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TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\n// Online C++ compiler to run C++ program online\n#include <iostream>\n#include <cstring>\n \nusing namespace std;\nlong long val;\nlong parent[200002],dp[200002][11],sizes[200002],ind[200002][11];\nlong par(long k){\n if(parent[k]==k)\n return k;\n return parent[k]=par(parent[k]);\n} \nvoid unionFind(long a,long b){\n a=par(a);\n b=par(b);\n if(a==b)\n return;\n if(sizes[b]>sizes[a]){\n swap(a,b);\n }\n \n parent[b]=a;\n sizes[a]+=sizes[b];\n \n return;\n}\nvoid solve(){\n long long n,m;\n cin>>n>>m;\n // cout<<n<<endl;\n val=n;\n for(int i=1;i<=n;i++){\n sizes[i]=1;\n parent[i]=i;\n for(int j=1;j<=10;j++)\n ind[i][j]=i;\n }\n \n memset(dp,0,sizeof(dp));\n \n for(int i=0;i<m;i++){\n long a,d,k;\n cin>>a>>d>>k;\n dp[a][d]++;\n long p=a+((k)*d);\n if(p>n+1)\n p=n+1;\n dp[p][d]--;\n }\n for(long i=1;i<=n;i++){\n for(long j=1;j<=10;j++){\n if(i-j>0){\n dp[i][j]=dp[i-j][j]+dp[i][j];\n \n \n if(dp[i-j][j]>0)\n unionFind(ind[i-j][j],i);\n ind[i][j]=ind[i-j][j];\n }\n \n }\n }\n val=0;\n for(int i=1;i<=n;i++){\n if(parent[i]==i)\n val++;\n }\n \n cout<<val<<\"\\n\";\n}\nint main() {\n // Write C++ code here\n long long t;\n ios_base::sync_with_stdio(0);\ncin.tie(0);\n cin>>t;\n for(int ii=0;ii<t;ii++){\n solve();\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 283861750,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define endl \"\\n\"\nusing namespace std;\n#define int long long\n \n \n#define mod 1000000007\n \nclass DSU{\npublic:\n\tint n, *par, *rank;\npublic:\n\tDSU(int n){\n\t\tthis->n=n;\n\t\tpar=(int*)malloc(sizeof(int)*(n+1));\n\t\trank=(int*)malloc(sizeof(int)*(n+1));\n\t\tmemset(rank, 0, sizeof(rank));\n\t\tfor(int i=1;i<=n;i++)par[i]=i;\n\t}\n\tint find(int x){\n\t\tif(par[x]==x)return x;\n\t\treturn find(par[x]);\n\t}\n\tvoid unionSet(int a, int b){\n\t\ta=find(a), b=find(b);\n\t\tif(a!=b){\n\t\t\tif(rank[a]<rank[b])swap(a, b);\n\t\t\tpar[b]=a;\n\t\t\tif(rank[a]==rank[b])rank[a]++;\n\t\t}\n\t}\n};\n \n/*\nINIT \nDSU dsu(len)\n \nFUNCS\nfind(x), unionSet(x, y)\n*/\n \n \nint32_t main(){\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n \n #ifndef ONLINE_JUDGE\n freopen(\"input.txt\",\"r\",stdin);\n freopen(\"output.txt\",\"w\",stdout);\n #endif\n \n int t; cin>>t;\n while(t--){\n \n \tint n,m; cin>>n>>m;\n \n \tDSU dsu(n);\n \tint dp[n+1][11];\n \n \tmemset(dp,0,sizeof(dp));\n \n \tfor(int i=0;i<m;i++){\n \t\tint a,d,k; cin>>a>>d>>k;\n \t\tfor(int j=1;j<=k;j++){\n \t\t\tif(dp[a+j*d][j])break;\n \t\t\tdp[a+j*d][j]=1;\n \t\t\tdsu.unionSet(a+j*d,a);\n \t\t}\n \n \t}\n \n \tset<int>st;\n \n \tfor(int i=1;i<=n;i++){\n \t\tst.insert(dsu.find(dsu.par[i]));\n \t}\n \n \tcout<<st.size()<<endl;\n \n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 285190316,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\n//libraries\n#include <bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\n \n \n \n \n//namespaces\nusing namespace std;\nusing namespace __gnu_pbds;\nusing namespace chrono;\n \n \n \n \n//utility variables\nlong long int _debug = 0, _nested = 0, _t = 1, _random_setup=0;\ntime_point <high_resolution_clock> _code_start, _code_end;\nduration<double> _code_duration;\n \n \n \n \n// macros-------------------------------------------------------------------------------\n#define ll long long\n#define int ll\n#define all(x) x.begin(), x.end()\n#define rep(i, x, n) for (int(i) = (x); (i) < (n); (i)++)\n#define endl \"\\n\"\n#define nl \\\n cout << endl; \\\n if (_debug) \\\n cerr << endl;\n \n \n//Random number generator\nint getRandomNumber(int start, int end) {\n\tstatic random_device rd; // Random device to generate a seed\n\tstatic mt19937 rng(rd()); // Mersenne Twister generator with random seed\n\tuniform_int_distribution<> dist(start, end); // Uniform distribution\n\treturn dist(rng); // Generate and return a random number\n}\n \n \n// print\n// helper-------------------------------------------------------------------------------\ntemplate <typename Primitive>\nvoid print(Primitive &&P)\n{\n\tcout << P << \" \";\n\tif (_debug)\n\t\tcerr << P << \" \";\n\tif (!_nested)\n\t{\n\t\tnl;\n\t}\n}\ntemplate <typename Complex>\nvoid print(vector<Complex> &Vector)\n{\n\t_nested = 1;\n\tfor (Complex &C : Vector)\n\t\tprint(C);\n\tnl;\n\t_nested = 0;\n}\ntemplate <typename Complex, typename... ComplexObjects>\nvoid print(Complex &&C, ComplexObjects &&...list)\n{\n\t_nested = 1, print(C);\n\t_nested = 0, print(list...);\n}\n \n \n \n \n// debug-------------------------------------------------------------------------------\n#define bug(...) __f(#__VA_ARGS__, __VA_ARGS__)\ntemplate <typename Arg1>\nvoid __f(const char *name, Arg1 &&arg1) {\n\tcout << name << \" : \" << arg1 << endl;\n}\ntemplate <typename Arg1, typename... Args>\nvoid __f(const char *names, Arg1 &&arg1, Args &&...args)\n{\n#ifndef ONLINE_JUDGE\n \n\tconst char *comma = strchr(names + 1, ',');\n\tcout.write(names, comma - names) << \":\" << arg1 << \"|\";\n\t__f(comma + 1, args...);\n \n#endif\n}\n \n \n \n// policy based data\n// structures---------------------------------------------------------------------------\n#define indexed_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>\n#define indexed_map tree<int, int, less<int>, rb_tree_tag, tree_order_statistics_node_update>\n \n \n \n \n// code----------------------------------------------------------------------------------\n \nstruct DSU {\n \n\tvector<int> parent;\n \n\tDSU(int n)\n\t{\n\t\tparent.resize(n+1,0);\n\t\tiota(parent.begin(), parent.end(), 0);\n\t}\n \n\tint find(int n)\n\t{\n\t\tif(parent[n]==n)return n;\n\t\treturn parent[n]=find(parent[n]);\n\t}\n \n \n\tbool unite(int a, int b)\n\t{\n\t\tauto p1 = find(a);\n\t\tauto p2 = find(b);\n \n\t\tif(p1==p2)\n\t\t{\n\t\t\treturn false;\n\t\t}\n \n\t\tif(p1<p2)\n\t\t{\n\t\t\tswap(p1,p2);\n\t\t}\n\t\tparent[p2]=p1;\n\t\treturn true;\n\t}\n \n};\n \n \n \nvoid solve()\n{\n\tint n, m;\n\tcin>>n>>m;\n \n\tvector<DSU> dus;\n\tfor(int i = 0; i<=10; i++)\n\t{\n\t\tDSU temp(n+1);\n\t\tdus.push_back(temp);\n\t}\n \n \n\tfor(int i = 0; i<m; i++)\n\t{\n\t\tint a, d, k;\n\t\tcin>>a>>d>>k;\n \n\t\tint last = a+k*d;\n\t\tint first = dus[d].parent[a];\n\t\tint second = first+d;\n \n\t\twhile(second<=last)\n\t\t{\n\t\t\tdus[d].unite(first, second);\n\t\t\tfirst = dus[d].parent[second];\n\t\t\tsecond = first+d;\n\t\t}\n\t}\n \n\tbool isRoot[n+1];\n\tmemset(isRoot, true, sizeof(isRoot));\n \n\tfor(int i = 1; i<=10; i++)\n\t{\n\t\tfor(int j = 1; j+i<=n; j++)\n\t\t{\n\t\t\tif(isRoot[j] && isRoot[j+i])\n\t\t\t{\n\t\t\t\tif(dus[i].find(j)==dus[i].find(j+i))\n\t\t\t\t{\n\t\t\t\t\tisRoot[j]=false;\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t}\n\t\n\tint ans =0;\n\tfor(int i = 1;i<=n;i++)\n\t{\n\t ans+= isRoot[i]==true;\n\t}\n\tprint(ans);\n \n \n}\n \nsigned main()\n{\n \n \n\t// _code_start =high_resolution_clock::now();\n// #ifndef ONLINE_JUDGE\n// freopen(\"input.txt\", \"r\", stdin);\n// freopen(\"output.txt\", \"w\", stdout);\n// #endif\n \n \n \n\tios_base::sync_with_stdio(false), cin.tie(NULL);\n\tcin >> _t;\n \n\tfor (int tc = 1; tc <= _t; tc++)\n\t{\n\t\t// cout << \"Case \" << tc << \":\"\n\t\t// << \" \";\n\t\tsolve();\n\t}\n \n \n \n\t// _code_end =high_resolution_clock::now();\n\t// _code_duration = _code_end - _code_start;\n\t// cerr << \"Code executed in \" << _code_duration.count() << \" seconds.\" <<endl;\n \n}\n \n// 1) cin does not move the input cursor to the next line, use cin.ignore() to\n// move it to the next line;\n \n// 2) GCD(a, b, c , d...) = GCD(a,b-a,c-a,d-a...);\n \n// 3) A reverse iterator moves backwards while doing it++\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 283666642,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nt = int(input())\nfor _ in range(t):\n n, m = map(int, input().split())\n qrs = []\n for _ in range(m):\n a, d, k = map(int, input().split())\n qrs.append((a, d, k))\n tab = [[0]*11 for _ in range(n + 1)]\n \n for i in range(m):\n a, d, k = qrs[i]\n tab[a][d] += 1\n tab[a + k * d][d] -= 1\n for j in range(1, 11):\n for z in range(0, j):\n cur = 0\n for i in range(1 + z, n + 1, j):\n if tab[i][j] == 0:\n tab[i][j] = 1 if cur > 0 else 0\n elif tab[i][j] > 0:\n cur += tab[i][j]\n tab[i][j] = 1 if cur > 0 else 0\n else:\n c = tab[i][j]\n tab[i][j] = 1 if cur > 0 else 0\n cur += c\n \n # for j in range(1, 11):\n # if sum(tab[i][j] for i in range(1, n + 1)) == 0:\n # continue\n # for i in range(1, n + 1):\n # print(tab[i][j], end=' ')\n # print()\n \n res = 0\n # find connected components with dfs:\n low = 1\n vis = [0] * (n + 1)\n vis_cn = 0\n while low < n + 1:\n stk = [low]\n while stk:\n x = stk.pop()\n vis[x] = 1\n vis_cn += 1\n for j in range(1, 11):\n if x + j < n + 1 and tab[x][j] == 1 and tab[x + j][j] == 1 and vis[x + j] == 0:\n stk.append(x + j)\n while low <=n and vis[low] == 1:\n low += 1\n res += 1\n \n print(res)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 284310017,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nt=int(input())\nfor _ in range(t):\n n,m = map(int,input().split(\" \"))\n lista = [i for i in range(1,n+1)]\n arch=0\n for __ in range(m):\n a,d,k = map(int,input().split(\" \"))\n allista = [a+i*d for i in range(k+1)]\n benne=True\n for v in allista:\n if v in lista:\n lista.remove(v)\n else:\n benne= False\n if benne:\n arch+=1\n print(arch+len(lista))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 283648986,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nfor _ in range(int(input())):\n n, m = map(int, input().split())\n queries = [[] for i in range(11)]\n for _ in range(m):\n a, d, k = map(int, input().split())\n queries[d].append([a, k])\n for i in range(11):\n queries[i].sort()\n nodes = [False for i in range(n+1)]\n \n ans = 0\n for i in range(1, 11):\n visited_till = [0 for i in range(i)]\n for j in queries[i]:\n a = j[0]\n k = j[1] + 1\n if a < visited_till[a%i]:\n k -= (visited_till[a%i] - a)//i\n a = visited_till[a%i]\n if k<=0:\n continue\n found = False\n while k>0:\n if nodes[a]:\n found = True\n else:\n nodes[a] = True\n k-=1\n a+=i\n if not found:\n ans += 1\n print(ans+nodes.count(False)-1)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 283630385,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nimport javax.print.attribute.HashPrintServiceAttributeSet;\nimport java.io.BufferedInputStream;\nimport java.io.IOException;\nimport java.util.ArrayList;\nimport java.util.Arrays;\nimport java.util.Collections;\nimport java.util.HashMap;\nimport java.util.*;\n \npublic class Main {\n public static void main(String[] args) {\n FastScanner in = new FastScanner();\n int testCount = in.nextInt();\n for (int i = 1; i <= testCount; i++) {\n Solver solver = new Solver();\n try {\n solver.solve(i, in, System.out);\n } catch (IOException e) {\n e.printStackTrace(); // Handle exception appropriately\n }\n }\n }\n \n static class Solver {\n int visited[];\n public int dfs(ArrayList<Integer> graph[]){\n int comp=0;\n for(int i=1;i<graph.length;i++){\n if(visited[i]==0){\n hlpr(graph,i);\n comp++;\n }\n }\n return comp;\n }\n public void hlpr(ArrayList<Integer> graph[],int i){\n Queue<Integer> q = new LinkedList<>();\n q.add(i);\n visited[i]=1;\n while(!q.isEmpty()){\n for(int temp:graph[q.remove()]){\n if(visited[temp]==0){\n visited[temp]=1;\n q.add(temp);\n }\n }\n }\n }\n public void solve(int testNumber, FastScanner sc, Appendable out) throws IOException {\n // Solution logic goes here\n int n=sc.nextInt();\n int m=sc.nextInt();\n int arr[][]=new int[n+1][11];\n visited=new int[n+1];\n ArrayList<Integer> graph[]=new ArrayList[n+1];\n for(int i=0;i<graph.length;i++){\n graph[i]=new ArrayList<Integer>();\n }\n \n for(int i=0;i<m;i++){\n int a=sc.nextInt();\n int d=sc.nextInt();\n int k=sc.nextInt();\n arr[a][d]=Math.max(k,arr[a][d]);\n }\n for(int i=1 ;i<=n;i++){\n for(int j=1;j<=10;j++){\n int iter=arr[i][j];\n arr[i][j]=0;\n int tempj=j;\n while(iter>0&&tempj<n){\n graph[i].add(i+tempj);\n graph[i+tempj].add(i);\n tempj+=j;\n iter=Math.max(iter,arr[tempj][j]);\n arr[tempj][j]=0;\n iter--;\n }\n }\n }\n System.out.println(dfs(graph));\n }\n }\n \n static class FastScanner {\n private int BS = 1 << 16;\n private char NC = (char) 0;\n private byte[] buf = new byte[BS];\n private int bId = 0, size = 0;\n private char c = NC;\n private double cnt = 1;\n private BufferedInputStream in;\n \n public FastScanner() {\n in = new BufferedInputStream(System.in, BS);\n }\n \n public int nextInt() {\n return (int) nextLong();\n }\n \n public int[] nextInts(int N) {\n int[] res = new int[N];\n for (int i = 0; i < N; i++) {\n res[i] = (int) nextLong();\n }\n return res;\n }\n \n public long nextLong() {\n cnt = 1;\n boolean neg = false;\n if (c == NC) c = getChar();\n for (; (c < '0' || c > '9'); c = getChar()) {\n if (c == '-') neg = true;\n }\n long res = 0;\n for (; c >= '0' && c <= '9'; c = getChar()) {\n res = (res << 3) + (res << 1) + c - '0';\n cnt *= 10;\n }\n return neg ? -res : res;\n }\n \n public double nextDouble() {\n double cur = nextLong();\n return c != '.' ? cur : cur + nextLong() / cnt;\n }\n \n public String next() {\n StringBuilder res = new StringBuilder();\n while (c <= 32) c = getChar();\n while (c > 32) {\n res.append(c);\n c = getChar();\n }\n return res.toString();\n }\n \n public String nextLine() {\n StringBuilder res = new StringBuilder();\n while (c <= 32) c = getChar();\n while (c != '\\n') {\n res.append(c);\n c = getChar();\n }\n return res.toString();\n }\n \n public boolean hasNext() {\n if (c > 32) return true;\n while (true) {\n c = getChar();\n if (c == NC) return false;\n else if (c > 32) return true;\n }\n }\n \n public int[] sortedIntArray(int n) {\n int[] arr = new int[n];\n for (int i = 0; i < n; i++) {\n arr[i] = nextInt();\n }\n Arrays.sort(arr);\n return arr;\n }\n \n public long[] sortedLongArray(int n) {\n long[] arr = new long[n];\n for (int i = 0; i < n; i++) {\n arr[i] = nextLong();\n }\n Arrays.sort(arr);\n return arr;\n }\n \n public long gcd(long a, long b) {\n while (b != 0) {\n long temp = b;\n b = a % b;\n a = temp;\n }\n return a;\n }\n \n public long lcm(long a, long b) {\n return (a / gcd(a, b)) * b;\n }\n \n public int[] prefixSum(int[] arr) {\n int n = arr.length;\n int[] prefixSum = new int[n];\n prefixSum[0] = arr[0];\n for (int i = 1; i < n; i++) {\n prefixSum[i] = prefixSum[i - 1] + arr[i];\n }\n return prefixSum;\n }\n public static void sort(int[] arr)\n {\n //because Arrays.sort() uses quicksort which is dumb\n //Collections.sort() uses merge sort\n ArrayList<Integer> ls = new ArrayList<Integer>();\n for(int x: arr)\n ls.add(x);\n Collections.sort(ls);\n for(int i=0; i < arr.length; i++)\n arr[i] = ls.get(i);\n }\n public static long power(long x, long y, long p)\n {\n //0^0 = 1\n long res = 1L;\n x = x%p;\n while(y > 0)\n {\n if((y&1)==1)\n res = (res*x)%p;\n y >>= 1;\n x = (x*x)%p;\n }\n return res;\n }\n public static void iniDP(int dp[][]){\n for(int arr[]:dp) Arrays.fill(arr,-1);\n }\n private char getChar() {\n while (bId == size) {\n try {\n size = in.read(buf);\n } catch (Exception e) {\n return NC;\n }\n if (size == -1) return NC;\n bId = 0;\n }\n return (char) buf[bId++];\n }\n }\n }\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 284185186,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class D {\n \n \n static class DSU {\n DSU(int n) {\n init(n);\n }\n \n private int[] p, sz;\n \n void init(int n) {\n p = new int[n];\n sz = new int[n];\n for (int i = 0; i < n; i++) {\n p[i] = i;\n sz[i] = 1;\n }\n }\n \n //用于连接两个连通分量\n void union(int x, int y) {\n x = find(x);\n y = find(y);\n if (x == y) return;\n sz[x] += sz[y];\n p[y] = x;\n }\n \n //用于查询某个元素所在连通分量的根节点\n int find(int x) {\n while (p[x] != x) {\n p[x] = p[p[x]];\n x = p[x];\n }\n return x;\n }\n \n //查询两个元素是否在同一连通分量中\n boolean same(int x, int y) {\n return find(x) == find(y);\n }\n \n //返回 x 所在连通分量的元素个数\n int size(int x) {\n return sz[find(x)];\n }\n }\n \n public static void main(String[] args) {\n int t = sc.nextInt();\n while (t-- > 0) {\n int n = sc.nextInt(), q = sc.nextInt();\n //a[i],d,k\n DSU dsu = new DSU(n);\n List<int[]>[][] query = new List[11][10];//d,idx%d\n for (int i = 0; i < 11; i++) {\n for (int j = 0; j < 10; j++) {\n query[i][j] = new ArrayList<>();\n }\n }\n for (int i = 0; i < q; i++) {\n int[] a = new int[3];\n a[0] = sc.nextInt() - 1;\n a[1] = sc.nextInt();\n a[2] = sc.nextInt();\n query[a[1]][a[0] % a[1]].add(a);\n }\n int[] diff = new int[n + 1];\n for (int i = 1; i < 11; i++) {//枚举 d\n for (int j = 0; j < i; j++) {//枚举 idx % d\n if (query[i][j].size() > 0) {\n Arrays.fill(diff, 0);\n for (int[] z : query[i][j]) {\n int ai = z[0], d = z[1], k = z[2];\n int idx = ai / d;\n diff[idx]++;\n diff[idx + k + 1]--;\n }\n for (int k = 1; k <= n; k++) {\n diff[k] += diff[k - 1];\n }\n for (int k = 1; k < n; k++) {\n if (diff[k] > 0 && diff[k - 1] > 0) {\n dsu.union(k * i + j, (k - 1) * i + j);\n }\n }\n \n }\n \n \n }\n }\n \n int ans = 0;\n for (int i = 0; i < n; i++) {\n if (dsu.find(i) == i) ans++;\n }\n out.println(ans);\n \n \n }\n out.close();\n }\n \n static Kattio sc = new Kattio();\n static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n \n static class Kattio {\n static BufferedReader r;\n static StringTokenizer st;\n \n public Kattio() {\n r = new BufferedReader(new InputStreamReader(System.in));\n }\n \n public String next() {\n try {\n while (st == null || !st.hasMoreTokens()) {\n st = new StringTokenizer(r.readLine());\n }\n return st.nextToken();\n } catch (Exception e) {\n return null;\n }\n }\n \n public int nextInt() {\n return Integer.parseInt(next());\n }\n \n public long nextLong() {\n return Long.parseLong(next());\n }\n \n public double nextDouble() {\n return Double.parseDouble(next());\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 283639731,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\n#include<iostream>\n#include<set>\n#include<queue>\n#include<set>\n#include<map>\n#include<algorithm>\n#include<unordered_map>\n#include<queue>\n \nusing namespace std;\n \nvector<int> dsu,ranked;\n \nint parent(int root){\n if(dsu[root]==root)return root;\n return dsu[root]= parent(dsu[root]);\n}\n \nvoid unite(int c1, int c2 ){\n int a = parent(c1);\n int b = parent(c2);\n if(a==b)return;\n if(ranked[b]>ranked[a])swap(a,b);\n ranked[a]+=ranked[b];\n dsu[b]=a;\n}\n \nvoid solve(){\n int n ,m;\n cin>>n>>m;\n dsu.resize(n+1),ranked.resize(n+1);\n for(int i = 0;i<=n;i++){\n dsu[i]=i;\n ranked[i]=1;\n }\n vector<vector<int> > q(m,vector<int>(3));\n for(int i = 0;i<m;i++){cin>>q[i][0]>>q[i][1]>>q[i][2];\n // q[i][0]--; \n }\n sort(q.begin(),q.end());\n vector<vector<int> > lst(10,vector<int>(10,-1)),p(10,vector<int>(10,-1));\n int idx = 0;\n for(int i = 0;i<=n;i++){\n \n \n for(int j=0;j<10;j++){\n int x = (i+j)%10, y = (10-j)%10;\n if(p[x][y]!=-1)\n if(lst[x][y]>=i){\n unite(i,p[x][y]);\n }\n }\n \n \n while(idx<m&&q[idx][0]==i){\n lst[i%10][q[idx][1]%10]=max(lst[i%10][q[idx][1]%10],q[idx][0]+q[idx][1]*q[idx][2]);\n p[i%10][q[idx][1]%10]=parent(i);\n for(int l=1;l<=q[idx][2]&&l<10&&q[idx][0]%10 != (q[idx][0] + q[idx][1]*l)%10;l++){\n int nxt = (q[idx][0] + q[idx][1]*l)%10;\n lst[nxt][q[idx][1]%10]=max(lst[nxt][q[idx][1]%10],q[idx][0]+q[idx][1]*q[idx][2]);\n if(p[nxt][q[idx][1]%10]!=-1)\n unite(p[nxt][q[idx][1]%10],i);\n p[nxt][q[idx][1]%10]=parent(i);\n }\n idx++;\n }\n \n }\n set<int> s;\n \n for(int i =1;i<=n;i++){\n s.insert(parent(i));\n }\n cout<<s.size()<<endl;\n \n}\n \nint main(){\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n #ifndef ONLINE_JUDGE\n freopen(\"input.txt\",\"r\",stdin);\n freopen(\"output.txt\",\"w\",stdout);\n #endif \n int t=1;\n cin>>t;\n while(t--){\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 284482787,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\nusing namespace std;\nusing namespace __gnu_pbds;\n \n#define print(x) for(auto it:(x)) cout<<it<<\" \"; cout<<endl;\n#define printp(x) for(auto it:(x)) cout<<it.first<<\" \"<<it.second<<endl;\n \ntypedef vector<int> vi;\ntypedef vector<pair<int,int>> vp;\ntypedef vector<pair<long long, long long>> vpll;\ntypedef vector<long long int> vll;\ntypedef vector<vector<int>> vvi;\ntypedef long long int ll;\ntypedef vector<vector<long long int>> vvll;\ntypedef vector<string> vs;\ntypedef pair<long long int, long long int> pll;\ntypedef tree <int , null_type , less<int> , rb_tree_tag , tree_order_statistics_node_update> ordered_set; //find_by order(k)->kth element of set(0 counting), order_of_key(k)-> number of elements strictly smaller than k\n \nll binpow(ll a, ll b, ll m) { // x^n % m\n a %= m;\n ll res = 1;\n while (b > 0) {\n if (b & 1)\n res = res * a % m;\n a = a * a % m;\n b >>= 1;\n }\n return res;\n}\n \nll n,m;\nvll parent(200005,0); // intialise in solve\nvll sizen(200005,1);\n \nll findUPar(ll node){\n\tif(parent[node] == node) return node;\n \n\treturn parent[node] = findUPar(parent[node]);\n}\n \nvoid unionBySize(ll a, ll b){\n\tll ulp_a = findUPar(a);\n\tll ulp_b = findUPar(b);\n\tif(ulp_a == ulp_b) return;\n\tif(sizen[ulp_a] > sizen[ulp_b]){\n\t\tparent[ulp_b] = ulp_a;\n\t\tsizen[ulp_a] += sizen[ulp_b];\n\t}else{\n\t\tparent[ulp_a] = ulp_b;\n\t\tsizen[ulp_b] += sizen[ulp_a];\n\t}\n\treturn;\n}\n \nvoid solve(){\n cin>>n>>m;\n for(ll i=1; i<=n; i++) parent[i] = i, sizen[i] = 1;;\n vector<vpll> darr(11); \n for(ll i=0; i<m; i++){\n ll a,d,k;\n cin>>a>>d>>k;\n darr[d].push_back({a,k});\n }\n for(ll i=0; i<11; i++){\n if(!darr[i].empty()) sort(darr[i].begin(),darr[i].end());\n }\n for(ll i=0; i<11; i++){\n if(darr[i].empty()) continue;\n // d value = i\n map<ll,ll> ub;\n for(auto it: darr[i]){\n ll st = it.first;\n ll en = it.first + it.second*i;\n ll j = max(st, ub[st%i]);\n ub[st%i] = en;\n while(j <= en){\n ub[j] = en;\n unionBySize(st,j);\n j+=i;\n }\n }\n }\n set<ll> ans;\n for(ll i=1; i<=n; i++) ans.insert(parent[i]);\n cout<<ans.size()<<endl;\n}\n \nint main(){\n ios::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n int t=1;\n cin>>t;\n while(t--){\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 284975165,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nimport sys\n \ndef find(u):\n\tif un[u] != u:\n\t\tun[u] = find(un[u])\n\treturn un[u]\n \ndef union(u, v):\n\tru = find(u)\n\trv = find(v)\n\tif ru == rv:\n\t\treturn 0\n\tun[rv] = ru\n\treturn 1\n \ninput = sys.stdin.buffer.readline\nfor _ in range(int(input())):\n\tn, m = map(int, input().split())\n\tgrp = [[[] for _ in range(10)] for _ in range(11)]\n\tun = list(range(n + 1))\n\tans = n\n\tfor _ in range(m):\n\t\ta, d, k = map(int, input().split())\n\t\tgrp[d][a % d].append([a, a + d * k])\n\tfor d in range(1, 10):\n\t\tfor mod in range(d):\n\t\t\tmx = -1\n\t\t\tgrp[d][mod].sort()\n\t\t\tfor l, r in grp[d][mod]:\n\t\t\t\tfor u in range(max(mx, l), r, d):\n\t\t\t\t\tans -= union(u, u + d)\n\t\t\t\tmx = max(r, mx)\n\tprint(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 291379123,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nfrom types import GeneratorType\ndef bootstrap(f, stack=[]):\n def wrappedfunc(*args, **kwargs):\n if stack:\n return f(*args, **kwargs)\n else:\n to = f(*args, **kwargs)\n while True:\n if type(to) is GeneratorType:\n stack.append(to)\n to = next(to)\n else:\n stack.pop()\n if not stack:\n break\n to = stack[-1].send(to)\n return to\n return wrappedfunc\n \ndef sol():\n n, m = map(int, input().split())\n \n # Dsu\n fa = [i for i in range(n + 1)]\n def fd(x: int) -> int:\n if fa[x] == x:\n return x\n fa[x] = fd(fa[x])\n return fa[x]\n def merge(x: int, y: int):\n fx = fd(x)\n fy = fd(y)\n if fx == fy:\n return\n fa[fx] = fy\n \n f = [[0 for _ in range(11)] for _ in range(n + 1)]\n \n @bootstrap\n def dfs(x: int, k: int, d: int):\n if f[x][d] >= k:\n yield\n lst = f[x][d]\n tmp = d\n now = 1\n while tmp <= 10:\n f[x][tmp] = max(f[x][tmp], k // now)\n now += 1\n tmp += d\n merge(x, x + (lst + 1) * d)\n yield (yield dfs(x + (lst + 1) * d, k - lst - 1, d))\n \n while m:\n m -= 1\n a, d, k = map(int, input().split())\n dfs(a, k, d)\n cnt = 0\n vis = [False for _ in range(n + 1)]\n for i in range(1, n + 1):\n x = fd(i)\n if not vis[x]:\n vis[x] = True\n cnt += 1\n print(cnt)\n \nt = int(input())\nwhile t:\n t -= 1\n sol()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 285402146,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\n \npublic class connectTheDots {\n static class DisJointSet {\n int[] parent;\n int[] rank;\n int N;\n int numOfComp = 0;\n \n DisJointSet(int n) {\n this.parent = new int[n];\n this.rank = new int[n];\n this.N = n;\n \n for (int i = 0; i < N; i++) {\n parent[i] = i;\n rank[i] = 1;\n }\n }\n \n int findParent(int node) {\n if (parent[node] == node) return node;\n return parent[node] = findParent(parent[node]);\n }\n \n void unionByRank(int u, int v) {\n int PU = findParent(u);\n int PV = findParent(v);\n \n if (rank[PU] > rank[PV]) {\n parent[PV] = PU;\n } else if (rank[PV] > rank[PU]) {\n parent[PU] = PV;\n } else {\n parent[PU] = PV;\n rank[PV]++;\n }\n }\n \n int components() {\n for (int i = 0; i < N; i++) {\n if (findParent(i) == i) numOfComp++;\n }\n return numOfComp-1;\n }\n }\n \n public static void main(String[] args) throws IOException {\n BufferedReader reader= new BufferedReader(new InputStreamReader(System.in));\n int t =Integer.parseInt(reader.readLine());\n \n while(t-->0){\n String[] inputs= reader.readLine().split(\" \");\n int n =Integer.parseInt(inputs[0]);\n int m =Integer.parseInt(inputs[1]);\n DisJointSet ds= new DisJointSet(n+1);\n for(int i=0;i<m;i++){\n String[] inputs2= reader.readLine().split(\" \");\n \n int a =Integer.parseInt(inputs2[0]);\n int d =Integer.parseInt(inputs2[1]);\n int k =Integer.parseInt(inputs2[2]);\n int an=a+(d*k);\n \n for(int j=an;j>=a;j-=d){\n if(ds.findParent(a)==ds.findParent(j)){\n for(int l=a+d;l<j;l+=d){\n if(ds.findParent(l)==ds.findParent(a)) break;\n ds.unionByRank(l,a);\n }\n break;\n }\n else ds.unionByRank(j,a);\n }\n \n }\n \n System.out.println(ds.components());\n \n }\n \n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 288346780,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\n/*\nhttps://codeforces.com/contest/2020/problem/D\n */\n \n \nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStream;\nimport java.io.InputStreamReader;\nimport java.util.HashSet;\nimport java.util.Set;\nimport java.util.StringTokenizer;\n \npublic class ConnectTheDots {\n \n public static void main(String[] args) throws IOException {\n Scanner scanner = new Scanner(System.in);\n int t = scanner.nextInt();\n while (t-- > 0) {\n int n = scanner.nextInt();\n int m = scanner.nextInt();\n DisjointSetDataStructure disjointSetDataStructure = new DisjointSetDataStructure(n);\n int count = 1;\n while (m-- > 0) {\n int a = scanner.nextInt();\n int d = scanner.nextInt();\n int k = scanner.nextInt();\n for (int i = 0; i <= k; i++) {\n int val = a + i * d;\n// if (!disjointSetDataStructure.union(val, a)) {\n// count++;\n// }\n disjointSetDataStructure.union(val, a);\n }\n }\n// System.out.println(count);\n Set<Integer> set = new HashSet<>();\n for (int i = 1; i < disjointSetDataStructure.parent.length; i++) {\n int parI = disjointSetDataStructure.parent[i];\n int rankI = disjointSetDataStructure.rank[i];\n if(rankI == 1) set.add(parI);\n }\n System.out.println(set.size());\n }\n }\n \n private static class DisjointSetDataStructure {\n \n int forests;\n int[] parent;\n int[] rank;\n \n public DisjointSetDataStructure(int n) {\n this.forests = n + 1;\n this.parent = new int[this.forests];\n this.rank = new int[this.forests];\n for (int i = 0; i < this.forests; i++) {\n parent[i] = i;\n rank[i] = 1;\n }\n }\n \n public int find(int i) {\n if (this.parent[i] == i) {\n return i;\n }\n int rep = find(parent[i]);\n this.parent[i] = rep;\n return rep;\n }\n \n public boolean union(int i, int j) {\n int irep = this.find(i);\n int jrep = this.find(j);\n if (irep == jrep) {\n return false;\n }\n int irank = rank[irep];\n int jrank = rank[jrep];\n if (irank > jrank) {\n this.parent[jrep] = irep;\n } else if (irank < jrank) {\n this.parent[irep] = jrep;\n } else {\n this.parent[irep] = jrep;\n rank[jrep]++;\n }\n return true;\n }\n }\n \n private static class Scanner {\n \n StringTokenizer st;\n \n BufferedReader br;\n \n public Scanner(InputStream is) {\n br = new BufferedReader(new InputStreamReader(is));\n }\n \n public String next() throws IOException {\n while (st == null || !st.hasMoreTokens()) {\n st = new StringTokenizer(br.readLine());\n }\n return st.nextToken();\n }\n \n public int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n \n public long nextLong() throws IOException {\n return Long.parseLong(next());\n }\n \n public double nextDouble() throws IOException {\n return Double.parseDouble(next());\n }\n \n public String nextLine() throws IOException {\n return br.readLine();\n }\n \n public boolean ready() throws IOException {\n return br.ready() || st.hasMoreTokens();\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 283635925,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\nusing namespace std;\n \nclass UnionFind {\npublic:\n vector<int> parent;\n vector<int> rank;\n int count;\n \n UnionFind(int size) {\n parent.resize(size);\n rank.resize(size, 0);\n for (int i = 0; i < size; ++i) {\n parent[i] = i;\n }\n count = size;\n }\n \n int find(int x) {\n if (parent[x] != x) {\n parent[x] = find(parent[x]); // Path compression\n }\n return parent[x];\n }\n \n void unionSet(int x, int y) {\n int rootX = find(x);\n int rootY = find(y);\n if (rootX != rootY) {\n if (rank[rootX] > rank[rootY]) {\n parent[rootY] = rootX;\n } else if (rank[rootX] < rank[rootY]) {\n parent[rootX] = rootY;\n } else {\n parent[rootY] = rootX;\n rank[rootX]++;\n }\n count--;\n }\n }\n};\n \nint main() {\n int t;\n cin >> t;\n while (t--) {\n int n, m;\n cin >> n >> m;\n UnionFind uf(n);\n for (int i = 0; i < m; ++i) {\n int a, d, k;\n cin >> a >> d >> k;\n for (int j = 0; j <= k; ++j) {\n uf.unionSet(a - 1, a - 1 + j * d);\n }\n }\n cout << uf.count << endl;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 293856178,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <unordered_set>\n \nusing namespace std;\n \nclass DSU {\npublic:\n vector<int> parent;\n vector<int> rank;\n \n DSU(int n) {\n parent.resize(n);\n rank.resize(n, 0);\n for (int i = 0; i < n; ++i) {\n parent[i] = i;\n }\n }\n \n int find(int x) {\n if (parent[x] != x) {\n parent[x] = find(parent[x]);\n }\n return parent[x];\n }\n \n void unionSets(int x, int y) {\n int rootX = find(x);\n int rootY = find(y);\n if (rootX != rootY) {\n if (rank[rootX] > rank[rootY]) {\n parent[rootY] = rootX;\n } else if (rank[rootX] < rank[rootY]) {\n parent[rootX] = rootY;\n } else {\n parent[rootY] = rootX;\n rank[rootX]++;\n }\n }\n }\n};\n \nint main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n \n int t;\n cin >> t;\n while (t--) {\n int n, m;\n cin >> n >> m;\n DSU dsu(n);\n \n for (int i = 0; i < m; ++i) {\n int a, d, k;\n cin >> a >> d >> k;\n for (int j = 0; j < k; ++j) {\n dsu.unionSets(a - 1, a - 1 + d * (j + 1));\n }\n }\n \n unordered_set<int> components;\n for (int i = 0; i < n; ++i) {\n components.insert(dsu.find(i));\n }\n \n cout << components.size() << '\\n';\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 294478500,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nclass DSU:\n def __init__(self, n):\n self.parents = list(range(n))\n self.sizes = [1] * n\n \n def find(self, node):\n while self.parents[node] != node:\n node = self.parents[node]\n return node\n \n def union(self, a, b):\n root_a = self.find(a)\n root_b = self.find(b)\n if root_a != root_b:\n if self.sizes[root_a] > self.sizes[root_b]:\n self.parents[root_b] = root_a\n self.sizes[root_a] += self.sizes[root_b]\n else:\n self.parents[root_a] = root_b\n self.sizes[root_b] += self.sizes[root_a]\n \ndef main():\n t = int(input())\n for _ in range(t):\n n, m = map(int, input().split())\n dsu = DSU(n+1)\n for _ in range(m):\n a, d, k = map(int, input().split())\n for i in range(k):\n dsu.union(a+i*d, a+(i+1)*d)\n print(len(set(dsu.find(i) for i in range(1, n+1))))\n \nif __name__ == \"__main__\":\n main()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 283627998,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nclass UnionFind:\n def __init__(self, n):\n self.parent = list(range(n))\n self.rank = [1] * n\n \n def find(self, x):\n if self.parent[x] != x:\n self.parent[x] = self.find(self.parent[x]) # Path compression\n return self.parent[x]\n \n def union(self, x, y):\n rootX = self.find(x)\n rootY = self.find(y)\n if rootX != rootY:\n # Union by rank\n if self.rank[rootX] > self.rank[rootY]:\n self.parent[rootY] = rootX\n elif self.rank[rootX] < self.rank[rootY]:\n self.parent[rootX] = rootY\n else:\n self.parent[rootY] = rootX\n self.rank[rootX] += 1\n \ndef solve():\n t = int(input()) # Number of test cases\n for _ in range(t):\n n, m = map(int, input().split()) # n points, m operations\n uf = UnionFind(n)\n \n for _ in range(m):\n a, d, k = map(int, input().split())\n # Points a, a+d, a+2d, ..., a+k*d are connected\n last = a - 1 # Convert to 0-indexed\n for i in range(1, k + 1):\n next_point = a + i * d - 1 # 0-indexed\n uf.union(last, next_point)\n last = next_point\n \n # Count unique connected components\n unique_components = set()\n for i in range(n):\n unique_components.add(uf.find(i))\n \n print(len(unique_components))\n \n# Reading input and running the solution\nsolve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 283616310,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class D_Connect_the_Dots {\n \n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n int t = Integer.parseInt(br.readLine().trim()); // Number of test cases\n \n while (t-- > 0) {\n StringTokenizer st = new StringTokenizer(br.readLine());\n int n = Integer.parseInt(st.nextToken()); // Number of physical nodes\n int m = Integer.parseInt(st.nextToken()); // Number of connections\n \n int[] arr = new int[n + 1]; // Stores the connection assigned to physical nodes\n PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] == b[0] ? b[2] - a[2] : a[0] - b[0]);\n boolean[][] vis = new boolean[n + 1][11]; // Visited array for each node and its distance\n \n // Reading connections and adding them to the priority queue\n for (int i = 0; i < m; i++) {\n st = new StringTokenizer(br.readLine());\n int a = Integer.parseInt(st.nextToken()); // Physical node\n int d = Integer.parseInt(st.nextToken()); // Distance\n int k = Integer.parseInt(st.nextToken()); // Steps\n pq.add(new int[]{a, d, k, i + 1}); // Add to queue with connection ID i+1\n }\n \n // Initialize graph for m connections\n List<Set<Integer>> graph = new ArrayList<>(m + 1);\n for (int i = 0; i <= m; i++) graph.add(new HashSet<>());\n \n // Processing priority queue\n while (!pq.isEmpty()) {\n int[] a = pq.poll(); // Get the top element from the priority queue\n int node = a[0], dist = a[1], steps = a[2], connID = a[3];\n \n // Skip if already visited with this distance\n if (vis[node][dist]) {\n graph.get(connID).add(arr[node]);\n graph.get(arr[node]).add(connID);\n continue;\n }\n \n // Add connections for all distances already visited for this node\n for (int i = 1; i <= 10; i++) {\n if (vis[node][i]) {\n graph.get(connID).add(arr[node]);\n graph.get(arr[node]).add(connID);\n }\n }\n \n // Mark the node with current distance as visited\n vis[node][dist] = true;\n arr[node] = connID; // Assign this connection to the node\n \n // If steps left, propagate to the next node\n if (steps > 0 && node + dist <= n) {\n pq.add(new int[]{node + dist, dist, steps - 1, connID});\n }\n }\n \n // Count connected components in the graph\n int numOfConnectedComponents = 0;\n boolean[] mark = new boolean[m + 1]; // Visited array for DFS on connections\n \n // Iterative DFS to count connected components\n for (int i = 1; i <= m; i++) {\n if (!mark[i]) {\n iterativeDFS(graph, i, mark);\n numOfConnectedComponents++; // Increment component count\n }\n }\n \n // Count the number of physical nodes connected to some connection\n int cnt = 0;\n for (int i = 1; i <= n; i++) {\n if (arr[i] != 0) cnt++;\n }\n \n // Output the result\n System.out.println(n - cnt + numOfConnectedComponents);\n }\n }\n \n // Iterative DFS to explore the connection graph\n public static void iterativeDFS(List<Set<Integer>> graph, int src, boolean[] vis) {\n Stack<Integer> stack = new Stack<>();\n stack.push(src);\n vis[src] = true;\n \n while (!stack.isEmpty()) {\n int node = stack.pop();\n for (int neighbor : graph.get(node)) {\n if (!vis[neighbor]) {\n vis[neighbor] = true;\n stack.push(neighbor);\n }\n }\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 283712003,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Connect the Dots\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: One fine evening, Alice sat down to play the classic game \"Connect the Dots\", but with a twist.\n\nTo play the game, Alice draws a straight line and marks $n$ points on it, indexed from $1$ to $n$. Initially, there are no arcs between the points, so they are all disjoint. After that, Alice performs $m$ operations of the following type:\n\n * She picks three integers $a_i$, $d_i$ ($1 \\le d_i \\le 10$), and $k_i$. * She selects points $a_i, a_i+d_i, a_i+2d_i, a_i+3d_i, \\ldots, a_i+k_i\\cdot d_i$ and connects each pair of these points with arcs. \n\nAfter performing all $m$ operations, she wants to know the number of connected components$^\\dagger$ these points form. Please help her find this number.\n\n$^\\dagger$ Two points are said to be in one connected component if there is a path between them via several (possibly zero) arcs and other points.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^5$). The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n \\le 2 \\cdot 10^5$, $1 \\le m \\le 2 \\cdot 10^5$).\n\nThe $i$-th of the following $m$ lines contains three integers $a_i$, $d_i$, and $k_i$ ($1 \\le a_i \\le a_i + k_i\\cdot d_i \\le n$, $1 \\le d_i \\le 10$, $0 \\le k_i \\le n$).\n\nIt is guaranteed that both the sum of $n$ and the sum of $m$ over all test cases do not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output the number of connected components.\n\nExamples:\ninput:\n3\n10 2\n1 2 4\n2 2 4\n100 1\n19 2 4\n100 3\n1 2 5\n7 2 6\n17 2 31\noutput:\n2\n96\n61\n\nNote: In the first test case, there are $n = 10$ points. The first operation joins the points $1$, $3$, $5$, $7$, and $9$. The second operation joins the points $2$, $4$, $6$, $8$, and $10$. There are thus two connected components: $\\\\{1, 3, 5, 7, 9\\\\}$ and $\\\\{2, 4, 6, 8, 10\\\\}$.\n\nIn the second test case, there are $n = 100$ points. The only operation joins the points $19$, $21$, $23$, $25$, and $27$. Now all of them form a single connected component of size $5$. The other $95$ points form single-point connected components. Thus, the answer is $1 + 95 = 96$.\n\nIn the third test case, there are $n = 100$ points. After the operations, all odd points from $1$ to $79$ will be in one connected component of size $40$. The other $60$ points form single-point connected components. Thus, the answer is $1 + 60 = 61$.\n\nBuggy Code:\nimport java.util.HashMap;\nimport java.util.Scanner;\n \npublic class D_Connect_the_Dots {\n \n public static class DisjointSet {\n HashMap<Integer, Integer> parent;\n HashMap<Integer, Integer> size;\n int count ;\n \n DisjointSet(int n) {\n parent = new HashMap<>();\n size = new HashMap<>();\n count = n ;\n }\n \n int findUlp(int node) {\n if (!parent.containsKey(node)) {\n parent.put(node, node);\n size.put(node, 1);\n }\n if (parent.get(node) == node) {\n return node;\n }\n int ulp = findUlp(parent.get(node));\n parent.put(node, ulp);\n return ulp;\n }\n \n void unionBySize(int u, int v) {\n int ulp_u = findUlp(u);\n int ulp_v = findUlp(v);\n \n if (ulp_u != ulp_v) {\n if (size.get(ulp_u) > size.get(ulp_v)) {\n parent.put(ulp_v, ulp_u);\n size.put(ulp_u, size.get(ulp_u) + size.get(ulp_v));\n } else {\n parent.put(ulp_u, ulp_v);\n size.put(ulp_v, size.get(ulp_u) + size.get(ulp_v));\n }\n \n count-- ;\n }\n }\n }\n \n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n while (t-- > 0) {\n int n = sc.nextInt();\n int m = sc.nextInt();\n \n DisjointSet ds = new DisjointSet(n);\n \n for (int i = 0; i < m; i++) {\n int a = sc.nextInt();\n int d = sc.nextInt();\n int k = sc.nextInt();\n \n int prev = a;\n ds.findUlp(prev);\n \n for (int j = 1; j <= k; j++) {\n int next = a + j * d;\n if (next > n) break;\n ds.findUlp(next);\n ds.unionBySize(prev, next);\n prev = next;\n }\n }\n \n \n \n System.out.println(ds.count);\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2020D",
"submission_id": 283618069,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define pb push_back\n#define ppb pop_back\n#define f first\n#define int long long int\n#define sec second\n#define MOD 1e9+7\n#define vi vector<int>\n#define pii pair<int, int>\n#define MOD1 1000000009\n#define MOD2 998244353\n#define PI 3.141592653589793238462\n#define set_bits __builtin_popcountll\nint dx[] = {-1, 0, 1, 0};\nint dy[] = {0, -1, 0, 1};\n#define for_(x) for(int i=0;i<x;i++)\nconst long long inf = 1e18;\ntypedef long long ll;\ntypedef unsigned long long ull;\ntypedef long double lld;\n#define all(v) v.begin(), v.end()\n#define yes cout << \"YES\" << endl\n#define no cout << \"NO\" << endl\n#define rall(x) (x).rbegin(),(x).rend()\n#define fastio() \\\n ios_base::sync_with_stdio(false); \\\n cin.tie(NULL);\n/*---------------------------------------------------------------------------------------------------------------------------*/\nll expo(ll a, ll b, ll mod) {ll res = 1; while (b > 0) {if (b & 1)res = (res * a) % mod; a = (a * a) % mod; b = b >> 1;} return res;}\nvoid extendgcd(ll a, ll b, ll*v) {if (b == 0) {v[0] = 1; v[1] = 0; v[2] = a; return ;} extendgcd(b, a % b, v); ll x = v[1]; v[1] = v[0] - v[1] * (a / b); v[0] = x; return;} //pass an arry of size1 3\nll mminv(ll a, ll b) {ll arr[3]; extendgcd(a, b, arr); return arr[0];} //for non prime b\nll mminvprime(ll a, ll b) {return expo(a, b - 2, b);}\nll combination(ll n, ll r, ll m, ll *fact, ll *ifact) {ll val1 = fact[n]; ll val2 = ifact[n - r]; ll val3 = ifact[r]; return (((val1 * val2) % m) * val3) % m;}\nll phin(ll n) {ll number = n; if (n % 2 == 0) {number /= 2; while (n % 2 == 0) n /= 2;} for (ll i = 3; i <= sqrt(n); i += 2) {if (n % i == 0) {while (n % i == 0)n /= i; number = (number / i * (i - 1));}} if (n > 1)number = (number / n * (n - 1)) ; return number;} //O(sqrt(N))\nll gcd(ll a, ll b) {if (b > a) {return gcd(b, a);} if (b == 0) {return a;} return gcd(b, a % b);}\nbool revsort(ll a, ll b) {return a > b;}\nvoid google(int t) {cout << \"Case #\" << t << \": \";}\nvector<ll> sieve(int n) {int*arr = new int[n + 1](); vector<ll> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}\nll mod_add(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}\nll mod_mul(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}\nll mod_div(ll a, ll b, ll m) {a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m;} \nll mod_sub(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}\nvector<int> decimalToBinary(int n) { vector<int> binary(32, 0); int index = 31; while (n > 0) { int remainder = n % 2; binary[index] = remainder; n /= 2; index--; } reverse(all(binary)); return binary;}\nll binpow(long long a, long long b) { if (b == 0) return 1; long long res = binpow(a, b / 2); if (b % 2)return res * res * a; else return res * res;}\nll binaryToDecimal(const std::vector<int>& binary) { int decimal = 0; int power = 1;for (int i = binary.size() - 1; i >= 0; --i) { decimal += binary[i] * power; power *= 2;}return decimal;}\nll binaryToDecimal(string binary) {long long decimal = 0;long long power = 1;for (long long i = binary.size() - 1; i >= 0; --i) {decimal += (binary[i]-'0') * power; power *= 2;}return decimal;}\n/*---------------------------------------------------------------------------------------------------------------------------*/\n \n#ifndef ONLINE_JUDGE\n#define debug(x) \\\n std::cerr << #x << \" \"; \\\n _print(x); \\\n cerr << endl;\n#else\n#define debug(x)\n \n#endif\nvoid _print(int t) { std::cerr << t; }\nvoid _print(string t) { std::cerr << t; }\nvoid _print(char t) { std::cerr << t; }\nvoid _print(double t) { std::cerr << t; }\nvoid _print(ull t) { std::cerr << t; }\ntemplate <class T, class V> void _print(pair <T, V> p) {cerr << \"{\"; _print(p.f); cerr << \",\"; _print(p.sec); cerr << \"}\";}\ntemplate <class T> void _print(vector <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T> void _print(set <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T> void _print(multiset <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T, class V> void _print(map <T, V> v) {cerr << \"[ \"; for (auto i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\n \n/*\ntry brute force\nuse clear naming\nwrite clear code\nrecheck the logic\nconcentrate\n*/\nvoid solve(){\n int n, c; cin >> n >> c;\n vector<int> vec(n);\n for_(n){ \n cin >> vec[i];\n }\n int idxMaxi = max_element(all(vec))-vec.begin();\n int sum = accumulate(all(vec), 0ll);\n int curr = 0;\n for_(n){\n if(i == 0){\n if(vec[idxMaxi] > vec[i]+c) cout << 1 << \" \";\n else cout << 0 << \" \";\n }\n else if(i == idxMaxi && vec[i] > vec[0]+c){\n cout << 0 << \" \";\n }\n else if((vec[i]+curr+c > vec[idxMaxi])||(vec[i]+curr+c == vec[idxMaxi] && i < idxMaxi)){\n cout << i << \" \";\n }\n else cout << i+1 << \" \";\n curr+=vec[i];\n }\n cout << endl;\n}\n \nsigned main(){\n\tfastio();\n#ifndef ONLINE_JUDGE\n\tfreopen(\"input.txt\", \"r\", stdin);\n\tfreopen(\"output.txt\", \"w\", stdout);\n\tfreopen(\"error.txt\", \"w\", stderr);\n#endif\n \nint n; cin >> n;\nwhile(n--)\n solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 289936998,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\n#include \"bits/stdc++.h\" \nusing namespace std; \n#define FOR(a, c) for (int(a) = 0; (a) < (c); (a)++) \n#define FORL(a, b, c) for (int(a) = (b); (a) <= (c); (a)++) \n#define FORR(a, b, c) for (int(a) = (b); (a) >= (c); (a)--) \n#define dbg(x) cout << #x << '=' << x << endl;\n#define int long long\n#define inc(v) sort(v.begin() , v.end()) \n#define dec(v) sort(v.rbegin() , v.rend()) \ntypedef long double ld; \ntypedef vector<int> vi; \ntypedef pair<int, int> pi; \n#define F first \n#define S second \n#define pb push_back \n#define pob pop_back \n#define mp make_pair \n \nvoid solve()\n{\n int n , k ;\n cin >> n >> k ;\n int x = -1 , j = -1 ;\n vi v(n);\n for(int i=0 ; i<n ; i++){\n cin >> v[i];\n int temp = x ;\n x = max(x,v[i]);\n if(x>temp){\n j = i ;\n }\n }\n vi a;\n v[0] += k ;\n if(v[0]>=x){\n a.pb(0);\n }else{\n a.pb(1);\n }\n int t = 1 ;\n for(int i=1 ; i<n ; i++){\n if(i == j){\n if(a[0] == 1){\n a.pb(0);\n }else{\n a.pb(t);\n }\n t++;\n continue;\n }\n a.pb(t);\n t++;\n }\n \n for(int i=0 ; i<n ; i++){\n cout << a[i] << \" \";\n }\n cout << '\\n';\n}\n \nint32_t main() \n{ \n ios::sync_with_stdio(0); \n cin.tie(0);\n cout.tie(0); \n int T; \n cin >> T; \n while (T--) { \n solve();\n } \n return 0; \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 297607230,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define fast \\\n ios_base::sync_with_stdio(false); \\\n cin.tie(NULL); \\\n cout.tie(NULL);\n#define int long long\n#define double long double\n#define pb push_back\n#define pii pair<int, int>\n#define vi vector<int>\n#define all(x) (x).begin(), (x).end()\n#define fr(i, n) for (int i = 1; i <= n; i++)\n \nconst int M = 1e9 + 7;\n \nvoid solve()\n{\n // Your code here\n int n, c;\n cin >> n >> c;\n vi nums(n + 1, 0);\n fr(i, n) cin >> nums[i];\n vi pre_sum(n + 1);\n nums[1] += c;\n fr(i, n) pre_sum[i] = pre_sum[i - 1] + nums[i];\n map<int, vi> mp;\n fr(i, n) mp[nums[i]].pb(i);\n int resev = *mp[*max_element(all(nums))].begin();\n int max_ = *max_element(all(nums));\n int a = lower_bound(all(pre_sum), max_) - pre_sum.begin();\n fr(i, n)\n {\n int ans = (i == resev) ? 0 : max(i - 1, a - 1);\n cout << ans << \" \";\n }\n cout << endl;\n}\n \nsigned main()\n{\n fast;\n int t;\n cin >> t;\n while (t--)\n {\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266067619,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\ndef ii():\n return int(input())\n \n \ndef I():\n return input()\n \n \ndef lmii():\n return list(map(int, input().split()))\n \n \ndef mii():\n return map(int, input().split())\n \n \ndef lmsi():\n return list(map(str, input().split()))\n \n \n# import copy\nimport math\n \n# M=1e9+7\n# import sys\nmod = 10**9 + 7\nM = 998244353\n \n# input = sys.stdin.readline\n \n \n# for _ in range(int(input())):\n# n, k = mii()\n# mx = 0\n# for i in range(1, n + 1):\n# mx += abs(n - i * 2 + 1)\n# if k % 2 or k > mx:\n# print(\"No\")\n# continue\n# res = [i for i in range(1, n + 1)]\n# sw = 2 * (n - 1)\n# n_swp = k // 2\n# j = n - 1\n# for i in range(n_swp):\n# if j <= 1:\n# j = n - 1\n# while j > 1 and n_swp > 0:\n# if abs(res[j] - j - 1) + abs(res[j - 1] - j) < abs(\n# res[j - 1] - j - 1\n# ) + abs(res[j] - j):\n# n_swp -= 1\n# res[j], res[j - 1] = res[j - 1], res[j]\n# j -= 1\n# print(\"Yes\")\n# print(*res)\n \nfor _ in range(ii()):\n n, c = mii()\n a = lmii()\n mx = max(a)\n ind = a.index(max(a))\n if ind == 0 or c + a[0] >= max(a):\n res = [i for i in range(n)]\n else:\n temp = c\n ind1 = -1\n for i in range(ind):\n temp += a[i]\n if temp >= mx:\n ind1 = i\n break\n # print(ind1, \"ind\")\n res = []\n if ind1 != -1:\n for i in range(ind1):\n res.append(i + 1)\n res.append(res[-1])\n for i in range(ind1 + 1, ind):\n res.append(ind - i + 1)\n res.append(0)\n for i in range(len(res), n):\n res.append(i)\n else:\n for i in range(ind):\n res.append(i + 1)\n res.append(0)\n for i in range(len(res), n):\n res.append(i)\n print(*res)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266070642,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\ndef solve():\n n, c = map(int, input().split())\n a = list(map(int,input().split()))\n if n == 1:\n print(0)\n return\n ans = []\n b = []\n for i in range(n):\n if a[i] == 0: continue\n b.append(a[i])\n ii = a.index(max(a))\n w = max(a)\n if min(b)+c >= w:\n ii = a.index(min(b))\n # print(w,ii)\n for i in range(n):\n if i == ii:\n ans.append(0)\n elif c + a[i] >= w:\n ans.append(i)\n else:\n ans.append(i+1)\n c+= a[i]\n print(*ans)\n \n \nfor _ in range(int(input())):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266212643,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\nimport os\nimport sys\nfrom io import BytesIO, IOBase\n \nBUFSIZE = 8192\n \n \nclass FastIO(IOBase):\n newlines = 0\n \n def __init__(self, file):\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n \n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n \n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n \n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\n \n \nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\n \n \nsys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\ninput = lambda: sys.stdin.readline().rstrip(\"\\r\\n\")\n#######################################\n# import sys,threading,math,functools\n# sys.setrecursionlimit(600000)\n# threading.stack_size(10**8)\ndef primes(n):\n ans=[]\n for i in range(2,int(n**0.5)+1):\n if n%i==0:\n ans.append(i)\n while(n%i==0):\n n//=i \n if n==1:\n break \n if n!=1:\n ans.append(n)\n return ans\n \ndef countSetBits(n):\n ans=0\n while(n):\n ans+=n%2\n n//=2\n return ans\n \ndef factors(n):\n ans=[]\n rev=[]\n for i in range(1,int(n**0.5)+1):\n if n%i==0:\n if i*i==n:\n ans.append(i)\n continue\n ans.append(i)\n rev.append(n//i)\n \n return ans+rev[::-1]\ndef factorial(x,mod):\n if x==1 or x==0 or x>=mod:\n return 1 \n ans=1\n for i in range(2,x+1):\n ans*=i \n if ans%mod==0:\n return 0\n ans%=mod \n return ans\n return ((x%mod)*factorial(x-1,mod))%mod \n \ndef primeFactorsRepeated(number):\n prime_factors=[]\n i=2\n while(i*i<=number and number!=1):\n while(number%i==0):\n # print(number,i)\n prime_factors.append(i)\n number//=i \n i+=1\n if number!=1:\n prime_factors.append(number)\n return prime_factors\n \ndef primeFactors(number):\n prime_factors=set()\n i=2\n while(i*i<=number and number!=1):\n while(number%i==0):\n # print(number,i)\n prime_factors.add(i)\n number//=i \n i+=1\n if number!=1:\n prime_factors.add(number)\n return list(prime_factors) \n \n \ndef primes(num):\n prime = [True for i in range(num+1)]\n p = 2\n while (p * p <= num):\n if (prime[p] == True):\n for i in range(p * p, num+1, p):\n prime[i] = False\n p += 1\n return prime\ndef ncr(n, r, p):\n num = den = 1\n for i in range(r):\n num = (num * (n - i)) % p\n den = (den * (i + 1)) % p\n return (num * pow(den, p - 2, p)) % p\ndef nonCoPrimes(n):\n res = n\n i = 2\n while i*i<=n:\n if n%i==0:\n res/=i\n res*=(i-1)\n \n while n%i==0:\n n/=i\n i+=1\n if n>1:\n res/=n\n res*=(n-1)\n return int(res)\n \n \ndef next_permutation(s):\n arr=[i for i in s]\n # print(arr)\n N=len(arr)\n ind = 0\n l = [i for i in arr]\n # step 1 : find the index from right from where changes will happen\n # it is where first the incresing order is breaking from right to left in last\n index=N-1\n while(index>0 and l[index]<=l[index-1]):\n index-=1\n # print(index)\n # now either index is greater than its before element or it is 0\n # if index==0: just reverse the string so wait for step-3\n if index:\n index-=1\n # step-2: find next greatest number from right and swap\n for temp in range(N-1,index,-1):\n if l[temp]>l[index]:\n break\n # print(temp,index)\n l[temp],l[index]=l[index],l[temp]\n left=index+1\n else:\n left=0\n # print(\"l: \",l)\n \n # step-3: reverse the decreasing order to get the smallest string after index\n right=N-1\n while(left<right):\n l[left],l[right]=l[right],l[left]\n left+=1\n right-=1\n # print(\"l2 : \",l)\n \n return l\n \nclass fenvik_tree:\n \n def __init__(self,n):\n self.arr=[0]*(n+1)\n self.len=n\n def add(self,ind,v):\n ind+=1\n while(ind<=self.len):\n self.arr[ind]+=v\n ind+= ind & (-ind)\n def query(self,ind):\n ind+=1\n ans=0\n while(ind):\n ans+=self.arr[ind]\n ind-= ind & (-ind)\n return ans\n \n \n# python a.py <inputf.in>outputf.in\n# list(map(int,input().split()))\n# from heapq import heappush, heappop\n \nimport math\nfrom collections import defaultdict\nfrom bisect import bisect_left,bisect_right\nmod=10**9+7\nfor _ in range(int(input())):\n n,c = map(int,input().split())\n arr = list(map(int,input().split()))\n arr[0]+=c\n dp=[arr[i] for i in range(n)]\n maxi,ind=arr[0],0\n for i in range(1,n):\n dp[i]+=dp[i-1]\n if arr[i]>maxi:\n maxi,ind=arr[i],i\n for i in range(n):\n if ind==i:\n print(0)\n elif ind<i:\n print(i)\n else:\n ans=i+1\n if dp[i]>maxi:\n ans=i\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 283378564,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\nimport java.util.*;\n \npublic class ProblemD {\n \n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n while (t-- > 0) {\n int n = sc.nextInt();\n long c = sc.nextLong();\n long arr[] = new long[n];\n for(int i=0;i<n;i++){\n arr[i] = sc.nextLong();\n }\n \n arr[0]+=c;\n \n long maxi = 0;\n long maxIndex = 0;\n for(int i=0;i<n;i++) {\n if(arr[i]>maxi){\n maxi = arr[i];\n maxIndex = i;\n }\n }\n \n long prefixSum = 0;\n for(int i=0;i<n;i++){\n prefixSum+=arr[i];\n \n if(i>maxIndex)\n System.out.print(i+\" \");\n else if(i==maxIndex)\n System.out.print(0+\" \");\n else{\n if(prefixSum>maxi)\n System.out.print(i +\" \");\n else\n System.out.print((i+1) +\" \");\n }\n }\n \n System.out.println();\n }\n }\n \n}\n \n/*\n 5,1,10,7\n if index<maxIndex\n if(prefixSum>max)\n ans = index\n else\n ans = index+1\n else if index>maxIndex\n ans = index+1\n else\n ans = 0\n \n 8,10,7,1\n 1,0,2,3\n \n 2,2,2,3,3,3\n 1,1\n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266787570,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\nimport java.util.*;\nimport java.awt.List;\nimport java.io.*;\n \npublic class Vikash_45{\n\tstatic FastScanner sc = new FastScanner();\n\tstatic PrintWriter out = new PrintWriter(System.out);\n\t\n\tpublic static void main(String[] args) {\n\t\tint t = sc.nextInt();\n\t while(t-->0) {\n\t \tvk45();\n\t \tout.flush();\n\t }\n\t out.close();\n\t}\n\t\n\t\n\t\n\t\n\t\n/*-----------------------------------------------------------------------------------------------------------------------------*/ \t\n\t\n \n\t\n\tstatic void vk45() {\n\t\tint n = sc.nextInt();\n\t\tint c = sc.nextInt();\n\t\t\n\t\tlong arr[] = sc.readArrayLong(n);\n\t\tarr[0]+=c;\n\t\t\n\t\tlong left[] = new long[n];\n\t\tleft[n-1] = arr[n-1];\n\t\tfor(int i=n-2; i>=0; i--) {\n\t\t\tleft[i] = Math.max(arr[i],left[i+1]);\n\t\t}\n\t\t\n//\t\tout.println(Arrays.toString(left));\n\t\t\n\t\tlong right[] = new long[n];\n\t\tright[0] = arr[0];\n\t\tfor(int i=1; i<n; i++) {\n\t\t\tright[i] = Math.max(arr[i],right[i-1]);\n\t\t}\n\t\t\n//\t\tout.println(Arrays.toString(right));\n\t\tlong sum = 0;\n\t\tfor(int i=0; i<n; i++) {\n\t\t\tsum+=arr[i];\n\t\t\tlong l = 0;\n\t\t\tlong r = 0;\n\t\t\tif(i-1>=0) {\n\t\t\t\tif(right[i-1]>=arr[i]) {\n\t\t\t\t\tr = 1;\n\t\t\t\t}\n\t\t\t}\n\t\t\tif(i+1<n) {\n\t\t\t\tif(left[i+1]>arr[i]) {\n\t\t\t\t\tl = 1;\n\t\t\t\t}\n\t\t\t}\n\t\t\t\n\t\t\tif(r==0) {\n\t\t\t\tif(l==0) {\n\t\t\t\t\tout.print(\"0 \");\n\t\t\t\t}else {\n\t\t\t\t\tout.print(\"1 \");\n\t\t\t\t}\n\t\t\t}else {\n\t\t\t\tif(i+1<n && sum<left[i+1]) {\n\t\t\t\t\tout.print((i+1)+\" \");\n\t\t\t\t}else {\n\t\t\t\t\tout.print(i+\" \");\n\t\t\t\t}\n\t\t\t\t\n\t\t\t}\n\t\t}\n\t\t\n\t\t\n\t\tout.println();\n\t\t\n\t}\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\tstatic class pair implements Comparable<pair>{\n\t\tint a;\n\t\tint b;\n\t\tint c;\n\t\tpublic pair(int a,int b,int c) {\n\t\t\tthis.a = a;\n\t\t\tthis.b = b;\n\t\t\tthis.c = c;\n\t\t}\n\t\t@Override\n\t\tpublic int compareTo(pair p) {\n\t\t\tif(this.a==p.a) {\n\t\t\t\treturn this.b-p.b;\n\t\t\t}\n\t\t\treturn this.a-p.a;\n\t\t}\n\t}\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n \n/*-----------------------------------------------------------------------------------------------------------------------------*/\n\t\n\tstatic class FastScanner {\n\t\tBufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n\t\tStringTokenizer st = new StringTokenizer(\"\");\n\t\tString next() {\n\t\t\twhile (!st.hasMoreTokens())\n\t\t\t\ttry {\n\t\t\t\t\tst = new StringTokenizer(br.readLine());\n\t\t\t\t} catch (IOException e) {\n\t\t\t\t\te.printStackTrace();\n\t\t\t\t}\n\t\t\treturn st.nextToken();\n\t\t}\n\t\tint nextInt() {\n\t\t\treturn Integer.parseInt(next());\n\t\t}\n\t\tint[] readArray(int n) {\n\t\t\tint[] a = new int[n];\n\t\t\tfor (int i = 0; i < n; i++)\n\t\t\t\ta[i] = nextInt();\n\t\t\treturn a;\n\t\t}\n\t\tlong[] readArrayLong(int n) {\n\t\t\tlong[] a = new long[n];\n\t\t\tfor (int i = 0; i < n; i++)\n\t\t\t\ta[i] = nextLong();\n\t\t\treturn a;\n\t\t}\n\t\tlong nextLong() {\n\t\t\treturn Long.parseLong(next());\n\t\t}\n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266465818,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define FOR(a, c) for (int(a) = 0; (a) < (c); (a)++)\n#define FORL(a, b, c) for (int(a) = (b); (a) <= (c); (a)++)\n#define FORR(a, b, c) for (int(a) = (b); (a) >= (c); (a)--)\ntypedef long long int ll;\ntypedef vector<int> vi;\ntypedef vector<long long> vl;\ntypedef pair<int, int> pi;\n#define read(v) FOR(i,v.size()){cin>>v[i];}\nint main()\n{\n ios::sync_with_stdio(0);\n cin.tie(0);\n int t;\n cin>>t;\n while(t--){\n int n;\n ll c;\n cin>>n>>c;\n vl a(n);\n read(a);\n ll mx=*max_element(a.begin()+1,a.end());\n ll mxc=max(a[0]+c,mx);\n ll win,sum=c;\n if(mxc==a[0]+c){\n win=0;\n }\n else{\n win=find(a.begin()+1,a.end(),mx)-a.begin();\n }\n FOR(i,n){\n int ans;\n if(i==win){\n ans=0;\n }\n else if(sum+a[i]>=mx){\n ans=i;\n }\n else{\n ans=i+1;\n }\n cout<<ans<<\" \";\n sum+=a[i];\n }\n cout<<endl;\n \n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266365628,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define endl '\\n'\n#define ll long long\n#define int ll\nusing namespace std;\nconstexpr int N=2e5+7;\nconstexpr int M=2e3+7;\nint n,c;\nint a[N],f[N][20];;\nint ans[N];\nint presum[N];\nint sufmax[N];\nint query(int l,int r)\n{\n\tint len=r-l+1;\n\tint k=log(len)/log(2);\n\tint res=max(f[l][k],f[r-(1<<k)+1][k]);\n\treturn res;\n}\nbool check(int mid,int idx)\n{\n\tif(mid==idx) return 0;\n\tif(query(mid,idx-1)>=a[idx]) return 1;\n\t//if(presum[mid-1]+c+a[mid]>=a[idx]) return 1;\n\treturn 0;\n}\nbool check2(int mid,int idx)\n{\n\tif(presum[mid]>a[idx]) return 0;\n\tif(presum[mid]==a[idx]) return 1;\n\treturn 1;\n}\nvoid solve()\n{\n\tcin>>n>>c;\n\tfor(int i=1;i<=n;i++)\n\t{\n\t\tcin>>a[i];\n\t\tpresum[i]=presum[i-1]+a[i];\n\t}\n\tfor(int i=1;i<=n;i++)\n\t{\n\t\tpresum[i]+=c;\n\t}\n\tsufmax[n+1]=0;\n\tfor(int i=n;i>=1;i--)\n\t{\n\t\tsufmax[i]=max(sufmax[i+1],a[i]);\n\t}\n\tfor(int j=0;j<20;j++)\n\t{\n\t\tfor(int i=1;i+(1<<j)-1<=n;i++)\n\t\t{\n\t\t\tif(j==0) f[i][j]=a[i];\n\t\t\telse f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);\n\t\t}\n\t}\n\t\n\tfor(int i=1;i<=n;i++)\n\t{\n\t\tint l=1,r=i;\n\t\t// 找小于a[i]的位置\n\t\twhile(l<r)\n\t\t{\n\t\t\tint mid=(l+r)>>1;\n\t\t\tif(check(mid,i)) l=mid+1;\n\t\t\telse r=mid;\n\t\t}\n\t\t//cout<<\"l: \"<<l<<endl;\n \n\t\tif(l==i)\n\t\t{\n\t\t\t//cout<<l<<' '<<i<<endl;\n\t\t\tans[i]=i-1;\n\t\t\t//if(i==1) cout<<ans[i]<<\"******\"<<endl;\n\t\t}\n\t\telse\n\t\t{\n\t\t\t// 找比a[i]小的位置中 [l,i-1] 中小于a[i]的presum的值\n\t\t\tint j=1;\n\t\t\tfor(j=l;j<=i-1;j++)\n\t\t\t{\n\t\t\t\tif(presum[j]<a[i]) break;\n\t\t\t\t//if(i==3) cout<<\"j: \"<<j<<endl;\n\t\t\t}\n\t\t\tans[i]=j-1;\n\t\t\t//if(i==3) cout<<\"j: \"<<j<<\"l: \"<<l<<endl;\n\t\t}\n \n \n\t\tint sum=0;\n\t\tif(ans[i]==i-1)\n\t\t{\n\t\t\tsum=(a[i]+presum[i-1]);\n\t\t\tif(i==1) sum+=c;\n\t\t}\n\t\telse sum=a[i];\n\t\tif(sufmax[i+1]<=sum) continue;\n\t\tans[i]++;\n\t}\n \n\tfor(int i=1;i<=n;i++) cout<<ans[i]<<' ';\n\tcout<<endl;\n}\nsigned main()\n{\n\tios::sync_with_stdio(0);cin.tie(0);cout.tie(0);\n\tint T=1; cin>>T;\n\twhile(T--){solve();}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 292895058,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\nimport bisect\nfrom collections import Counter\n \nfor _ in range(int(input())):\n n, c = map(int, input().split())\n a = list(map(int, input().split()))\n ans = []\n a_sort = a.copy()\n a_sort.sort()\n a_sort_pre = [0]\n for i in range(n):\n a_sort_pre.append(a_sort_pre[i - 1] + a[i])\n cc = Counter()\n for i in range(n):\n if i == 0:\n if a[i] + c < a_sort[-1]:\n j = n - 1\n s = a[i] + c\n res = i\n while s < a_sort[j]:\n s += a_sort[j]\n j -= 1\n res += 1\n ans.append(res)\n else:\n ans.append(0)\n elif a[i] == a_sort[-1] and cc[a[i]] == 0 and a[0] + c < a_sort[-1]:\n ans.append(0)\n else:\n if a_sort_pre[i] + c + a[i] < a_sort[-1]:\n j = n - 1\n s = a_sort_pre[i] + c + a[i]\n res = i\n while s < a_sort[j] and j > i:\n s += a_sort[j]\n j -= 1\n res += 1\n ans.append(res)\n else:\n ans.append(i)\n cc[a[i]] += 1\n print(*ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266796346,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\nimport sys\nfrom itertools import accumulate\n \n# region fastio\ninput = lambda: sys.stdin.readline().rstrip()\nsint = lambda: int(input())\nmint = lambda: map(int, input().split())\nints = lambda: list(map(int, input().split()))\n# endregion fastio\n \n# MOD = 998_244_353\n# MOD = 10 ** 9 + 7\n# DIR4 = ((-1, 0), (0, 1), (1, 0), (0, -1)) #URDL\n# DIR8 = ((-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1))\n \ndef solve() -> None:\n n, c = mint()\n nums = ints()\n pres = list(accumulate(nums, initial=0))\n ans = [0] * n\n suff = [0] * (n + 1)\n for i in range(n - 1, -1, -1):\n suff[i] = max(nums[i], suff[i + 1])\n mx = -1\n for i, x in enumerate(nums):\n if i == 0:\n x += c\n y = x\n if y <= mx:\n ans[i] += i\n y += pres[i] + c\n if y < suff[i + 1]:\n ans[i] += 1\n mx = max(mx, x)\n print(*ans)\n \nfor _ in range(int(input())):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266011409,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\nimport java.util.*;\n \npublic class Elections {\n public static void main(String[] args) {\n int i = 0;\n List<String> ll = new ArrayList<>();\n Scanner sc = new Scanner(System.in);\n while (true) {\n if (i == 0) {\n i = Integer.valueOf(sc.nextLine()) * 2;\n } else {\n ll.add(sc.nextLine());\n if (--i == 0) {\n break;\n }\n }\n }\n int p = ll.size();\n for (int k = 0; k < p; k += 2) {\n int c = Integer.valueOf(ll.get(k).split(\" \")[1]);\n String[] ss = ll.get(k + 1).split(\" \");\n int[] arr = new int[ss.length];\n for (int q = 0; q < ss.length; q++) {\n arr[q] = Integer.valueOf(ss[q]);\n }\n elections(arr, c);\n }\n }\n \n static void elections(int[] nums, long c) {\n int n = nums.length;\n if (n == 1) {\n System.out.println(0);\n return;\n }\n TreeSet<int[]> set = new TreeSet<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);\n int[][] arr = new int[n][2];\n for (int i = 0; i < n; i++) {\n arr[i][0] = nums[i];\n arr[i][1] = i;\n set.add(arr[i]);\n }\n long sum = 0;\n TreeSet<int[]> set1 = new TreeSet<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);\n for (int i = 0; i < n; i++) {\n set.remove(arr[i]);\n if (set.size() == 0) {\n if (arr[i][0] > sum + arr[0][0] + c) {\n System.out.print(0 + \" \");\n } else {\n System.out.print(i + \" \");\n }\n break;\n }\n int[] fir = set.first();\n if (i == 0) {\n if (arr[i][0] + c >= fir[0]) {\n System.out.print(0 + \" \");\n } else {\n System.out.print(1 + \" \");\n }\n } else if (i == 1) {\n if (arr[i][0] >= fir[0]) {\n if (arr[i][0] <= arr[0][0] + c) {\n System.out.print(1 + \" \");\n } else {\n System.out.print(0 + \" \");\n }\n } else {\n if (arr[i][0] + arr[0][0] + c >= fir[0]) {\n System.out.print(1 + \" \");\n } else {\n System.out.print(2 + \" \");\n }\n }\n set1.add(arr[i]);\n sum += arr[i][0];\n } else {\n int[] fir1 = set1.first();\n if (arr[i][0] >= fir[0]) {\n if (arr[i][0] <= fir1[0] || arr[i][0] <= arr[0][0] + c) {\n System.out.print(i + \" \");\n } else {\n System.out.print(0 + \" \");\n }\n } else {\n if (arr[i][0] + sum + arr[0][0] + c >= fir[0]) {\n System.out.print(i + \" \");\n } else {\n System.out.print(i + 1 + \" \");\n }\n }\n set1.add(arr[i]);\n sum += arr[i][0];\n }\n }\n System.out.println();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 287767125,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\n// package problems;\n \n \nimport java.util.*;\n \npublic class CommonTemplate {\n static class Pair {\n \n // Pair attributes\n public long index;\n public long value;\n \n // Constructor to initialize pair\n public Pair(long index, long value)\n {\n // This keyword refers to current instance\n this.index = index;\n this.value = value;\n }\n }\n \n static int lcmOfTwoNumber(int a,int b){\n // Checking for the largest\n // Number between them\n int ans = Math.max(a, b);\n \n // Checking for smallest number that\n // can de divided by both numbers\n while (ans % a != 0 || ans % b != 0) {\n ans++;\n }\n return ans;\n }\n static ArrayList<Integer> primeFactors(int n)\n {\n // Print the number of 2s that divide n\n ArrayList<Integer> l1 = new ArrayList<>();\n while (n % 2 == 0) {\n// System.out.print(2 + \" \");\n l1.add(2);\n n /= 2;\n }\n \n // n must be odd at this point. So we can\n // skip one element (Note i = i +2)\n for (int i = 3; i <= Math.sqrt(n); i += 2) {\n // While i divides n, print i and divide n\n while (n % i == 0) {\n// System.out.print(i + \" \");\n l1.add(i);\n n /= i;\n }\n }\n \n // This condition is to handle the case when\n // n is a prime number greater than 2\n if (n > 2)l1.add(n);\n// System.out.print(n);\n return l1;\n }\n \n static Integer[] sortedIndices(long[] array){\n // Create an array of indices\n Integer[] indices = new Integer[array.length];\n for (int i = 0; i < array.length; i++) {\n indices[i] = i;\n }\n \n // Sort indices based on the values in the original array\n Arrays.sort(indices, Comparator.comparingLong(a -> array[a]));\n \n return indices;\n }\n static long gcd(long a, long b)\n {\n return b == 0? a:gcd(b, a % b);\n }\n \n static long LcmOfArray(int[] arr, int idx)\n {\n if (idx == arr.length - 1){\n return arr[idx];\n }\n long a = arr[idx];\n long b = LcmOfArray(arr, idx+1);\n return (a*b/gcd(a,b)); //\n }\n static int mod(int idx,int n){\n if (idx>=n){\n idx-=n;\n }\n if (idx<0){\n idx+=n;\n }\n return idx;\n }\n \n private static int countDigits(long l) {\n if (l >= 1000000000000000000L) return 19;\n if (l >= 100000000000000000L) return 18;\n if (l >= 10000000000000000L) return 17;\n if (l >= 1000000000000000L) return 16;\n if (l >= 100000000000000L) return 15;\n if (l >= 10000000000000L) return 14;\n if (l >= 1000000000000L) return 13;\n if (l >= 100000000000L) return 12;\n if (l >= 10000000000L) return 11;\n if (l >= 1000000000L) return 10;\n if (l >= 100000000L) return 9;\n if (l >= 10000000L) return 8;\n if (l >= 1000000L) return 7;\n if (l >= 100000L) return 6;\n if (l >= 10000L) return 5;\n if (l >= 1000L) return 4;\n if (l >= 100L) return 3;\n if (l >= 10L) return 2;\n return 1;\n }\n static int[] numToNumArray(long x){\n StringBuilder str = new StringBuilder(String.valueOf(x));\n int[] arr = new int[str.length()];\n for (int i = 0; i < str.length(); i++) {\n int ch = str.charAt(i)-'0';\n arr[i] = ch;\n }\n return arr;\n }\n static int trailingZero(long n){\n int ct = 0;\n while (n%10==0){\n ct++;\n n/=10;\n \n }\n return ct;\n }\n static Scanner sc = new Scanner(System.in);\n public static void main(String[] args) {\n int q = sc.nextInt();\n loop1:\n while (q-->0){\n solve();\n }\n }\n public static void solve(){\n int n = sc.nextInt();\n boolean flag1 = true;\n long c = sc.nextLong(), max = Integer.MIN_VALUE;\n long[] arr = new long[n];\n HashMap<Integer,Long> h1= new HashMap<>();\n for(int i=0; i<n;i++) {\n \tarr[i] = sc.nextLong();\n if (i==0){\n h1.put(i,0L);\n }else {\n h1.put(i,max);\n }\n \tmax = Math.max(max,arr[i]);\n \n }\n int idx = 0, jdx = 0;\n for(int i=0; i<n;i++) {\n if (arr[i]==max){\n idx = i;\n }\n if (arr[i]==max&&flag1){\n jdx = i;\n flag1 = false;\n }\n }\n long[] ps = new long[n];\n ps[0] = arr[0];\n for (int i = 1; i <n; i++) {\n ps[i] = ps[i-1]+arr[i];\n }\n long[] brr = new long[n];\n for (int i = 0; i < n; i++) {\n long ct = 0;\n if (i==0||i==jdx){\n if (i==jdx){\n if (arr[i]>h1.get(i)+c){\n ct = 0;\n }else {\n ct = i;\n }\n }else {\n if (ps[i]+c>=max){\n ct+= i;\n }else {\n ct+= i+1;\n }\n }\n }else {\n if (i<jdx){\n if (ps[i]+c>=max){\n ct+= i;\n }else {\n ct+= i+1;\n }\n }else {\n ct = i;\n }\n }\n brr[i] = ct;\n }\n// System.out.println(h1);\n// System.out.println(idx+\" \"+jdx);\n for (long e:brr){\n System.out.print(e+\" \");\n }\n System.out.println();\n }\n public static int binarySearch(long[] arr, long x) {\n // System.out.println(x);\n int left = 0;\n int right = arr.length - 1;\n int result = -1; // Innitialize result to -1\n \n while (left <= right) {\n int mid = left + (right - left) / 2;\n \n // If arr[mid] is less than or equal to x, update result and search in right half\n if (arr[mid] <= x) {\n result = mid;\n left = mid + 1;\n } else {\n // If arr[mid] is greater than x, search in left half\n right = mid - 1;\n }\n }\n \n return result;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266171146,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\nt = int(input())\n \nwhile(t):\n n, c = map(int, input().split())\n a = list(map(int, input().split()[:n]))\n \n if(n == 1):\n print(0)\n elif(n == 2):\n if(a[1] > a[0] + c):\n print(*[1, 0])\n else:\n print(*[0, 1])\n else:\n res = []\n max_val = max(a)\n \n if(a[0] + c >= max_val):\n res.append(0)\n else:\n res.append(1)\n \n if(a[1] == max_val and a[0] < max_val):\n res.append(0)\n else:\n if(a[1] + a[0] + c >= max_val):\n res.append(1)\n else:\n res.append(2)\n \n \n cur_sum = a[0] + a[1]\n cur_max = max(a[0], a[1])\n for i in range(2, n):\n if(a[i] == max_val and a[i] > cur_max + c):\n res.append(0)\n else:\n to_add = i\n if(cur_sum + c + a[i] >= max_val):\n res.append(to_add)\n else:\n res.append(to_add + 1)\n cur_sum += a[i]\n cur_max = max(cur_max, a[i])\n \n \n print(*res)\n \n \n \n \n t -= 1\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266777565,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\nimport sys\nfrom itertools import accumulate\n \n# region fastio\ninput = lambda: sys.stdin.readline().rstrip()\nsint = lambda: int(input())\nmint = lambda: map(int, input().split())\nints = lambda: list(map(int, input().split()))\n# endregion fastio\n \n# MOD = 998_244_353\n# MOD = 10 ** 9 + 7\n# DIR4 = ((-1, 0), (0, 1), (1, 0), (0, -1)) #URDL\n# DIR8 = ((-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1))\n \ndef solve() -> None:\n n, c = mint()\n nums = ints()\n pres = list(accumulate(nums, initial=0))\n pres[1] += c\n ans = [0] * n\n suff = [0] * (n + 1)\n for i in range(n - 1, -1, -1):\n suff[i] = max(nums[i], suff[i + 1])\n pref = [-1] * (n + 1)\n for i, x in enumerate(nums):\n if i == 0:\n x += c\n pref[i + 1] = max(pref[i], x)\n if x <= pref[i] or x < suff[i + 1]:\n ans[i] += i\n x += pres[i]\n if x < suff[i + 1]:\n ans[i] += 1\n s = 0\n stk = []\n for i in range(n - 1, 0, -1):\n while stk and stk[-1] <= nums[i]:\n s -= stk.pop()\n if s + nums[0] + c < nums[i] and pref[i] < nums[i] and len(stk) < ans[i]:\n ans[i] = len(stk)\n stk.append(nums[i])\n s += nums[i]\n print(*ans)\n \nfor _ in range(int(input())):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266022665,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n static PrintWriter pw = new PrintWriter(System.out);\n static Scanner sc = new Scanner(System.in);\n \n public static void solve() throws IOException {\n int n = sc.nextInt(), c = sc.nextInt();\n if(n == 200000){\n while (true){\n \n }\n// pw.println();\n// return;\n }\n Integer a [] = sc.nextIntegerArray(n);\n Integer b [] = a.clone();\n int ans [] = new int [n];\n Arrays.sort(b);\n long prefix [] = new long [n + 1];\n for(int i = 1; i <= n ; i++)\n prefix[i] = prefix[i - 1] + b[i - 1];\n long Max = b[n - 1];\n long cnt = 0;\n for(int i = 0; i < n ; i++){\n int low = 1, high = n;\n if(0l + a[i] + cnt + c >= Max){\n ans[i] = i;\n }\n else {\n while (low <= high) {\n int mid = (low + high) / 2;\n if (0l + a[i] + prefix[n] - prefix[mid - 1] + cnt + c >= Max)\n low = mid + 1;\n else\n high = mid - 1;\n }\n ans[i] = i + n - high + 1;\n }\n cnt += a[i];\n }\n for(int i = 0 ; i < n ; i++){\n if(a[i] + c >= Max){\n ans[i] = 0;\n break;\n }\n }\n for(int i : ans)\n pw.print(i+\" \");\n pw.println();\n }\n \n \n public static void main(String[] args) throws IOException{\n int t = sc.nextInt();\n while(t -- > 0)\n solve();\n pw.flush();\n }\n \n static class Scanner {\n StringTokenizer st;\n BufferedReader br;\n \n public Scanner(InputStream s) {\n br = new BufferedReader(new InputStreamReader(s));\n }\n \n public Scanner(String s) throws FileNotFoundException {\n br = new BufferedReader(new FileReader(s));\n }\n \n public String next() throws IOException {\n while (st == null || !st.hasMoreTokens())\n st = new StringTokenizer(br.readLine());\n return st.nextToken();\n }\n \n public int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n \n public long nextLong() throws IOException {\n return Long.parseLong(next());\n }\n \n public String nextLine() throws IOException {\n return br.readLine();\n }\n \n public double nextDouble() throws IOException {\n String x = next();\n StringBuilder sb = new StringBuilder(\"0\");\n double res = 0, f = 1;\n boolean dec = false, neg = false;\n int start = 0;\n if (x.charAt(0) == '-') {\n neg = true;\n start++;\n }\n for (int i = start; i < x.length(); i++)\n if (x.charAt(i) == '.') {\n res = Long.parseLong(sb.toString());\n sb = new StringBuilder(\"0\");\n dec = true;\n } else {\n sb.append(x.charAt(i));\n if (dec)\n f *= 10;\n }\n res += Long.parseLong(sb.toString()) / f;\n return res * (neg ? -1 : 1);\n }\n \n public long[] nextlongArray(int n) throws IOException {\n long[] a = new long[n];\n for (int i = 0; i < n; i++)\n a[i] = nextLong();\n return a;\n }\n \n public Long[] nextLongArray(int n) throws IOException {\n Long[] a = new Long[n];\n for (int i = 0; i < n; i++)\n a[i] = nextLong();\n return a;\n }\n \n public int[] nextIntArray(int n) throws IOException {\n int[] a = new int[n];\n for (int i = 0; i < n; i++)\n a[i] = nextInt();\n return a;\n }\n \n public Integer[] nextIntegerArray(int n) throws IOException {\n Integer[] a = new Integer[n];\n for (int i = 0; i < n; i++)\n a[i] = nextInt();\n return a;\n }\n \n public boolean ready() throws IOException {\n return br.ready();\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266003679,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.util.*;\n \npublic class b2\n{\n static long MOD = 998244353;\n public static void main (String[] args) throws IOException {\n FastReader sc = new FastReader();\n int t = sc.nextInt();\n \n loop: for (int z = 0; z < t; z++) {\n int n = sc.nextInt();\n long c = sc.nextInt();\n long[] a = new long[n];\n long[] sort = new long[n];\n for (int i = 0; i < n; i++) {\n a[i] = sc.nextLong();\n sort[i] = a[i];\n }\n \n TreeMap<Long, Integer> map = new TreeMap<>();\n Arrays.sort(sort);\n ArrayList<Integer> count = new ArrayList<>();\n count.add(1);\n map.put(sort[0], 0);\n for (int i = 1; i < n; i++) {\n if(sort[i] == sort[i-1]){\n count.set(count.size()-1, count.get(count.size()-1) + 1);\n }else{\n map.put(sort[i], count.size());\n count.add(1);\n }\n }\n map.put(Long.MAX_VALUE, Integer.MAX_VALUE);\n \n long[] counts = new long[count.size()];\n for (int i = 0; i < count.size(); i++) {\n counts[i] = count.get(i);\n }\n \n// map.forEach((i,j) -> System.out.println(i + \" ; \" + j));\n \n FenwickTree ft = new FenwickTree(counts);\n \n long sumFromPrev = c;\n long mostPrev = a[0] + c;\n StringBuilder ans = new StringBuilder();\n for (int i = 0; i < n; i++) {\n long kos2 = Integer.MAX_VALUE, kos1 = Integer.MAX_VALUE;\n if(a[i] > mostPrev){\n int ind = map.higherEntry(a[i]).getValue();\n if(ind == Integer.MAX_VALUE){\n kos1 = 0;\n }else {\n kos1 = ft.query(ind + 1, n);\n }\n }\n \n kos2 = i;\n long votes = a[i] + sumFromPrev;\n// System.out.println(\"votes = \" + votes);\n int ind = map.higherEntry(votes).getValue();\n if(ind != Integer.MAX_VALUE && ft.query(ind + 1, counts.length) > 0){\n kos2 ++;\n }\n \n \n ft.update(map.get(a[i]) + 1, -1);\n sumFromPrev += a[i];\n mostPrev = Math.max(mostPrev, a[i]);\n ans.append(Math.min(kos1, kos2) + \" \");\n }\n \n System.out.println(ans);\n }\n }\n //One Indexed, Also the update just adds and subtracts\n static class FenwickTree {\n int size;\n long ft[];\n public FenwickTree(int n) {\n this.ft = new long[n+1];\n this.size = n;\n }\n public FenwickTree(long[] arr) {\n this(arr.length);\n for (int i = 0; i < arr.length; i++) {\n update(i+1, arr[i]);\n }\n }\n \n void update(int p, long val) {\n for (; p <= size; p += p&(-p))\n ft[p] += val;\n }\n \n long query(int p) {\n long sum = 0;\n for (; p > 0; p -= p&(-p)) sum += ft[p];\n return sum;\n }\n \n long query(int a, int b) {\n return query(b) - query(a-1);\n }\n }\n static final long mod_inverse(long num) {\n return binExp(num, MOD - 2);\n }\n \n static final long binExp(long base, long exp) {\n base %= MOD;\n long result = 1;\n while (exp > 0) {\n if ((exp & 1) > 0) {\n result *= base;\n result %= MOD;\n }\n base *= base;\n base %= MOD;\n exp >>= 1;\n }\n return result;\n }\n \n static class FastReader{\n BufferedReader br;\n StringTokenizer st;\n \n public FastReader(){\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() throws IOException {\n while(st == null || !st.hasMoreElements()){\n st = new StringTokenizer(br.readLine());\n }\n return st.nextToken();\n }\n \n int nextInt() throws IOException {return Integer.parseInt(next());}\n long nextLong() throws IOException {return Long.parseLong(next());}\n double nextDouble() throws IOException {return Double.parseDouble(next());}\n \n String nextLine() throws IOException {\n String str = \"\";\n if(st.hasMoreTokens()){\n str = st.nextToken(\"\\n\");\n }\n else{\n str = br.readLine();\n }\n return str;\n }\n \n int[] intArray(int n) throws IOException {\n int[] arr = new int[n];\n \n for (int i = 0; i < n; i++) {\n arr[i] = nextInt();\n }\n \n return arr;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266046174,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define Paddi ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)\nconst int N = 5e5 + 10;\nconst int LOG = 20;\nint s[N];\nint pre[N];\nint mapre[N];\nint masuf[N];\nint n, c;\nvoid init()\n{\n for (int i = 0; i <= n + 5; i++)\n s[i] = 0, pre[i] = 0, mapre[i] = 0, masuf[i] = 0;\n}\nsigned main()\n{\n // Paddi;\n int T;\n cin >> T;\n while (T--)\n {\n cin >> n >> c;\n init();\n for (int i = 1; i <= n; i++)\n cin >> s[i];\n for (int i = 1; i <= n; i++)\n pre[i] = pre[i - 1] + s[i];\n for (int i = 1; i <= n; i++)\n mapre[i] = max(mapre[i - 1], s[i]);\n for (int i = n; i >= 1; i--)\n masuf[i] = max(masuf[i + 1], s[i]);\n vector<int> ans(n + 10);\n for (int i = 1; i <= n; i++)\n {\n int pre_max = mapre[i - 1];\n int suf_max = masuf[i + 1];\n int pre_sum = pre[i - 1];\n if (i == 1)\n {\n if (c + s[1] >= suf_max)\n ans[1] = 0;\n else\n ans[1] = 1;\n }\n else\n {\n if (pre_max < s[i])\n {\n pre_max = max(pre_max, s[1] + c);\n if (pre_max < s[i])\n {\n if (suf_max <= s[i])\n ans[i] = 0;\n else\n {\n int t = pre_sum + c + s[i];\n if (t >= suf_max)\n ans[i] = i - 1;\n else\n ans[i] = i;\n }\n }\n else\n {\n int t = pre_sum + c + s[i];\n if (t >= suf_max)\n ans[i] = i - 1;\n else\n ans[i] = i;\n }\n }\n else\n {\n int t = pre_sum + c + s[i];\n if (t >= suf_max)\n ans[i] = i - 1;\n else\n ans[i] = i;\n }\n }\n }\n for (int i = 1; i <= n; i++)\n cout << ans[i] << \" \";\n cout << endl;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 272239569,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Elections\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.\n\nThe candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.\n\nYou found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.\n\nYou are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 2 \\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains two integers $n$ and $c$ ($1 \\le n \\le 2 \\cdot 10^5$, $0 \\le c \\le 10^9$) — the number of candidates in the elections and the number of undecided people.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($0 \\le a_i \\le 10^9$) — the number of fans for each candidate.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.\n\nExamples:\ninput:\n5\n3 1\n2 0 3\n2 3\n0 10\n5 3\n5 4 3 2 1\n4 5\n3 10 7 1\n6 0\n2 2 2 3 3 3\noutput:\n0 1 2\n1 0\n0 1 2 3 4\n1 0 2 3\n1 1 2 0 4 5\n\nNote: In the first test case:\n\n * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$. * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$. * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.\n\nIn the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.\n\nBuggy Code:\n#include <iostream>\n#include <cstring>\n#include <algorithm>\n \nusing namespace std;\ntypedef long long ll;\nconst int N = 3e5;\nconst int M = 25;\nconst int INF = 0x3f3f3f3f;\n \nint n, m, ans ,_;\nint w[N] ,f[N];\nbool cmp ( int x ,int y){ return x>y;}\nint main() {\n ios_base::sync_with_stdio(0),cin.tie(0);\n cin>>_;\n while(_--)\n {\n cin>>n>>m;\n memset(f,0,sizeof f);\n memset(w,0,sizeof w);\n w[1] = m;\n \n int mav = 0;\n for(int i=1;i<=n;i++)\n {\n int p;\n cin>>p;\n w[i] += p;\n if(mav <w[i]){\n mav = w[i];\n }\n f[i] += w[i] + f[i-1];\n }\n bool flag = true;\n for(int i=1;i<=n;i++)\n {\n if(w[i]==mav && flag){\n cout<<0<< \" \";\n flag =false;\n }else{\n if(f[i] >= mav){\n cout<<i-1<<\" \";\n }else{\n cout<<i<<\" \";\n }\n }\n \n }\n cout<<endl;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1978D",
"submission_id": 266806817,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include <cstdio>\n#include <algorithm>\n#include <initializer_list>\nusing ll=long long;\nconstexpr int N=200010,V=2010;\nconstexpr ll INF=0x3fffffffffffffff;\nint T,Ch,n,m,Q,i,j,x,a[N];\nll c[N],f[V][V],ans[V];\nchar op;\ntemplate<typename T>\ninline void r(T&x){\n\twhile((Ch=getchar())<48);\n\tx=Ch&15;\n\twhile((Ch=getchar())>47)x=(x<<1)+(x<<3)+(Ch&15);\n}\ninline void r(char&x){while((x=getchar())<33);}\ntemplate<typename...Ar>\ninline void r(Ar&&...x){std::initializer_list<int>{(r(x),0)...};}\ntemplate<typename T>\ninline void w(T x){\n\tstatic char stk[64],*top=stk;\n\tdo{\n\t\t*++top=x%10;\n\t\tx/=10;\n\t}while(x);\n\tdo putchar(*top--|48);while(top!=stk);\n\tputchar(32);\n}\ntemplate<typename T,int N>\nclass BIT{\n\tT t[N];\n\tint n;\npublic:\n\tvoid init(const int&x){\n\t\tn=x;\n\t\tstd::fill(t+1,t+n+1,-INF);\n\t}\n\tT query(int x)const{\n\t\tT ans=-INF;\n\t\twhile(x){\n\t\t\tans=std::max(ans,t[x]);\n\t\t\tx&=x-1;\n\t\t}\n\t\treturn ans;\n\t}\n\tvoid modify(int x,const T&v){\n\t\twhile(x<=n){\n\t\t\tt[x]=std::max(t[x],v);\n\t\t\tx+=x&-x;\n\t\t}\n\t}\n};\nBIT<ll,V>t1[V],t2[V];\nint main(){\n\tr(T);\n\twhile(T--){\n\t\tr(n,Q,m);\n\t\tfor(i=1;i<=n;i++){\n\t\t\tr(c[i]);\n\t\t\ta[i]=0;\n\t\t}\n\t\twhile(Q--){\n\t\t\tr(op,i);\n\t\t\tif(op=='L'){\n\t\t\t\ta[1]++;\n\t\t\t\ta[i+1]--;\n\t\t\t}\n\t\t\telse a[i]++;\n\t\t}\n\t\tfor(i=1;i<=n;i++)a[i]+=a[i-1];\n\t\tx=0;\n\t\tfor(i=1;i<=n;i++){\n\t\t\tif(a[i]>m)continue;\n\t\t\tif(a[i]==a[x])c[x]+=c[i];\n\t\t\telse{\n\t\t\t\ta[++x]=a[i];\n\t\t\t\tc[x]=c[i];\n\t\t\t}\n\t\t}\n\t\tn=x;\n\t\tfor(i=0;i<=m;i++){\n\t\t\tt1[i].init(m+1);\n\t\t\tt1[i].modify(m+1,0);\n\t\t\tt2[i].init(m+1);\n\t\t\tt2[i].modify(1,0);\n\t\t\tans[i]=0;\n\t\t}\n\t\tfor(i=1;i<=n;i++){\n\t\t\tfor(j=a[i];j<=m;j++){\n\t\t\t\t// for(int k=0;k<i;k++)\n\t\t\t\t// \tif(a[k]<a[i]){if(j-a[i]+a[k]>=0)f[i][j]=std::max(f[i][j],f[k][j-a[i]+a[k]]);}\n\t\t\t\t// \telse f[i][j]=std::max(f[i][j],f[k][j]);\n\t\t\t\tf[i][j]=std::max(t1[j].query(m-a[i]+1),t2[j-a[i]].query(a[i]+1))+c[i];\n\t\t\t\tans[j]=std::max(ans[j],f[i][j]);\n\t\t\t}\n\t\t\tfor(j=a[i];j<=m;j++){\n\t\t\t\tt1[j].modify(m-a[i]+1,f[i][j]);\n\t\t\t\tt2[j-a[i]].modify(a[i]+1,f[i][j]);\n\t\t\t}\n\t\t}\n\t\tfor(i=1;i<=m;i++){\n\t\t\tans[i]=std::max(ans[i],ans[i-1]);\n\t\t\tw(ans[i]);\n\t\t}\n\t\tputchar(10);\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 291549350,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define F first\n#define S second \n#define sz(x) (int)x.size()\n#define all(x) x.begin(), x.end()\n#define make_unique(x) sort(all(x)), x.erase(unique(all(x)), x.end())\n \nmt19937 rng(chrono::steady_clock::now().time_since_epoch().count());\nmt19937_64 Rng(chrono::steady_clock::now().time_since_epoch().count());\n \ntypedef long long ll;\ntypedef long double ld;\ntypedef unsigned long long ull;\n \nconst int N = 2e3 + 200;\nconst int M = 2e5;\nconst int inf = 2e9 + 3;\nconst ll INF = 1e18;\n \nint T;\nint n, m, V;\nint was[2][N], Was[2][N];\nlong long dp[2 * N][N], ans[N];\nlong long c[N], C[N];\n \nvoid solve() {\n cin >> n >> m >> V;\n for (int i = 0; i <= n + 1; ++i) {\n was[0][i] = was[1][i] = 0;\n }\n for (int i = 1; i <= n; ++i) {\n cin >> c[i];\n }\n for (int i = 1; i <= m; ++i) {\n char c;\n int x;\n cin >> c >> x;\n was[c == 'R'][x]++;\n }\n for (int i = 1; i <= n; ++i) {\n was[1][i] += was[1][i - 1];\n }\n for (int i = n; i >= 1; --i) {\n was[0][i] += was[0][i + 1];\n }\n int sz = 0;\n long long sum = 0;\n for (int i = 1; i <= n; ++i) {\n if (was[0][i] + was[1][i] > V) \n continue;\n if (i == 1 || was[0][i] != was[0][i - 1] || was[1][i] != was[1][i - 1]) {\n ++sz;\n C[sz] = c[i];\n Was[0][sz] = was[0][i];\n Was[1][sz] = was[1][i];\n }\n else {\n C[sz] += c[i];\n }\n }\n n = sz;\n /* for (int i = 1; i <= n; ++i) { */\n /* cout << C[i] << \" \" << Was[0][i] << \" \" << Was[1][i] << '\\n'; */\n /* } */\n assert(n <= 2 * V);\n for (int i = 1; i <= n; ++i) {\n for (int k = 0; k <= V; ++k) {\n dp[i][k] = -INF;\n }\n }\n for (int i = 1; i <= V; ++i)\n ans[i] = 0;\n for (int i = 1; i <= n; ++i) {\n dp[i][Was[1][i]] = max(dp[i][Was[1][i]], C[i]);\n for (int k = 0; k <= V; ++k) {\n if (k + Was[0][i] <= V)\n ans[k + Was[0][i]] = max(ans[k + Was[0][i]], dp[i][k]);\n for (int j = i + 1; j <= n; ++j) {\n int nk = k + max(Was[0][i] - Was[0][j], Was[1][j] - Was[1][i]);\n if (nk <= V)\n dp[j][nk] = max(dp[j][nk], dp[i][k] + C[j]);\n }\n }\n }\n for (int i = 1; i <= V; ++i) {\n ans[i] = max(ans[i], ans[i - 1]);\n cout << ans[i] << \" \\n\"[i == V];\n }\n}\n \nint main() {\n cin.tie(0)->sync_with_stdio(0);\n cin >> T;\n while (T--) solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 293147123,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define lowbit(x) (x&(-x))\nusing namespace std;\ntypedef long long ll;\nconst ll N = 2e5+10,M = 4010,INF = 1e18;\ntemplate<typename T>void tomin(T& x,T y){x = min(x,y);}\ntemplate<typename T>void tomax(T& x,T y){x = max(x,y);}\n//\nint T,n,m,V,c[N],a[N],L,R;\n \nll va[M],vc[M],cnt,ans[M];\n \nll dp[M][M];\n \nstruct BIT{\n ll t[M];\n void clear(){\n for(int i=1;i<=V;i++)t[i] = -INF;\n }\n void mdf(int x,ll v){\n for(;x<=V;x+=lowbit(x))tomax(t[x],v);\n }\n ll qry(ll x,ll S=-INF){\n for(;x;x-=lowbit(x))tomax(S,t[x]);\n return S;\n }\n}t1[M],t2[M];\n \nvoid solve(){\n cin>>n>>m>>V;\n for(int i=1;i<=n;i++)cin>>c[i]; fill(a+1,a+1+n,0);\n for(int i=1;i<=m;i++){\n char s; int x; cin>>s>>x;\n if(s=='L')a[1]++,a[x+1]--;\n else a[x]++;\n }\n for(int i=1;i<=n;i++)a[i] += a[i-1];\n L = 1,R = n;\n while(L<=n && a[L] > V)L++; while(R>=1 && a[R] > V)R--;\n if(L>R){\n for(int i=1;i<=V;i++)cout<<\"0 \"; cout<<\"\\n\";\n return;\n }\n cnt = 0;\n for(int i=L,j;i<=R;i=j+1){\n j=i; while(j<R && a[j+1] == a[j])j++;\n va[++cnt] = a[j]; vc[cnt] = 0; for(int k=i;k<=j;k++)vc[cnt] += c[k];\n }\n //\n for(int v=1;v<=V;v++)ans[v] = 0;\n for(int v=1;v<=V;v++)t1[v].clear(); for(int v=0;v<=2*V;v++)t2[v].clear();\n for(int i=1;i<=cnt;i++)for(int j=0;j<=V;j++)dp[i][j] = -INF;\n for(int i=1;i<=cnt;i++){\n if(va[i] <= V)tomax(dp[i][va[i]],vc[i]);\n for(int j=1;j<=V;j++){\n ll mx = max(t1[j].qry(V-va[i]+1),t2[j-va[i]+V].qry(va[i]));\n tomax(dp[i][j],mx + vc[i]);\n }\n for(int j=1;j<=V;j++){\n tomax(ans[j],dp[i][j]);\n t1[j].mdf(V-va[i]+1,dp[i][j]);\n t2[j-va[i]+V].mdf(va[i],dp[i][j]);\n }\n }\n for(int v=1;v<=V;v++)tomax(ans[v],ans[v-1]);\n for(int v=1;v<=V;v++)cout<<ans[v]<<\" \"; cout<<\"\\n\";\n}\n \nint main(){\n ios::sync_with_stdio(false); cin.tie(0);\n cin>>T;\n while(T--)solve();\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 290882780,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\nN = int(2e5 + 2)\nM = 4002\n \ndef upd(x, y, z, v):\n x += 1\n y += 1\n for i in range(x, v + 2):\n for j in range(y, v + 2):\n t[i][j] = max(t[i][j], z)\n \ndef qry(x, y, v):\n x += 1\n y += 1\n z = 0\n for i in range(x, 0, -1):\n for j in range(y, 0, -1):\n z = max(z, t[i][j])\n return z\n \nT = int(input())\nfor _ in range(T):\n n, m, v = map(int, input().split())\n c = list(map(int, input().split()))\n s = [0] * (n + 1)\n \n for _ in range(m):\n op, x = input().split()\n x = int(x)\n if op == 'L':\n s[1] += 1\n if x + 1 <= n:\n s[x + 1] -= 1\n else:\n s[x] += 1\n \n for i in range(1, n + 1):\n s[i] += s[i - 1]\n \n g = []\n i = 1\n while i <= n:\n C = 0\n j = i\n while j <= n and s[j] == s[i]:\n C += c[j - 1]\n j += 1\n if s[i] <= v:\n g.append((s[i], C))\n i = j\n \n t = [[0] * (M) for _ in range(M)]\n f = [0] * (M)\n \n for x, val in g:\n for i in range(v + 1 - x):\n f[i] = qry(i, i + x, v)\n for i in range(v + 1 - x):\n upd(i, i + x, f[i] + val, v)\n \n for i in range(1, v + 1):\n print(qry(v, i, v), end=' ')\n print()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 297705225,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\nN = int(2e5 + 2)\nM = 4002\n \ndef upd(x, y, z):\n x += 1\n y += 1\n for i in range(x, v + 2):\n for j in range(y, v + 2):\n t[i][j] = max(t[i][j], z)\n \ndef qry(x, y):\n x += 1\n y += 1\n z = 0\n for i in range(x, 0, -1):\n for j in range(y, 0, -1):\n z = max(z, t[i][j])\n return z\n \nT = int(input())\nfor _ in range(T):\n n, m, v = map(int, input().split())\n c = list(map(int, input().split()))\n s = [0] * (n + 1)\n \n for _ in range(m):\n op, x = input().split()\n x = int(x)\n if op == 'L':\n s[1] += 1\n if x + 1 <= n:\n s[x + 1] -= 1\n else:\n s[x] += 1\n \n for i in range(1, n + 1):\n s[i] += s[i - 1]\n \n g = []\n i = 1\n while i <= n:\n C = 0\n j = i\n while j <= n and s[j] == s[i]:\n C += c[j - 1]\n j += 1\n if s[i] <= v:\n g.append((s[i], C))\n i = j\n \n t = [[0] * (M) for _ in range(M)]\n f = [0] * (M)\n \n for x, val in g:\n for i in range(v + 1 - x):\n f[i] = qry(i, i + x)\n for i in range(v + 1 - x):\n upd(i, i + x, f[i] + val)\n \n for i in range(1, v + 1):\n print(qry(v, i), end=' ')\n print()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 297704669,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define For(i,x,y) for (int i=(x);i<=(y);i++)\n#define FOR(i,x,y) for (int i=(x);i<(y);i++)\n#define Dow(i,x,y) for (int i=(x);i>=(y);i--)\n#define mp make_pair\n#define fi first\n#define se second\n#define pb push_back\n#define ep emplace_back\n#define siz(x) ((int)x.size())\n#define all(x) x.begin(),x.end()\n#define fil(a,b) memset(a,b,sizeof a)\nusing namespace std;\ntypedef long long ll;\ntypedef unsigned long long ull;\ntypedef pair<int,int> pa;\ntypedef pair<ll,ll> PA;\ntypedef vector<int> poly;\ninline ll read(){\n\tll x=0,f=1;char c=getchar();\n\twhile ((c<'0'||c>'9')&&(c!='-')) c=getchar();\n\tif (c=='-') f=-1,c=getchar();\n\twhile (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();\n\treturn x*f;\n}\n \nconst int N = 2e5+10, V = 2010;\nint n,m,v,cnt,a[N],s[N],b[V<<1];\nll c[V<<1],f[V<<1][V],ans[V];\n \nll t[V<<1][V];\ninline void Add(int k,int x,ll y){\n\t++x;\n\tfor (;x<=v+1;x+=x&-x) t[k][x]=max(t[k][x],y);\n}\ninline ll Query(int k,ll x){\n\t++x;\n\tll ret=0;\n\tfor (;x;x-=x&-x) ret=max(ret,t[k][x]);\n\treturn ret;\n}\n \ninline void solve(){\n\tn=read(),m=read(),v=read();\n\tFor(i,1,n) a[i]=read();\n\tFor(i,1,n) s[i]=0;\n\tFor(i,1,m){\n\t\tchar op[4];\n\t\tscanf(\"%s\",op);\n\t\tint x=read();\n\t\tif (op[0]=='L'){\n\t\t\ts[1]++;\n\t\t\ts[x+1]--;\n\t\t} else {\n\t\t\ts[x]++;\n\t\t\ts[n+1]--;\n\t\t}\n\t}\n\tFor(i,1,n) s[i]+=s[i-1];\n\tint cnt=0;\n\tfor (int l=1,r=1;l<=n;l=++r){\n\t\tll sa=a[r];\n\t\twhile (r<n&&s[r+1]==s[r]) ++r,sa+=a[r];\n\t\tif (s[l]<=v){\n\t\t\tb[++cnt]=s[l];\n\t\t\tc[cnt]=sa;\n\t\t}\n\t}\n\tFor(i,1,cnt) For(j,0,v) f[i][j]=0;\n\tFor(i,1,v) ans[i]=0;\n\tFor(i,0,v+1) For(j,0,v+1) t[i][j]=0;\n\tDow(i,cnt,1){\n\t\tf[i][0]=max(f[i][0],c[i]);\n\t\tFor(j,0,v){\n\t\t\tif (b[i]+j>v) break;\n\t\t\tf[i][j]=max(f[i][j],Query(j,b[i])+c[i]);\n\t\t\tif (j) f[i][j]=max(f[i][j],f[i][j-1]);\n\t\t\tans[b[i]+j]=max(ans[b[i]+j],f[i][j]);\n\t\t}\n\t\tDow(k,i-1,1) if (b[k]<=b[i]){\n\t\t\tFor(j,0,v-(b[i]-b[k])) f[k][j+b[i]-b[k]]=max(f[k][j+b[i]-b[k]],f[i][j]+c[k]);\n\t\t\tbreak;\n\t\t}\n\t\tFor(j,0,v){\n\t\t\tif (b[i]+j>v) break;\n\t\t\tAdd(j,b[i],f[i][j]);\n\t\t}\n\t}\n\tFor(i,1,v) printf(\"%lld \",ans[i]);\n\tputs(\"\");\n}\n \nint main(){\n\tint T=read();\n\twhile (T--) solve();\n}\n/*\n1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 291792634,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define int long long\n#define rep(i,n) for(int i=0;i<n;i++)\n#define pb push_back\n#define fr first\n#define se second\n#define ford(i,n) for(int i=n-1;i>=0;i--)\n#define forn(i,a,n) for(int i=a;i<n;i++)\n#define foreach(i,c) for(__typeof(c.begin())i=(c.begin());i!=(c).end();i++)\n#define pii pair<int,int>\n#define vi vector<int>\n#define ll long long\n#define sz(s) (int)(s.size())\n#define all(s) s.begin(),s.end()\n#define vii vector<pair<int,int> >\n#define re return\n#define mod 998244353\nusing namespace std;\n \nll q_p(ll x,ll y){\n ll res=1;\n x%=mod;\n while(y){\n if(y%2){\n res=res*x;\n res%=mod;\n }\n y/=2;\n x=x*x;\n x%=mod;\n }\n re res;\n}\nint dx[4]={0,-1,0,1},dy[4]={1,0,-1,0};\nint n,m,v;\nint a[200001],c[200001];\nstruct BIT{\n int lowbit(int x){\n re x&(-x);\n }\n int bit[2001];\n void init(){\n rep(i,v+1)\n bit[i]=0;\n }\n int qry(int x){\n int res=0;\n while(x){\n res=max(res,bit[x]);\n x-=lowbit(x);\n }\n re res;\n }\n void add(int x,int v){\n while(x<=n){\n bit[x]=max(bit[x],v);\n x+=lowbit(x);\n }\n }\n}BIT[2001][2];\nint ans[2001];\nsigned main(){\n ios::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n int t;\n cin>>t;\n while(t--){\n cin>>n>>m>>v;\n rep(i,n)\n cin>>c[i];\n rep(j,v+1)\n rep(i,2)\n BIT[j][i].init();\n rep(i,n)\n a[i]=0;\n rep(i,m){\n char op;\n int x;\n cin>>op>>x;\n if(op=='L'){\n a[0]++;\n a[x]--;\n }\n if(op=='R'){\n a[x-1]++;\n }\n }\n forn(i,1,n)\n a[i]+=a[i-1];\n vii d;\n d.clear();\n int sumc=0,sa=-1;\n rep(i,n){\n if(a[i]!=sa&&sa!=-1){\n d.pb({sa,sumc});\n sa=-1;\n sumc=0;\n }\n if(a[i]>v)\n continue;\n if(a[i]==sa){\n sumc+=c[i];\n }else{\n sa=a[i];\n sumc=c[i];\n }\n }\n if(sa!=-1)\n d.pb({sa,sumc});\n // dp vi,j=max dp k<vi,j-vi+k j'-k=j-vi\n // 1~V\n rep(i,v+1)\n ans[i]=0;\n for(pii p:d){\n int nv,w;\n nv=p.fr,w=p.se;\n forn(j,nv,v+1){\n int dpij=0;\n dpij=max(dpij,BIT[j-nv][0].qry(nv)+w);\n dpij=max(dpij,BIT[j][1].qry(v-nv+1)+w);\n BIT[j-nv][0].add(nv,dpij);\n BIT[j][1].add(v-nv+1,dpij);\n ans[j]=max(ans[j],dpij);\n }\n }\n forn(i,1,v+1)\n ans[i]=max(ans[i],ans[i-1]);\n if(t>50)\n cout<<sz(d)<<\"\\n\";\n if(t<=50){\n forn(i,1,v+1)\n cout<<ans[i]<<\" \";\n cout<<\"\\n\"; \n }\n \n }\n re 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 290820171,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n//��_G\n#include<bits/stdc++.h>\nusing namespace std;\ntypedef long long LL;\ntypedef double DB;\nconst int N = 222222;\nconst int V = 6666;\nconst LL inf = 1e18;\nint n,m,v,a[N],b[N];\nLL c[N],f[V][V],g[V][V],h[V][V],ans[V];\nchar ch[111];\nvoid cmax(LL&x,LL y){\n\tif(x<y)\n\t\tx=y;\n}\nvoid modify(LL*t,int i,LL x){\n\twhile(i<=v+1){\n\t\tcmax(t[i],x);\n\t\ti+=i&-i;\n\t}\n}\nLL query(LL*t,int i){\n\tLL x=-inf;\n\twhile(i){\n\t\tcmax(x,t[i]);\n\t\ti-=i&-i;\n\t}\n\treturn x;\n}\nint main(){\n\tint T,i,j,x;\n\tscanf(\"%d\",&T);\n\twhile(T--){\n\t\tscanf(\"%d%d%d\",&n,&m,&v);\n\t\tfor(i=1;i<=n;i++)\n\t\t\tscanf(\"%lld\",c+i);\n\t\tfor(i=0;i<=n+5;i++)\n\t\t\ta[i]=0,b[i]=0;\n\t\tfor(i=1;i<=m;i++){\n\t\t\tscanf(\"%s%d\",ch,&x);\n\t\t\tif(ch[0]=='L')\n\t\t\t\ta[x]++;\n\t\t\telse\n\t\t\t\tb[x]++;\n\t\t}\n\t\tfor(i=n-1;i>=1;i--)\n\t\t\ta[i]+=a[i+1];\n\t\tfor(i=2;i<=n;i++)\n\t\t\tb[i]+=b[i-1];\n\t\tfor(i=1;i<=n;i++)\n\t\t\ta[i]+=b[i];\n\t\tj=0;\n\t\tfor(i=1;i<=n;i++){\n\t\t\tif(a[i]<=v){\n\t\t\t\tif(j&&a[j]==a[i])\n\t\t\t\t\tc[j]+=c[i];\n\t\t\t\telse\n\t\t\t\t\tj++,a[j]=a[i],c[j]=c[i];\n\t\t\t}\n\t\t}\n\t\tfor(i=j+1;i<=n;i++)\n\t\t\ta[i]=0,b[i]=0,c[i]=0;\n\t\tn=j;\n\t\tfor(i=0;i<=v*3;i++)\n\t\t\tfor(j=0;j<=v*3;j++)\n\t\t\t\tf[i][j]=-inf;\n\t\tfor(i=0;i<=v*3;i++)\n\t\t\tfor(j=0;j<=v*3;j++)\n\t\t\t\tg[i][j]=-inf,h[i][j]=-inf;\n\t\tfor(i=0;i<=v*3;i++)\n\t\t\tans[i]=0;\n\t\tfor(i=1;i<=n;i++){\n\t\t\tcmax(f[i][a[i]],c[i]);\n\t\t\tfor(j=0;j<=v*2;j++){\n\t\t\t\tcmax(f[i][j],c[i]+query(g[j-a[i]+v],1+a[i]));\n\t\t\t\tcmax(f[i][j],c[i]+query(h[j+a[i]],v+1-a[i]));\n\t\t\t}\n\t\t\tfor(j=0;j<=v*2;j++)\n\t\t\t\tcmax(ans[(j+a[i]+1)/2],f[i][j]);\n\t\t\tfor(j=0;j<=v*2;j++){\n\t\t\t\tmodify(g[j-a[i]+v],1+a[i],f[i][j]);\n\t\t\t\tmodify(h[j+a[i]],v+1-a[i],f[i][j]);\n\t\t\t}\n\t\t\t//for(j=0;j<=v*2;j++)\n\t\t\t//\tif(f[i][j]>=0)\n\t\t\t//\t\tcout<<i<<' '<<j<<' '<<f[i][j]<<endl;\n\t\t}\n\t\tfor(i=2;i<=v;i++)\n\t\t\tans[i]=max(ans[i],ans[i-1]);\n\t\tfor(i=1;i<=v;i++)\n\t\t\tprintf(\"%lld%c\",ans[i],\" \\n\"[i==v]);\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 290757812,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define f first\n#define s second\n#define int li\n \nusing namespace std;\nusing li = long long;\nusing ld = long double;\nconst int N = 4005;\nconst int INF = 1e18;\n \nint t1[N][N], t2[N][N];\n \nint get(int t[N][N], int j, int r)\n{\n int ans = -INF;\n \n for(; r >= 0; r = (r & r + 1) - 1)\n ans = max(ans, t[j][r]);\n \n return ans;\n}\n \nvoid upd(int t[N][N], int j, int i, int x)\n{\n for(; i < N; i |= i + 1)\n t[j][i] = max(t[j][i], x);\n}\n \nint32_t main()\n{\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n \n int t;\n cin >> t;\n \n while(t--)\n {\n int n, m, v;\n cin >> n >> m >> v;\n \n vector <int> c(n);\n \n for(int i = 0; i < n; i++)\n cin >> c[i];\n \n vector <int> dif(n + 1);\n \n for(int i = 0; i < m; i++)\n {\n char c;\n int x;\n cin >> c >> x;\n x--;\n \n if(c == 'L')\n {\n dif[0]++;\n dif[x + 1]--;\n }\n else\n {\n dif[x]++;\n }\n }\n \n for(int i = 1; i < n; i++)\n dif[i] += dif[i - 1];\n \n vector <int> b, cb;\n \n for(int i = 0; i < n; i++)\n if(dif[i] <= v)\n {\n b.push_back(v - dif[i]);\n cb.push_back(c[i]);\n }\n \n vector <int> nb, ncb;\n nb.push_back(v);\n ncb.push_back(0);\n \n// for(int i = 0; i < b.size(); i++)\n// {\n// nb.push_back(b[i]);\n// ncb.push_back(cb[i]);\n// }\n \n int curc = 0;\n \n for(int i = 0; i < b.size(); i++)\n {\n if(i > 0 && b[i] != b[i - 1])\n {\n nb.push_back(b[i - 1]);\n ncb.push_back(curc);\n curc = cb[i];\n }\n else curc += cb[i];\n }\n \n if(b.size())\n {\n nb.push_back(b.back());\n ncb.push_back(curc);\n }\n \n// cout << \"!!!!\" << endl;\n//\n// for(auto x:nb)\n// cout << x << ' ';\n//\n// cout << endl;\n//\n// for(auto x:ncb)\n// cout << x << ' ';\n//\n// cout << endl;\n \n n = nb.size();\n b = nb;\n c = ncb;\n \n int dp[n][v + 1], dp2[n][v + 1];\n \n for(int i = 0; i < n; i++)\n for(int j = 0; j <= v; j++)\n dp[i][j] = dp2[i][j] = -INF;\n \n dp[0][0] = 0;\n \n vector <int> ans(v + 1);\n \n for(int i = 0; i <= v; i++)\n for(int j = 0; j <= 2 * v; j++)\n t1[i][j] = t2[i][j] = -INF;\n \n for(int i = 0; i < n; i++)\n {\n for(int j = 0; j <= v; j++)\n {\n dp[i][j] = max(dp[i][j], get(t1, j, b[i]) + c[i]);\n dp[i][j] = max(dp[i][j], get(t2, j + b[i], v - b[i]) + c[i]);\n upd(t1, j, b[i], dp[i][j]);\n upd(t2, j + b[i], v - b[i], dp[i][j]);\n \n// for(int ni = i + 1; ni < n; ni++)\n// {\n// if(b[ni] >= b[i])\n// dp[ni][j] = max(dp[ni][j], dp[i][j] + c[ni]);\n// if(b[ni] < b[i] && j + b[i] - b[ni] <= v)\n// {\n// int nj = j + b[i] - b[ni];\n// dp[ni][nj] = max(dp[ni][nj], dp[i][j] + c[ni]);\n// }\n// }\n \n ans[j] = max(ans[j], dp[i][j]);\n }\n }\n \n for(int i = 1; i <= v; i++)\n ans[i] = max(ans[i], ans[i - 1]);\n \n for(int i = 1; i <= v; i++)\n cout << ans[i] << ' ';\n \n cout << \"\\n\";\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 290771211,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing std::cin;\nusing std::cout;\nusing std::string;\ntypedef long long lli;\n \nconst int N=2e5,M=5e3;\nint n,m,V,rc[N+5];\nint cf[N+5];\n \nint va[N+5];lli vc[N+5];int vt;\n \nstruct FTree {\n\tlli mx[M+5];\n\tinline void ins(int x,lli z) {\n\t\tfor(;x<=V;x+=x&-x)mx[x]=std::max(mx[x],z);\n\t}\n\tinline lli qry(int x) {\n\t\tlli z=0;\n\t\tfor(;x;x-=x&-x)z=std::max(z,mx[x]);\n\t\treturn z;\n\t}\n\tinline void clear() {\n\t\tstd::fill(mx+1,mx+V+1,0);\n\t}\n}ft1[M+5],ft2[M+5];\nlli f[M+5][M+5];\n \ninline void solve() {\n\tcin>>n>>m>>V;\n\tfor(int i=1;i<=n;++i)cin>>rc[i];\n\tfor(int i=1;i<=m;++i) {\n\t\tstring o;int x;\n\t\tcin>>o>>x;\n\t\tif(o==\"L\")++cf[1],--cf[x+1];\n\t\telse ++cf[x];\n\t}\n \n\tfor(int i=1,nv=0;i<=n;++i) {\n\t\tnv+=cf[i];\n\t\tif(nv<=V) {\n\t\t\tif(vt==0||nv!=va[vt])++vt,va[vt]=va[vt-1]+cf[i],vc[vt]=rc[i];\n\t\t\telse vc[vt]+=rc[i];\n\t\t}\n\t}\n\tfor(int i=1;i<=vt;++i)++va[i];++V;\n \n\tfor(int i=1;i<=vt;++i) {\n\t\tfor(int j=1;j<=V;++j) {\n\t\t\tf[i][j]=ft1[j].qry(V+1-va[i])+vc[i];\n\t\t\tif(j>=va[i])f[i][j]=std::max(f[i][j],ft2[j-va[i]].qry(va[i])+vc[i]);\n\t\t\tf[i][j]=std::max(f[i][j],f[i][j-1]);\n\t\t}\n\t\tfor(int j=va[i];j<=V;++j)\n\t\t\tft1[j].ins(V+1-va[i],f[i][j]),ft2[j-va[i]].ins(va[i],f[i][j]);\n\t}\n \n\tlli a=0;\n\tfor(int i=1;i<=V;++i) {\n\t\ta=std::max(a,ft1[i].qry(V));\n\t\tif(i>1)cout<<a<<' ';\n\t}\n\tcout<<'\\n';\n}\ninline void clear() {\n\tfor(int i=1;i<=vt;++i)std::fill(f[i]+1,f[i]+V+1,0);\n\tfor(int i=0;i<=V;++i)ft1[i].clear(),ft2[i].clear();\n\tstd::fill(cf+1,cf+n+2,0);\n\tvt=0;\n}\n \nint main() {\n\tstd::ios::sync_with_stdio(0);\n\tcin.tie(0),cout.tie(0);\n \n\tint t;\n\tcin>>t;\n\twhile(t--) {\n\t\tif(t==900)return 0;\n\t\tsolve();\n\t\tclear();\n\t}\n \n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 295498306,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\nimport sys\nfrom collections import defaultdict\n \ninput = sys.stdin.read\nN = int(2e5 + 2)\nM = 4002\n \ndef upd(x, y, z):\n x += 1\n y += 1\n for i in range(x, v + 2):\n for j in range(y, v + 2):\n t[i][j] = max(t[i][j], z)\n \ndef qry(x, y):\n x += 1\n y += 1\n z = 0\n for i in range(x, 0, -1):\n for j in range(y, 0, -1):\n z = max(z, t[i][j])\n return z\n \ndata = input().split()\nindex = 0\nT = int(data[index])\nindex += 1\n \nfor _ in range(T):\n n = int(data[index])\n m = int(data[index + 1])\n v = int(data[index + 2])\n index += 3\n \n c = [0] * (n + 1)\n s = [0] * (n + 1)\n for i in range(1, n + 1):\n c[i] = int(data[index])\n s[i] = 0\n index += 1\n \n for _ in range(m):\n op = data[index]\n x = int(data[index + 1])\n index += 2\n if op == 'L':\n s[1] += 1\n if x + 1 <= n:\n s[x + 1] -= 1\n else:\n s[x] += 1\n \n for i in range(1, n + 1):\n s[i] += s[i - 1]\n \n g = []\n i = 1\n while i <= n:\n j = i\n C = 0\n while j <= n and s[j] == s[i]:\n C += c[j]\n j += 1\n if s[i] <= v:\n g.append((s[i], C))\n i = j\n \n t = [[0] * (M) for _ in range(M)]\n f = [0] * (M)\n \n for x, val in g:\n for i in range(0, v + 1 - x):\n f[i] = qry(i, i + x)\n for i in range(0, v + 1 - x):\n upd(i, i + x, f[i] + val)\n \n result = []\n for i in range(1, v + 1):\n result.append(qry(v, i))\n \n print(' '.join(map(str, result)))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 297703675,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define F first\n#define S second \n#define sz(x) (int)x.size()\n#define all(x) x.begin(), x.end()\n#define make_unique(x) sort(all(x)), x.erase(unique(all(x)), x.end())\n \nmt19937 rng(chrono::steady_clock::now().time_since_epoch().count());\nmt19937_64 Rng(chrono::steady_clock::now().time_since_epoch().count());\n \ntypedef long long ll;\ntypedef long double ld;\ntypedef unsigned long long ull;\n \nconst int N = 2e3 + 200;\nconst int M = 2e5;\nconst int inf = 2e9 + 3;\nconst ll INF = 1e18;\n \nint T;\nint n, m, V;\nint was[2][N], Was[2][N];\nlong long dp[2 * N][N], ans[N];\nlong long c[N], C[N];\n \nvoid solve() {\n cin >> n >> m >> V;\n for (int i = 0; i <= n + 1; ++i) {\n was[0][i] = was[1][i] = 0;\n }\n for (int i = 1; i <= n; ++i) {\n cin >> c[i];\n }\n for (int i = 1; i <= m; ++i) {\n char c;\n int x;\n cin >> c >> x;\n was[c == 'R'][x]++;\n }\n for (int i = 1; i <= n; ++i) {\n was[1][i] += was[1][i - 1];\n }\n for (int i = n; i >= 1; --i) {\n was[0][i] += was[0][i + 1];\n }\n int sz = 0;\n long long sum = 0;\n for (int i = 1; i <= n; ++i) {\n if (was[0][i] + was[1][i] > V) \n continue;\n if (i == 1 || was[0][i] != was[0][i - 1] || was[1][i] != was[1][i - 1]) {\n ++sz;\n C[sz] = c[i];\n Was[0][sz] = was[0][i];\n Was[1][sz] = was[1][i];\n }\n else {\n C[sz] += c[i];\n }\n }\n n = sz;\n /* for (int i = 1; i <= n; ++i) { */\n /* cout << C[i] << \" \" << Was[0][i] << \" \" << Was[1][i] << '\\n'; */\n /* } */\n assert(n <= 2 * V);\n for (int i = 1; i <= n; ++i) {\n for (int k = 0; k <= V; ++k) {\n dp[i][k] = -INF;\n }\n }\n for (int i = 1; i <= V; ++i)\n ans[i] = 0;\n for (int i = 1; i <= n; ++i) {\n dp[i][Was[1][i]] = max(dp[i][Was[1][i]], C[i]);\n for (int k = 0; k <= V; ++k) {\n if (k + Was[0][i] <= V)\n ans[k + Was[0][i]] = max(ans[k + Was[0][i]], dp[i][k]);\n for (int j = i + 1; j <= n; ++j) {\n int nk = k + max(Was[0][i] - Was[0][j], Was[1][j] - Was[1][i]);\n dp[j][nk] = max(dp[j][nk], dp[i][k] + C[j]);\n }\n }\n }\n for (int i = 1; i <= V; ++i) {\n ans[i] = max(ans[i], ans[i - 1]);\n cout << ans[i] << \" \\n\"[i == V];\n }\n}\n \nint main() {\n cin.tie(0)->sync_with_stdio(0);\n cin >> T;\n while (T--) solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 293146973,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n//��_G\n#include<bits/stdc++.h>\nusing namespace std;\ntypedef long long LL;\ntypedef double DB;\nconst int N = 222222;\nconst int V = 6666;\nconst LL inf = 1e18;\nint n,m,v,a[N],b[N];\nLL c[N],f[V][V],g[V][V],h[V][V],ans[V];\nchar ch[111];\nvoid cmax(LL&x,LL y){\n\tif(x<y)\n\t\tx=y;\n}\nvoid modify(LL*t,int i,LL x){\n\twhile(i<=v+1){\n\t\tcmax(t[i],x);\n\t\ti+=i&-i;\n\t}\n}\nLL query(LL*t,int i){\n\tLL x=-inf;\n\twhile(i){\n\t\tcmax(x,t[i]);\n\t\ti-=i&-i;\n\t}\n\treturn x;\n}\nint main(){\n\tint T,i,j,x;\n\tscanf(\"%d\",&T);\n\twhile(T--){\n\t\tscanf(\"%d%d%d\",&n,&m,&v);\n\t\tfor(i=1;i<=n;i++)\n\t\t\tscanf(\"%lld\",c+i);\n\t\tfor(i=1;i<=n;i++)\n\t\t\ta[i]=0,b[i]=0;\n\t\tfor(i=1;i<=m;i++){\n\t\t\tscanf(\"%s%d\",ch,&x);\n\t\t\tif(ch[0]=='L')\n\t\t\t\ta[x]++;\n\t\t\telse\n\t\t\t\tb[x]++;\n\t\t}\n\t\tfor(i=n-1;i>=1;i--)\n\t\t\ta[i]+=a[i+1];\n\t\tfor(i=2;i<=n;i++)\n\t\t\tb[i]+=b[i-1];\n\t\tfor(i=1;i<=n;i++)\n\t\t\ta[i]+=b[i];\n\t\tj=0;\n\t\tfor(i=1;i<=n;i++){\n\t\t\tif(a[i]<=v){\n\t\t\t\tif(j&&a[j]==a[i])\n\t\t\t\t\tc[j]+=c[i];\n\t\t\t\telse\n\t\t\t\t\tj++,a[j]=a[i],c[j]=c[i];\n\t\t\t}\n\t\t}\n\t\tn=j;\n\t\tfor(i=1;i<=n;i++)\n\t\t\tfor(j=0;j<=v*2;j++)\n\t\t\t\tf[i][j]=-inf;\n\t\tfor(i=0;i<=v*3;i++)\n\t\t\tfor(j=1;j<=v+1;j++)\n\t\t\t\tg[i][j]=-inf,h[i][j]=-inf;\n\t\tfor(i=1;i<=v;i++)\n\t\t\tans[i]=0;\n\t\tfor(i=1;i<=n;i++){\n\t\t\tcmax(f[i][a[i]],c[i]);\n\t\t\tfor(j=0;j<=v*2;j++){\n\t\t\t\tcmax(f[i][j],c[i]+query(g[j-a[i]+v],1+a[i]));\n\t\t\t\tcmax(f[i][j],c[i]+query(h[j+a[i]],v+1-a[i]));\n\t\t\t}\n\t\t\tfor(j=0;j<=v*2;j++)\n\t\t\t\tcmax(ans[(j+a[i]+1)/2],f[i][j]);\n\t\t\tfor(j=0;j<=v*2;j++){\n\t\t\t\tmodify(g[j-a[i]+v],1+a[i],f[i][j]);\n\t\t\t\tmodify(h[j+a[i]],v+1-a[i],f[i][j]);\n\t\t\t}\n\t\t\t//for(j=0;j<=v*2;j++)\n\t\t\t//\tif(f[i][j]>=0)\n\t\t\t//\t\tcout<<i<<' '<<j<<' '<<f[i][j]<<endl;\n\t\t}\n\t\tfor(i=2;i<=v;i++)\n\t\t\tans[i]=max(ans[i],ans[i-1]);\n\t\tfor(i=1;i<=v;i++)\n\t\t\tprintf(\"%lld%c\",ans[i],\" \\n\"[i==v]);\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 290748519,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define ll long long\nusing namespace std;\nconst int maxn=200010,maxv=2010;\nint i,j,k,n,m,v,a[maxn];\nll val[maxn],len[maxn];\nint tmp[maxn],cnt;\nint det[maxn];\nll dp[maxv*2][maxv],Max[maxv*2];\nll tMax[maxv][maxv],tS[maxv][2*maxv];\nll tree[2][maxv][maxv*2];\nint lb(int x){return x&(-x);}\nvoid change(int x,int id0,int id1,ll s){\n for(;x<=v+1;x+=lb(x))tree[id0][id1][x]=max(tree[id0][id1][x],s);\n}\nll find(int x,int id0,int id1){\n ll s=0;for(;x;x-=lb(x))s=max(s,tree[id0][id1][x]);return s;\n}\nint main(){\n int T;cin>>T;\n int b=(T==1000);\n while(T--){\n scanf(\"%d%d%d\",&n,&m,&v);\n for(int i=1;i<=n;i++)a[i]=len[i]=val[i]=tmp[i]=det[i]=0;\n for(int i=1;i<=n;i++)scanf(\"%lld\",&val[i]);\n for(int i=1;i<=m;i++){\n char c;\n do c=getchar();\n while(c!='L' && c!='R');\n int x;scanf(\"%d\",&x);\n if(c=='L')a[1]++,a[x+1]--;\n else a[x]++;\n }\n for(int i=1;i<=n;i++)a[i]+=a[i-1];\n cnt=0;\n for(int i=1;i<=n;i++){\n if(a[i]<=v){\n for(int j=i;j<=n;){\n for(int k=j+1;k<=n+1;k++)\n if(k==n+1 || a[k]!=a[j]){\n if(a[j]<=v){\n ++cnt;\n tmp[cnt]=a[j];\n for(int h=j;h<k;h++)len[cnt]+=val[h];\n }\n j=k;\n }\n }\n break;\n }\n }\n for(int i=1;i<=cnt;i++)det[i]=tmp[i]-tmp[i-1];\n if(!b)\n for(int i=cnt;i>=1;i--){\n for(int j=0;j<=v;j++)dp[i][j]=len[i];\n for(int k=0;k<=v;k++){\n dp[i][k]=max(dp[i][k],find(v+1-tmp[i],1,k+tmp[i])+len[i]);\n dp[i][k]=max(dp[i][k],find(tmp[i],0,k)+len[i]);\n }\n for(int j=0;j<=v;j++){\n change(tmp[i],0,j,dp[i][j]);\n }\n for(int j=0;j<=v;j++){\n change(v+1-tmp[i],1,j+tmp[i],dp[i][j]);\n }\n }\n for(int i=0;i<=v+v;i++)\n for(int j=0;j<=v+1;j++)\n tree[0][i][j]=tree[1][i][j]=0;\n \n for(int r=1;r<=v;r++){\n ll ans=0;\n for(int i=1;i<=cnt;i++){\n int y=r-tmp[i];\n if(y>=0)ans=max(ans,dp[i][y]);\n }\n printf(\"%lld \",ans);\n }puts(\"\");\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 290795371,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define F first\n#define S second \n#define sz(x) (int)x.size()\n#define all(x) x.begin(), x.end()\n#define make_unique(x) sort(all(x)), x.erase(unique(all(x)), x.end())\n \nmt19937 rng(chrono::steady_clock::now().time_since_epoch().count());\nmt19937_64 Rng(chrono::steady_clock::now().time_since_epoch().count());\n \ntypedef long long ll;\ntypedef long double ld;\ntypedef unsigned long long ull;\n \nconst int N = 2e3 + 200;\nconst int M = 2e5 + 200;\nconst int inf = 2e9 + 3;\nconst ll INF = 1e18;\n \nint T;\nint n, m, V;\nint was[2][M], Was[2][N];\nlong long dp[2 * N][N], ans[N];\nlong long c[M], C[2 * N];\n \nvoid solve() {\n cin >> n >> m >> V;\n for (int i = 0; i <= n + 1; ++i) {\n was[0][i] = was[1][i] = 0;\n }\n for (int i = 1; i <= n; ++i) {\n cin >> c[i];\n }\n for (int i = 1; i <= m; ++i) {\n char c;\n int x;\n cin >> c >> x;\n was[c == 'R'][x]++;\n }\n for (int i = 1; i <= n; ++i) {\n was[1][i] += was[1][i - 1];\n }\n for (int i = n; i >= 1; --i) {\n was[0][i] += was[0][i + 1];\n }\n int sz = 0;\n for (int i = 1, j = 0; i <= n; ++i) {\n if (was[0][i] + was[1][i] > V) \n continue;\n if (j == 0 || was[0][i] != was[0][j] || was[1][i] != was[1][j]) {\n ++sz;\n C[sz] = c[i];\n Was[0][sz] = was[0][i];\n Was[1][sz] = was[1][i];\n }\n else {\n C[sz] += c[i];\n }\n j = i;\n }\n n = sz;\n assert(n <= 2 * V);\n /* for (int i = 1; i <= n; ++i) { */\n /* cout << C[i] << \" \" << Was[0][i] << \" \" << Was[1][i] << '\\n'; */\n /* } */\n for (int i = 1; i <= n; ++i) {\n for (int k = 0; k <= V; ++k) {\n dp[i][k] = -INF;\n }\n }\n for (int i = 0; i <= V; ++i)\n ans[i] = 0;\n for (int i = 1; i <= n; ++i) {\n dp[i][Was[1][i]] = max(dp[i][Was[1][i]], C[i]);\n for (int k = 0; k <= V; ++k) {\n if (k + Was[0][i] <= V)\n ans[k + Was[0][i]] = max(ans[k + Was[0][i]], dp[i][k]);\n for (int j = i + 1; j <= n; ++j) {\n int nk = k + max(Was[0][i] - Was[0][j], Was[1][j] - Was[1][i]);\n if (nk <= V)\n dp[j][nk] = max(dp[j][nk], dp[i][k] + C[j]);\n }\n }\n }\n for (int i = 1; i <= V; ++i) {\n ans[i] = max(ans[i], ans[i - 1]);\n cout << ans[i] << \" \\n\"[i == V];\n }\n}\n \nint main() {\n cin.tie(0)->sync_with_stdio(0);\n cin >> T;\n while (T--) solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 293148745,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define int long long\nusing namespace std;\nint n,m,v,test,c[200010],a[200010],tmpa[200010],tmpc[200010],f[5010][5010];\nstruct fenwick{\n\tint tr[5010];\n\tvoid add(int x,int y)\n\t{\n\t\tfor(;x<=5000;x+=(x&(-x)))\n\t\t\ttr[x]=max(tr[x],y);\n\t}\n\tint ask(int x)\n\t{\n\t\tint maxv=-2147483647;\n\t\tfor(;x;x-=(x&(-x)))\n\t\t\tmaxv=max(maxv,tr[x]);\n\t\treturn maxv;\n\t}\n}c1[5010],c2[5010];\nsigned main()\n{\n\tscanf(\"%lld\",&test);\n\tfor(;test--;)\n\t{\n\t\tscanf(\"%lld%lld%lld\",&n,&m,&v);\n\t\tfor(int i=1;i<=n;i++) scanf(\"%lld\",&c[i]),a[i]=0;\n\t\tfor(int i=1;i<=m;i++)\n\t\t{\n\t\t\tchar op;\n\t\t\tint x;\n\t\t\tscanf(\"%*c%c%lld\",&op,&x);\n\t\t\tif(op=='L')\n\t\t\t\ta[1]++,a[x+1]--;\n\t\t\telse\n\t\t\t\ta[x]++;\n\t\t}\n\t\tfor(int i=2;i<=n;i++) a[i]+=a[i-1];\n\t\tint cnt=0;\n\t\tfor(int i=1;i<=n;i++)\n\t\t\tif(a[i]<=v)\n\t\t\t\ttmpa[++cnt]=a[i],tmpc[cnt]=c[i];\n\t\tn=cnt;\n\t\tcnt=0;\n\t\tfor(int i=1;i<=n;i++)\n\t\t\tif(cnt==0||tmpa[i]!=a[cnt])\n\t\t\t\ta[++cnt]=tmpa[i],c[cnt]=tmpc[i];\n\t\t\telse\n\t\t\t\tc[cnt]+=tmpc[i];\n\t\tn=cnt;\n\t\tint value=0;\n\t\tfor(int i=1;i<=n;i++) if(a[i]<=1) value+=c[i];\n\t\tif(test==0&&m==4000&&v==2000) printf(\"%lld\\n\",value);\n\t\tfor(int i=0;i<=n;i++)\n\t\t\tfor(int j=0;j<=v;j++)\n\t\t\t\tf[i][j]=-1e18;\n\t\tf[0][0]=0;\n\t\tfor(int i=0;i<=2*v+10;i++)\n\t\t\tfor(int j=0;j<=2*v+10;j++)\n\t\t\t\tc1[i].tr[j]=c2[i].tr[j]=-1e18;\n\t\tfor(int i=0;i<=n;i++)\n\t\t\tfor(int j=0;j<=v;j++)\n\t\t\t{\n//\t\t\t\tprintf(\"%lld %lld\\n\",i,j);\n\t\t\t\tif(i)\n\t\t\t\t{\n\t\t\t\t\tf[i][j]=max(f[i][j],c2[j-a[i]+v+1].ask(a[i]+1)+c[i]);\n\t\t\t\t\tf[i][j]=max(f[i][j],c1[j].ask(v-a[i]+1)+c[i]);\n\t\t\t\t}\n\t\t\t\tc1[j].add(v-a[i]+1,f[i][j]);\n\t\t\t\tc2[j-a[i]+v+1].add(a[i]+1,f[i][j]);\n\t\t\t}\n\t\tint maxv=-2147483647;\n\t\tfor(int j=0;j<=n;j++) maxv=max(maxv,f[j][0]);\n\t\tfor(int i=1;i<=v;i++)\n\t\t{\n\t\t\tfor(int j=0;j<=n;j++) maxv=max(maxv,f[j][i]);\n\t\t\tprintf(\"%lld \",maxv);\n\t\t}\n\t\tprintf(\"\\n\");\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 291518806,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define int long long\nusing namespace std;\nint n,m,v,test,c[200010],a[200010],tmpa[200010],tmpc[200010],f[5010][5010];\nstruct fenwick{\n\tint tr[5010],maxn;\n\tvoid add(int x,int y)\n\t{\n\t\tfor(;x<=maxn;x+=(x&(-x)))\n\t\t\ttr[x]=max(tr[x],y);\n\t}\n\tint ask(int x)\n\t{\n\t\tint maxv=-2147483647;\n\t\tfor(;x;x-=(x&(-x)))\n\t\t\tmaxv=max(maxv,tr[x]);\n\t\treturn maxv;\n\t}\n}c1[5010],c2[5010];\nsigned main()\n{\n\tscanf(\"%lld\",&test);\n\tfor(;test--;)\n\t{\n\t\tscanf(\"%lld%lld%lld\",&n,&m,&v);\n\t\tfor(int i=1;i<=n;i++) scanf(\"%lld\",&c[i]),a[i]=0;\n\t\tfor(int i=1;i<=m;i++)\n\t\t{\n\t\t\tchar op;\n\t\t\tint x;\n\t\t\tscanf(\"%*c%c%lld\",&op,&x);\n\t\t\tif(op=='L')\n\t\t\t\ta[1]++,a[x+1]--;\n\t\t\telse\n\t\t\t\ta[x]++;\n\t\t}\n\t\tfor(int i=2;i<=n;i++) a[i]+=a[i-1];\n\t\tint cnt=0;\n\t\tfor(int i=1;i<=n;i++)\n\t\t\tif(a[i]<=v)\n\t\t\t\ttmpa[++cnt]=a[i],tmpc[cnt]=c[i];\n\t\tn=cnt;\n\t\tcnt=0;\n\t\tfor(int i=1;i<=n;i++)\n\t\t\tif(cnt==0||tmpa[i]!=a[cnt])\n\t\t\t\ta[++cnt]=tmpa[i],c[cnt]=tmpc[i];\n\t\t\telse\n\t\t\t\tc[cnt]+=tmpc[i];\n\t\tn=cnt;\n\t\tint value=0;\n\t\tfor(int i=1;i<=n;i++) if(a[i]<=1) value+=c[i];\n//\t\tif(test==0&&m==4000&&v==2000) printf(\"%lld\\n\",value);\n\t\tfor(int i=0;i<=n;i++)\n\t\t\tfor(int j=0;j<=v;j++)\n\t\t\t\tf[i][j]=-1e18;\n\t\tf[0][0]=0;\n\t\tfor(int i=0;i<=2*v+10;i++)\n\t\t\tfor(int j=0;j<=2*v+10;j++)\n\t\t\t\tc1[i].tr[j]=c2[i].tr[j]=-1e18,c1[i].maxn=c2[i].maxn=2*v+5;\n\t\tfor(int i=0;i<=n;i++)\n\t\t\tfor(int j=0;j<=v;j++)\n\t\t\t{\n//\t\t\t\tprintf(\"%lld %lld\\n\",i,j);\n\t\t\t\tif(i)\n\t\t\t\t{\n\t\t\t\t\tf[i][j]=max(f[i][j],c2[j-a[i]+v+1].ask(a[i]+1)+c[i]);\n\t\t\t\t\tif(!(m==4000&&v==2000)) f[i][j]=max(f[i][j],c1[j].ask(v-a[i]+1)+c[i]);\n\t\t\t\t}\n\t\t\t\tc1[j].add(v-a[i]+1,f[i][j]);\n\t\t\t\tc2[j-a[i]+v+1].add(a[i]+1,f[i][j]);\n\t\t\t}\n\t\tint maxv=-2147483647;\n\t\tfor(int j=0;j<=n;j++) maxv=max(maxv,f[j][0]);\n\t\tif(test==0&&m==4000&&v==2000) printf(\"%lld\\n\",maxv);\n\t\tfor(int i=1;i<=v;i++)\n\t\t{\n\t\t\tfor(int j=0;j<=n;j++) maxv=max(maxv,f[j][i]);\n\t\t\tprintf(\"%lld \",maxv);\n\t\t}\n\t\tprintf(\"\\n\");\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 291520822,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define int long long\nusing namespace std;\nint n,m,v,test,c[200010],a[200010],tmpa[200010],tmpc[200010],f[5010][5010];\nstruct fenwick{\n\tint tr[5010],maxn;\n\tvoid add(int x,int y)\n\t{\n\t\tfor(;x<=maxn;x+=(x&(-x)))\n\t\t\ttr[x]=max(tr[x],y);\n\t}\n\tint ask(int x)\n\t{\n\t\tint maxv=-2147483647;\n\t\tfor(;x;x-=(x&(-x)))\n\t\t\tmaxv=max(maxv,tr[x]);\n\t\treturn maxv;\n\t}\n}c1[5010],c2[5010];\nsigned main()\n{\n\tscanf(\"%lld\",&test);\n\tfor(;test--;)\n\t{\n\t\tscanf(\"%lld%lld%lld\",&n,&m,&v);\n\t\tfor(int i=1;i<=n;i++) scanf(\"%lld\",&c[i]),a[i]=0;\n\t\tfor(int i=1;i<=m;i++)\n\t\t{\n\t\t\tchar op;\n\t\t\tint x;\n\t\t\tscanf(\"%*c%c%lld\",&op,&x);\n\t\t\tif(op=='L')\n\t\t\t\ta[1]++,a[x+1]--;\n\t\t\telse\n\t\t\t\ta[x]++;\n\t\t}\n\t\tfor(int i=2;i<=n;i++) a[i]+=a[i-1];\n\t\tint cnt=0;\n\t\tfor(int i=1;i<=n;i++)\n\t\t\tif(a[i]<=v)\n\t\t\t\ttmpa[++cnt]=a[i],tmpc[cnt]=c[i];\n\t\tn=cnt;\n\t\tcnt=0;\n\t\tfor(int i=1;i<=n;i++)\n\t\t\tif(cnt==0||tmpa[i]!=a[cnt])\n\t\t\t\ta[++cnt]=tmpa[i],c[cnt]=tmpc[i];\n\t\t\telse\n\t\t\t\tc[cnt]+=tmpc[i];\n\t\tn=cnt;\n//\t\tint value=0;\n//\t\tfor(int i=1;i<=n;i++) if(a[i]<=1) value+=c[i];\n//\t\tif(test==0&&m==4000&&v==2000) printf(\"%lld\\n\",value);\n\t\tfor(int i=0;i<=n;i++)\n\t\t\tfor(int j=0;j<=v;j++)\n\t\t\t\tf[i][j]=-1e18;\n\t\tf[0][0]=0;\n\t\tfor(int i=0;i<=2*v+10;i++)\n\t\t\tfor(int j=0;j<=2*v+10;j++)\n\t\t\t\tc1[i].tr[j]=c2[i].tr[j]=-1e18,c1[i].maxn=c2[i].maxn=2*v+5;\n\t\tfor(int i=1;i<=n;i++)\n\t\t{\n\t\t\tif(m==4000&&v==2000)\n\t\t\t{\n\t\t\t\tassert(a[i]>1);\n\t\t\t}\n\t\t}\n\t\tfor(int i=0;i<=n;i++)\n\t\t\tfor(int j=0;j<=v;j++)\n\t\t\t{\n\t\t\t\tif(i)\n\t\t\t\t{\n\t\t\t\t\tf[i][j]=max(f[i][j],c2[j-a[i]+v+1].ask(a[i]+1)+c[i]);\n\t\t\t\t\tf[i][j]=max(f[i][j],c1[j].ask(v-a[i]+1)+c[i]);\n\t\t\t\t}\n\t\t\t\tc1[j].add(v-a[i]+1,f[i][j]);\n\t\t\t\tc2[j-a[i]+v+1].add(a[i]+1,f[i][j]);\n\t\t\t}\n\t\tint maxv=-2147483647;\n\t\tfor(int j=0;j<=n;j++) maxv=max(maxv,f[j][0]);\n//\t\tif(test==0&&m==4000&&v==2000) printf(\"%lld\\n\",maxv);\n\t\tfor(int i=1;i<=v;i++)\n\t\t{\n\t\t\tfor(int j=0;j<=n;j++) maxv=max(maxv,f[j][i]);\n\t\t\tprintf(\"%lld \",maxv);\n\t\t}\n\t\tprintf(\"\\n\");\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 291549644,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define pi pair<int,int>\n#define pb push_back\n#define vec vector<int>\n#define M 998244353\n#define int long long\nusing namespace std;\nvoid in(int &x){\n int f=1;x=0;\n char c=getchar();\n while(!isdigit(c)&&c!='-')c=getchar();\n if(c=='-')f=-1,c=getchar();\n while(isdigit(c))x=x*10+c-'0',c=getchar();\n x*=f;\n}\nint t,n,m,k,x,y,b[200005],d[200005],a[4005],c[4005],res[4005];\nchar s[205];\nstruct BIT{\n int c[2005];\n void clear(){for(int i=0;i<=k+1;i++)c[i]=-1e9;}\n void upd(int x,int y){x++;while(x<=k+1)c[x]=max(c[x],y),x+=(x&(-x));}\n int query(int x){x++;int ans=-1e9;while(x>=1)ans=max(ans,c[x]),x^=(x&(-x));return ans;}\n}T[4005],B[2005];\nvoid solve(){\n in(n);in(m);in(k);\n for(int i=0;i<=n;i++)b[i]=0;\n for(int i=1;i<=n;i++)in(d[i]);\n for(int i=1;i<=m;i++){\n scanf(\"%s\",s+1);in(x);\n if(s[1]=='L')b[1]++,b[x+1]--;\n else b[x]++;\n }\n for(int i=1;i<=n;i++)b[i]+=b[i-1];\n m=0;\n for(int i=1;i<=n;){\n int j=i;while(j<n&&b[j]==b[j+1])j++;\n int sum=0;for(int o=i;o<=j;o++)sum+=d[o];\n if(b[j]<=k)a[++m]=b[j],c[m]=sum;\n i=j+1;\n }\n for(int i=0;i<=2*k;i++)T[i].clear();\n for(int i=0;i<=k;i++)B[i].clear();\n T[k].upd(0,0);B[0].upd(k,0);\n for(int i=0;i<=k+1;i++)res[i]=0;\n for(int i=1;i<=m;i++){\n for(int j=0;j<=k;j++){\n int f=-1e9;\n f=max(f,T[j-a[i]+k].query(a[i])+c[i]);\n f=max(f,B[j].query(k-a[i]-1)+c[i]);\n T[j-a[i]+k].upd(a[i],f);\n B[j].upd(k-a[i],f);\n res[j]=max(res[j],f);\n }\n }\n for(int i=1;i<=k;i++)res[i]=max(res[i],res[i-1]);\n for(int i=1;i<=k;i++)printf(\"%lld \",res[i]);puts(\"\");\n}\nsigned main(){\n in(t);\n while(t--)solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 293497086,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Balanced Problem\ntime_limit_ms: 3000\nmemory_limit_mb: 1024\nDescription: There is an array $a$ consisting of $n$ integers. Initially, all elements of $a$ are equal to $0$.\n\nKevin can perform several operations on the array. Each operation is one of the following two types:\n\n * Prefix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $1\\le j\\le x$, increases $a_j$ by $1$; * Suffix addition — Kevin first selects an index $x$ ($1\\le x\\le n$), and then for each $x\\le j\\le n$, increases $a_j$ by $1$.\n\nIn the country of KDOI, people think that the integer $v$ is balanced. Thus, Iris gives Kevin an array $c$ consisting of $n$ integers and defines the beauty of the array $a$ as follows:\n\n * Initially, set $b=0$; * For each $1\\le i\\le n$, if $a_i=v$, add $c_i$ to $b$; * The beauty of $a$ is the final value of $b$.\n\nKevin wants to maximize the beauty of $a$ after all the operations. However, he had already performed $m$ operations when he was sleepy. Now, he can perform an arbitrary number (possibly zero) of new operations.\n\nYou have to help Kevin find the maximum possible beauty if he optimally performs the new operations.\n\nHowever, to make sure that you are not just rolling the dice, Kevin gives you an integer $V$, and you need to solve the problem for each $1\\le v\\le V$.\n\nInput: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\\le t\\le 1000$) — the number of test cases. The description of test cases follows.\n\nThe first line of each test case contains three integers $n$, $m$, and $V$ ($1\\le n, m\\le 2\\cdot 10^5$, $1\\le V\\le 2000$) — the length of the array $a$, the number of initial operations, and the number that Kevin gives you.\n\nThe second line contains $n$ integers $c_1, c_2, \\ldots, c_n$ ($1\\le c_i\\le 10^9$) — the elements in the array $c$.\n\nThen $m$ lines follow, the $i$-th line containing a character $op$ and an integer $x$ ($op=\\mathtt{L}$ or $\\mathtt{R}$, $1\\le x\\le n$) — the type of the $i$-th operation and the selected index.\n\n * If $op=\\mathtt{L}$, this operation is a prefix addition on index $x$; * If $op=\\mathtt{R}$, this operation is a suffix addition on index $x$. \n\nIt is guaranteed that:\n\n * the sum of $n$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $m$ over all test cases does not exceed $2\\cdot 10^5$; * the sum of $V^2$ over all test cases does not exceed $4\\cdot 10^6$.\n\nOutput: For each test case, output $V$ integers in a single line, the $i$-th integer denoting the maximum possible beauty after Kevin performs some new operations when $v=i$.\n\nExamples:\ninput:\n5\n3 3 2\n1 2 4\nL 3\nR 3\nL 1\n3 3 2\n5 1 4\nL 3\nR 3\nL 1\n5 4 5\n1 1 1 1 1\nL 3\nR 2\nL 5\nL 4\n10 12 9\n10 9 8 7 6 5 4 3 2 1\nL 2\nL 4\nR 4\nR 4\nL 6\nR 8\nL 3\nL 2\nR 1\nR 10\nL 8\nL 1\n1 1 4\n1000000000\nL 1\noutput:\n2 6\n1 9\n0 1 3 5 5\n0 0 0 6 25 32 35 44 51\n1000000000 1000000000 1000000000 1000000000\n\nNote: In the first test case, the array $a$ changes as follows for the initial operations: $[0, 0, 0] \\xrightarrow{\\mathtt{L}\\ 3} [1, 1, 1] \\xrightarrow{\\mathtt{R}\\ 3} [1, 1, 2] \\xrightarrow{\\mathtt{L}\\ 1} [2, 1, 2]$.\n\n * For $v=1$, it is optimal to not perform any new operations, and the beauty is $b=c_2=2$; * For $v=2$, it is optimal to perform a prefix addition operation on index $2$. After that, $a$ becomes $[3,2,2]$, and the beauty is $b=c_2+c_3=6$. \n\nIn the second test case, for both $v=1$ and $v=2$, it is optimal to not perform any new operations.\n\nBuggy Code:\n#pragma GCC optimize(\"Ofast\",\"unroll-all-loops\",\"fast-math\",\"no-stack-protector\")\n#include <bits/stdc++.h>\nusing namespace std;\n#define fi first\n#define se second\n#if 1\nvoid __print(int x) {cerr << x;}\nvoid __print(long x) {cerr << x;}\nvoid __print(long long x) {cerr << x;}\nvoid __print(unsigned x) {cerr << x;}\nvoid __print(unsigned long x) {cerr << x;}\nvoid __print(unsigned long long x) {cerr << x;}\nvoid __print(float x) {cerr << x;}\nvoid __print(double x) {cerr << x;}\nvoid __print(long double x) {cerr << x;}\nvoid __print(char x) {cerr << '\\'' << x << '\\'';}\nvoid __print(const char *x) {cerr << '\\\"' << x << '\\\"';}\nvoid __print(const string &x) {cerr << '\\\"' << x << '\\\"';}\nvoid __print(bool x) {cerr << (x ? \"true\" : \"false\");}\ntemplate<typename T, typename V>\nvoid __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}\ntemplate<typename T>\nvoid __print(const T &x) {int f = 0; cerr << '['; for (auto &i: x) cerr << (f++ ? \",\" : \"\"), __print(i); cerr << \"]\";}\nvoid _print() {cerr << endl << flush;}\ntemplate <typename T, typename... V>\nvoid _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << \", \"; _print(v...);}\n#define DEBUG(x...) cerr << \"*[\"<<__LINE__<<\"]\\t\"<< #x << \" = \"; _print(x)\ntemplate<class T>\nistream& operator>>(istream& is, vector<T>& v) {\n\tint n; is>>n; v.resize(n);\n\tfor(auto &i:v) is>>i;\n\treturn is;\n}\ntemplate<class T1,class T2>\nistream& operator>>(istream& is, pair<T1,T2>& p) {\n\tis>>p.fi>>p.se;\n\treturn is;\n}\ntemplate<class T>\nostream& operator<<(ostream& os, const vector<T>& v) {\n\tfor(const auto &i:v) os<<i<<' ';\n\treturn os;\n}\ntemplate<class T1,class T2>\nostream& operator<<(ostream& os, const pair<T1,T2>& p) {\n\tos<<p.fi<<' '<<p.se; return os;\n}\n#endif\n \n#define ll long long\n#define mp make_pair\n#define pb push_back\n#define vi vector<ll>\n#define pi pair<int, int>\n#define mod 998244353\ntemplate<typename T> bool chkmin(T &a, T b){return (b < a) ? a = b, 1 : 0;}\ntemplate<typename T> bool chkmax(T &a, T b){return (b > a) ? a = b, 1 : 0;}\nll ksm(ll a, ll b) {if (b == 0) return 1; ll ns = ksm(a, b >> 1); ns = ns * ns % mod; if (b & 1) ns = ns * a % mod; return ns;}\n \nconst int maxn = 200005;\nint val[maxn];\nconst int V = 6005;\nint cur[maxn];\nint cnt = 0;\nint need[V];\nll sumv[V];\nll bst[V][V]; // best when last = c, sumdiff = h\nll dp[V][V];\nint main() {\n int t;\n cin >> t;\n while (t--) {\n int n, m, v;\n cin >> n >> m >> v;\n for (int i = 1; i <= n; i++) cin >> val[i];\n memset(cur, 0, sizeof(cur));\n for (int i = 1; i <= m; i++) {\n char tp[3]; int v;\n cin >> tp >> v;\n if (tp[0] == 'L') cur[1] += 1, cur[v + 1] -= 1;\n else cur[v] += 1;\n }\n for (int i = 1; i <= n; i++) cur[i] += cur[i - 1];\n // for (int i = 1; i <= n; i++) cout << cur[i] << ' '; cout << endl;\n cnt = 0;\n for (int i = 1; i <= n; ) {\n if (cur[i] > v) {\n i += 1;\n continue;\n }\n int j = i; ll sum = 0;\n while (j <= n && cur[j] == cur[i]) sum += val[j], j += 1;\n need[++cnt] = v - cur[i];\n sumv[cnt] = sum;\n i = j;\n }\n // for (int i = 1; i <= cnt; i++) cout << need[i] << ' ' << sumv[i] << endl;\n \n for (int i = 0; i <= v; i++) \n for (int j = 0; j <= v; j++) bst[i][j] = -1e18;\n for (int i = 1; i <= cnt; i++) {\n for (int d = 0; d <= v; d++) {\n dp[i][d] = sumv[i];\n for (int lst = 0; lst <= v; lst++) {\n int nxt = d; \n if (lst < need[i]) nxt -= need[i] - lst; // nxt = d + lst - need[i]\n if (nxt >= 0) \n chkmax(dp[i][d], bst[lst][nxt] + sumv[i]);\n }\n // max(lst > v; bst[lst][d])\n // or : max(lst <= v; bst[lst][d + lst - need[i]])\n }\n for (int d = 1; d <= v; d++) chkmax(dp[i][d], dp[i][d - 1]);\n for (int d = 0; d <= v; d++) chkmax(bst[need[i]][d], dp[i][d]);\n }\n for (int i = 1; i <= v; i++) {\n ll ans = 0;\n int minus = v - i;\n for (int j = 1; j <= cnt; j++) {\n // need[j] -> need[j]\n if (need[j] >= minus) chkmax(ans, dp[j][need[j] - minus]);\n }\n cout << ans << ' ';\n }\n cout << endl;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2029G",
"submission_id": 290762841,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\n// Problem: F. Expected Median\n// Contest: Codeforces - Codeforces Round 964 (Div. 4)\n// URL: https://codeforces.com/contest/1999/problem/F\n// Memory Limit: 256 MB\n// Time Limit: 3000 ms\n// \n// Powered by CP Editor (https://cpeditor.org)\n \n#include <bits/stdc++.h>\nusing namespace std;\n \n/*#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\nusing namespace __gnu_pbds;\ntemplate<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\ntemplate<class T> using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;\n// order_of_key(k): Gives the count of elements smaller than k. � O(log n)\n// find_by_order(k): Returns the iterator for the kth element (use k = 0 for the first element). � O(log n)*/\n#define int unsigned long long\n#define endl '\\n'\n#define thokkukuntopovale ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);\n#define vi vector<int>\n#define pb push_back\n#define vii vector<vector<int>>\n#define ch custom_hash\nint mod = 1e9+7;\nint inf = 1e18;\nvoid TxtIO(string name){\n\tfreopen((name + \".in\").c_str(), \"r\", stdin);\n\tfreopen((name + \".out\").c_str(), \"w\", stdout);\n}\nmt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());\nint fact[1<<20];\n \nint power(int a,int b){\n\tint res = 1;\n\ta = a%mod;\n\tif(a==0)return 0;\n\twhile(b>0){\n\t\tif(b%2){\n\t\t\tres = (res*a)%mod;\n\t\t}\n\t\tb = b/2;\n\t\ta = (a*a)%mod;\n\t}\n\treturn res%mod;\n}\n \nint ncr(int n,int r){\n\treturn (fact[n]*power((fact[n-r]*fact[r])%mod,mod-2))%mod;\n}\n \nvoid solve() {\n int n,m,k,i,j,ans=0;\n cin>>n>>k;\n int cnt0=0,cnt1=0;\n for(i=0;i<n;i++){\n \tcin>>j;\n \tif(j)cnt1++;\n \telse cnt0++;\n }\n if(cnt1<(k+1)/2){\n \tcout<<0<<endl;\n \treturn;\n }\n \n for(i=(k+1)/2;i<=k;i++){\n \tj = k - i;\n \tif(j>cnt0)continue;\n \tans=(ans+(ncr(cnt0,j)*ncr(cnt1,i))%mod)%mod;\n }\n cout<<ans<<endl;\n}\n \nint32_t main() {\n \n thokkukuntopovale;\n \n int t = 1;\n cin >> t;\n fact[0]=1;\n int n = 2e5+5;\n for(int i=1;i<=n;i++){\n \tfact[i] = (fact[i-1]*i)%mod;\n }\n while(t--) {\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 275183262,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\n// time-limit: 3000\n// problem-url: https://codeforces.com/contest/1999/problem/F\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define ll long long\n#define ld long double\n#define ar array\n \n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\nusing namespace __gnu_pbds;\n \ntemplate <typename T>\nusing oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\n \n#define vt vector\n#define pb push_back\n#define inf 0x3f3f3f3f\n#define infll 0x3f3f3f3f3f3f3f3f\n#define all(x) (x).begin(), (x).end()\n#define allr(x) (x).rbegin(), (x).rend()\n#define sz(x) (int)(x).size()\n#define nl '\\n'\n \n#define F_OR(i, a, b, s) for (int i = (a); (s) > 0 ? i < (b) : i > (b); i += (s))\n#define F_OR1(e) F_OR(i, 0, e, 1)\n#define F_OR2(i, e) F_OR(i, 0, e, 1)\n#define F_OR3(i, b, e) F_OR(i, b, e, 1)\n#define F_OR4(i, b, e, s) F_OR(i, b, e, s)\n#define GET5(a, b, c, d, e, ...) e\n#define F_ORC(...) GET5(__VA_ARGS__, F_OR4, F_OR3, F_OR2, F_OR1)\n#define FOR(...) F_ORC(__VA_ARGS__)(__VA_ARGS__)\n#define each(x, a) for (auto& x : a)\n \ntemplate <class T>\nbool cmin(T& a, const T& b) {\n return b < a ? a = b, 1 : 0;\n}\ntemplate <class T>\nbool cmax(T& a, const T& b) {\n return a < b ? a = b, 1 : 0;\n}\n \nvoid __print(int x) { cerr << x; }\nvoid __print(long x) { cerr << x; }\nvoid __print(long long x) { cerr << x; }\nvoid __print(unsigned x) { cerr << x; }\nvoid __print(unsigned long x) { cerr << x; }\nvoid __print(unsigned long long x) { cerr << x; }\nvoid __print(float x) { cerr << x; }\nvoid __print(double x) { cerr << x; }\nvoid __print(long double x) { cerr << x; }\nvoid __print(char x) { cerr << '\\'' << x << '\\''; }\nvoid __print(const char* x) { cerr << '\\\"' << x << '\\\"'; }\nvoid __print(const string& x) { cerr << '\\\"' << x << '\\\"'; }\nvoid __print(bool x) { cerr << (x ? \"true\" : \"false\"); }\n \ntemplate <typename T, typename V>\nvoid __print(const pair<T, V>& x) {\n cerr << '{';\n __print(x.first);\n cerr << ',';\n __print(x.second);\n cerr << '}';\n}\ntemplate <typename T>\nvoid __print(const T& x) {\n int f = 0;\n cerr << '{';\n for (auto& i : x) cerr << (f++ ? \",\" : \"\"), __print(i);\n cerr << \"}\";\n}\nvoid _print() { cerr << \"]\\n\"; }\ntemplate <typename T, typename... V>\nvoid _print(T t, V... v) {\n __print(t);\n if (sizeof...(v)) cerr << \", \";\n _print(v...);\n}\n#ifndef ONLINE_JUDGE\n#define debug(x...) \\\n cerr << \"[\" << #x << \"] = [\"; \\\n _print(x)\n#else\n#define debug(x...)\n#endif\n \ntemplate <class A>\nvoid read(vt<A>& v);\ntemplate <class A, size_t S>\nvoid read(ar<A, S>& a);\ntemplate <class T>\nvoid read(T& x) {\n cin >> x;\n}\nvoid read(double& d) {\n string t;\n read(t);\n d = stod(t);\n}\nvoid read(long double& d) {\n string t;\n read(t);\n d = stold(t);\n}\ntemplate <class H, class... T>\nvoid read(H& h, T&... t) {\n read(h);\n read(t...);\n}\ntemplate <class A>\nvoid read(vt<A>& x) {\n each(a, x)\n read(a);\n}\ntemplate <class A, size_t S>\nvoid read(array<A, S>& x) {\n each(a, x)\n read(a);\n}\n \nstring to_string(char c) {\n return string(1, c);\n}\nstring to_string(bool b) {\n return b ? \"true\" : \"false\";\n}\nstring to_string(const char* s) {\n return string(s);\n}\nstring to_string(string s) {\n return s;\n}\nstring to_string(vt<bool> v) {\n string res;\n FOR(sz(v))\n res += char('0' + v[i]);\n return res;\n}\n \ntemplate <size_t S>\nstring to_string(bitset<S> b) {\n string res;\n FOR(S)\n res += char('0' + b[i]);\n return res;\n}\ntemplate <class T>\nstring to_string(T v) {\n bool f = 1;\n string res;\n each(x, v) {\n if (!f)\n res += ' ';\n f = 0;\n res += to_string(x);\n }\n return res;\n}\n \ntemplate <class A>\nvoid write(A x) {\n cout << to_string(x);\n}\ntemplate <class H, class... T>\nvoid write(const H& h, const T&... t) {\n write(h);\n write(t...);\n}\nvoid print() {\n write(\"\\n\");\n}\ntemplate <class H, class... T>\nvoid print(const H& h, const T&... t) {\n write(h);\n if (sizeof...(t))\n write(' ');\n print(t...);\n}\n \nconst ll MAXN = 2e5 + 7;\nconst ll MOD = 1e9 + 7;\n \nll fact[MAXN], invFact[MAXN];\n \nll pow(ll a, ll b) {\n if (!b)\n return 1LL;\n \n ll ans = pow(a, b >> 1LL);\n ans = (ans * ans) % MOD;\n \n if (b & 1LL)\n ans = (ans * a) % MOD;\n \n return ans;\n}\n \nll ncr(ll n, ll r) {\n ll ans = fact[n];\n ans = (ans * invFact[r]) % MOD;\n ans = (ans * invFact[n - r]) % MOD;\n return ans;\n}\n \nvoid solve() {\n ll N,K; read(N,K);\n vt<ll> A(N); read(A);\n ll ones = 0;\n each(a,A) ones += a == 1LL;\n ll zeros = N - ones;\n ll need_zeros = min(K/2,zeros);\n ll ans = ncr(need_zeros + ones, K) % MOD;\n ans = (ans * ncr(zeros,need_zeros)) % MOD;\n // print(\"needz z \", need_zeros, zeros);\n // print(\"A \", A);\n if (need_zeros < zeros && K <= ones) {\n ans -= ncr(zeros,need_zeros) * ncr(ones,K);\n ans += ncr(ones,K);\n }\n print(ans);\n}\n \nint main() {\n ios::sync_with_stdio(0);\n cin.tie(0);\n \n fact[0] = invFact[0] = 1LL;\n FOR(i, 1, 1e5 + 7) {\n fact[i] = (fact[i - 1] * i) % MOD;\n invFact[i] = pow(fact[i], MOD - 2LL);\n }\n \n int t;\n read(t);\n FOR(t) {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 274923901,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n \nusing namespace std;\n#define int long long\n#define ll long long\n#define endl \"\\n\"\n#define INF 0x3f3f3f3f3f3f3f3f\n#define cy cout<<\"Yes\"<<endl\n#define cn cout<<\"No\"<<endl\n#define debug(x) std::cout<<#x<<\" = \"<<x<<endl;\n#define x first\n#define y second\ntypedef pair<int,int> PII;\ntypedef pair<int,pair<int,int> > PIII;\n \nbool cmp0(int a,int b) {return a>b;}\ntemplate<typename T> T gcd(T a,T b) {return b==0?a:gcd(b,a%b);}\ntemplate<typename T> T lcm(T a,T b) {return a*b/gcd(a,b);}\n \nconst int N=2e5+10,mod=1e9+7;\nint qkp(int base,int power){\n\tint ret=1;\n\twhile(power){\n\t\tif(power%2) ret=ret*base%mod;\n\t\tbase=base*base%mod;\n\t\tpower/=2;\n\t}\n\treturn ret%mod;\n} \n \nint n,m,k,p,h;\nint a[N];\nvoid solve()\n{\t\n\tcin>>n>>k;\n\tint sum=0;\n\tfor(int i=1;i<=n;i++)\n\t{\n\t\tcin>>a[i];\n\t\tif(a[i]==1)sum++;\n\t}\n\tint zw=(k+1)/2;\n\tif(sum<zw)\n\t{\n\t\tcout<<0<<endl;\n\t\treturn ;\n\t}\n\telse\n\t{\n\t\tint ans=1;\n\t\t//c[n][k]-c[n-sum+zw-1][k]*c[sum][zw-1]\n\t\tint tmp=n;\n\t\tint ct=k;\n\t\twhile(ct--)\n\t\t{\n\t\t\tans=ans*tmp%mod;\n\t\t\ttmp--;\n\t\t}\n\t\tct=k;\n\t\twhile(ct)\n\t\t{\n\t\t\tans=ans*qkp(ct,mod-2)%mod;\n\t\t\tct--;\n\t\t}\n\t\tint res=1;\n\t\ttmp=n-sum+zw-1;\n\t\tct=k;\n\t\twhile(ct--)\n\t\t{\n\t\t\tres=res*tmp%mod;\n\t\t\ttmp--;\n\t\t}\n\t\tct=k;\n\t\twhile(ct)\n\t\t{\n\t\t\tres=res*qkp(ct,mod-2)%mod;\n\t\t\tct--;\n\t\t}\n\t\ttmp=sum;\n\t\tct=zw-1;\n\t\twhile(ct--)\n\t\t{\n\t\t\tres=res*tmp%mod;\n\t\t\ttmp--;\n\t\t}\n\t\tct=zw-1;\n\t\twhile(ct)\n\t\t{\n\t\t\tres=res*qkp(ct,mod-2)%mod;\n\t\t\tct--;\n\t\t}\n\t\tcout<<(ans-res+2*mod)%mod<<endl;\n\t}\n}\nsigned main()\n{\n\tios::sync_with_stdio(false);\n\tcin.tie(0);\n\tint t=1;\n\tcin>>t;\n\twhile(t--)solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 274923561,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\ndef comb(fact, Mod, n, r):\n if n < r:\n return 0\n num = fact[n]\n den = (fact[n-r] * fact[r]) % Mod\n den = pow(den, Mod-2, Mod)\n return (num * den) % Mod\n \nt = int(input())\nwhile t:\n t -= 1\n n, k = map(int, input().split())\n l = []\n s = input()\n count_0=0\n count_1=0\n \n for i in range(0,2*n,2):\n x = int(s[i])\n l.append(int(x))\n if x == 0 :\n count_0+=1\n else:\n count_1+=1\n \n #print(l)\n \n Mod = int(1e9 + 7)\n fact = [1] * (int(2e5) + 1)\n for i in range(1, int(2e5) + 1):\n fact[i] = fact[i-1] * i % Mod\n \n \n \n mid = (k + 1) // 2\n ans = 0\n \n for i in range(mid):\n x = k - i\n comb0 = comb(fact, Mod, count_0, i)\n comb1 = comb(fact, Mod, count_1, x)\n ans = (ans + comb0 * comb1 % Mod) % Mod\n \n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 275369620,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\n\"\"\"\n思路：对于这种 在长度为 n 的数组里 选 m 个子序列，一般要用到组合数\n我们首先可以得出一个结论，即这个数组是一个二进制数组，也就是说里面的元素无非就是1和0，然后题意说取中位数就是说排序之\n的第 (数组长度+1) / 2位元素就是中位数，但是这个是二进制数组，只需要保证1的数量大于0，这个数组的中位数就是1，这里就\n省略证明了\n然后我们可以先求出组合数，然后 - 那些配对的不合法的1和0的数量\nf = C(1的数量,m // 2) * C(0的数量,向上取整(m/2))\nans = C(n,m) - f\n\"\"\"\nimport math\ndef invC(n,m,MOD):\n def exgcd(a,b):\n if b == 0:\n return a,1,0\n d,x,y = exgcd(b,a%b)\n return d,y,x - (a//b)* y\n def cor(a,b,m):\n gcd,x,y = exgcd(a,m)\n if b % gcd != 0:return False\n if b == 1:return (x % m + m) % m\n else:\n return x * (b/gcd)\n inv = cor((math.factorial(m) * math.factorial(n - m)),1,MOD)\n return math.factorial(n) * inv\n \ndef solve():\n n, m = map(int,input().split())\n li = list(map(int,input().split()))\n MOD = 10**9 + 7\n cnt0,cnt1 = 0,0\n for i in li:\n if i == 0:cnt0+=1\n if i == 1:cnt1+=1\n if cnt1 < m//2:return 0\n if cnt0 < math.ceil(m/2):f = 0\n else:\n f = (invC(cnt1, m // 2,MOD) * invC(cnt0, math.ceil(m / 2),MOD))% MOD\n return int((invC(n, m,MOD) - f) % MOD)\n \n \nfor i in range(int(input())):print(solve())\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 275220426,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nfrom math import perm, factorial, comb\nmod = pow(10, 9) + 7\n \nfor _ in range(int(input())):\n n, k = map(int, input().split())\n a = list(map(int, input().split()))\n zeros = a.count(0)\n ones = a.count(1)\n m = comb(n, k)\n m1 = comb(zeros, k // 2 + 1)\n m2 = comb(ones, k // 2) if ones > 0 else 1\n print((m - (m2 * m1)) % mod)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 275013789,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \n \npublic class Main {\n static int mod = (int) (1e9 + 7);\n static long[] jc;\n \n \n public static void main(String[] args) {\n Kattio sc = new Kattio();\n int t = sc.nextInt();\n jc = new long[(int) (3e5 + 1)];\n jc[0] = 1;\n for (int i = 1; i < jc.length; i++) {\n jc[i] = jc[i - 1] * i % mod;\n jc[i] %= mod;\n }\n while (t-- > 0) solve(sc);\n }\n \n private static void solve(Kattio sc) {\n int n = sc.nextInt();\n int k = sc.nextInt();\n int[] a = new int[2];\n for (int i = 0; i < n; i++) {\n a[sc.nextInt()]++;\n }\n long ans = 0;\n if (a[1] >= k / 2 + 1) {\n if (a[0] != 0) {\n for (int i = k / 2 + 1; i <=k&&i<=a[1]&&a[0]>=k-i&&k-i>=0; i++) {\n ans += (C(a[1], i) % mod * C(a[0], k - i) % mod) % mod;\n ans %= mod;\n }\n }else {\n ans=C(a[1], Math.min(k,a[1])) % mod;\n }\n System.out.println(ans);\n } else {\n System.out.println(0);\n }\n \n \n }\n \n private static long C(int n, int dx) { //组合数\n if (n == 0) return 1;\n long x = jc[n] % mod;\n long y = jc[dx] * jc[n - dx] % mod;\n return x * fastPow(y, mod - 2, mod) % mod;\n }\n \n private static long fastPow(long n, int m, int p) {\n if (m <= 0) return 1;\n long t = fastPow(n, m >> 1, p) % p;\n if (m % 2 == 0) return (t % p * t % p) % p;\n return (n * t % p * t % p) % p;\n }\n \n static class Kattio extends PrintWriter {\n private BufferedReader r;\n private StringTokenizer st;\n \n public Kattio() {\n this(System.in, System.out);\n }\n \n public Kattio(InputStream i, OutputStream o) {\n super(o);\n r = new BufferedReader(new InputStreamReader(i));\n }\n \n public Kattio(String intput, String output) throws IOException {\n super(output);\n r = new BufferedReader(new FileReader(intput));\n }\n \n public String next() {\n try {\n while (st == null || !st.hasMoreTokens())\n st = new StringTokenizer(r.readLine());\n return st.nextToken();\n } catch (Exception e) {\n }\n return null;\n }\n \n public int nextInt() {\n return Integer.parseInt(next());\n }\n \n public double nextDouble() {\n return Double.parseDouble(next());\n }\n \n public long nextLong() {\n return Long.parseLong(next());\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 274944026,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.math.BigInteger;\nimport java.text.DecimalFormat;\nimport java.util.*;\nimport java.util.stream.Collectors;\n \nimport javax.net.ssl.HandshakeCompletedEvent;\nimport javax.xml.transform.Source;\n \npublic class problem {\n \n static int mod=1000000007;\n public static void main(String[] args) {\n FastReader sc = new FastReader();\n\t\tint t=sc.nextInt();\n \n\t\tl1:\n\t\twhile(t-->0){\n\t\t \n \n \n int n=sc.nextInt();\n int k=sc.nextInt();\n \n int[] arr=new int[n];\n int z=0,o=0;\n for(int i=0; i<n; i++){\n arr[i]=sc.nextInt();\n if(arr[i]==0)z++;\n else o++;\n }\n if(o<=k/2){\n System.out.println(0);\n continue;\n }\n if(k==1){\n System.out.println(o);\n continue;\n }\n long ans=combination(n, k);\n int req=(k/2)+1;\n if(req>z){\n System.out.println(ans%mod);\n }else{\n long com=combination(z, req);\n ans-=com*o;\n System.out.println(ans%mod);\n }\n }\n }\n \n \n \n public static int temp(int a,int b,int c,int d){\n int count1=0;\n int count2=0;\n \n if(a>b){\n count1++;\n }else if(a!=b){\n count2++;\n }\n \n if(c>d){\n count1++;\n }else if(c!=d){\n count2++;\n }\n return count1>count2?1:0;\n }\n \n \n \n public static boolean[] prime(int limit) {\n boolean[] isPrime=new boolean[limit];\n Arrays.fill(isPrime, true);\n isPrime[0] = false;\n isPrime[1] = false;\n \n for (int i = 2; i * i <= limit; i++) {\n if (isPrime[i]) {\n for (int j = i * i; j <= limit; j += i) {\n isPrime[j] = false;\n }\n }\n }\n return isPrime;\n \n \n }\n \n public static boolean isPrime(int num) {\n if (num <= 1) {\n return false;\n }\n if (num <= 3) {\n return true;\n }\n if (num % 2 == 0 || num % 3 == 0) {\n return false; \n }\n for (int i = 5; i * i <= num; i += 6) {\n if (num % i == 0 || num % (i + 2) == 0) {\n return false; \n }\n }\n return true; \n }\n \n public static int gcd(int a, int b) {\n while (b != 0) {\n int temp = b;\n b = a % b;\n a = temp;\n }\n return a;\n }\n \n public static int findGCD(int[] arr) {\n int result = arr[0];\n for (int i = 1; i < arr.length; i++) {\n result = gcd(result, arr[i]);\n if (result == 1) {\n return 1;\n }\n }\n return result;\n }\n \n public static long combination(int n, int k) {\n if (k > n) {\n return 0;\n }\n if (k == 0 || k == n) {\n return 1;\n }\n k = Math.min(k, n - k);\n long c = 1;\n for (int i = 0; i < k; i++) {\n c = c * (n - i) / (i + 1);\n }\n return c;\n }\n \n static class FastReader { \n BufferedReader br; \n StringTokenizer st; \n \n public FastReader() \n { \n br = new BufferedReader( \n new InputStreamReader(System.in)); \n } \n \n String next() \n { \n while (st == null || !st.hasMoreElements()) { \n try { \n st = new StringTokenizer(br.readLine()); \n } \n catch (IOException e) { \n e.printStackTrace(); \n } \n } \n return st.nextToken(); \n } \n \n int nextInt() { return Integer.parseInt(next()); } \n \n long nextLong() { return Long.parseLong(next()); } \n \n double nextDouble() \n { \n return Double.parseDouble(next()); \n } \n \n String nextLine() \n { \n String str = \"\"; \n try { \n if(st.hasMoreTokens()){ \n str = st.nextToken(\"\\n\"); \n } \n else{ \n str = br.readLine(); \n } \n } \n catch (IOException e) { \n e.printStackTrace(); \n } \n return str; \n } \n } \n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 275056133,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define problem \"test\"\n#define ll long long\n#define pii pair<int,int>\n#define pll pair<ll,ll>\n#define fi first\n#define se second\n#define ALL(v) (v).begin(),(v).end()\n#define UNIQUE(v) (v).resize(unique(ALL(v)) - (v).begin())\n#define BIT(x,i) (((x) >> (i)) & 1)\n#define MASK(i) (1LL << (i))\nusing namespace std;\nconst int N = 2e5;\nconst ll mod = 1e9 + 7;\n \nint n,k;\nint numOne,numZero;\nll fact[N + 5],inv[N + 5];\n \nll binaryExpo(ll base,ll k){\n ll res = 1;\n for (;k > 0;k >>= 1){\n if (k & 1) res = (res * base) % mod;\n base = (base * base) % mod;\n }\n return res;\n}\n \nll Com(int r,int i){\n if (i > r) return 0;\n if (i == 0 || i == r) return 1;\n return (((1LL*fact[r]*inv[i]) % mod)*inv[r - i]) % mod;\n}\n \nint main(){\n ios_base :: sync_with_stdio(false);\n cin.tie(NULL);\tcout.tie(NULL);\n // freopen(problem\".inp\",\"r\",stdin);\n // freopen(problem\".out\",\"w\",stdout);\n int numTest; cin>>numTest;\n \n fact[0] = 1;\n for (int i=1;i<= N;i++ )\n fact[i] = (1LL*fact[i - 1]*i) % mod;\n inv[N] = binaryExpo(fact[N],mod - 2);\n for (int i=N - 1;i>= 1;i-- )\n inv[i] = (1LL*inv[i + 1]*(i + 1)) % mod;\n \n while (numTest-- ){\n cin>>n>>k;\n numZero = numOne = 0;\n for (int i=1;i<= n;i++ ){\n int x; cin>>x;\n (x == 1 ? numOne++ : numZero++);\n }\n \n int need = (k + 1) >> 1;\n if (numOne < need)\n cout<<0<<'\\n';\n else{\n ll res = 0;\n for (int i=need;i<= min(numOne,k);i++ )\n res += (1LL*Com(numOne,i)*Com(numZero,k - i))% mod;\n cout<<res<<'\\n';\n }\n \n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 274908172,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\n#include <iostream>\n#include <cmath>\n#include <algorithm>\nusing namespace std;\n \nlong long mod = 1e9+7;\n \n// Returns factorial of n\ndouble fact(int n)\n{\n if(n==0)\n return 1;\n double res = 1;\n for (int i = 2; i <= n; i++)\n res = res * i;\n return res;\n}\ndouble nCR(int n, int r)\n{\n return fact(n) / (fact(r) * fact(n - r));\n}\n \n \n \nint main() {\n int t; cin >> t;\n while (t-->0){\n int n, k; cin >> n >> k;\n int numones = 0; int numzero = 0;\n for(int i=0; i<n; i++) {\n int x; cin >> x;\n if(x==0) numzero++; else numones++;\n }\n int choseones = min(k, numones); int choosezero = k-choseones;\n long long sum = 0;\n if(choseones==k){\n double val = nCR(numones,choseones);\n sum = (long long)fmod(val,mod);\n choseones--; choosezero++;\n }\n while (choseones>k/2){\n if(numones>=choseones && numzero>=choosezero){\n double val = nCR(numones,choseones)*nCR(numzero,choosezero);\n sum += (long long)(fmod(val,mod)); sum=(long long)fmod(sum,mod);\n choseones--; choosezero++;}\n else break;\n }\n cout << sum << '\\n';\n }\n \n \n \n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 274981487,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nimport sys\ninput = sys.stdin.readline\n \n \nM = 10**9+7\nN = 2*10**5\nd = [1]\nfor i in range(1, N+1):\n d.append(d[-1]*i%M)\n \n \ndef comb(k, n):\n return d[n] * pow((d[k]*d[n-k]), -1, M) % M\n \n \nfor _ in range(int(input())):\n n, k = map(int, input().split())\n w = input()\n a = w.count('1')\n b = n-a\n c = 0\n x = (k+1)//2\n for i in range(x, min(a, k)+1):\n j = k-i\n if j <= b:\n c += comb(i, a) * comb(j, b)\n print(c)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 280467961,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nimport os\nimport sys\nfrom io import BytesIO, IOBase\n \nBUFSIZE = 8192\n \n \nclass FastIO(IOBase):\n newlines = 0\n \n def __init__(self, file):\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n \n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n \n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n \n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\n \n \nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\n \n \nsys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\ninput = lambda: sys.stdin.readline().rstrip(\"\\r\\n\")\n \nimport decimal\n \ndecimal.getcontext().prec = 10000\n \n \ndef solve():\n n, k = map(int, input().split())\n arr = list(map(int, input().split()))\n x = arr.count(0)\n y = n - x\n if y < (k + 1) // 2:\n print(0)\n return\n min_num_of_zero = max(k - y, 0)\n min_num_of_1 = max((k + 1) // 2, k - x)\n temp1 = decimal.Decimal(1)\n temp2 = decimal.Decimal(1)\n for i in range(min_num_of_zero):\n temp1 *= decimal.Decimal((x - i)) / decimal.Decimal((min_num_of_zero - i))\n for i in range(k - min_num_of_zero):\n temp2 *= (decimal.Decimal(y - i)) / (decimal.Decimal(k - min_num_of_zero - i))\n temp = (int(temp1) % (1000000007)) * (int(temp2) % (1000000007)) % (1000000007)\n ans = temp\n for i in range(min_num_of_zero + 1, k - min_num_of_1 + 1):\n # C_x^i * C_y^(k-i)\n j = y - (k - i)\n temp = int((decimal.Decimal(temp) * decimal.Decimal((y - j + 1))) / decimal.Decimal(j)) % (1000000007)\n temp = int((decimal.Decimal(temp) * decimal.Decimal((x - i + 1))) / decimal.Decimal(i)) % (1000000007)\n # print(\"Uytr\", i, x, j, y, temp)\n ans = (ans + temp) % (1000000007)\n print(int(ans))\n \n \nfor _ in range(int(input())):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 274978949,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n \n static long MOD = (long) 1e9 + 7;\n static int MAXN = (int) 2e5;\n static long[] invFact = new long[MAXN + 1];\n static long[] fact = new long[MAXN + 1];\n \n public static void main(String[] args) throws IOException {\n FastIO io = new FastIO();\n int T = io.nextInt();\n \n fact[0] = 1;\n invFact[0] = inv(1);\n for (int i = 1; i <= MAXN; i++) {\n fact[i] = (fact[i - 1] * i) % MOD;\n invFact[i] = inv(fact[i]);\n }\n \n for (int t = 0; t < T; t++) {\n int N = io.nextInt();\n int K = io.nextInt();\n \n int[] arr = new int[N];\n int tot = 0;\n \n for (int i = 0; i < N; i++) {\n arr[i] = io.nextInt();\n tot += arr[i];\n }\n \n long ans = 0;\n for (int ones = (K + 1) / 2; ones <= Math.min(tot, K); ones++) {\n int zeros = K - ones;\n ans += (choose(tot, ones) * choose(N - tot, zeros)) % MOD;\n }\n \n io.println(ans);\n }\n io.close();\n }\n \n static long choose(int n, int k) {\n if (n < k) return 0;\n return ((fact[n] * invFact[k]) % MOD * invFact[n - k]) % MOD;\n }\n \n static long exp(long num, long x) {\n if (x == 0) return 1;\n long res = exp(num, x / 2);\n long prod = (res * res) % MOD;\n if (x % 2 == 1) return (prod * num) % MOD;\n else return prod;\n }\n \n static long inv(long x) {\n return exp(x, MOD - 2);\n }\n}\nclass FastIO extends PrintWriter {\n \n\tprivate InputStream stream;\n \n\tprivate byte[] buf = new byte[1 << 16];\n \n\tprivate int curChar;\n \n\tprivate int numChars;\n \n \n\t// standard input\n \n\tpublic FastIO() { this(System.in, System.out); }\n \n \n\tpublic FastIO(InputStream i, OutputStream o) {\n \n\t\tsuper(o);\n \n\t\tstream = i;\n \n\t}\n \n \n\t// file input\n \n\tpublic FastIO(String i, String o) throws IOException {\n \n\t\tsuper(new FileWriter(o));\n \n\t\tstream = new FileInputStream(i);\n \n\t}\n \n \n\t// throws InputMismatchException() if previously detected end of file\n \n\tprivate int nextByte() {\n \n\t\tif (numChars == -1) { throw new InputMismatchException(); }\n \n\t\tif (curChar >= numChars) {\n \n\t\t\tcurChar = 0;\n \n\t\t\ttry {\n \n\t\t\t\tnumChars = stream.read(buf);\n \n\t\t\t} catch (IOException e) { throw new InputMismatchException(); }\n \n\t\t\tif (numChars == -1) {\n \n\t\t\t\treturn -1; // end of file\n \n\t\t\t}\n \n\t\t}\n \n\t\treturn buf[curChar++];\n \n\t}\n \n \n\t// to read in entire lines, replace c <= ' '\n \n\t// with a function that checks whether c is a line break\n \n\tpublic String next() {\n \n\t\tint c;\n \n\t\tdo { c = nextByte(); } while (c <= ' ');\n \n \n\t\tStringBuilder res = new StringBuilder();\n \n\t\tdo {\n \n\t\t\tres.appendCodePoint(c);\n \n\t\t\tc = nextByte();\n \n\t\t} while (c > ' ');\n \n\t\treturn res.toString();\n \n\t}\n \n \n\tpublic int nextInt() { // nextLong() would be implemented similarly\n \n\t\tint c;\n \n\t\tdo { c = nextByte(); } while (c <= ' ');\n \n \n\t\tint sgn = 1;\n \n\t\tif (c == '-') {\n \n\t\t\tsgn = -1;\n \n\t\t\tc = nextByte();\n \n\t\t}\n \n \n\t\tint res = 0;\n \n\t\tdo {\n \n\t\t\tif (c < '0' || c > '9') { throw new InputMismatchException(); }\n \n\t\t\tres = 10 * res + c - '0';\n \n\t\t\tc = nextByte();\n \n\t\t} while (c > ' ');\n \n\t\treturn res * sgn;\n \n\t}\n \n\tpublic long nextLong() { // nextLong() would be implemented similarly\n \n\t\tint c;\n \n\t\tdo { c = nextByte(); } while (c <= ' ');\n \n \n\t\tint sgn = 1;\n \n\t\tif (c == '-') {\n \n\t\t\tsgn = -1;\n \n\t\t\tc = nextByte();\n \n\t\t}\n \n \n\t\tlong res = 0;\n \n\t\tdo {\n \n\t\t\tif (c < '0' || c > '9') { throw new InputMismatchException(); }\n \n\t\t\tres = 10 * res + c - '0';\n \n\t\t\tc = nextByte();\n \n\t\t} while (c > ' ');\n \n\t\treturn res * sgn;\n \n\t} \n \n\tpublic double nextDouble() { return Double.parseDouble(next()); }\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 276321698,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nimport java.util.Arrays;\nimport java.util.Scanner;\n \npublic class F{\n static int mod=(int)1e9+7;\n private static long combi(int n,int k){\n long ans=1;\n k=k>n-k?n-k:k;\n int j=1;\n for(;j<=k;j++,n--)\n {\n if(n%j==0)\n {\n ans*=n/j;\n }else\n if(ans%j==0)\n {\n ans=ans/j*n;\n }else\n {\n ans=(ans*n)/j;\n }\n }\n return ans%mod;\n }\n \n public static void main(String[] args) {\n Scanner in = new Scanner(System.in);\n int ti=1;\n ti=in.nextInt();\n \n while((ti--)>0){\n int n=in.nextInt();\n int k=in.nextInt();\n int[] a=new int[n];\n for(int i=0;i<n;i++){\n a[i]=in.nextInt();\n }\n Arrays.sort(a);\n k/=2;\n int ans=0;\n for(int i=k;i<n-k;i++){\n if(a[i]==1){\n ans = (int)((ans+((combi(i, k)*combi(n-i-1, k))%mod))%mod);\n }\n }\n System.out.println(ans);\n }\n in.close();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 274882185,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nimport sys\nimport math\nfrom itertools import combinations\nfrom functools import lru_cache\n \nMOD = 1000000007\n \n# Returns factorial of n using memoization for efficiency\n@lru_cache(None)\ndef fact(n):\n if n == 0:\n return 1\n res = 1\n for i in range(2, n + 1):\n res = res * i\n return res\n \n# Returns nCr (combination of n items taken r at a time)\ndef nCr(n, r):\n return fact(n) // (fact(r) * fact(n - r))\n \ndef solve():\n n, k = map(int, input().split())\n a = list(map(int, input().split()))\n b = list(range(n))\n ans = 0\n half = k // 2\n \n b.sort(key=lambda x: a[x])\n a.sort()\n \n pos = {}\n for i in range(n):\n pos[b[i]] = (i, n - i - 1)\n \n for i in range(n):\n x = a[i]\n curr = 0\n if i >= half and n - i > half:\n curr = (nCr(i, half) * nCr(n - 1 - i, half)) % MOD\n ans = (curr * x + ans) % MOD\n \n print(ans % MOD)\n \ndef main():\n t = int(input())\n for _ in range(t):\n solve()\n \nif __name__ == \"__main__\":\n main()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 278944813,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nimport sys\nfrom collections import defaultdict,deque,Counter\nfrom heapq import heappush,heappop,heapify\nfrom math import inf,comb\nfrom functools import cache\nfrom itertools import accumulate,combinations,permutations\ninput=lambda:sys.stdin.readline().rstrip()\ninp=lambda:map(int,input().split())\nt=int(input())\nmod=10**9+7\nfor _ in range(t):\n n,k=inp()\n *s,=inp()\n \n occ=[0,0]\n o=0\n for e in s:\n occ[e]+=1\n for i in range((k+1)//2,k+1):\n o+=comb(occ[1],i)*comb(occ[0],k-i)\n o%=mod\n print(o)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 274852014,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.math.BigInteger;\nimport java.text.DecimalFormat;\nimport java.util.*;\nimport java.util.stream.Collectors;\n \nimport javax.net.ssl.HandshakeCompletedEvent;\nimport javax.xml.transform.Source;\n \npublic class problem {\n \n static BigInteger mod=new BigInteger(\"1000000007\");\n public static void main(String[] args) {\n FastReader sc = new FastReader();\n\t\tint t=sc.nextInt();\n \n\t\tl1:\n\t\twhile(t-->0){\n\t\t \n \n \n int n=sc.nextInt();\n int k=sc.nextInt();\n \n int[] arr=new int[n];\n int z=0,o=0;\n for(int i=0; i<n; i++){\n arr[i]=sc.nextInt();\n if(arr[i]==0)z++;\n else o++;\n }\n if(o<=k/2){\n System.out.println(0);\n continue;\n }\n if(k==1){\n System.out.println(o);\n continue;\n }\n BigInteger ans=combination(n, k);\n int req=(k/2)+1;\n int req1=k-req;\n if(req>z){\n System.out.println(ans.mod(mod));\n }else{\n BigInteger temp=BigInteger.ZERO;\n while(req<=z && req<=k){\n BigInteger com1=combination(z, req);\n BigInteger com2=combination(o, req1);\n temp=temp.add(com1.multiply(com2));\n req++;\n if(req1>0)\n req1--;\n }\n ans=ans.subtract(temp).mod(mod);\n System.out.println(ans);\n }\n }\n }\n \n \n \n public static int temp(int a,int b,int c,int d){\n int count1=0;\n int count2=0;\n \n if(a>b){\n count1++;\n }else if(a!=b){\n count2++;\n }\n \n if(c>d){\n count1++;\n }else if(c!=d){\n count2++;\n }\n return count1>count2?1:0;\n }\n \n \n \n public static boolean[] prime(int limit) {\n boolean[] isPrime=new boolean[limit];\n Arrays.fill(isPrime, true);\n isPrime[0] = false;\n isPrime[1] = false;\n \n for (int i = 2; i * i <= limit; i++) {\n if (isPrime[i]) {\n for (int j = i * i; j <= limit; j += i) {\n isPrime[j] = false;\n }\n }\n }\n return isPrime;\n \n \n }\n \n public static boolean isPrime(int num) {\n if (num <= 1) {\n return false;\n }\n if (num <= 3) {\n return true;\n }\n if (num % 2 == 0 || num % 3 == 0) {\n return false; \n }\n for (int i = 5; i * i <= num; i += 6) {\n if (num % i == 0 || num % (i + 2) == 0) {\n return false; \n }\n }\n return true; \n }\n \n public static int gcd(int a, int b) {\n while (b != 0) {\n int temp = b;\n b = a % b;\n a = temp;\n }\n return a;\n }\n \n public static int findGCD(int[] arr) {\n int result = arr[0];\n for (int i = 1; i < arr.length; i++) {\n result = gcd(result, arr[i]);\n if (result == 1) {\n return 1;\n }\n }\n return result;\n }\n \n public static BigInteger combination(int n, int k) {\n if (k > n) {\n return BigInteger.ZERO;\n }\n if (k == 0 || k == n) {\n return BigInteger.ONE;\n }\n k = Math.min(k, n - k); // Take advantage of symmetry\n BigInteger c = BigInteger.ONE;\n for (int i = 0; i < k; i++) {\n c = c.multiply(BigInteger.valueOf(n - i))\n .divide(BigInteger.valueOf(i + 1));\n }\n return c;\n }\n \n static class FastReader { \n BufferedReader br; \n StringTokenizer st; \n \n public FastReader() \n { \n br = new BufferedReader( \n new InputStreamReader(System.in)); \n } \n \n String next() \n { \n while (st == null || !st.hasMoreElements()) { \n try { \n st = new StringTokenizer(br.readLine()); \n } \n catch (IOException e) { \n e.printStackTrace(); \n } \n } \n return st.nextToken(); \n } \n \n int nextInt() { return Integer.parseInt(next()); } \n \n long nextLong() { return Long.parseLong(next()); } \n \n double nextDouble() \n { \n return Double.parseDouble(next()); \n } \n \n String nextLine() \n { \n String str = \"\"; \n try { \n if(st.hasMoreTokens()){ \n str = st.nextToken(\"\\n\"); \n } \n else{ \n str = br.readLine(); \n } \n } \n catch (IOException e) { \n e.printStackTrace(); \n } \n return str; \n } \n } \n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 275062972,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\nusing namespace std;\n \nconst int mod = 1000000007;\n \nlong long factorial(long long n, long long mod) {\n long long result = 1;\n for (long long i = 2; i <= n; ++i) {\n result = (result * i) % mod;\n }\n return result;\n}\n \nlong long mod_inverse(long long x, long long mod) {\n long long result = 1, base = x;\n long long exp = mod - 2;\n while (exp > 0) {\n if (exp % 2 == 1) {\n result = (result * base) % mod;\n }\n base = (base * base) % mod;\n exp /= 2;\n }\n return result;\n}\n \nlong long binomial_coefficient(long long n, long long k, long long mod) {\n if (k > n) return 0;\n if (k == 0 || k == n) return 1;\n long long numerator = factorial(n, mod);\n long long denominator = (factorial(k, mod) * factorial(n - k, mod)) % mod;\n return (numerator * mod_inverse(denominator, mod)) % mod;\n}\n \nvoid solve() {\n // Your code goes here\n int n, k;\n cin >> n >> k;\n vector<int> a(n);\n int c = 0, d = 0;\n for (int i = 0; i < n; ++i) {\n cin >> a[i];\n if (a[i] == 0) c++;\n else d++;\n }\n int ans = 0;\n for (int i = (k / 2) + 1; i <= min(d, k); i++) {\nans = (ans + ((binomial_coefficient(d, i, mod) % mod) * (binomial_coefficient(c, k - i, mod) % mod)) % mod) % mod; }\n cout << ans << endl;\n}\n \nint main() {\n cin.tie(NULL)->sync_with_stdio(false);\n int t = 1;\n cin >> t;\n // cout<<binomial_coefficient(200000,100000,mod);\n while (t--) solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 274936530,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ll long long\n#define int ll\n#define ull unsigned long long\n#define ld long double\n#define eb emplace_back\n#define pii pair<int, int>\n#define tpl tuple<int, int, int>\n#define OO LLONG_MAX / 2\n#define oo INT_MAX / 2\nusing namespace std;\n \n \nconst int MOD=1e9+7;\n \nint lgpow(int x,int y)\n{\n int z = 1;\n while (y != 0)\n {\n if (y % 2 == 1)\n z = 1LL*z * x % MOD;\n x = 1LL*x * x % MOD;\n y /= 2;\n }\n return z;\n}\n \nint fact(int n)\n{\n int p=1;\n for(int i=2;i<=n;i++)\n p*=i,p%=MOD;\n return p;\n}\n \nint combi(int n,int k)\n{\n if(k>n)\n return 1;\n return fact(n)*lgpow(fact(n-k),MOD-2)%MOD*lgpow(fact(k),MOD-2)%MOD;\n}\n \nvoid solve()\n{\n int n,k;\n cin>>n>>k;\n \n int _1=0,_0=0;\n \n for(int i=1;i<=n;i++)\n {\n int x;\n cin>>x;\n _1+=x==1;\n _0+=x==0;\n }\n \n int med=(k+1)/2;\n int ans=0;\n //cout<<max(1LL,k-_1)<<' '<<min(med-1,_0)<<'\\n';\n for(int i=max(1LL,k-_1);i<=min(med-1,_0);i++)\n {\n ///cout<<fact[2]*invfact[2]%MOD<<'\\n';\n //cout<<combi(_1,k-i)<<' '<<combi(_0,i)<<'\\n';\n ans=(ans+combi(_1,k-i)*combi(_0,i)%MOD)%MOD;\n }\n \n if(_1>=k)\n {\n ans+=combi(_1,k);\n ans%=MOD;\n }\n cout<<ans<<'\\n';\n}\n \n \n \nsigned main()\n{\n ios_base::sync_with_stdio(false);\n cin.tie(0);\n cout.tie(0);\n //init();\n int q = 1;\n cin >> q;\n while (q--)\n solve();\n exit(0);\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 274816873,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Expected Median\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Arul has a binary array$^{\\text{∗}}$ $a$ of length $n$.\n\nHe will take all subsequences$^{\\text{†}}$ of length $k$ ($k$ is odd) of this array and find their median.$^{\\text{‡}}$\n\nWhat is the sum of all these values?\n\nAs this sum can be very large, output it modulo $10^9 + 7$. In other words, print the remainder of this sum when divided by $10^9 + 7$.\n\n$^{\\text{∗}}$A binary array is an array consisting only of zeros and ones.\n\n$^{\\text{†}}$An array $b$ is a subsequence of an array $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements. Subsequences don't have to be contiguous.\n\n$^{\\text{‡}}$The median of an array of odd length $k$ is the $\\frac{k+1}{2}$-th element when sorted.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $k$ ($1 \\leq k \\leq n \\leq 2 \\cdot 10^5$, $k$ is odd) — the length of the array and the length of the subsequence, respectively.\n\nThe second line of each test case contains $n$ integers $a_i$ ($0 \\leq a_i \\leq 1$) — the elements of the array.\n\nIt is guaranteed that sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print the sum modulo $10^9 + 7$.\n\nExamples:\ninput:\n8\n4 3\n1 0 0 1\n5 1\n1 1 1 1 1\n5 5\n0 1 0 1 0\n6 3\n1 0 1 0 1 1\n4 3\n1 0 1 1\n5 3\n1 0 1 1 0\n2 1\n0 0\n34 17\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\noutput:\n2\n5\n0\n16\n4\n7\n0\n333606206\n\nNote: In the first test case, there are four subsequences of $[1,0,0,1]$ with length $k=3$:\n\n * $[1,0,0]$: median $= 0$. * $[1,0,1]$: median $= 1$. * $[1,0,1]$: median $= 1$. * $[0,0,1]$: median $= 0$. \n\nThe sum of the results is $0+1+1+0=2$.\n\nIn the second test case, all subsequences of length $1$ have median $1$, so the answer is $5$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\npublic class Main {\n public static final int MOD = 1_000_000_007;\n static PrintWriter out;\n static FastReader in;\n \n public static void main(String[] args) {\n in = new FastReader();\n out = new PrintWriter(System.out);\n int t = in.nextInt();\n while(t-->0){\n int n = in.nextInt();\n int k = in.nextInt();\n int[] b = new int[n];\n int a = 0;\n for(int i=0;i<n;i++){\n b[i]=in.nextInt();\n if(b[i]==1){\n a++;\n }\n }\n long ans = 0;\n if(a>=(k+1)/2){\n for(int i=(k+1)/2;i<=Math.min(a,k);i++){\n ans = (ans + (nCk(a, i) * nCk(n - a, k - i)) % MOD) % MOD;\n }\n }\n out.println(ans);\n }\n \n out.flush();\n out.close();\n }\n \n \n \n \n public static long nCk(int n, int k) {\n if (k > n) return 0;\n return (fact(n) * modInverse(fact(k), MOD) % MOD * modInverse(fact(n - k), MOD) % MOD) % MOD;\n }\n public static long fact(int n) {\n long result = 1;\n for (int i = 2; i <= n; i++) {\n result = (result * i) % MOD;\n }\n return result;\n }\n \n public static long modInverse(long a, int mod) {\n return power(a, mod - 2, mod);\n }\n \n public static long power(long base, long exp, int mod) {\n long result = 1;\n while (exp > 0) {\n if ((exp & 1) == 1) {\n result = (result * base) % mod;\n }\n base = (base * base) % mod;\n exp >>= 1;\n }\n return result;\n }\n public static int binSearch(int[] arr, int key){\n int l = 0;\n int r = arr.length-1;\n while (l<=r){\n int mid = (l+r)/2;\n if (arr[mid]==key){\n return mid;\n }\n if(arr[l]<arr[mid]){\n if (key>=arr[l]&&key<arr[mid]){\n r=mid-1;\n }\n else {\n l=mid+1;\n }\n }\n else {\n if (key>arr[mid]&&key<=arr[r]){\n l=mid+1;\n }\n else {\n r=mid-1;\n }\n }\n }\n return -1;\n }\n public static int lower_bound(int[] arr,int key){\n int l = 0;\n int r = arr.length-1;\n int mid;\n while(r-l>1){\n mid = (r+l)/2;\n if(key>arr[mid]){\n l=mid+1;\n }\n else{\n r=mid;\n }\n }\n if(arr[l]>=key){\n return l;\n }\n else if(arr[r]>=key){\n return r;\n }\n return -1;\n }\n \n public static int upper_bound(int[] arr,int key){\n int l = 0;\n int r = arr.length-1;\n int mid;\n while(r-l>1){\n mid = (r+l)/2;\n if(key>=arr[mid]){\n l=mid+1;\n }\n else{\n r=mid;\n }\n }\n if(arr[l]>key){\n return l;\n }\n else if(arr[r]>key){\n return r;\n }\n return -1;\n }\n \n public static boolean inRange(int[] arr, int L, int R) {\n Arrays.sort(arr);\n long product = 1;\n \n for (int i = 0; i < arr.length; i++) {\n for (int j = i + 1; j < arr.length; j++) {\n int xor = arr[i] ^ arr[j];\n product *= xor;\n \n if (product >= R || product==0) {\n return false;\n }\n }\n }\n return product>L&&product<R;\n }\n public static int findIndex(long[] arr, int target) {\n int low = 0, high = arr.length - 1;\n \n while (low <= high) {\n int mid = low + (high - low) / 2;\n \n if (arr[mid] == target) {\n return mid; // Target found\n } else if (arr[mid] < target) {\n low = mid + 1;\n } else {\n high = mid - 1;\n }\n }\n \n return low;\n }\n public static int maxOcurrences(int[] arr) {\n HashMap<Integer, Integer> frequencyMap = new HashMap<>();\n \n // Count the frequency of each number\n for (int num : arr) {\n frequencyMap.put(num, frequencyMap.getOrDefault(num, 0) + 1);\n }\n \n // Find the number with the highest frequency\n int maxCount = 0;\n int maxNumber = -1; // Default value (if the array is empty)\n for (HashMap.Entry<Integer, Integer> entry : frequencyMap.entrySet()) {\n if (entry.getValue() > maxCount) {\n maxCount = entry.getValue();\n maxNumber = entry.getKey();\n }\n }\n \n return maxCount;\n }\n public static String changeCharAt(String str, int index, char newChar) {\n StringBuilder sb = new StringBuilder(str);\n sb.setCharAt(index, newChar);\n return sb.toString();\n }\n static int gcd(int a, int b){\n if(b==0){\n return a;\n }\n return gcd(b,a%b);\n }\n \n static Boolean[] isPrime;\n static{\n int k = 10_000_010;\n isPrime = new Boolean[k];\n Arrays.fill(isPrime, true);\n isPrime[0]=isPrime[1]=false;\n for(int i=2;i<k;i++){\n if(isPrime[i]){\n for(int j=2*i;j<k;j+=i){\n isPrime[j]=false;\n }\n }\n }\n }\n static long exp(int a, int b){\n long ans = 1;\n long base = a;\n while(b>0){\n if((b&1)!=0){\n ans*=base;\n }\n base*=base;\n b>>=1;\n }\n return ans;\n }\n static long[] L_arr(int n) {\n long[] array = new long[n];\n for (int i = 0; i < n; i++) {\n array[i] = in.nextLong();\n }\n return array;\n }\n static <T> ArrayList<T> Al(int n) {\n ArrayList<T> list = new ArrayList<>();\n for (int i = 0; i < n; i++) {\n T element = (T) in.next();\n list.add(element);\n }\n return list;\n }\n \n static <K, V> HashMap<K, V> map() {\n return new HashMap<>();\n }\n static <T> ArrayList<T> list() {\n return new ArrayList<>();\n }\n static <K, V> Pair<K, V> pair(K key, V value) {\n return new Pair<>(key, value);\n }\n \n static class FastReader {\n BufferedReader br;\n StringTokenizer st;\n \n public FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() {\n while (st == null || !st.hasMoreElements()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n }\n \n static class Pair<K, V> extends AbstractMap.SimpleEntry<K, V> {\n Pair(K key, V value) {\n super(key, value);\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999F",
"submission_id": 296260176,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define int long long\nconst int maxn = 200000;\npair<int, int> arrpii[maxn];\nmap<int, vector<int>> mii;\nbitset<maxn> bs;\nsigned main() {\n cin.tie(0)->sync_with_stdio(0);\n cin.exceptions(cin.failbit);\n int tests;\n cin >> tests;\n while (tests--) {\n int n, m, k;\n cin >> n >> m >> k;\n mii.clear();\n for (int i = 0; i < k; i++) {\n cin >> arrpii[i].first >> arrpii[i].second;\n mii[arrpii[i].second].push_back(i);\n }\n int lowest = 0;\n vector<int> vi;\n bs.reset();\n for (auto &a : mii) {\n sort(a.second.begin(), a.second.end(),\n [](int a, int b) { return arrpii[a].first < arrpii[b].first; });\n if (arrpii[a.second.back()].first > lowest) {\n vi.push_back(a.second.back());\n lowest = arrpii[a.second.back()].first;\n bs[a.second.back()] = 1;\n }\n }\n int all = n * m;\n int last = 0;\n vector<int> prev;\n for (auto a : vi) {\n prev.push_back(last);\n pair<int, int> pii = arrpii[a];\n all -= (pii.first - last) * (m - pii.second + 1);\n last = pii.first;\n }\n cout << all << \"\\n\";\n \n for (int i = 0; i < k; i++) {\n if (!bs[i]) {\n cout << \"0 \";\n continue;\n }\n int itvi = lower_bound(vi.begin(), vi.end(), i) - vi.begin();\n int deficit = (arrpii[i].first - prev[itvi]) * (m - arrpii[i].second + 1);\n auto it = mii.find(arrpii[i].second);\n int last = prev[itvi];\n if (it->second.size() >= 2) {\n pair<int, int> coord = arrpii[it->second[it->second.size() - 2]];\n deficit -= (coord.first - last) * (m - coord.second + 1);\n last = coord.first;\n }\n it++;\n while (it != mii.end()) {\n int in = it->second.back();\n if (arrpii[in].first > last) {\n deficit -= (arrpii[in].first - last) * (m - arrpii[in].second + 1);\n last = arrpii[in].first;\n }\n if (bs[in]) {\n deficit +=\n (arrpii[in].first - prev[itvi + 1]) * (m - arrpii[in].second + 1);\n break;\n }\n \n it++;\n }\n cout << deficit << \" \";\n }\n cout << \"\\n\";\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 270822414,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n/**\n * https://codeforces.com/contest/1980/submission/267827508\n *\n * Copyright (c) 2024 Diego Sogari\n */\n#include <bits/stdc++.h>\n \nusing namespace std;\nusing filesystem::path;\nusing i64 = int64_t;\nusing f64 = double;\n \nvoid solve(int t) {\n int n, m, k;\n cin >> n >> m >> k;\n if (t == 16) {\n cout << n << ',' << m << ',' << k << ';';\n }\n map<int, map<int, int>> frows;\n for (int i = 0, r, c; i < k; i++) { // O(k*log k)\n cin >> r >> c;\n if (t == 16) {\n cout << r << ',' << c << ';';\n }\n auto &rows = frows[c];\n rows[r] = i;\n if (rows.size() > 2) {\n rows.erase(rows.begin());\n }\n }\n if (t == 16) {\n return;\n }\n vector<int> ans(k);\n i64 a = 0;\n int pra = 0, prb = 0, pca = 1, pcb = 1, pi = 0;\n auto f = [&](int c) { a += i64(n - prb) * (c - pcb); };\n auto g = [&](int c) { ans[pi] += (prb - pra) * (c - max(pca, pcb)); };\n for (auto &[c, map] : frows) { // O(k)\n auto [rb, i] = *map.rbegin();\n if (rb > prb) {\n f(c), g(c);\n auto ra = map.size() > 1 ? map.begin()->first : 0;\n pra = ra > pra ? ra : prb;\n pca = ra > pra ? c : pcb;\n prb = rb, pcb = c, pi = i;\n } else if (rb > pra) {\n g(c);\n pra = rb, pca = c;\n }\n }\n f(m + 1), g(m + 1);\n cout << a << endl;\n for (auto &&ai : ans) {\n cout << ai << ' ';\n }\n cout << endl;\n}\n \nint main() {\n#ifdef LOCAL\n freopen(path(__FILE__).replace_filename(\"input.txt\").c_str(), \"r\", stdin);\n#endif\n cin.tie(nullptr)->tie(nullptr)->sync_with_stdio(false);\n int t;\n cin >> t;\n for (int i = 1; i <= t; ++i) {\n solve(i);\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 267830731,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nusing i32 = int32_t;\nusing i64 = int64_t;\n \nusing pi32 = pair<i32, i32>;\nusing pi64 = pair<i64, i64>;\n \n#define MP make_pair\n#define MT make_tuple\n#define F first\n#define S second\n \nusing vi32 = vector<i32>;\nusing vi64 = vector<i64>;\n \nusing vpi32 = vector<pi32>;\nusing vpi64 = vector<pi64>;\n \nusing vvi32 = vector<vi32>;\nusing vvi64 = vector<vi64>;\n \nusing vvpi32 = vector<vpi32>;\nusing vvpi64 = vector<vpi64>;\n \n#define SZ(x) static_cast<i32>((x).size())\n#define RSZ(x, n) x.resize(n)\n#define ME(x, a) memset(x, a, sizeof(x))\n \n#define ALL(x) x.begin(), x.end()\n#define RALL(x) x.rbegin(), x.rend()\n \n#define PB push_back\n#define PF push_front\n#define PPB pop_back\n#define PPF pop_front\n#define EB emplace_back\n \n#define INS insert\n \n#define F0R(i, n) for(int i = 0; i < n; i ++) \n#define FOR(i, a, b) for(int i = a; i < b; i ++)\n#define R0F(i, n) for(int i = n - 1; i >= 0; i --) \n#define ROF(i, a, b) for(int i = b - 1; i >= a; i --)\n#define AUTO(a, x) for(auto &a : x)\n \nconst i64 INF = 1e18;\nconst i64 MOD = 1e9 + 7; // 998244353;\n \nvoid Main() {\n\ti64 n, m, k;\n\tcin >> n >> m >> k;\n\tvector<tuple<i64, i64, i64>> f(k + 1);\n\tF0R(i, k) {\n\t\ti64 r, c;\n\t\tcin >> r >> c;\n\t\tf[i] = MT(c, r, i);\n\t}\n\tf[k] = MT(m + 1, n, -1);\n\tsort(ALL(f));\n\ti64 res = 0;\n\ti64 prev = 0;\n\tvi64 ans(k);\n\tF0R(i, k + 1) {\n\t\tif(get<2>(f[i]) == -1) {\n\t\t\tres += (n - prev) * m;\n\t\t\tcontinue;\n\t\t}\n\t\tif(get<1>(f[i]) > prev) {\n\t\t\ti32 j = i + 1;\n\t\t\ti64 tmp = 0, tmp_prev = prev;\n\t\t\tbool next = true;\n\t\t\tif(get<0>(f[i + 1]) > get<0>(f[i])) {\n\t\t\t\twhile(get<1>(f[j]) < get<1>(f[i])) {\n\t\t\t\t\tnext = false;\n\t\t\t\t\tif(get<1>(f[j]) > tmp_prev) {\n\t\t\t\t\t\ttmp += (get<1>(f[j]) - tmp_prev) * ((get<0>(f[j])) - (get<0>(f[i]) - 1));\n\t\t\t\t\t\ttmp_prev = max(tmp_prev, get<1>(f[j]));\n\t\t\t\t\t}\n\t\t\t\t\tj ++;\n\t\t\t\t}\n\t\t\t\tif(next) {\n\t\t\t\t\tans[get<2>(f[i])] = (get<1>(f[i]) - prev) * ((get<0>(f[j]) - 1) - (get<0>(f[i]) - 1));\n\t\t\t\t}\n\t\t\t\telse {\n\t\t\t\t\tans[get<2>(f[i])] = (get<1>(f[i]) - prev) * ((get<0>(f[j]) - 1) - (get<0>(f[i]) - 1)) - tmp;\n\t\t\t\t}\n\t\t\t}\n\t\t\tres += (get<0>(f[i]) - 1) * (get<1>(f[i]) - prev);\n\t\t\tprev = get<1>(f[i]);\n\t\t}\n\t}\n\tcout << res << '\\n';\n\tF0R(i, k) {\n\t\tcout << ans[i] << ' '; \n\t}\n\tcout << '\\n';\n}\n \ni32 main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n //freopen(\"input.txt\", \"r\", stdin);\n //freopen(\"output.txt\", \"w\", stdout);\n i64 tc = 1ll;\n cin >> tc;\n while(tc--) {\n Main();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 266450426,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\nimport sys\nfrom collections import defaultdict,deque,Counter\nfrom heapq import heappush,heappop,heapify\nfrom math import inf,comb,ceil,log,gcd\nfrom functools import cache\nfrom itertools import accumulate,combinations,permutations\nfrom bisect import bisect_left,bisect_right,insort_left\nfrom random import randint\nxor=randint(0,2**32-1)\ninput=lambda:sys.stdin.readline().rstrip()\ninp=lambda:map(int,input().split())\nt=int(input())\nfor _ in range(t):\n n,m,k=inp()\n X=[[0,0]]\n for _ in range(k):\n X+=[*inp()][::-1],\n X+=[m+1,n+1],\n oX=X.copy()\n X=sorted([(e,i)for i,e in enumerate(X)],key=lambda t:(t[0][0],-t[0][1]))\n #print(X)\n h=-1\n ar=0\n s=[0]\n for (x,y),i in X:\n \n x-=1\n y-=1\n #print(\"xyi\",x,y,i,la,h)\n #if i==k+1:print(x,y)\n if y>h:\n ar+=(m-x)*(y-h)\n h=y\n s.append(i)\n \n s1=[]\n s2=deque()\n ftb=[*range(k+2)]\n stb=[*range(k+2)]\n for (x,y),i in X:\n #print(x,y,i)\n while s2 and (y,x)>(s2[-1][1],s2[-1][0]):\n cur=s2.pop()\n stb[cur[2]]=i\n while s1 and (y,x)>(s1[-1][1],s1[-1][0]):\n #print(s1,i)\n cur=s1.pop()\n s2.appendleft(cur)\n ftb[cur[2]]=i\n \n s1.append((x,y,i))\n #print([oX[i]for i in ftb])\n def rect(i1,i2,i3):\n return (oX[i2][0]-oX[i1][0])*(oX[i1][1]-oX[i3][1])\n res=[0]*k\n for i in s:\n el=ftb[i]\n el2=ftb[ftb[i]]\n el3=stb[i]\n if el-1<k:\n #print(oX[i],i-1,oX[el],el-1,oX[el2],el2-1,oX[el3],el3-1)\n #print(rect(el3,el2,i))\n res[el-1]=abs(rect(el,el2,i)-rect(el3,el2,i))\n #print(ar)\n #print(s)\n print(n*m-ar)\n print(*res)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 281118299,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n# import numpy as np\n \n# def getComponent(coor,mat):\n# tmp = mat\n# succ = getSuccessors(coor,tmp)\n# for s in succ:\n# tmp[s]='-'\n# getComponent(s,tmp)\n# componentLength = np.count_nonzero(tmp == '-')\n# return 1 if not componentLength else componentLength\n# def getMaxComponent(matrix,l,c):\n# mat = np.copy(matrix)\n# mat[l,:]='#'\n# mat[:,c]='#'\n# ls = list()\n# X,Y = np.where(mat == '#')[0],np.where(mat == '#')[1]\n# indexes = [(X[i],Y[i]) for i in range(np.shape(X)[0])]\n# for coor in indexes: \n# ls.append(getComponent(coor,np.copy(mat)))\n# return max(ls)\n \n# def getSuccessors(cell,mat):\n# successors = list()\n# if cell[0] != 0 and mat[cell[0]-1,cell[1]] == '#':\n# successors.append((cell[0]-1,cell[1]))\n# if cell[0] != np.shape(mat)[0]-1 and mat[cell[0]+1,cell[1]] == '#':\n# successors.append((cell[0]+1,cell[1]))\n# if cell[1] != 0 and mat[cell[0],cell[1]-1] == '#':\n# successors.append((cell[0],cell[1]-1))\n# if cell[1] != np.shape(mat)[1]-1 and mat[cell[0],cell[1]+1] == '#':\n# successors.append((cell[0],cell[1]+1))\n# return successors\n \n \n# t = int(input())\n# result = list()\n# for _ in range(t):\n# line = str(input()).split(' ')\n# n = int(line[0].strip())\n# m = int(line[1].strip())\n# m = np.ndarray(shape=(n,m),dtype='<U9')\n# for i in range (n):\n# row = input()\n# m[i]=[char for char in row]\n# lengthes = list()\n# for l in range (np.shape(m)[0]):\n# for c in range (np.shape(m)[1]):\n# lengthes.append(getMaxComponent(m,l,c))\n# result.append(max(lengthes))\n# for num in result:\n# print(num)\n# def calculateMinTurns(boss,attacks,cooldowns):\n# minTurns=0\n# tmpCooldown = [0]*len(cooldowns)\n# while(boss>0):\n# totDamage=0\n# for i in range(len(attacks)):\n# if(tmpCooldown[i] != 0):\n# tmpCooldown[i]+=1\n# if(tmpCooldown[i] == 0):\n# totDamage += attacks[i]\n# tmpCooldown[i] = 0 - cooldowns[i]\n \n# boss-=totDamage\n# minTurns+=1\n# return minTurns\n \n \n# t=int(input())\n# totScores = list()\n# for _ in range(t):\n# firstLine = input().split(' ')\n# bossHealth = int(firstLine[0])\n# numAttacks = int(firstLine[1])\n# attacks = [int(a) for a in input().split(' ')]\n# cooldowns = [int(c) for c in input().split(' ')]\n# totScores.append(calculateMinTurns(bossHealth,attacks,cooldowns))\n# for _ in totScores:\n# print(_)\n \n \n \n# t=int(input())\n# words=list()\n# for _ in range(t):\n# word1, word2 = input().split(' ')\n# words.append((word1,word2))\n# for elem in words:\n# print(elem[1][0]+elem[0][1:],elem[0][0]+elem[1][1:])\n# def process_test_case(ls):\n# n = len(ls)\n# iter_count = 1 << n # Equivalent to 2^n\n# max_sum = float('-inf')\n# count_max_sum = 0\n \n# for i in range(iter_count):\n# c = 0\n# for j in range(n):\n# if i & (1 << j): # Check if j-th bit is set in i\n# c = abs(c + ls[j])\n# else:\n# c += ls[j]\n# if c > max_sum:\n# max_sum = c\n# count_max_sum = 1\n# elif c == max_sum:\n# count_max_sum += 1\n \n# return count_max_sum\n \n# # Read input and process\n# t = int(input())\n# results = []\n# for _ in range(t):\n# input() # Skip the first line of each test case\n# ls = list(map(int, input().split()))\n# results.append(process_test_case(ls))\n \n# # Print results\n# for result in results:\n# print(result)\n# def generate_combinations(lst):\n# # Create a new list with elements as (value, index)\n# indexed_lst = [(value, idx) for idx, value in enumerate(lst)]\n \n# def combinations_recursive(current, remaining):\n# if current and remaining:\n# result.append((current, remaining))\n# for i in range(len(remaining)):\n# combinations_recursive(current + [remaining[i]], remaining[:i] + remaining[i+1:])\n \n# result = []\n# combinations_recursive([], indexed_lst)\n# return result\n# def getRange(ls):\n# min = 2**31-1\n# max = -min\n# for elem in ls:\n# if(elem[0]>max):\n# max = elem[0]\n# if(elem[0]<min):\n# min = elem[0]\n# return max - min\n \n# def getOutput(lst):\n# combinations_list = generate_combinations(lst)\n# for i in range(len(combinations_list)):\n# if(getRange(combinations_list[i][0]) != getRange(combinations_list[i][1])):\n# result = ['R']*len(lst)\n# for _ in combinations_list[i][0]:\n# result[_[1]]='B'\n# return ''.join(result)\n \n \n# t = int(input())\n# results = []\n# for _ in range(t):\n# input()\n# lst = [int(x) for x in input().split(' ')]\n# results.append(getOutput(lst))\n# for result in results:\n# if result == None:\n# print(\"NO\")\n# continue\n# print(f\"YES\\n{result}\")\nN=0\nM=0\nK=0\nimport gc\n \ndef getAliceCellsInRowCount(row):\n s = 0\n for cell in row:\n if not cell:\n break\n s += 1\n return s\n \ndef getOutput(kmatrix):\n matrix = [[True for _ in range(M)] for _ in range(N)]\n for elem in kmatrix:\n matrix[elem[0]][elem[1]:] = [False] * (M - elem[1])\n for i in range(elem[0] + 1):\n matrix[i][elem[1]] = False\n \n return sum(getAliceCellsInRowCount(row) for row in matrix)\n \nt = int(input())\nresults = []\nfor _ in range(t):\n possibilities = []\n kmatrix = []\n N, M, K = map(int, input().split())\n for _ in range(K):\n x, y = map(int, input().split())\n kmatrix.append((x - 1, y - 1))\n possibilities.append(getOutput(kmatrix))\n for k in range(K):\n possibilities.append(getOutput(kmatrix[:k] + kmatrix[k+1:]))\n \n results.append(possibilities[0])\n rr = \" \".join(str(res - possibilities[0]) for res in possibilities[1:])\n results.append(rr)\n \n # Clear memory for the current iteration\n del kmatrix\n del possibilities\n gc.collect()\n \nfor result in results:\n print(result)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 266192208,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\nimport math,random \nfrom collections import defaultdict\nfrom bisect import bisect_left, bisect_right\nimport random \nrd = random.randint(10**8,10**9) \nclass mydict:\n def __init__(self, func=lambda: 0):\n self.random = random.randint(0, 1 << 32)\n self.default = func\n self.dict = {}\n def __getitem__(self, key):\n mykey = self.random ^ key\n if mykey not in self.dict:\n self.dict[mykey] = self.default()\n return self.dict[mykey]\n def get(self, key, default):\n mykey = self.random ^ key\n if mykey not in self.dict:\n return default\n return self.dict[mykey]\n def __setitem__(self, key, item):\n mykey = self.random ^ key\n self.dict[mykey] = item\n def getkeys(self):\n return [self.random ^ i for i in self.dict]\n def __str__(self):\n return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'\nt = int(input())\nfor _ in range(t):\n n,m,k = map(int,input().split())\n a = []\n d = {}\n d2 = {}\n for i in range(k):\n r,c = map(int,input().split())\n a.append([r,c,i])\n \n a.sort(key = lambda x: (-x[0],x[1]))\n r = n+1\n rows = []\n cols = []\n d3 = mydict(list)\n rl = -1\n for i in range(k):\n d3[a[i][0]].append(a[i][1])\n if a[i][0]!=r:\n r = a[i][0]\n rows.append(a[i][0])\n cols.append(a[i][1])\n d2[(a[i][0],a[i][1])] = a[i][2]\n r = n+1\n c = m \n ans=0\n rows.append(0)\n cols.append(0)\n #print(rows)\n res = [0]*k\n for i in range(len(rows)):\n if i==0:\n x = r - rows[i]\n y = cols[i]-1\n ans+= y \n ans+=(x-1)*(c)\n #print(ans)\n #if fountain a[i] can be taken=>\n \n p = c-y\n if len(d3[rows[i]])>1:\n ll = list(d3[rows[i]])\n ll.sort()\n p = ll[1]-y-1\n rnext=0\n cnext=0\n if i<len(rows)-1: \n rnext = rows[i+1] \n cnext = cols[i+1]\n q = rows[i] - rnext\n temp = p*q\n if cnext>cols[i]: temp+=cnext-cols[i]\n res[d2[(rows[i],cols[i])]]=temp\n d[rows[i]]=temp \n #print(i,temp)\n prevc = cols[i] \n else:\n x = r - rows[i]\n ans+=(x-1)*(prevc-1)\n if rows[i]==0: break \n pprevc = prevc\n prevc = min(prevc,cols[i])\n y = prevc-1 \n ans+=y \n #print(ans,x,prevc,rows[i])\n #if fountain a[i] can be taken \n if cols[i]<pprevc:\n rnext = rows[i+1]\n p = rows[i]-rnext\n q=0\n if cols[i+1]>cols[i]: q = rows[i]-rows[i+1]\n d[rows[i]] = (p+q)*(pprevc-cols[i])\n #print(rows[i],d[rows[i]],p,q,pprevc)\n res[d2[(rows[i],cols[i])]]= d[rows[i]]\n else:\n d[rows[i]] = 0\n #print(ans)\n r = rows[i]\n \n print(ans)\n print(*res)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 264599753,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n#include <bits/stdc++.h>\n// #define LOOP_TEST_ABFS\n#define FROM_FILE_IO\n \n#if (defined XqAiyee) && (!defined LOOP_TEST_ABFS)\n#include <cur.hpp>\n#include <chronos.hpp>\n#define debug(...) (XQ::out_s(XQ::Not_Char{'\\0'}, XQ::Not_Str{#__VA_ARGS__ }, XQ::Not_Str{\":\"}, __VA_ARGS__) << std::endl)\n#else\n#define debug(...) (NULL)\n#endif\n \nusing i64 = int64_t;\nusing u64 = uint64_t;\nusing std::cin, std::cout, std::endl;\n \nconstexpr int MAXN = 2e5, N = MAXN + 2, MOD = 1e9 + 7, INF = 0x3f3f3f3f;\n \ni64 pre[N];\ni64 dp[N];\nint tr[N * 2];\n \nvoid solve(char opt = '\\0') {\n\tint n, m, k, id = 0;\n cin >> n >> m >> k;\n // col row id\n std::vector<std::array<int, 3>> cr(k);\n std::vector<std::pair<i64, std::array<int, 2>>> res(k + 2);\n for (auto &i : cr) cin >> i[1] >> i[0], i[2] = ++id;\n std::sort(cr.begin(), cr.end(), [](auto &a, auto &b) -> bool {\n if (a[0] != b[0]) return a[0] < b[0];\n return a[1] > b[1];\n });\n std::function<void (int, int, int)> build = [&](int u, int L, int R) -> void {\n if (L == R) return tr[u] = L, void();\n int mid = L + R >> 1, y = u + (mid - L + 1 << 1);\n build(u + 1, L, mid);\n build(y, mid + 1, R);\n if (cr[tr[u + 1]][1] >= cr[tr[y]][1]) tr[u] = tr[u + 1];\n else tr[u] = tr[y];\n };\n cr.push_back({m, 1, k + 1});\n build(0, 0, k);\n std::stack<std::array<int, 3>> stk;\n stk.push({m, 1, k + 1}), dp[k + 1] = 0;\n for (int i = k - 1; i >= 0; -- i) {\n while (stk.size() > 1 && stk.top()[1] <= cr[i][1]) stk.pop();\n i64 calc = i64(stk.top()[0] - cr[i][0] - 1) * (n - cr[i][1]);\n // debug(stk.top(), cr[i]);\n calc += n - std::max(cr[i][1], stk.top()[1]);\n dp[cr[i][2]] = calc + dp[stk.top()[2]];\n stk.push(cr[i]);\n }\n // for (int i = 0; i <= k; ++ i) cout << dp[i] << '|';\n // cout << endl;\n int y = 0, x = 0;\n res[0] = {0, {y, x}};\n for (int j = 0; j <= k; ++ j) {\n auto &i = cr[j];\n i64 ans = res[j].first;\n ans += i64(i[0] - x - 1) * (n - y);\n y = std::max(y, i[1]);\n ans += n - y;\n x = i[0];\n // debug(ans, y, x, i);\n res[j + 1] = {ans, {y, x}};\n }\n cout << res[k + 1].first << '\\n';\n std::function<int (int, int, int, int, int)> check = [&](int u, int L, int R, int l, int v) {\n if (l <= L) return tr[u];\n int mid = L + R >> 1, y = u + (mid - L + 1 << 1);\n int pos;\n if (l <= mid) {\n pos = check(u + 1, L, mid, l, v);\n if (cr[pos][1] >= v) return pos;\n return tr[y];\n }\n return check(y, mid + 1, R, l, v);\n };\n std::vector<i64> oo(k + 1);\n for (int j = 1; j <= k; ++ j) {\n i64 ans = res[j - 1].first;\n auto [y, x] = res[j - 1].second;\n auto &i = cr[j];\n debug(j, i, cr[j - 1], res[j - 1]);\n if (y <= i[1]) {\n ans += i64(i[0] - x - 1) * (n - y);\n oo[cr[j - 1][2]] = ans + n - i[1] + dp[cr[j][2]];\n } else if (y >= cr[j - 1][1]) {\n oo[cr[j - 1][2]] = res[k + 1].first;\n } else {\n int pos = check(0, 0, k, j, y);\n ans += i64(cr[pos][0] - x - 1) * (n - y);\n ans += n - std::max(y, cr[pos][1]);\n oo[cr[j - 1][2]] = ans + dp[cr[pos][2]];\n }\n }\n for (int i = 1; i <= k; ++ i) {\n cout << oo[i] - res[k + 1].first << \"\\n \"[i != k];\n }\n}\n \nsigned main() {\n#if (defined XqAiyee && !defined LOOP_TEST_ABFS)\n system(\"chcp 65001 >nul\");\n#ifdef FROM_FILE_IO\n freopen(MY_IN_FILE, \"r\", stdin);\n#endif\n SHOW_USED_TIME_BEG\n#endif\n int T = 1, n = 0;\n std::cin.tie(0)->sync_with_stdio(false);\n std::cin >> T;\n while (T --) {\n solve();\n }\n#if (defined XqAiyee && !defined LOOP_TEST_ABFS)\n SHOW_USED_TIME_END\n#ifdef FROM_FILE_IO\n freopen(\"CON\", \"r\", stdin);\n#endif\n\tstd::cout << std::endl;\n\tsystem(\"pause\");\n#endif\n return exit(0), 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 264631202,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define int long long\nstruct node {\n int x, y, id;\n} a[200005];\nint mx[200005];\nbool cmp(node a, node b) {\n if (a.x != b.x) return a.x < b.x;\n return a.y < b.y;\n}\nint val[200005];\nsigned main() {\n int t;\n cin >> t;\n while (t--) {\n int n, m, k;\n cin >> n >> m >> k;\n for (int i = 1; i <= k; i++) {\n cin >> a[i].x >> a[i].y;\n a[i].id = i;\n val[i] = mx[i] = 0;\n }\n sort(a + 1, a + k + 1, cmp);\n mx[k + 1] = m + 1;\n for (int i = k; i >= 1; i--) {\n mx[i] = min(mx[i + 1], a[i].y);\n // cout << mx[k] << endl;\n }\n int ans = 0, now = 0, last = 0;\n for (int i = 1; i <= k; i++) {\n // cout << a[i].x << \" \" << last << \" \" << mx[i] << endl;\n ans += 1ll * (a[i].x - last) * (mx[i] - 1);\n last = a[i].x;\n }\n ans += 1ll * (n - last) * m;\n cout << ans << endl;\n a[0].x = 0, a[0].y = m + 1;\n a[k + 1].x = n + 1, a[k + 1].y = m + 1;\n for (int i = k; i >= 1; i--) {\n if (a[i].y == a[i + 1].y || a[i].x == a[i - 1].x) {\n continue;\n }\n int l = i, mn = m + 1, r = i - 1, pos;\n // 删除一个点的价值\n // a[i] < a[j] min(mn[i+1 - j-1], mx[j+1]) * (a[i].x - last)\n \n while (r > 0) {\n if (a[r].y <= a[i].y)\n break;\n else {\n if (a[r].y < mn) {\n mn = min(mn, a[r].y);\n pos = r;\n }\n r--;\n }\n }\n // printf(\"id = %d r = %d pos= %d last = %d mn = %d mxid + 1 = %d\\n\", a[i].id, r,\n // pos,\n // last, mn, mx[i + 1]);\n if (mx[i + 1] <= mn) {\n val[a[i].id] = (abs(a[i].x - a[r].x)) * abs(a[i].y - min(mn, mx[i + 1]));\n } else {\n val[a[i].id] += (abs(a[pos].x - a[r].x)) * abs(a[i].y - mn);\n val[a[i].id] += (abs(a[pos].x - a[l].x)) * abs(a[i].y - mx[i + 1]);\n }\n i = r + 1;\n }\n for (int i = 1; i <= k; i++) cout << val[i] << \" \";\n cout << endl;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 264099526,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define ll long long\nint t, n, m, q;\nint pos[200005];\nll ans[200005];\nstruct node {\n\tint x, y, id;\n\tbool operator<(const node& p)const {\n\t\tif (x == p.x) return y > p.y;\n\t\treturn x < p.x;\n\t}\n}a[200005];\nvoid test() {\n\tcin >> n >> m >> q;\n\tfor (int i = 1; i <= q; i++) {\n\t\tcin >> a[i].x >> a[i].y;\n\t\ta[i].id = i;\n\t}\n\ta[0].x = 0;\n\tpos[0] = 0;\n\ta[q + 1].x = n;\n\tpos[q + 1] = m + 1;\n\tsort(a + 1, a + q + 1);\n\tpos[q] = a[q].y;\n\tfor (int i = q - 1; i >= 1; i--) {\n\t\tpos[i] = min(pos[i + 1], a[i].y);\n\t}\n\tll sum = 0;\n\tfor (int i = 1; i <= q + 1; i++) {\n\t\tsum += 1ll * (a[i].x - a[i - 1].x) * (pos[i] - 1);\n\t}\n\tcout << sum << endl;\n\tfor (int i = 1; i <= q; i++) {\n\t\tif (i == q || (i < q && a[i].y == pos[i] && pos[i + 1] != pos[i])) {\n\t\t\tll num1 = 1ll * (a[i].x - a[i - 1].x) * (pos[i] - 1);\n\t\t\tll num2 = 1ll * (a[i].x - a[i - 1].x) * (pos[i + 1] - 1);\n\t\t\t//cout << num2 << endl;\n\t\t\tint r = i - 1, mi = pos[i + 1];\n\t\t\twhile (r >= 0 && pos[i + 1] > pos[i] && a[r].y > pos[i]) {\n\t\t\t\tmi = min(mi, a[r].y);\n\t\t\t\tr--;\n\t\t\t}\n\t\t\tif (r != i - 1) {\n\t\t\t\t//cout << a[i].id << \" \" << 1 << endl;\n\t\t\t\t//cout << a[i - 1].x - a[r + 1].x + 1 << \" \" << mi - pos[i] << endl;\n\t\t\t\tnum2 += 1ll * (a[i - 1].x - a[r + 1].x + 1) * (mi - pos[i]);\n\t\t\t}\n\t\t\tans[a[i].id] = num2 - num1;\n\t\t\t//cout << num1 << \" \" << num2 << endl;\n\t\t}\n\t\telse ans[a[i].id] = 0;\n\t}\n\tfor (int i = 1; i <= q; i++)cout << ans[i] << \" \";\n\tcout << endl;\n}\nsigned main() {\n\tcin.tie(0);\n\tcout.tie(0);\n\tios::sync_with_stdio(false);\n\tcin >> t;\n\twhile (t--) {\n\t\ttest();\n\t}\n\t//test();\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 276417524,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nmt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());\ntypedef long long LL;\ntypedef unsigned long long ULL;\ntypedef pair<int,int> PII;\nconst int _MAX = 0x3f3f3f3f;\nconst int N = 2e5 + 10,mod = 1e9 + 7;\nstruct node{\n int x, y;\n}point[N];\nvoid solve() {\n int n, m, k;\n cin >> n >> m >> k;\n map<int,vector<int>> hh;\n vector<int> tt;\n map<int,int> st;\n map<int,int> change;\n vector<int> chan;\n map<int,int> id;\n for (int i = 1; i <= k; i++) {\n int x, y;\n cin >> x >> y;\n point[i].x = x, point[i].y = y;\n hh[y].push_back(x);\n if (st[y] == 0) {\n tt.push_back(y);\n st[y] = 1;\n }\n }\n sort(tt.begin(),tt.end());\n LL ans = 0;\n int tmplow = 0;\n int last_y = 1;\n for (int i = 0; i < tt.size(); i++) {\n sort(hh[tt[i]].begin(),hh[tt[i]].end());\n if (hh[tt[i]].back() > tmplow) {\n change[tt[i]] = tmplow + 1; //特意加一，后面再减一，防止tmplow = 0;\n chan.push_back(i);\n id[tt[i]] = i;\n ans += (LL)(n - tmplow) * (tt[i] - last_y);\n tmplow = hh[tt[i]].back();\n last_y = tt[i];\n }\n }\n ans += (LL)(n - tmplow) * (m - last_y + 1);\n cout << ans << '\\n';\n int idx = -1;\n for (int i = 0; i < tt.size(); i++) {\n if (tt[i] > last_y) {\n idx = i;\n break;\n }\n }\n for (int i = 1; i <= k; i++) {\n int x = point[i].x, y = point[i].y;\n if (y == last_y && x == hh[y].back()) {\n int llaasstt_yy = last_y;\n if (idx == -1) {\n if (hh[y].size() > 1) {\n int kk = max(change[y] - 1,hh[y][hh[y].size() - 2]);\n cout << (LL)(hh[y].back() - kk) * (m - last_y + 1) << ' ';\n }\n else {\n cout << (LL)(hh[y].back() - change[y] + 1) * (m - last_y + 1) << ' ';\n }\n }\n else {\n LL tmpans = 0;\n int tttmplow;\n if (hh[y].size() > 1) {\n tttmplow = max(change[y] - 1,hh[y][hh[y].size() - 2]);\n }\n else {\n tttmplow = change[y] - 1;\n }\n for (int i = idx; i < tt.size(); i++) {\n if (hh[tt[i]].back() > tttmplow) {\n tmpans += (LL)(tmplow - tttmplow) * (tt[i] - llaasstt_yy);\n tttmplow = hh[tt[i]].back();\n llaasstt_yy = tt[i];\n }\n }\n tmpans += (LL)(tmplow - tttmplow) * (m - llaasstt_yy + 1);\n cout << tmpans << ' ';\n }\n continue;\n }\n if (change[y] && x == hh[y].back()) {\n LL tmpans = 0;\n int tttmplow;\n if (hh[y].size() > 1) {\n tttmplow = max(change[y] - 1,hh[y][hh[y].size() - 2]);\n }\n else tttmplow = change[y] - 1;\n int last = y;\n int l = 0, r = chan.size() - 1;\n while (l < r) {\n int mid = (l + r) / 2;\n if (hh[tt[chan[mid]]].back() > tttmplow) r = mid;\n else l = mid + 1;\n }\n l++;\n while (l < r) {\n int mid = (l + r) / 2;\n if (hh[tt[chan[mid]]].back() > tttmplow) r = mid;\n else l = mid + 1;\n }\n for (int i = id[y]; i < chan[l]; i++) {\n if (hh[tt[i]].back() > tttmplow) {\n tmpans += (hh[tt[i]].back() - tttmplow) * (tt[i] - last + 1);\n tttmplow = hh[tt[i]].back();\n last = tt[i];\n }\n }\n cout << tmpans << ' ';\n } \n else cout << 0 << ' ';\n }\n cout << '\\n';\n}\nint main() {\n ios::sync_with_stdio(false);\n cin.tie(0);\n cout.tie(0);\n int t;\n cin>>t;\n while (t--) {\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 264205512,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\nimport sys\nfrom collections import defaultdict,deque,Counter\nfrom heapq import heappush,heappop,heapify\nfrom math import inf,comb,ceil,log,gcd\nfrom functools import cache\nfrom itertools import accumulate,combinations,permutations\nfrom bisect import bisect_left,bisect_right,insort_left\nfrom random import randint\nxor=randint(0,2**32-1)\ninput=lambda:sys.stdin.readline().rstrip()\ninp=lambda:map(int,input().split())\nt=int(input())\nfor _ in range(t):\n n,m,k=inp()\n X=[[0,0]]\n for _ in range(k):\n X+=[*inp()][::-1],\n X+=[m+1,n+1],\n oX=X.copy()\n X=sorted([(e,i)for i,e in enumerate(X)],key=lambda t:(t[0][0],-t[0][1]))\n #print(X)\n h=-1\n ar=0\n s=set()\n for (x,y),i in X:\n \n x-=1\n y-=1\n #print(\"xyi\",x,y,i,la,h)\n #if i==k+1:print(x,y)\n if y>h:\n ar+=(m-x)*(y-h)\n h=y\n s.add(i)\n s1=[]\n s2=deque()\n ftb=[*range(k+2)]\n stb=[*range(k+2)]\n for (x,y),i in X:\n #print(x,y,i)\n while s2 and y>s2[-1][1]:\n cur=s2.pop()\n stb[cur[2]]=i\n while s1 and y>s1[-1][1]:\n #print(s1,i)\n cur=s1.pop()\n s2.appendleft(cur)\n ftb[cur[2]]=i\n \n s1.append((x,y,i))\n #print([oX[i]for i in ftb])\n def rect(i1,i2,i3):\n return (oX[i2][0]-oX[i1][0])*(oX[i1][1]-oX[i3][1])\n res=[0]*k\n for i in range(k+1):\n el=ftb[i]\n el2=ftb[ftb[i]]\n el3=stb[i]\n if el not in s:continue\n if el-1<k:\n #print(oX[i],i-1,oX[el],el-1,oX[el2],el2-1,oX[el3],el3-1)\n res[el-1]=abs(rect(el,el2,i)-rect(el3,el2,i))\n #print(ar)\n #print(s)\n print(n*m-ar)\n print(*res)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 281115538,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\nimport math,random \nfrom collections import defaultdict\nfrom bisect import bisect_left, bisect_right\nimport random \nrd = random.randint(10**8,10**9) \nclass mydict:\n def __init__(self, func=lambda: 0):\n self.random = random.randint(0, 1 << 32)\n self.default = func\n self.dict = {}\n def __getitem__(self, key):\n mykey = self.random ^ key\n if mykey not in self.dict:\n self.dict[mykey] = self.default()\n return self.dict[mykey]\n def get(self, key, default):\n mykey = self.random ^ key\n if mykey not in self.dict:\n return default\n return self.dict[mykey]\n def __setitem__(self, key, item):\n mykey = self.random ^ key\n self.dict[mykey] = item\n def getkeys(self):\n return [self.random ^ i for i in self.dict]\n def __str__(self):\n return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'\nt = int(input())\nfor _ in range(t):\n n,m,k = map(int,input().split())\n a = []\n d = {}\n d2 = {}\n for i in range(k):\n r,c = map(int,input().split())\n a.append([r,c,i])\n \n a.sort(key = lambda x: (-x[0],x[1]))\n r = n+1\n rows = []\n cols = []\n d3 = mydict(list)\n rl = -1\n for i in range(k):\n d3[a[i][0]].append(a[i][1])\n if a[i][0]!=r:\n r = a[i][0]\n rows.append(a[i][0])\n cols.append(a[i][1])\n d2[(a[i][0],a[i][1])] = a[i][2]\n r = n+1\n c = m \n ans=0\n rows.append(0)\n cols.append(0)\n #print(rows)\n res = [0]*k\n for i in range(len(rows)):\n if i==0:\n x = r - rows[i]\n y = cols[i]-1\n ans+= y \n ans+=(x-1)*(c)\n #print(ans)\n #if fountain a[i] can be taken=>\n \n p = c-y\n if len(d3[rows[i]])>1:\n ll = list(d3[rows[i]])\n ll.sort()\n p = ll[1]-y-1\n rnext=0\n cnext=0\n if i<len(rows)-1: \n rnext = rows[i+1] \n cnext = cols[i+1]\n q = rows[i] - rnext\n temp = p*q\n if cnext>cols[i]: temp+=cnext-cols[i]\n res[d2[(rows[i],cols[i])]]=temp\n d[rows[i]]=temp \n #print(i,temp)\n prevc = cols[i] \n else:\n x = r - rows[i]\n ans+=(x-1)*(prevc-1)\n if rows[i]==0: break \n pprevc = prevc\n prevc = min(prevc,cols[i])\n y = prevc-1 \n ans+=y \n #print(ans,x,prevc,rows[i])\n #if fountain a[i] can be taken \n if cols[i]<pprevc:\n rnext = rows[i+1]\n p = rows[i]-rnext\n q=0\n if cols[i+1]>cols[i]: q = rows[i]-rows[i+1]\n temp = p*(pprevc-cols[i]) + q*(min(pprevc,cols[i+1])-1)\n #print(temp,)\n d[rows[i]] = temp\n #print(rows[i],d[rows[i]],p,q,pprevc)\n res[d2[(rows[i],cols[i])]]= d[rows[i]]\n else:\n d[rows[i]] = 0\n #print(ans)\n r = rows[i]\n \n print(ans)\n print(*res)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 264600913,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n// Source: https://usaco.guide/general/io\n \nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n\tpublic static void main(String[] args) throws IOException {\n\t\tBufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n\t\tint T = Integer.parseInt(br.readLine());\n\t\tfor(int ii = 1; ii <= T; ii++){\n\t\t\tString[] s = br.readLine().split(\" \");\n\t\t\tlong N = Integer.parseInt(s[0]);\n\t\t\tlong M = Integer.parseInt(s[1]);\n\t\t\tint K = Integer.parseInt(s[2]); // Number of fountains\n\t\t\tTreeMap<Long,TreeSet<Long>> coordsByCol = new TreeMap<Long,TreeSet<Long>>();\n\t\t\tTreeMap<Long,TreeSet<Long>> coordsByRow = new TreeMap<Long,TreeSet<Long>>();\n\t\t\tSet<String> maxPairsDown = new HashSet<String>();\n\t\t\tSet<String> maxPairsRight = new HashSet<String>();\n\t\t\tString[] p = new String[K];\n\t\t\tfor(int j = 1; j <= K; j++){\n\t\t\t\tString[] rc = br.readLine().split(\" \");\n\t\t\t\tlong r = Long.parseLong(rc[0]);\n\t\t\t\tlong c = Long.parseLong(rc[1]);\n\t\t\t\tp[j-1] = r + \" \" + c;\n\t\t\t\tif(!coordsByCol.containsKey(c)) coordsByCol.put(c, new TreeSet<Long>());\n\t\t\t\tcoordsByCol.get(c).add(r);\n\t\t\t\tif(!coordsByRow.containsKey(r)) coordsByRow.put(r, new TreeSet<Long>());\n\t\t\t\tcoordsByRow.get(r).add(c);\n\t\t\t}\n\t\t\t//System.out.println(coordsByCol);\n\t\t\tlong sumOfDownMaxes = 0;\n\t\t\tlong prevCol = 1;\n\t\t\tlong prevMax = 0;\n\t\t\tlong partialSum = 0;\n\t\t\tlong prevVal = 0;\n\t\t\tMap<Long,Long> deltas = new HashMap<Long,Long>();\n\t\t\t//List<Long> prevVals = new ArrayList<Long>();\n\t\t\t//System.out.println(coordsByCol);\n\t\t\tList<Long> stuff = new ArrayList<Long>();\n\t\t\tfor(Long col : coordsByCol.keySet()){\n\t\t\t\t\n\t\t\t\t//System.out.println(\"Adding \" + prevMax + \" for \" + prevCol + \" thru \" + (col - 1));\n\t\t\t\tsumOfDownMaxes += (col - prevCol) * prevMax;\n\t\t\t\tpartialSum += prevVal * (col - prevCol);\n\t\t\t\tdeltas.put(prevCol,-partialSum + sumOfDownMaxes);\n\t\t\t\t//System.out.println(\"Now prevMax is \" + prevMax);\n\t\t\t\t//System.out.println(\"When adding col \" + col + \" real sum becomes \" + sumOfDownMaxes + \" while partialSum \" + partialSum);\n\t\t\t\tprevCol = col;\n\t\t\t\tlong value = coordsByCol.get(col).last();\n\t\t\t\tif(value > prevVal){\n\t\t\t\t\tprevVal = Math.max(prevVal,value);\n\t\t\t\t\t\n\t\t\t\t}\n\t\t\t\tif(value > prevMax){\n\t\t\t\t\tstuff.add(col);\n\t\t\t\t\t//System.out.println(\"For col \" + col + \" prevVal is now \" + prevVal);\n\t\t\t\t\t\n\t\t\t\t\tpartialSum = sumOfDownMaxes;\n\t\t\t\t\t// Setting prevVal to the value that would be max here by removing this max value\n\t\t\t\t\tcoordsByCol.get(col).remove(coordsByCol.get(col).last());\n\t\t\t\t\tprevVal = (coordsByCol.get(col).size() >= 1 ? coordsByCol.get(col).last() : prevMax);\n\t\t\t\t\tprevMax = Math.max(prevMax,value);\n\t\t\t\t\tmaxPairsDown.add(value + \" \" + col);\n\t\t\t\t\t\n\t\t\t\t}\n\t\t\t}\n\t\t\t//System.out.println(\"Adding \" + prevMax + \" for \" + prevCol + \" thru \" + M);\n\t\t\tsumOfDownMaxes += (M - prevCol + 1) * prevMax;\n\t\t\tpartialSum += prevVal * (M - prevCol + 1);\n\t\t\tdeltas.put(prevCol, -partialSum + sumOfDownMaxes);\n\t\t\tdeltas.put(stuff.get(stuff.size() - 1), -partialSum + sumOfDownMaxes);\n\t\t\tlong alpha = N * M - sumOfDownMaxes;\n\t\t\tSystem.out.println(alpha);\n\t\t\tlong[] ans = new long[K];\n\t\t\tfor(int i = 0; i < p.length; i++){\n\t\t\t\tif(maxPairsDown.contains(p[i])){\n\t\t\t\t\tans[i] = N * M - sumOfDownMaxes + deltas.get(Long.parseLong(p[i].split(\" \")[1])) - alpha;\n\t\t\t\t}\n\t\t\t\telse{\n\t\t\t\t\tans[i] = 0;\n\t\t\t\t}\n \n\t\t\t\t\n\t\t\t}\n\t\t\t//System.out.println((N * M - sumOfDownMaxes) + \" \" + (N * M - sumOfRightMaxes));\n\t\t\tfor(int i = 0; i < ans.length; i++){\n\t\t\t\tif(i < K-1) System.out.print(ans[i] + \" \");\n\t\t\t\telse System.out.println(ans[i]);\n\t\t\t} \n\t\t}\n\t}\n}\nclass Pair{\n\tlong x;\n\tlong y;\n\tPair(long x, long y){\n\t\tthis.x = x;\n\t\tthis.y = y;\n\t}\n\tpublic String toString(){\n\t\treturn \"(\" + x + \",\" + y + \")\";\n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 265925918,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n// Source: https://usaco.guide/general/io\n \nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n\tpublic static void main(String[] args) throws IOException {\n\t\tBufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n\t\tint T = Integer.parseInt(br.readLine());\n\t\tfor(int ii = 1; ii <= T; ii++){\n\t\t\tString[] s = br.readLine().split(\" \");\n\t\t\tlong N = Integer.parseInt(s[0]);\n\t\t\tlong M = Integer.parseInt(s[1]);\n\t\t\tint K = Integer.parseInt(s[2]); // Number of fountains\n\t\t\tTreeMap<Long,TreeSet<Long>> coordsByCol = new TreeMap<Long,TreeSet<Long>>();\n\t\t\tTreeMap<Long,TreeSet<Long>> coordsByRow = new TreeMap<Long,TreeSet<Long>>();\n\t\t\tSet<String> maxPairsDown = new HashSet<String>();\n\t\t\tSet<String> maxPairsRight = new HashSet<String>();\n\t\t\tString[] p = new String[K];\n\t\t\tfor(int j = 1; j <= K; j++){\n\t\t\t\tString[] rc = br.readLine().split(\" \");\n\t\t\t\tlong r = Long.parseLong(rc[0]);\n\t\t\t\tlong c = Long.parseLong(rc[1]);\n\t\t\t\tp[j-1] = r + \" \" + c;\n\t\t\t\tif(!coordsByCol.containsKey(c)) coordsByCol.put(c, new TreeSet<Long>());\n\t\t\t\tcoordsByCol.get(c).add(r);\n\t\t\t\tif(!coordsByRow.containsKey(r)) coordsByRow.put(r, new TreeSet<Long>());\n\t\t\t\tcoordsByRow.get(r).add(c);\n\t\t\t}\n\t\t\t//System.out.println(coordsByCol);\n\t\t\tlong sumOfDownMaxes = 0;\n\t\t\tlong prevCol = 1;\n\t\t\tlong prevMax = 0;\n\t\t\tlong partialSum = 0;\n\t\t\tlong prevVal = 0;\n\t\t\tMap<Long,Long> deltas = new HashMap<Long,Long>();\n\t\t\t//List<Long> prevVals = new ArrayList<Long>();\n\t\t\t//System.out.println(coordsByCol);\n\t\t\tList<Long> stuff = new ArrayList<Long>();\n\t\t\tfor(Long col : coordsByCol.keySet()){\n\t\t\t\t\n\t\t\t\t//System.out.println(\"Adding \" + prevMax + \" for \" + prevCol + \" thru \" + (col - 1));\n\t\t\t\tsumOfDownMaxes += (col - prevCol) * prevMax;\n\t\t\t\tpartialSum += prevVal * (col - prevCol);\n\t\t\t\tdeltas.put(prevCol,-partialSum + sumOfDownMaxes);\n\t\t\t\t//System.out.println(\"Now prevMax is \" + prevMax);\n\t\t\t\t//System.out.println(\"When adding col \" + col + \" real sum becomes \" + sumOfDownMaxes + \" while partialSum \" + partialSum);\n\t\t\t\tprevCol = col;\n\t\t\t\tlong value = coordsByCol.get(col).last();\n\t\t\t\tif(value > prevVal){\n\t\t\t\t\tprevVal = Math.max(prevVal,value);\n\t\t\t\t\t\n\t\t\t\t}\n\t\t\t\tif(value > prevMax){\n\t\t\t\t\tstuff.add(col);\n\t\t\t\t\t//System.out.println(\"For col \" + col + \" prevVal is now \" + prevVal);\n\t\t\t\t\t\n\t\t\t\t\tpartialSum = sumOfDownMaxes;\n\t\t\t\t\t// Setting prevVal to the value that would be max here by removing this max value\n\t\t\t\t\tcoordsByCol.get(col).remove(coordsByCol.get(col).last());\n\t\t\t\t\tprevVal = (coordsByCol.get(col).size() >= 1 ? coordsByCol.get(col).last() : prevMax);\n\t\t\t\t\tprevMax = Math.max(prevMax,value);\n\t\t\t\t\tmaxPairsDown.add(value + \" \" + col);\n\t\t\t\t\t\n\t\t\t\t}\n\t\t\t}\n\t\t\t//System.out.println(\"Adding \" + prevMax + \" for \" + prevCol + \" thru \" + M);\n\t\t\tsumOfDownMaxes += (M - prevCol + 1) * prevMax;\n\t\t\tpartialSum += prevVal * (M - prevCol + 1);\n\t\t\tdeltas.put(stuff.get(stuff.size() - 1), -partialSum + sumOfDownMaxes);\n\t\t\tlong alpha = N * M - sumOfDownMaxes;\n\t\t\tSystem.out.println(alpha);\n\t\t\tlong[] ans = new long[K];\n\t\t\tfor(int i = 0; i < p.length; i++){\n\t\t\t\tif(maxPairsDown.contains(p[i])){\n\t\t\t\t\tans[i] = N * M - sumOfDownMaxes + deltas.get(Long.parseLong(p[i].split(\" \")[1])) - alpha;\n\t\t\t\t}\n\t\t\t\telse{\n\t\t\t\t\tans[i] = 0;\n\t\t\t\t}\n \n\t\t\t\t\n\t\t\t}\n\t\t\t//System.out.println((N * M - sumOfDownMaxes) + \" \" + (N * M - sumOfRightMaxes));\n\t\t\tfor(int i = 0; i < ans.length; i++){\n\t\t\t\tif(i < K-1) System.out.print(ans[i] + \" \");\n\t\t\t\telse System.out.println(ans[i]);\n\t\t\t} \n\t\t}\n\t}\n}\nclass Pair{\n\tlong x;\n\tlong y;\n\tPair(long x, long y){\n\t\tthis.x = x;\n\t\tthis.y = y;\n\t}\n\tpublic String toString(){\n\t\treturn \"(\" + x + \",\" + y + \")\";\n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 265925713,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n#include<iostream>\n#include<algorithm>\n#include<vector>\n#include<array>\n#define SZ(X) ((int)(X).size())\nusing namespace std;\nvoid solve() {\n int n, m, k;\n cin >> n >> m >> k;\n vector<array<int, 3>> p;\n vector<int> val(k + 2), an(k + 1);\n for(int i = 1; i <= k; i++) {\n int r, c;\n cin >> r >> c;\n p.push_back({-r, c, i});\n }\n p.push_back({-n - 1, m + 1, 0});\n p.push_back({0, 0, k + 1});\n sort(p.begin(), p.end());\n int lt_r = n + 1;\n int lt_c = m + 1;\n long long alpha = 0;\n val[0] = 1;\n for(int i = 1; i <= k; i++) {\n int r = -p[i][0];\n int c = p[i][1];\n int id = p[i][2];\n if(c < lt_c) {\n val[i] = (lt_c - 1ll) * (lt_r - r);\n alpha += val[i];\n lt_c = c;\n lt_r = r;\n }\n }\n val[k + 1] = (lt_c - 1ll) * lt_r + 1;\n alpha += val[k + 1];\n cout << alpha - m - 1 << '\\n';\n for(int i = 1; i <= k; i++) {\n if(val[i]) {\n int id = p[i][2];\n int ll = i - 1;\n while(!val[ll]) ll--;\n int rr = i + 1;\n while(!val[rr]) rr++;\n an[id] = - val[i] - val[rr];\n lt_r = -p[ll][0];\n lt_c = p[ll][1];\n for(int j = i + 1; j <= rr; j++) {\n int r = -p[j][0];\n int c = p[j][1];\n if(c < lt_c) {\n an[id] += (lt_c - 1ll) * (lt_r - r);\n lt_c = c;\n lt_r = r;\n }\n }\n if(rr > k) an[id]++;\n }\n }\n for(int i = 1; i <= k; i++) cout << an[i] << (i < k ? ' ' : '\\n');\n}\nint main() {\n cin.tie(0);\n ios_base::sync_with_stdio(false);\n int t;\n cin >> t;\n while(t--) solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 264046337,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n/* === INCLUDES === ---------------------------------------------------------- */\n#include <algorithm>\n#include <bitset>\n#include <cmath>\n#include <iostream>\n#include <map>\n#include <numeric>\n#include <queue>\n#include <random>\n#include <set>\n#include <stack>\n#include <string>\n#include <unordered_map>\n#include <unordered_set>\n#include <utility>\n#include <vector>\nusing namespace std;\n \n/* === UTILITY === ----------------------------------------------------------- */\n#ifdef _LOCAL_\n#define SHOW(x) \\\n { \\\n cerr << \"[\" #x \"]: \" << (x) << \"\\n\"; \\\n }\n#define SHOW_EACH(x) \\\n { \\\n bool _head = true; \\\n cerr << \"[\" #x \"]: {\"; \\\n for (auto _iter : (x)) { \\\n if (_head) { \\\n _head = false; \\\n } else { \\\n cerr << \", \"; \\\n } \\\n cerr << _iter; \\\n } \\\n cerr << \"}\" << \"\\n\"; \\\n }\n#define SHOW_RANGE(x, l, r) \\\n { \\\n bool _head = true; \\\n cerr << \"[\" #x \"]: {\"; \\\n for (auto _iter = (l); _iter <= (r); ++_iter) { \\\n if (_head) { \\\n _head = false; \\\n } else { \\\n cerr << \", \"; \\\n } \\\n cerr << (x)[_iter]; \\\n } \\\n cerr << \"}\" << \"\\n\"; \\\n }\n#else\n#define SHOW(x)\n#define SHOW_EACH(x)\n#define SHOW_RANGE(x, l, r)\n#endif // _LOCAL_\n \n/* === MACROS === -------------------------------------------------------------*/\n#define FASTIO \\\n ios::sync_with_stdio(false); \\\n cin.tie(0), cout.tie(0)\n#define MULITCASE \\\n int _T; \\\n cin >> _T; \\\n while (_T--)\n#define LL long long\n#define IMAX 2147483647\n#define LLMAX 9223372036854775807LL\n#define endl \"\\n\"\n \n////////////////////////////////// ┏━━━━━━━━━┓ //////////////////////////////////\n////////////////////////////////// ┣━╸ ┏━━━╸ ┃ //////////////////////////////////\n////////////////////////////////// ┃ ┏━┻━━━━━┫ //////////////////////////////////\n////////////////////////////////// ┃ ┣━━━━━╸ ┃ //////////////////////////////////\n////////////////////////////////// ┃ ┃ ╺━━━━━┫ //////////////////////////////////\n////////////////////////////////// ┣━┻━━━━━┳━┫ //////////////////////////////////\n////////////////////////////////// ┃ ╻ ╻ ╺━┫ ┃ //////////////////////////////////\n////////////////////////////////// ┣━╋━╋━┳━┫ ┃ //////////////////////////////////\n////////////////////////////////// ┃ ╹ ╹ ┃ ┃ ┃ //////////////////////////////////\n////////////////////////////////// ┃ ╻ ╻ ┃ ┃ ┃ //////////////////////////////////\n////////////////////////////////// ┣━┻━┻━┻━┻━┫ //////////////////////////////////\n////////////////////////////////// ┣━╸ ╺┳╸ ╺━┫ //////////////////////////////////\n////////////////////////////////// ┣━╸ ╺┻╸ ╺━┫ //////////////////////////////////\n////////////////////////////////// ┣━┳━┳━┳━┳━┫ //////////////////////////////////\n////////////////////////////////// ┣━┛ ┗━┛ ┗━┫ //////////////////////////////////\n////////////////////////////////// ┣━━━━━━━━━┫ //////////////////////////////////\n////////////////////////////////// ┣╸ ╺━━━╸ ╺┫ //////////////////////////////////\n////////////////////////////////// ┣━━━┓ ┏━━━┫ //////////////////////////////////\n////////////////////////////////// ┃ ╻ ╹ ╹ ╻ ┃ //////////////////////////////////\n////////////////////////////////// ┣━┻━━━━━┻━┫ //////////////////////////////////\n////////////////////////////////// ┣━━━╸ ┏━━━┫ //////////////////////////////////\n////////////////////////////////// ┃ ╺━━━┛ ╻ ┃ //////////////////////////////////\n////////////////////////////////// ┗━━━━━━━┻━┛ //////////////////////////////////\n \n \n \nint n, m, k;\nvector<tuple<int, int, int>> viii;\n \nvoid ChromicChlorideHexahydrate()\n{\n viii.clear();\n cin >> n >> m >> k;\n for (int i = 1; i <= k; i++) {\n int r, c;\n cin >> r >> c;\n viii.push_back({r, c, i - 1});\n }\n viii.push_back({0, 0, k++});\n sort(viii.begin(), viii.end(), [](auto l, auto r) {\n auto [lr, lc, li] = l;\n auto [rr, rc, ri] = r;\n if (lr == rr) {\n return lc < rc;\n } else {\n return lr > rr;\n }\n });\n \n vector<int> ans(k), area(k + 5), lastcol(k + 5, m + 1),\n lastrow(k + 5, n);\n for (int i = 1; i <= k; i++) {\n auto [r, c, id] = viii[i - 1];\n area[i] = area[i - 1];\n lastcol[i] = lastcol[i - 1];\n lastrow[i] = lastrow[i - 1];\n if (lastcol[i] > c) {\n ans[id] = 1;\n area[i] += (lastcol[i] - 1) * (lastrow[i] - r);\n lastcol[i] = c;\n lastrow[i] = r;\n }\n }\n cout << area[k] << endl;\n for (int i = 1; i <= k; i++) {\n auto [r, c, id] = viii[i - 1];\n if (ans[id] == 0) {\n continue;\n }\n int narea = area[i - 1];\n int nlcol = lastcol[i - 1];\n int nlrow = lastrow[i - 1];\n for (int j = i + 1; j <= k; j++) {\n auto [nr, nc, nid] = viii[j - 1];\n if (nlcol > nc) {\n narea += (nlcol - 1) * (nlrow - nr);\n nlcol = nc;\n nlrow = nr;\n }\n if (ans[nid] == 1) {\n ans[id] = narea - area[j];\n break;\n }\n }\n }\n ans.pop_back();\n for (auto it : ans) {\n cout << it << \" \";\n }\n cout << endl;\n}\n \nsigned main()\n{\n FASTIO;\n MULITCASE ChromicChlorideHexahydrate();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 264259990,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n# import numpy as np\n \n# def getComponent(coor,mat):\n# tmp = mat\n# succ = getSuccessors(coor,tmp)\n# for s in succ:\n# tmp[s]='-'\n# getComponent(s,tmp)\n# componentLength = np.count_nonzero(tmp == '-')\n# return 1 if not componentLength else componentLength\n# def getMaxComponent(matrix,l,c):\n# mat = np.copy(matrix)\n# mat[l,:]='#'\n# mat[:,c]='#'\n# ls = list()\n# X,Y = np.where(mat == '#')[0],np.where(mat == '#')[1]\n# indexes = [(X[i],Y[i]) for i in range(np.shape(X)[0])]\n# for coor in indexes: \n# ls.append(getComponent(coor,np.copy(mat)))\n# return max(ls)\n \n# def getSuccessors(cell,mat):\n# successors = list()\n# if cell[0] != 0 and mat[cell[0]-1,cell[1]] == '#':\n# successors.append((cell[0]-1,cell[1]))\n# if cell[0] != np.shape(mat)[0]-1 and mat[cell[0]+1,cell[1]] == '#':\n# successors.append((cell[0]+1,cell[1]))\n# if cell[1] != 0 and mat[cell[0],cell[1]-1] == '#':\n# successors.append((cell[0],cell[1]-1))\n# if cell[1] != np.shape(mat)[1]-1 and mat[cell[0],cell[1]+1] == '#':\n# successors.append((cell[0],cell[1]+1))\n# return successors\n \n \n# t = int(input())\n# result = list()\n# for _ in range(t):\n# line = str(input()).split(' ')\n# n = int(line[0].strip())\n# m = int(line[1].strip())\n# m = np.ndarray(shape=(n,m),dtype='<U9')\n# for i in range (n):\n# row = input()\n# m[i]=[char for char in row]\n# lengthes = list()\n# for l in range (np.shape(m)[0]):\n# for c in range (np.shape(m)[1]):\n# lengthes.append(getMaxComponent(m,l,c))\n# result.append(max(lengthes))\n# for num in result:\n# print(num)\n# def calculateMinTurns(boss,attacks,cooldowns):\n# minTurns=0\n# tmpCooldown = [0]*len(cooldowns)\n# while(boss>0):\n# totDamage=0\n# for i in range(len(attacks)):\n# if(tmpCooldown[i] != 0):\n# tmpCooldown[i]+=1\n# if(tmpCooldown[i] == 0):\n# totDamage += attacks[i]\n# tmpCooldown[i] = 0 - cooldowns[i]\n \n# boss-=totDamage\n# minTurns+=1\n# return minTurns\n \n \n# t=int(input())\n# totScores = list()\n# for _ in range(t):\n# firstLine = input().split(' ')\n# bossHealth = int(firstLine[0])\n# numAttacks = int(firstLine[1])\n# attacks = [int(a) for a in input().split(' ')]\n# cooldowns = [int(c) for c in input().split(' ')]\n# totScores.append(calculateMinTurns(bossHealth,attacks,cooldowns))\n# for _ in totScores:\n# print(_)\n \n \n \n# t=int(input())\n# words=list()\n# for _ in range(t):\n# word1, word2 = input().split(' ')\n# words.append((word1,word2))\n# for elem in words:\n# print(elem[1][0]+elem[0][1:],elem[0][0]+elem[1][1:])\n# def process_test_case(ls):\n# n = len(ls)\n# iter_count = 1 << n # Equivalent to 2^n\n# max_sum = float('-inf')\n# count_max_sum = 0\n \n# for i in range(iter_count):\n# c = 0\n# for j in range(n):\n# if i & (1 << j): # Check if j-th bit is set in i\n# c = abs(c + ls[j])\n# else:\n# c += ls[j]\n# if c > max_sum:\n# max_sum = c\n# count_max_sum = 1\n# elif c == max_sum:\n# count_max_sum += 1\n \n# return count_max_sum\n \n# # Read input and process\n# t = int(input())\n# results = []\n# for _ in range(t):\n# input() # Skip the first line of each test case\n# ls = list(map(int, input().split()))\n# results.append(process_test_case(ls))\n \n# # Print results\n# for result in results:\n# print(result)\n# def generate_combinations(lst):\n# # Create a new list with elements as (value, index)\n# indexed_lst = [(value, idx) for idx, value in enumerate(lst)]\n \n# def combinations_recursive(current, remaining):\n# if current and remaining:\n# result.append((current, remaining))\n# for i in range(len(remaining)):\n# combinations_recursive(current + [remaining[i]], remaining[:i] + remaining[i+1:])\n \n# result = []\n# combinations_recursive([], indexed_lst)\n# return result\n# def getRange(ls):\n# min = 2**31-1\n# max = -min\n# for elem in ls:\n# if(elem[0]>max):\n# max = elem[0]\n# if(elem[0]<min):\n# min = elem[0]\n# return max - min\n \n# def getOutput(lst):\n# combinations_list = generate_combinations(lst)\n# for i in range(len(combinations_list)):\n# if(getRange(combinations_list[i][0]) != getRange(combinations_list[i][1])):\n# result = ['R']*len(lst)\n# for _ in combinations_list[i][0]:\n# result[_[1]]='B'\n# return ''.join(result)\n \n \n# t = int(input())\n# results = []\n# for _ in range(t):\n# input()\n# lst = [int(x) for x in input().split(' ')]\n# results.append(getOutput(lst))\n# for result in results:\n# if result == None:\n# print(\"NO\")\n# continue\n# print(f\"YES\\n{result}\")\nN=0\nM=0\nK=0\ndef getAliceCellsInRowCount(row):\n s = 0\n for cell in row:\n if not cell:\n break\n s += 1\n return s\n \ndef getOutput(kmatrix, N, M):\n matrix = [[True for _ in range(M)] for _ in range(N)]\n for elem in kmatrix:\n matrix[elem[0]][elem[1]:] = [False] * (M - elem[1])\n for i in range(elem[0] + 1):\n matrix[i][elem[1]] = False\n \n return sum(getAliceCellsInRowCount(row) for row in matrix)\n \nt = int(input())\nresults = []\nfor _ in range(t):\n possibilities = []\n N, M, K = map(int, input().split())\n kmatrix = [tuple(map(lambda x: int(x) - 1, input().split())) for _ in range(K)]\n \n possibilities.append(getOutput(kmatrix, N, M))\n for k in range(K):\n possibilities.append(getOutput(kmatrix[:k] + kmatrix[k+1:], N, M))\n \n results.append(possibilities[0])\n rr = \" \".join(str(res - possibilities[0]) for res in possibilities[1:])\n results.append(rr)\n \nfor result in results:\n print(result)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 266191902,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\n# import numpy as np\n \n# def getComponent(coor,mat):\n# tmp = mat\n# succ = getSuccessors(coor,tmp)\n# for s in succ:\n# tmp[s]='-'\n# getComponent(s,tmp)\n# componentLength = np.count_nonzero(tmp == '-')\n# return 1 if not componentLength else componentLength\n# def getMaxComponent(matrix,l,c):\n# mat = np.copy(matrix)\n# mat[l,:]='#'\n# mat[:,c]='#'\n# ls = list()\n# X,Y = np.where(mat == '#')[0],np.where(mat == '#')[1]\n# indexes = [(X[i],Y[i]) for i in range(np.shape(X)[0])]\n# for coor in indexes: \n# ls.append(getComponent(coor,np.copy(mat)))\n# return max(ls)\n \n# def getSuccessors(cell,mat):\n# successors = list()\n# if cell[0] != 0 and mat[cell[0]-1,cell[1]] == '#':\n# successors.append((cell[0]-1,cell[1]))\n# if cell[0] != np.shape(mat)[0]-1 and mat[cell[0]+1,cell[1]] == '#':\n# successors.append((cell[0]+1,cell[1]))\n# if cell[1] != 0 and mat[cell[0],cell[1]-1] == '#':\n# successors.append((cell[0],cell[1]-1))\n# if cell[1] != np.shape(mat)[1]-1 and mat[cell[0],cell[1]+1] == '#':\n# successors.append((cell[0],cell[1]+1))\n# return successors\n \n \n# t = int(input())\n# result = list()\n# for _ in range(t):\n# line = str(input()).split(' ')\n# n = int(line[0].strip())\n# m = int(line[1].strip())\n# m = np.ndarray(shape=(n,m),dtype='<U9')\n# for i in range (n):\n# row = input()\n# m[i]=[char for char in row]\n# lengthes = list()\n# for l in range (np.shape(m)[0]):\n# for c in range (np.shape(m)[1]):\n# lengthes.append(getMaxComponent(m,l,c))\n# result.append(max(lengthes))\n# for num in result:\n# print(num)\n# def calculateMinTurns(boss,attacks,cooldowns):\n# minTurns=0\n# tmpCooldown = [0]*len(cooldowns)\n# while(boss>0):\n# totDamage=0\n# for i in range(len(attacks)):\n# if(tmpCooldown[i] != 0):\n# tmpCooldown[i]+=1\n# if(tmpCooldown[i] == 0):\n# totDamage += attacks[i]\n# tmpCooldown[i] = 0 - cooldowns[i]\n \n# boss-=totDamage\n# minTurns+=1\n# return minTurns\n \n \n# t=int(input())\n# totScores = list()\n# for _ in range(t):\n# firstLine = input().split(' ')\n# bossHealth = int(firstLine[0])\n# numAttacks = int(firstLine[1])\n# attacks = [int(a) for a in input().split(' ')]\n# cooldowns = [int(c) for c in input().split(' ')]\n# totScores.append(calculateMinTurns(bossHealth,attacks,cooldowns))\n# for _ in totScores:\n# print(_)\n \n \n \n# t=int(input())\n# words=list()\n# for _ in range(t):\n# word1, word2 = input().split(' ')\n# words.append((word1,word2))\n# for elem in words:\n# print(elem[1][0]+elem[0][1:],elem[0][0]+elem[1][1:])\n# def process_test_case(ls):\n# n = len(ls)\n# iter_count = 1 << n # Equivalent to 2^n\n# max_sum = float('-inf')\n# count_max_sum = 0\n \n# for i in range(iter_count):\n# c = 0\n# for j in range(n):\n# if i & (1 << j): # Check if j-th bit is set in i\n# c = abs(c + ls[j])\n# else:\n# c += ls[j]\n# if c > max_sum:\n# max_sum = c\n# count_max_sum = 1\n# elif c == max_sum:\n# count_max_sum += 1\n \n# return count_max_sum\n \n# # Read input and process\n# t = int(input())\n# results = []\n# for _ in range(t):\n# input() # Skip the first line of each test case\n# ls = list(map(int, input().split()))\n# results.append(process_test_case(ls))\n \n# # Print results\n# for result in results:\n# print(result)\n# def generate_combinations(lst):\n# # Create a new list with elements as (value, index)\n# indexed_lst = [(value, idx) for idx, value in enumerate(lst)]\n \n# def combinations_recursive(current, remaining):\n# if current and remaining:\n# result.append((current, remaining))\n# for i in range(len(remaining)):\n# combinations_recursive(current + [remaining[i]], remaining[:i] + remaining[i+1:])\n \n# result = []\n# combinations_recursive([], indexed_lst)\n# return result\n# def getRange(ls):\n# min = 2**31-1\n# max = -min\n# for elem in ls:\n# if(elem[0]>max):\n# max = elem[0]\n# if(elem[0]<min):\n# min = elem[0]\n# return max - min\n \n# def getOutput(lst):\n# combinations_list = generate_combinations(lst)\n# for i in range(len(combinations_list)):\n# if(getRange(combinations_list[i][0]) != getRange(combinations_list[i][1])):\n# result = ['R']*len(lst)\n# for _ in combinations_list[i][0]:\n# result[_[1]]='B'\n# return ''.join(result)\n \n \n# t = int(input())\n# results = []\n# for _ in range(t):\n# input()\n# lst = [int(x) for x in input().split(' ')]\n# results.append(getOutput(lst))\n# for result in results:\n# if result == None:\n# print(\"NO\")\n# continue\n# print(f\"YES\\n{result}\")\nN=0\nM=0\nK=0\ndef getAliceCellsInRowCount(row):\n s=0\n for _ in row:\n if(not _):\n break\n s+=1\n return s\ndef getOutput(kmatrix):\n matrix = [[True for _ in range(M)] for _ in range(N)]\n for elem in kmatrix:\n matrix[elem[0]][elem[1]:]=[False]*(M-elem[1])\n for i in range(0,elem[0]+1):\n matrix[i][elem[1]]=False\n \n return sum(getAliceCellsInRowCount(row) for row in matrix)\nt = int(input())\nresults=list()\nfor _ in range(t):\n possibilities = list()\n kmatrix = list()\n N,M,K = [int(x) for x in input().split(' ')]\n for _ in range(K):\n x,y=input().split(' ')\n kmatrix.append((int(x)-1,int(y)-1))\n possibilities.append(getOutput(kmatrix))\n for k in range(K):\n possibilities.append(getOutput(kmatrix[:k]+kmatrix[k+1:]))\n results.append(possibilities[0])\n rr=\"\"\n for res in possibilities[1:]:\n rr+=f\"{str(res-possibilities[0])} \"\n results.append(rr.strip())\nfor result in results:\n print(result)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 266191418,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \nimport static java.lang.Math.min;\n \npublic class F_1980 {\n \n private static final int MAX = 200005;\n \n public static void main(String[] args) throws IOException {\n final var cin = new Reader();\n final var sb = new StringBuilder();\n int t = cin.nextInt();\n while (t-- > 0) {\n int n = cin.nextInt();\n int m = cin.nextInt();\n int k = cin.nextInt();\n ArrayList<Pair>[] tower = new ArrayList[n];\n for (int i = 0; i < n; i++) {\n tower[i] = new ArrayList<>();\n }\n \n for (int i = 0; i < k; i++) {\n int x = cin.nextInt();\n int y = cin.nextInt();\n tower[x-1].add(new Pair(y, i));\n }\n \n for (int i = 0; i < n; i++) {\n tower[i].sort(Pair::compareTo);\n }\n \n Integer current = m;\n int[] width = new int[n];\n int total = 0;\n int[] res = new int[k];\n Pair last = null;\n int lastAcc = 0;\n int lastWidth = 0;\n \n for (int i = n - 1; i >= 0; i--) {\n if (!tower[i].isEmpty()) {\n Pair rowTower = tower[i].get(0);\n int rowWidth = rowTower.x - 1;\n if (current > rowWidth) {\n if (last != null) {\n res[last.y] = lastAcc;\n }\n last = rowTower;\n if (tower[i].size() != 1) {\n lastWidth = min(current, tower[i].get(1).x - 1) - rowWidth;\n } else {\n lastWidth = current - rowWidth;\n }\n lastAcc = lastWidth;\n \n current = rowWidth; // required for total\n } else {\n lastWidth = min(rowWidth - current, lastWidth);\n lastAcc += lastWidth;\n }\n } else {\n lastAcc += lastWidth;\n }\n width[i] = current;\n total += current;\n }\n \n if (last != null) {\n res[last.y] = lastAcc;\n }\n \n sb.append(total);\n sb.append(\"\\n\");\n for (int i = 0; i < k; i++) {\n sb.append(res[i]).append(\" \");\n }\n sb.append(\"\\n\");\n \n }\n System.out.println(sb);\n }\n \n static class Reader {\n final private int BUFFER_SIZE = 1 << 16;\n private DataInputStream din;\n private byte[] buffer;\n private int bufferPointer, bytesRead;\n \n public Reader() {\n din = new DataInputStream(System.in);\n buffer = new byte[BUFFER_SIZE];\n bufferPointer = bytesRead = 0;\n }\n \n public Reader(String file_name) throws IOException {\n din = new DataInputStream(\n new FileInputStream(file_name));\n buffer = new byte[BUFFER_SIZE];\n bufferPointer = bytesRead = 0;\n }\n \n public String readLine() throws IOException {\n byte[] buf = new byte[64]; // line length\n int cnt = 0, c;\n while ((c = read()) != -1) {\n if (c == '\\n') {\n if (cnt != 0) {\n break;\n } else {\n continue;\n }\n }\n buf[cnt++] = (byte) c;\n }\n return new String(buf, 0, cnt);\n }\n \n public int nextInt() throws IOException {\n int ret = 0;\n byte c = read();\n while (c <= ' ') {\n c = read();\n }\n boolean neg = (c == '-');\n if (neg)\n c = read();\n do {\n ret = ret * 10 + c - '0';\n } while ((c = read()) >= '0' && c <= '9');\n \n if (neg)\n return -ret;\n return ret;\n }\n \n public long nextLong() throws IOException {\n long ret = 0;\n byte c = read();\n while (c <= ' ')\n c = read();\n boolean neg = (c == '-');\n if (neg)\n c = read();\n do {\n ret = ret * 10 + c - '0';\n } while ((c = read()) >= '0' && c <= '9');\n if (neg)\n return -ret;\n return ret;\n }\n \n public double nextDouble() throws IOException {\n double ret = 0, div = 1;\n byte c = read();\n while (c <= ' ')\n c = read();\n boolean neg = (c == '-');\n if (neg)\n c = read();\n \n do {\n ret = ret * 10 + c - '0';\n } while ((c = read()) >= '0' && c <= '9');\n \n if (c == '.') {\n while ((c = read()) >= '0' && c <= '9') {\n ret += (c - '0') / (div *= 10);\n }\n }\n \n if (neg)\n return -ret;\n return ret;\n }\n \n private void fillBuffer() throws IOException {\n bytesRead = din.read(buffer, bufferPointer = 0,\n BUFFER_SIZE);\n if (bytesRead == -1)\n buffer[0] = -1;\n }\n \n private byte read() throws IOException {\n if (bufferPointer == bytesRead)\n fillBuffer();\n return buffer[bufferPointer++];\n }\n \n public void close() throws IOException {\n if (din == null)\n return;\n din.close();\n }\n }\n \n static final class Pair implements Comparable<Pair> {\n int x;\n int y;\n \n Pair(int x, int y) {\n this.x = x;\n this.y = y;\n }\n \n @Override\n public String toString() {\n return \"(\" + x + \", \" + y + \")\";\n }\n \n @Override\n public int hashCode() {\n return x + y + (y << 16);\n }\n \n @Override\n public boolean equals(Object obj) {\n if (this == obj)\n return true;\n if (obj == null)\n return false;\n if (getClass() != obj.getClass())\n return false;\n Pair other = (Pair) obj;\n return x == other.x && y == other.y;\n }\n \n @Override\n public int compareTo(Pair o) {\n if (this.x < o.x) return -1;\n if (this.x > o.x) return 1;\n if (this.y < o.y) return -1;\n if (this.y > o.y) return 1;\n return 0;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 265871346,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Field Division (hard version)\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.\n\nAlice and Bob are dividing the field. The field is a rectangle of size $n \\times m$ ($2 \\le n, m \\le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).\n\nBob has $k$ ($2 \\le k \\le 2 \\cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:\n\n * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field. * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob. * Alice will own the part that includes the cell ($n, 1$). * Bob will own the part that includes the cell ($1, m$). \n\nAlice wants to divide the field in such a way as to get as many cells as possible.\n\nBob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).\n\nThen output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains three integers $n$, $m$, and $k$ ($2 \\le n, m \\le 10^9$, $2 \\le k \\le 2 \\cdot 10^5$) — the field sizes and the number of fountains, respectively.\n\nThen follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \\le r_i \\le n$, $1 \\le c_i \\le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).\n\nIt is guaranteed that the sum of $k$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, first output $\\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \\dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\\alpha + a_i$.\n\nExamples:\ninput:\n5\n2 2 3\n1 1\n1 2\n2 2\n5 5 4\n1 2\n2 2\n3 4\n4 3\n2 5 9\n1 2\n1 5\n1 1\n2 2\n2 4\n2 5\n1 4\n2 3\n1 3\n6 4 4\n6 2\n1 3\n1 4\n1 2\n3 4 5\n2 1\n3 2\n1 4\n1 3\n2 4\noutput:\n1\n1 0 1 \n11\n0 1 0 4 \n1\n0 0 1 1 0 0 0 0 0 \n6\n15 0 0 0 \n1\n2 3 0 0 0\n\nNote: Below are the images for the second example:\n\n The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.\n\nNote that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.Arrays;\nimport java.util.Comparator;\nimport java.util.StringTokenizer;\n \npublic class Pro7 {\n \n public static void solve() throws IOException{\n int n = in.nextInt();\n int m = in.nextInt();\n int k = in.nextInt();\n int[][] points = new int[k][3];\n for (int i = 0; i < k; i++) {\n points[i][0] = i;\n points[i][1] = in.nextInt();\n points[i][2] = in.nextInt() - 1;\n }\n Arrays.sort(points, new Comparator<int[]>() {\n @Override\n public int compare(int[] o1, int[] o2) {\n if (o1[2] == o2[2]) return Integer.compare(o1[1], o2[1]);\n return Integer.compare(o1[2], o2[2]);\n }\n });\n long sum = 0, row = 0;\n boolean[] vis = new boolean[k];\n for (int i = 0; i < k; i++) {\n if (row >= points[i][1]) continue;\n sum += (points[i][1] - row) * points[i][2];\n row = points[i][1];\n vis[points[i][0]] = true;\n }\n sum += (n - row) * m;\n out.println(sum);\n long[] ans = new long[k];\n for (int i = 0; i < k; i++) {\n if (!vis[points[i][0]]) continue;\n int L = i - 1, R = i + 1;\n while (L >= 0 && !vis[points[L][0]]) L--;\n while (R < k && !vis[points[R][0]]) R++;\n int x1 = L < 0 ? 0 : points[L][1];\n int y1 = L < 0 ? 0 : points[L][2];\n int x2 = R >= k ? n : points[R][1];\n int y2 = R >= k ? m : points[R][2];\n long res = 0;\n res -= (points[i][1] - x1) * points[i][2];\n res -= (x2 - points[i][1]) * y2;\n for (int j = i + 1; j < R; j++) {\n if (x1 >= points[j][1]) continue;\n res += (points[j][1] - x1) * points[j][2];\n x1 = points[j][1];\n }\n res += (x2 - x1) * y2;\n ans[points[i][0]] = res;\n }\n for (long v : ans) out.print(v + \" \");\n out.println();\n }\n \n static boolean MULTI_CASE = true;\n \n public static void main(String[] args) throws IOException {\n if (MULTI_CASE) {\n int T = in.nextInt();\n for (int i = 0; i < T; ++i) {\n solve();\n }\n } else {\n solve();\n }\n out.close();\n }\n \n static InputReader in = new InputReader();\n static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));\n \n static class InputReader {\n private StringTokenizer st;\n private BufferedReader bf;\n \n public InputReader() {\n bf = new BufferedReader(new InputStreamReader(System.in));\n st = null;\n }\n \n public String next() throws IOException {\n while (st == null || !st.hasMoreTokens()) {\n st = new StringTokenizer(bf.readLine());\n }\n return st.nextToken();\n }\n \n public String nextLine() throws IOException {\n return bf.readLine();\n }\n \n public int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n \n public long nextLong() throws IOException {\n return Long.parseLong(next());\n }\n \n public double nextDouble() throws IOException {\n return Double.parseDouble(next());\n }\n }\n}\n \n/*\n \n2 2 1 2 3 2 2 3\n1 2 2 2 2 2 3 3\n \n \n */\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1980F2",
"submission_id": 264089421,
"task_type": "TestCase-Eval-Task2-DO"
} |
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