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TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\ntypedef long long ll;\ntypedef long double ld;\n#define endl '\\n'\nconst ll N = 2e5 + 10;\nconst ll M = 1e9 + 7;\nconst ll mod = 998244353;\nconst ll INF = 9e18;\n \n \n \nint main()\n{\n ios_base::sync_with_stdio(false);\n cin.tie(0);\n cout.tie(0);\n \n ll tt;\n cin >> tt;\n while (tt--)\n {\n int n,m;cin>>n>>m;\n vector<vector<char>> a(n,vector<char>(m));\n for(int i=0;i<n;i++){\n for(int j=0;j<m;j++)\n cin>>a[i][j];\n }\n int count=0;\n for(int i=0;i<n;i++){\n for(int j=0;j<m;j++){\n if(i-1>=0 and j-2 >=0 and a[i][j]=='1' and a[i][j-1]=='5' and a[i][j-2]=='4' and a[i-1][j-2]=='3') count++;\n if(i+2<n and j-1>=0 and a[i][j]=='1' and a[i+1][j]=='5' and a[i+2][j]=='4' and a[i+2][j-1]=='3') count++;\n // if(i-3>=0 and a[i][j]=='1' and a[i-1][j]=='5' and a[i-2][j]=='4' and a[i-3][j]=='3') count++;\n if(i-2>=0 and j+1<m and a[i][j]=='1' and a[i-1][j]=='5' and a[i-2][j]=='4' and a[i-2][j+1]=='3') count++;\n if(i-1>=0 and j+2<m and a[i][j]=='1' and a[i-1][j]=='5' and a[i-1][j+1]=='4' and a[i-1][j+2]=='3') count++;\n if(j+3<m and a[i][j]=='1' and a[i][j+1]=='5' and a[i][j+2]=='4' and a[i][j+3]=='3') count++;\n if(i+1<n and j+2<m and a[i][j]=='1' and a[i][j+1]=='5' and a[i][j+2]=='4' and a[i+1][j+2]=='3') count++;\n if(i+2<n and j+1<m and a[i][j]=='1' and a[i][j+1]=='5' and a[i+1][j+1]=='4' and a[i+2][j+1]=='3') count++;\n // if(i+3<n and a[i][j]=='1' and a[i+1][j]=='5' and a[i+2][j]=='4' and a[i+3][j]=='3') count++;\n if(i+2<n and j-1>=0 and a[i][j]=='1' and a[i+1][j]=='5' and a[i+2][j]=='4' and a[i+2][j-1]=='3') count++;\n if(i+1<n and j-2>=0 and a[i][j]=='1' and a[i+1][j]=='5' and a[i+1][j-1]=='4' and a[i+1][j-2]=='3') count++;\n if(j+1<m and i+1<n and a[i][j]=='1' and a[i][j+1]=='5' and a[i+1][j+1]=='4' and a[i+1][j]=='3') count++;\n if(j-1>=0 and i+1<n and a[i][j]=='1' and a[i+1][j]=='5' and a[i+1][j-1]=='4' and a[i][j-1]=='3') count++;\n if(i-1>=0 and j-1>=0 and a[i][j]=='1' and a[i][j-1]=='5' and a[i-1][j-1]=='4' and a[i-1][j]=='3') count++;\n if(i-1>=0 and j+1<m and a[i][j]=='1' and a[i-1][j]=='5' and a[i-1][j+1]=='4' and a[i][j+1]=='3') count++;\n }\n }\n cout<<count<<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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289587537,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\ntypedef long long ll;\n#define debug(X) cerr << #X << \" : \" << X << '\\n';\nconst int N = 1010;\nint a[N][N], f[4];\nll ans = 0;\nint ii, jj;\nbool u, d, l, r;\nchar xx;\nvoid change (int min_x, int min_y, int max_x, int max_y) {\n\tif (u) {\n\t\tif (ii - 1 >= max_x) {\n\t\t\tii --;\n\t\t}\n\t\telse {\n\t\t\tu = 0;\n\t\t\tr = 1;\n\t\t}\n\t}\n\tif (r) {\n\t\tif (jj + 1 <= max_y) {\n\t\t\tjj ++;\n\t\t}\n\t\telse {\n\t\t\tr = 0;\n\t\t\td = 1;\n\t\t}\n\t}\n\tif (d) {\n\t\tif (ii + 1 <= min_x) {\n\t\t\tii ++;\n\t\t}\n\t\telse {\n\t\t\td = 0;\n\t\t\tl = 1;\n\t\t}\n\t}\n\tif (l) {\n\t\tif (jj - 1 >= min_y) {\n\t\t\tjj --;\n\t\t}\n\t\telse {\n\t\t\tl = 0;\n\t\t\tu = 1;\n\t\t\tchange (min_x, min_y, max_x, max_y);\n\t\t}\n\t}\n\treturn;\n}\nll navar (int min_x, int min_y, int max_x, int max_y) {\n\tbool first = 1;\n\tii = min_x;\n\tjj = min_y;\n\tu = 1;\n\td = 0;\n\tl = 0;\n\tr = 0;\n\tll ret = 0;\n\tint x = 0;\n\twhile (ii != min_x || jj != min_y || first) {\n\t\tif (first) {\n\t\t\tfirst = 0;\n\t\t}\n\t\tif (a[ii][jj] == f[x]) {\n\t\t\tx ++;\n\t\t}\n\t\telse {\n\t\t\tx = 0;\n\t\t}\n\t\tif (x == 4) {\n\t\t\tret ++;\n\t\t\tx = 0;\n\t\t}\n\t\tchange (min_x, min_y, max_x, max_y);\n\t}\n\tii = min_x;\n\tjj = min_y;\n\tu = 1;\n\td = 0;\n\tl = 0;\n\tr = 0;\n\twhile (x != 0 && x != 4) {\n\t\tif (a[ii][jj] == f[x]) {\n\t\t\tx ++;\n\t\t}\n\t\telse {\n\t\t\tx = 0;\n\t\t}\n\t\tchange (min_x, min_y, max_x, max_y);\n\t}\n\tif (x == 4) {\n\t\tret ++;\n\t}\n\treturn ret;\n}\nint main () {\n\tint t, n, m;\n\tf[0] = 1;\n\tf[1] = 5;\n\tf[2] = 4;\n\tf[3] = 3;\n\tcin >> t;\n\twhile (t --) {\n\t\tcin >> n >> m;\n\t\tans = 0;\n\t\tfor (int i = 1; i <= n; i ++) {\n\t\t\tfor (int j = 1; j <= m; j ++) {\n\t\t\t\tcin >> xx;\n\t\t\t\ta[i][j] = xx - '0';\n\t\t\t}\n\t\t}\n\t\tint mn = min (n / 2, m / 2);\n\t\tfor (int i = 1; i <= mn; i ++) {\n\t\t\tans += navar (n - i + 1, i, i, m - i + 1);\n\t\t}\n\t\tcout << ans << '\\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289605128,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\n//author : Sajid Ayman\n// problem :\n#include <bits/stdc++.h>\n#define ll long long\n#define ld long double\n#define all(x) (x).begin(), (x).end()\n#define nl '\\n'\n#define tt int tc;cin >> tc;while (tc--)\n//if(i<1||j<1||i>n||j>m)\nusing namespace std;\nconst double PI=3.14159265359;\nconst int dx[8]={0, 0, 1, -1, 1, -1, -1, 1};\nconst int dy[8]={1, -1, 0, 0, 1, -1, 1, -1};\nconst int N = 1e5+5, Mod = 1e9+7;\nconst ll inf=1e18;\n#include <ext/pb_ds/tree_policy.hpp>\n#include <ext/pb_ds/assoc_container.hpp>\nusing namespace __gnu_pbds;\n// indexed set\ntemplate<typename T>\nusing o_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\n// multi ordered set\ntemplate<typename T>\nusing mo_set = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;\n// fast power method\nll power(ll a,ll b){\n if(b==0)return 1;\n if(b==1)return a;\n ll ans;\n ans=power(a,b/2);\n ans*=ans;\n// ans%=Mod;\n if(b%2)ans*=a;\n// return ans%Mod;\n return ans;\n}\n// solution starts here\nvoid sol(){\n int n,m;cin>>n>>m;\n string a[n];\n for(int i=0;i<n;i++) cin>>a[i];\n vector<vector<char>>v(min(n,m)/2+1);\n for(int i=0;i<min(n,m);i++) {\n for (int j = i; j < m - i; ++j)v[i].push_back(a[i][j]);\n for (int j = i + 1; j < n - i; ++j)v[i].push_back(a[j][m -i-1]);\n for (int j = m - i - 2; j >=0+i;--j)v[i].push_back(a[n - i-1][j]);\n for (int j = n - i - 2; j >= i + 1; --j)v[i].push_back(a[j][i]);\n }\n ll res=0;\n for(int i=0;i<min(n,m)/2+1;i++){\n if(v[i].empty())break;\n int idx=0;\n while(v[i][idx]!='1'&&idx<v[i].size()-1)idx++;\n if(idx!=v[i].size()){\n string s=\"\";\n while(v[i][idx]!='*'){\n s+=v[i][idx];\n if(s==\"1\"||s==\"15\"||s==\"154\"||s==\"1543\");\n else s=\"\";\n v[i][idx]='*',idx++;\n if(idx==v[i].size())idx=0;\n if(s==\"1543\")res++,s=\"\";\n// cout<<s<<nl;\n }\n }\n }\n cout<<res<<nl;\n}\nint main() {\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n tt sol();\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289689553,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\ndef fun():\n n, m = map(int, input().split())\n mx = []\n for i in range(n):\n mx.append(list(input()))\n \n a = [\"1\", \"5\", \"4\", \"3\"]\n cnt = 0\n \n num = min(n, m) // 2\n \n for idx in range(num):\n top, bottom = idx, n - idx - 1\n left, right = idx, m - idx - 1\n layer = []\n \n for i in range(left, right + 1):\n layer.append(mx[top][i])\n \n for i in range(top + 1, bottom + 1):\n layer.append(mx[i][right])\n \n if top < bottom:\n for i in range(right - 1, left - 1, -1):\n layer.append(mx[bottom][i])\n \n if left < right:\n for i in range(bottom - 1, top, -1):\n layer.append(mx[i][left])\n \n cur = 0\n for i in range(len(layer)):\n if layer[i] == a[cur]:\n cur += 1\n if cur == 4:\n cnt += 1\n cur = 0\n else:\n cur = 0\n \n if len(layer) >= (2 * (m + n - 2 - 4 * idx)):\n lay2 = layer[-3:] + layer[:3]\n cur = 0\n for i in range(len(lay2)):\n if lay2[i] == a[cur]:\n cur += 1\n if cur == len(a):\n cnt += 1\n cur = 0\n else:\n cur = 0\n \n print(cnt)\n \n# t = 1\nt = int(input())\nit = input\not = print\nitint = lambda: int(input())\nfor _ in range(t):fun()\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289560350,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\ndef get_matrix_rings(matrix):\n if not matrix or not matrix[0]:\n return []\n \n rows, cols = len(matrix), len(matrix[0])\n \n top = 0\n bottom = rows - 1\n left = 0\n right = cols - 1\n cnt = 0\n while top <= bottom and left <= right:\n ring = []\n \n for j in range(left, right + 1):\n ring.append(matrix[top][j])\n \n for i in range(top + 1, bottom + 1):\n ring.append(matrix[i][right])\n \n if top < bottom:\n for j in range(right - 1, left - 1, -1):\n ring.append(matrix[bottom][j])\n \n if left < right:\n for i in range(bottom - 1, top, -1):\n ring.append(matrix[i][left])\n \n i = 0\n while i < len(ring) and (ring[i] == 5 or ring[i] == 4 or ring[i] == 3):\n i += 1\n # print(ring, i)\n if i == len(ring):\n return 0\n reqd = [1, 5, 4, 3]\n curr = 0\n for j in range(i, i + len(ring)):\n j = j % len(ring)\n # print(j, curr)\n if ring[j] == reqd[curr]:\n curr += 1\n else:\n curr = 0\n if curr == len(reqd):\n cnt += 1\n curr = 0\n top += 1\n bottom -= 1\n left += 1\n right -= 1\n \n return cnt\n \nt = int(input())\nfor k in range(t):\n n, m = map(int, input().split())\n matrix = []\n for _ in range(n):\n matrix.append([int(i) for i in input()])\n # print(matrix)\n print(get_matrix_rings(matrix))\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289588381,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\nimport sys\nfrom itertools import chain, islice, repeat\n \ninput = sys.stdin.readline\n \nnum = [1, 5, 4, 3]\n \nfor _ in range(int(input())):\n gy, gx = map(int, input().split())\n ys = (0, gy - 1)\n xs = (0, gx - 1)\n grid = [list(map(int, input().rstrip())) for _ in range(gy)]\n \n times = 0\n while ys[0] < ys[1] and xs[0] < xs[1]:\n layer = [\n (y, x)\n for y, x in chain(\n zip(repeat(ys[0]), range(xs[0], xs[1] + 1)),\n zip(range(ys[0] + 1, ys[1]), repeat(xs[1])),\n zip(repeat(ys[1]), range(xs[1], xs[0] - 1, -1)),\n zip(range(ys[1] - 1, ys[0], -1), repeat(xs[0])),\n )\n ]\n layer.extend(layer[:3])\n \n have = 0\n for y, x in layer:\n if grid[y][x] == num[have]:\n have += 1\n if have == len(num):\n times += 1\n have = 0\n else:\n have = 0\n \n ys = (ys[0] + 1, ys[1] - 1)\n xs = (xs[0] + 1, xs[1] - 1)\n \n print(times)\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 290032326,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\npublic class ques4 {\n private static final int mod = 1000000007;\n // Custom class to hold pairs\n static class Pair {\n int value;\n int index;\n Pair(int value, int index) {\n this.value = value;\n this.index = index;\n }\n }\n // Euclidean algorithm for finding GCD for int type\n public static int gcdInt(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 // Euclidean algorithm for finding GCD for long type\n public static long gcdLong(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 // Binary exponentiation for int type\n public static int power_Int(int base, int exponent) {\n int result = 1;\n while (exponent > 0) {\n if ((exponent & 1) == 1) {\n result *= base;\n }\n base *= base;\n exponent >>= 1;\n }\n return result;\n }\n // Binary exponentiation for long type\n public static long power_Long(long base, long exponent) {\n long result = 1;\n while (exponent > 0) {\n if ((exponent & 1) == 1) {\n result *= base;\n }\n base *= base;\n exponent >>= 1;\n }\n return result;\n }\n // Modular exponentiation for int type with MOD = 1e9 + 7\n public static int power_modInt(int base, int exponent) {\n int result = 1;\n base %= mod;\n while (exponent > 0) {\n if ((exponent & 1) == 1) {\n result = (int) ((result * (long) base) % mod);\n }\n base = (int) ((base * (long) base) % mod);\n exponent >>= 1;\n }\n return result;\n }\n // greater than or equal to key\n public static long long_lowerBound(long[] arr, long key) {\n long low = 0, high = arr.length - 1;\n long ans = arr.length;\n while (low <= high) {\n long mid = (low + high) / 2;\n if (arr[(int)mid] >= key) {\n ans = mid;\n high = mid - 1;\n } else {\n low = mid + 1;\n }\n }\n return ans;\n }\n // greater than or equal to key\n public static long long_upperBound(long[] arr, long key) {\n long low = 0, high = arr.length - 1;\n long ans = arr.length;\n while (low <= high) {\n long mid = (low + high) / 2;\n if (arr[(int)mid] > key) {\n ans = mid;\n high = mid - 1;\n } else {\n low = mid + 1;\n }\n }\n return ans;\n }\n // greater than equal to key\n public static int int_lowerBound(int[] arr, int key) {\n int low = 0, high = arr.length - 1;\n int ans = arr.length;\n while (low <= high) {\n int mid = (low + high) / 2;\n if (arr[mid] >= key) {\n ans = mid;\n high = mid - 1;\n } else {\n low = mid + 1;\n }\n }\n return ans;\n }\n // greater than key\n public static int int_upperBound(int[] arr, int key) {\n int low = 0, high = arr.length - 1;\n int ans = arr.length;\n while (low <= high) {\n int mid = (low + high) / 2;\n if (arr[mid] > key) {\n ans = mid;\n high = mid - 1;\n } else {\n low = mid + 1;\n }\n }\n return ans;\n }\n // solve\n public static void main(String args[]) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n PrintWriter out = new PrintWriter(System.out);\n String test = br.readLine();\n int t = Integer.parseInt(test);\n while (t-- > 0) {\n String[] nk = br.readLine().split(\" \");\n int n = Integer.parseInt(nk[0]);\n int m = Integer.parseInt(nk[1]);\n int[][] a = new int[n][m];\n int[][] cal = new int[n][m];\n for (int i = 0; i < n; i++) {\n for(int j = 0; j<m; j++){\n cal[i][j]=0;\n }\n }\n for (int i = 0; i < n; i++) {\n String s = br.readLine();\n int val = Integer.parseInt(s);\n for(int j = 0; j<m; j++){\n a[i][m-j-1]=val%10;\n val/=10;\n }\n }\n long ans = 0;\n for (int i = 0; i < n; i++) {\n for(int j = 0; j<m; j++){\n if(a[i][j]==1){\n if(j+1<m && a[i][j+1]==5){\n if(j+2<m && a[i][j+2]==4){\n if(j+3<m && a[i][j+3]==3){\n ans++;\n }\n if(i+1<n && a[i+1][j+2]==3){\n ans++;\n }\n }\n if(i+1<n && a[i+1][j+1]==4){\n if(i+2<n && a[i+2][j+1]==3){\n ans++;\n }\n if(a[i+1][j]==3){\n ans++;\n }\n }\n }\n if(i+1<n && a[i+1][j]==5){\n if(i+2<n && a[i+2][j]==4){\n if(j-1>=0 && a[i+2][j-1]==3){\n ans++;\n }\n }\n if(j-1>=0 && a[i+1][j-1]==4){\n if(a[i][j-1]==3){\n ans++;\n }\n if(j-2>=0 && a[i+1][j-2]==3){\n ans++;\n }\n }\n }\n if(i-1>=0 && a[i-1][j]==5){\n if(j+1<m && a[i-1][j+1]==4){\n if(a[i][j+1]==3){\n ans++;\n }\n if(j+2<m && a[i-1][j+2]==3){\n ans++;\n }\n }\n if(i-2>=0 && a[i-2][j]==4){\n if(j+1<m && a[i-2][j+1]==3){\n ans++;\n }\n if(i-3>=0 && a[i-3][j]==3){\n ans++;\n }\n }\n }\n if(j-1>=0 && a[i][j-1]==5){\n if(j-2>=0 && i-1>=0 && a[i][j-2]==4 && a[i-1][j-2]==3){\n ans++;\n }\n if(i-1>=0 && a[i-1][j-1]==4){\n if(i-2>=0 && a[i-2][j-1]==3){\n ans++;\n }\n if(a[i-1][j]==3){\n ans++;\n }\n }\n }\n }\n }\n }\n out.println(ans);\n }\n out.flush();\n out.close();\n br.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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289781590,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class challenge {\n static String target = \"1543\";\n static HashSet<String> hash = new HashSet<>();\n \n // 1= ul to ur;\n // 2= ur to lr\n // 3= lr to ll\n // 4= ll to ul\n static int solve(String[] k, int x, int y, int type, int j, int total, int minX, int minY, int maxX, int maxY,\n int c) {\n if (c == 2) {\n return total;\n }\n int r = total;\n if (k[y].charAt(x) == target.charAt(j)) {\n j++;\n j %= 4;\n if (!hash.contains(x + \",\" + y) && j == 0) {\n hash.add(x + \",\" + y);\n r += j == 0 ? 1 : 0;\n }\n } else {\n \n j = 0;\n }\n switch (type) {\n case 1:\n if (x + 1 >= maxX) {\n r = solve(k, x, y + 1, 2, j, r, minX, minY, maxX, maxY, c);\n break;\n }\n r = solve(k, x + 1, y, 1, j, r, minX, minY, maxX, maxY, c);\n break;\n case 2:\n if (y + 1 >= maxY) {\n r = solve(k, x - 1, y, 3, j, r, minX, minY, maxX, maxY, c);\n break;\n }\n r = solve(k, x, y + 1, 2, j, r, minX, minY, maxX, maxY, c);\n break;\n case 3:\n if (x - 1 < minX) {\n r = solve(k, x, y - 1, 4, j, r, minX, minY, maxX, maxY, c);\n break;\n }\n r = solve(k, x - 1, y, 3, j, r, minX, minY, maxX, maxY, c);\n break;\n case 4:\n if (y - 1 < minY) {\n return r = solve(k, x + 1, y, 1, j, r, minX, minY, maxX, maxY, c + 1);\n }\n r = solve(k, x, y - 1, 4, j, r, minX, minY, maxX, maxY, c);\n break;\n default:\n break;\n }\n return r;\n }\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());\n \n outer: for (int q = 0; q < t; q++) {\n String p[] = br.readLine().split(\" \");\n int n = Integer.parseInt(p[0]);\n int m = Integer.parseInt(p[1]);\n hash = new HashSet<>();\n \n String[] k = new String[n];\n for (int i = 0; i < n; i++) {\n k[i] = br.readLine();\n }\n \n int min = Math.min(n, m) / 2;\n int r = 0;\n for (int i = 0; i < min; i++) {\n r += solve(k, i, i, 1, 0, 0, i, i, m - i, n - i, 0);\n }\n System.out.println(r);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 291186975,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\nt = int(input())\n \ndef check(n, m, matrix, layer):\n ans = 0\n sc = 0\n seq = '1543'\n for i in range(m-2*layer):\n if matrix[layer][layer+i] == seq[sc]:\n sc += 1\n else:\n sc = 0\n \n if sc == 4:\n ans += 1\n sc = 0\n \n if sc > 0:\n sc -= 1\n \n for i in range(n-2*layer):\n # print(layer+i, m-1-layer)\n if matrix[layer+i][m-1-layer] == seq[sc]:\n sc += 1\n else:\n sc = 0\n \n if sc == 4:\n ans += 1\n sc = 0\n \n if sc > 0:\n sc -= 1\n \n \n for i in range(m-2*layer):\n # print(n-1-layer, m-1-layer-i)\n if matrix[n-1-layer][m-1-layer-i] == seq[sc]:\n sc += 1\n else:\n sc = 0\n \n if sc == 4:\n ans += 1\n sc = 0\n \n if sc > 0:\n sc -= 1\n \n for i in range(n-2*layer):\n # print(n-1-layer-i, layer)\n if matrix[n-1-layer-i][layer] == seq[sc]:\n sc += 1\n else:\n sc = 0\n \n if sc == 4:\n ans += 1\n sc = 0\n \n if sc > 0:\n sc -= 1\n \n # print()\n \n if m - 2*layer > 2:\n for i in range(3):\n # print(layer, layer+i)\n if matrix[layer][layer+i] == seq[sc]:\n sc += 1\n else:\n sc = 0\n \n if sc == 4:\n ans += 1\n sc = 0\n else:\n # print(\"here\", sc)\n if sc == 0:\n pass\n elif sc == 1:\n if matrix[layer][layer] == seq[1] and matrix[layer][layer+1] == seq[2] and matrix[layer+1][layer+1] == seq[3]:\n ans += 1\n \n elif sc == 2:\n if matrix[layer][layer] == seq[2] and matrix[layer][layer+1] == seq[3]:\n ans += 1\n elif sc == 3:\n if matrix[layer][layer] == seq[3]:\n ans += 1\n \n return ans\n \nfor _ in range(t):\n n, m = map(int,input().split())\n matrix = []\n \n for i in range(n):\n a = input()\n matrix.append(a)\n \n ans = 0\n \n \n for layer in range(min(n, m)//2 + 1):\n #print(\"layer\", layer)\n x = check(n, m, matrix, layer)\n # print(\"layer\", layer, x)\n ans += x\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289541525,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\ndef solve():\n n, m = list(map(int, input().split()))\n arr = []\n for _ in range(n):\n arr.append([int(e) for e in list(input())])\n w1 = m - 1\n w2 = n - 1\n w3 = 0\n w4 = 0\n dontdec = False\n if arr[0][0] == 1:\n i, j = 0, 0\n dire = 0\n elif arr[1][0] == 1:\n i, j = 1, 0\n dire = 3\n dontdec = True\n elif n > 2 and arr[2][0] == 1:\n i, j = 2, 0\n dire = 3\n dontdec = True\n elif arr[1][1] == 1:\n i, j = 1, 1\n dire = 2\n dontdec = True\n elif m > 2 and arr[1][2] == 1:\n i, j = 1, 2\n dire = 2\n dontdec = True\n else:\n i, j = 0, 0\n dire = 0\n num = []\n cnt = 0\n while w1 >= w3 and w2 >= w4:\n if dire == 0:\n num.append(arr[i][j])\n if len(num) > 4:\n num.pop(0)\n if j == w1:\n w1 -= 1\n dire = 1\n j -= 1\n i += 1\n if dontdec:\n dontdec = False\n w1 += 1\n if num == [1, 5, 4, 3]:\n cnt += 1\n j += 1\n elif dire == 1:\n num.append(arr[i][j])\n if len(num) > 4:\n num.pop(0)\n if i == w2:\n w2 -= 1\n dire = 2\n i -= 1\n j -= 1\n if dontdec:\n dontdec = False\n w2 += 1\n if num == [1, 5, 4, 3]:\n cnt += 1\n i += 1\n elif dire == 2:\n num.append(arr[i][j])\n if len(num) > 4:\n num.pop(0)\n if j == w3:\n w3 += 1\n dire = 3\n j += 1\n i -= 1\n if num == [1, 5, 4, 3]:\n cnt += 1\n j -= 1\n elif dire == 3:\n num.append(arr[i][j])\n if len(num) > 4:\n num.pop(0)\n if i == w4:\n w4 += 1\n dire = 0\n i += 1\n j += 1\n if dontdec:\n w4 -= 1\n dontdec = False\n if num == [1, 5, 4, 3]:\n cnt += 1\n i -= 1\n # print(num, w3, dire, i, j, w3, w1, w2, w4)\n print(cnt)\nt = int(input())\nfor _ in range(t):\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289571362,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\nimport java.util.Scanner;\npublic class ilove1543 {\n public static void main(String[] args) {\n Scanner in = new Scanner(System.in);\n int t = in.nextInt();\n for (int k = 0; k < t; k++) {\n int n = in.nextInt();\n int m = in.nextInt();\n int[][] mat = new int[n][m];\n for (int i = 0; i < n; i++) {\n \n String line = in.next();\n for (int j = 0; j < m; j++) {\n mat[i][j] = line.charAt(j) - '0';\n }\n }\n \n \n }\n System.out.println(1);\n System.out.println(1);\n System.out.println(0);\n System.out.println(1);\n System.out.println(0);\n System.out.println(2);\n System.out.println(2);\n System.out.println(2);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 292107697,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.Locale;\nimport java.util.StringTokenizer;\n \n \npublic class Template {\n public static void main(String[] args) throws FileNotFoundException {\n Locale.setDefault(Locale.ROOT);\n InputStream inputStream = System.in;\n OutputStream outputStream = System.out;\n InputReader in = new InputReader(inputStream);\n PrintWriter out = new PrintWriter(outputStream);\n// InputReader in = new InputReader(new FileInputStream(\"sum.in\"));\n// PrintWriter out = new PrintWriter(new FileOutputStream(\"sum.out\"));\n solve(in, out);\n out.close();\n }\n \n \n public static void solve(InputReader in, PrintWriter out) {\n int t = in.nextInt();\n char[] kk = new char[]{'1', '5', '4', '3'};\n for (int i = 0; i < t; i++) {\n int n = in.nextInt();\n int m = in.nextInt();\n char[][] matr = new char[n][m];\n int ans = 0;\n \n for (int j = 0; j < n; j++) {\n matr[j] = in.next().toCharArray();\n }\n \n for (int layer = 0; layer < Math.min(n, m) / 2; layer++) {\n int lookup = 0;\n for (int j = layer; j < m - layer; j++) {\n if (matr[layer][j] == kk[lookup]) {\n lookup++;\n if (lookup == 4) {\n ans++;\n lookup = 0;\n }\n } else {\n lookup = 0;\n }\n }\n \n \n for (int j = layer + 1; j < n - layer; j++) {\n if (matr[j][m - layer - 1] == kk[lookup]) {\n lookup++;\n if (lookup == 4) {\n ans++;\n lookup = 0;\n }\n } else {\n lookup = 0;\n }\n }\n \n \n for (int j = m - layer - 2; j >= layer; j--) {\n if (matr[n - layer - 1][j] == kk[lookup]) {\n lookup++;\n if (lookup == 4) {\n ans++;\n lookup = 0;\n }\n } else {\n lookup = 0;\n }\n }\n \n \n for (int j = n - layer - 2; j > layer; j--) {\n if (matr[j][layer] == kk[lookup]) {\n lookup++;\n if (lookup == 4) {\n ans++;\n lookup = 0;\n }\n } else {\n lookup = 0;\n }\n }\n \n //++\n if (lookup > 0) {\n int lookup1 = lookup;\n for (int j = layer; j < Math.min(5 - lookup1, m - layer); j++) {\n if (matr[layer][j] == kk[lookup]) {\n lookup++;\n if (lookup == 4) {\n ans++;\n lookup = 0;\n }\n } else {\n lookup = 0;\n }\n }\n \n if ((m - layer - layer == 2 || n - layer - layer == 2) && lookup > 0) {\n lookup1 = lookup;\n for (int j = layer + 1; j < Math.min(5 - lookup1, n - layer); j++) {\n if (matr[j][m - layer - 1] == kk[lookup]) {\n lookup++;\n if (lookup == 4) {\n ans++;\n lookup = 0;\n }\n } else {\n lookup = 0;\n }\n }\n \n if (lookup > 0) {\n lookup1 = lookup;\n for (int j = m - layer - 2; j >= Math.min(5 - lookup1, layer); j--) {\n if (matr[n - layer - 1][j] == kk[lookup]) {\n lookup++;\n if (lookup == 4) {\n ans++;\n lookup = 0;\n }\n } else {\n lookup = 0;\n }\n }\n \n \n }\n \n }\n }\n \n }\n out.println(ans);\n }\n }\n \n \n static class InputReader {\n public BufferedReader reader;\n public StringTokenizer tokenizer;\n \n public InputReader(InputStream stream) {\n reader = new BufferedReader(new InputStreamReader(stream), 32768);\n tokenizer = null;\n }\n \n public String next() {\n while (tokenizer == null || !tokenizer.hasMoreTokens()) {\n try {\n tokenizer = new StringTokenizer(reader.readLine());\n } catch (IOException e) {\n throw new RuntimeException(e);\n }\n }\n return tokenizer.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine() {\n try {\n return reader.readLine();\n } catch (IOException e) {\n throw new RuntimeException(e);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289553442,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\n#define _USE_MATH_DEFINES\n#include <unordered_set>\n#include <unordered_map>\n#include <algorithm>\n#include <iostream>\n#include <sstream>\n#include <fstream>\n#include <iomanip>\n#include <cstring>\n#include <fstream>\n#include <utility>\n#include <cstdlib>\n#include <vector>\n#include <string>\n#include <limits>\n#include <cmath>\n#include <stack>\n#include <queue>\n#include <regex>\n#include <ctime>\n#include <set>\n#include <map>\nint main()\n{\n int t;\n std::cin >> t;\n while (t) {\n --t;\n int n, m;\n std::string s;\n std::cin >> n >> m;\n getline(std::cin, s);\n int** a = new int* [n];\n for (int i = 0; i < n; ++i) {\n a[i] = new int[m];\n }\n for (int i = 0; i < n; ++i) {\n getline(std::cin, s);\n for (int j = 0; j < m; ++j) {\n a[i][j] = s[j]-'0';\n }\n }\n int q = 0,d=0,r=0;\n while (q < n * m) {\n int* b = new int[2 * n + 2 * m + 10];\n for (int i = d; i < m-d; ++i) {\n if (q >= n * m)break;\n b[q] = a[d][i];\n //std::cout << b[q] << \" \";\n ++q;\n }\n for (int j = d+1; j < n-d; ++j) {\n if (q >= n * m)break;\n b[q] = a[j][m-d-1];\n //std::cout << b[q] << \" \";\n ++q;\n }\n for (int i = d+1; i < m-d; ++i) {\n if (q >= n * m)break;\n b[q] = a[n-d-1][m-1-i];\n //std::cout << b[q] << \" \";\n ++q;\n }\n for (int j = d+1; j<n-d-1; ++j) {\n if (q >= n * m)break;\n b[q] = a[n-j-1][d];\n //std::cout << b[q] << \" \";\n ++q;\n }\n int y = q;\n for (int i = d; i <std::min(3, m - d); ++i) {\n b[y] = a[d][i];\n //std::cout << b[y] << \" \";\n ++y;\n }\n if (m - d == 2) {\n for (int j = d + 1; j < n - d; ++j) {\n b[y] = a[j][m - d - 1];\n //std::cout << b[q] << \" \";\n ++y;\n break;\n }\n }\n for (int i = 3; i < y; ++i) {\n if (b[i - 3] == 1 && b[i - 2] == 5 && b[i - 1] == 4 && b[i] == 3)++r;\n }\n ++d;\n }\n std::cout << r << \"\\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289573697,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std ;\n#define ll long long \n \nint main(){\n\tint t;\n\tcin>>t;\n\twhile(t--){\n\t\tint n,m ;\n\t\tcin >> n >> m ;\n \n\t\tvector<vector<int>> arr(n,vector<int>(m,0));\n\t\tfor(int i=0 ; i<n ; i++){\n\t\t\tint num ; cin >> num ;\n\t\t\tfor(int j=m-1 ; j>=0 ; j--){\n\t\t\t\tarr[i][j]= num%10 ;\n\t\t\t\tnum=num/10;\n\t\t\t\t// cout << arr[i][j] << \" \" ;\n\t\t\t}\n\t\t\t// cout << endl;\n\t\t}\n \n \n\t\tint top=0, bottom = n-1 , left=0, right = m-1 ;\n\t\tint ans = 0 ;\n \n\t\twhile(top < bottom && left < right) {\n\t\t\tvector<int> temp ;\n \n\t\t\tfor(int i=left ; i<=right ; i++){\n\t\t\t\ttemp.push_back(arr[top][i]) ;\n\t\t\t}\n \n\t\t\ttop++ ;\n\t\t\tfor(int i=top ; i<=bottom ; i++){\n\t\t\t\ttemp.push_back(arr[i][right]);\n\t\t\t}\n\t\t\tright-- ;\n \n\t\t\tif(top<=bottom)\n\t\t\t{for(int i=right ; i>=left ; i--){\n\t\t\t\t\t\t\ttemp.push_back(arr[bottom][i]);\n\t\t\t\t\t\t}\n\t\t\t\t\t\tbottom--;}\n \n\t\t\tif (left <= right) \n\t\t\t{for(int i=bottom; i>=top ; i--){\n\t\t\t\t\t\t\ttemp.push_back(arr[i][left]);\n\t\t\t\t\t\t}\n\t\t\t\t\t\tleft++;}\n \n\t\t\t// cout << \"TEMP\\n\" ;\n\t\t\t// for(auto i : temp)\n\t\t\t// \tcout << i<< \"\\t\" ;\n\t\t\t// cout << endl ;\n \n\t\t\tint n_ = temp.size();\n\t\t\tfor(int i=0 ; i< n_ - 3 ; i++){\n\t\t\t\tif(temp[i]==1 && temp[i+1]==5 && temp[i+2]==4 && temp[i+3]==3)ans++;\n\t\t\t}\n \n\t\t\tif(n_ > 3)\n\t\t\t{\n\t\t\t\tif(temp[n_-3]==1 && temp[n_-2]==5 && temp[n_-1]==4 && temp[0]==3) ans++;\n\t\t\t\telse if(temp[1]==3 && temp[n_-2]==1 && temp[n_-1]==5 && temp[0]==4) ans++;\n\t\t\t\telse if(temp[1]==4 && temp[2]==3 && temp[n_-1]==1 && temp[0]==5) ans++;\n\t\t\t}\n\t\t}\n\t\tcout << ans << endl;\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289552697,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\nt = int(input())\n \nfor ii in range(t):\n n,m = list(map(int,input().split()))\n t = [[0]*m for i in range(n)]\n for jj in range(n):\n s = int(input())\n for kk in range(m):\n t[jj][kk] = s // 10**(m-kk-1) % 10\n bound = min(n//2,m//2)\n count = 0\n rotate = 0\n for jj in range(bound):\n upperBound = jj\n rightBound = m - jj - 1\n lowerBound = n - jj - 1\n leftBound = jj\n hash = set()\n numList = []\n currentX = jj\n currentY = jj\n nextX = currentX\n nextY = currentY + 1\n direction = 0\n hash.add((currentX,currentY))\n numList.append(t[currentX][currentY])\n while (nextX,nextY) not in hash:\n currentX = nextX\n currentY = nextY\n hash.add((currentX,currentY))\n numList.append(t[currentX][currentY])\n if direction == 0 and currentY != rightBound:\n nextX = currentX\n nextY = currentY + 1\n elif direction == 0 and currentY == rightBound:\n nextX = currentX + 1\n nextY = currentY\n direction = 1\n elif direction == 1 and currentX != lowerBound:\n nextX = currentX + 1\n nextY = currentY\n elif direction == 1 and currentX == lowerBound:\n nextX = currentX\n nextY = currentY - 1\n direction = 2\n elif direction == 2 and currentY != leftBound:\n nextX = currentX\n nextY = currentY - 1\n elif direction == 2 and currentY == leftBound:\n nextX = currentX - 1\n nextY = currentY\n elif direction == 3 and currentX != upperBound:\n nextX = currentX - 1\n nextY = currentY\n direction = 0\n lengthList = len(numList)\n \n for kk in range(lengthList):\n rotate = rotate + 1\n if numList[kk % lengthList] == 1 and numList[(kk + 1) % lengthList] == 5 and numList[(kk + 2) % lengthList] == 4 and numList[(kk + 3) % lengthList] == 3:\n count = count + 1\n print(count)\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289562858,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\nfrom collections import defaultdict\nfrom sys import stdin\n \ndef I(): return int(stdin.readline().strip())\n \ndef II(): return map(int, stdin.readline().strip().split())\n \ndef IL(): return list(map(int, stdin.readline().strip().split()))\n \ndef SIL(): return sorted(map(int, stdin.readline().strip().split()),)\n \ndef S() : return stdin.readline().strip()\n \ndef SL() : return list(stdin.readline().strip().split())\n \ndef count(part):\n ans = 0\n for i in range(len(part) - 3):\n if part[i:i + 4] == \"1543\":\n ans += 1\n return ans\n \ndef solve():\n n, m = II()\n mat = []\n for _ in range(n):\n mat.append(list(S()))\n result = []\n while mat:\n result += mat.pop(0)\n mat = (list(zip(*mat)))[::-1]\n \n broken = []\n i = 0\n j = i + (2 * n + 2 * m - 4)\n while i < len(result):\n broken.append(result[i:j])\n n -= 2\n m -= 2\n i = j\n j = i + (2 * n + 2 * m - 4)\n \n for i in range(len(broken)):\n broken[i] = \"\".join(broken[i] + broken[i][0:(3 if len(broken[i]) >= 4 else 0)])\n ans = 0\n for part in broken:\n ans += count(part)\n \n print(ans)\n \n \n \n \nT = I()\nfor _ in range(T):\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289586736,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\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#define op() \\\n ios_base::sync_with_stdio(0); \\\n cin.tie(0); \\\n cout.tie(0);\n#define srt(a) sort(a.begin(), a.end())\n#define rsrt(a) sort(a.begin(), a.end(), greater<>())\n#define fo(n) for (int i = 0; i < n; i++)\n#define fo1(n) for (int i = 1; i <= n; i++)\n#define all(a) a.begin(), a.end()\n#define pb push_back\n#define ll long long\n#define Inf 10000000000000000\n#define pi 3.1415926536\n#define gcd __gcd\n#define vi vector<int>\n#define vl vector<ll>\n#define nl '\\n'\n#define min_priority priority_queue<ll, vector<ll>, greater<ll>>\n#define ordered_set tree<pair<ll, ll>, null_type, greater<pair<ll, ll>>, rb_tree_tag, tree_order_statistics_node_update>\n#define show(a) \\\n for (auto e : a) \\\n cout << e << \" \"; \\\n cout << nl;\n#define rY \\\n cout << \"YES\" << nl; \\\n return;\n#define rN \\\n cout << \"NO\" << nl; \\\n return;\n#define rr return\nconst ll mod = 1e9 + 7;\n \nvoid Solve()\n{\n ll n, m, x, y, z, i, j, k, l, r;\n cin >> n >> m;\n char a[n + 3][m + 2];\n for (i = 1; i <= n; i++)\n {\n for (j = 1; j <= m; j++)\n cin >> a[i][j];\n }\n \n ll n1 = n, m1 = m;\n l = 1, r = 1;\n ll ans = 0;\n while (l < n1)\n {\n string s;\n for (i = r; i <= m1; i++)\n s.push_back(a[l][i]);\n for (i = l + 1; i <= n1; i++)\n s.push_back(a[i][m1]);\n for (i = m1 - 1; i >= r; i--)\n s.push_back(a[n1][i]);\n for (i = n1 - 1; i > l; i--)\n s.push_back(a[i][r]);\n l++, r++, m1--, n1--;\n for (i = 0; i < min((int)s.size(), 3); i++)\n s.push_back(s[i]);\n for (i = 0; i < s.size() - 3; i++)\n {\n if (s[i] == '1' && s[i + 1] == '5' && s[i + 2] == '4' && s[i + 3] == '3')\n ans++;\n }\n }\n cout << ans << nl;\n}\nint main()\n{\n op();\n int test = 1, c = 1;\n cin >> test;\n while (test--)\n {\n // cin.ignore();\n // cout << \"Case \" << c++ << \":\\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289795854,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\n#include<iostream>\n#include<vector>\nusing namespace std;\n int n,m;\n void print(vector<int>&layer)\n {\n for(auto&i:layer)\n {\n cout<<i<<\" \";\n }\n cout<<endl;\n \n }\nvoid strips(int w,vector<int>&layer,vector<vector<int>>&arr)\n{\n layer.clear();\n for(int i=w,j=w;j<=m-w;j++)\n {\n layer.push_back(arr[i][j]);\n }\n for(int i=w+1,j=m-w;i<=n-w;i++)\n {\n layer.push_back(arr[i][j]);\n }\n for(int i=n-w,j=m-w-1;j>=w;j--)\n {\n layer.push_back(arr[i][j]);\n }\n for(int i=n-w-1, j=w;i>w;i--)\n {\n layer.push_back(arr[i][j]);\n }\n}\n \nint main()\n{\n int t;\n cin>>t;\n while(t--)\n {\n \n cin>>n>>m;\n \n vector<vector<int>>arr(n,vector<int>(m));\n for(int i=0;i<n;i++)\n {\n for(int j=0;j<m;j++)\n {\n char c;\n cin>>c;\n arr[i][j]=c-'0';\n }\n }\n n--;m--;\n vector<int>layer;\n int w=0;\n long long ans=0;\n vector<int>tar={1,5,4,3};\n while(w<=n&&w<=m)\n {\n strips(w,layer,arr);\n int yoyo=0;\n for(int i=0;i<(int)layer.size();i++)\n {\n if(layer[i]==tar[yoyo])\n {\n yoyo++;\n if(yoyo==4)\n {\n yoyo=0;ans++;\n }\n }\n else {yoyo=0;if(layer[i]==tar[yoyo])yoyo++;}\n }\n if(yoyo!=0)\n {\n for(int i=0;i<(int)layer.size();i++)\n {\n if(layer[i]==tar[yoyo])\n {\n yoyo++;\n if(yoyo==4) {yoyo=0;ans++; break; }\n }\n else {yoyo=0;break;}\n }\n }\n w++;\n }\n cout<<ans<<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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 291239522,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\nimport java.util.StringTokenizer;\n \npublic class D_I_Love_1543{\npublic static int mod = (int)(1e9 + 7);\n public static void main(String[] args) {\n FastScanner sc = new FastScanner(System.in);\n PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n int t = sc.nextInt();\n while (t-- > 0)\n solve(sc, out);\n out.close();\n }\n \n public static void solve(FastScanner sc, PrintWriter out) {\n int n = sc.nextInt();\n int m = sc.nextInt();\n char[][] carpet = new char[n][m];\n \n for (int i = 0; i < n; i++) {\n carpet[i] = sc.next().toCharArray();\n }\n \n int ans = 0;\n String target = \"1543\";\n int r = 0, c = 1;\n while (n >= c && m >= r) {\n StringBuilder layer = new StringBuilder();\n \n int count = 0;\n String str = \"\";\n for (int j = r; j < m; j++) {\n layer.append(carpet[c-1][j]);\n if(count < 3){\n str += \"\" + carpet[c-1][j];\n count++;\n }\n }\n for (int i = c; i < n; i++) {\n layer.append(carpet[i][m - 1]);\n if(count < 3){\n str += \"\" + carpet[i][m - 1];\n count++;\n }\n }\n \n for (int j = m - 2; j >= r; j--) {\n layer.append(carpet[n - 1][j]);\n }\n for (int i = n - 2; i >= c; i--) {\n layer.append(carpet[i][r]);\n }\n layer.append(str);\n ans += countSub(layer.toString(), target);\n n -= 1; r++;\n m -= 1; c++;\n }\n out.println(ans);\n }\n private static int countSub(String str, String target) {\n int count = 0;\n for (int i = 0; i <= str.length() - target.length(); i++) {\n if (str.startsWith(target, i)) {\n count++;\n }\n }\n return count;\n }\n public static class Pair implements Comparable<Pair>{\n int first;\n int second;\n \n public Pair(int first, int second) {\n this.first = first;\n this.second = second;\n }\n \n @Override\n public boolean equals(Object obj) {\n if (this == obj) return true;\n if (obj == null || getClass() != obj.getClass()) return false;\n \n Pair pair = (Pair) obj;\n \n if (first != pair.first) return false;\n return second == pair.second;\n }\n \n @Override\n public int hashCode() {\n int result = first;\n result = 31 * result + second;\n return result;\n }\n @Override\npublic int compareTo(Pair other) {\n if (this.first != other.first) {\n return Integer.compare(this.first, other.first); // Compare by 'first'\n } else {\n return Integer.compare(this.second, other.second); // If 'first' values are equal, compare by 'second'\n}\n}\n}\nprivate static long gcd(long a, long b){\n while(a != 0 && b != 0){\n if(a > b) a = a % b;\n else b = b % a;\n }\n if(a == 0) return b;\n return a;\n}\n \n \n static class FastScanner {\n BufferedReader br;\n StringTokenizer st;\n \n public FastScanner(InputStream in) {\n br = new BufferedReader(new InputStreamReader(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 int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine() {\n String str = \"\";\n try {\n str = br.readLine();\n } catch (IOException e) {\n e.printStackTrace();\n }\n return str;\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289781436,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: I Love 1543\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.\n\nThe first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.\n\nPolycarp became curious about how many times the number $1543$ would appear in all layers$^{\\text{β}}$ of the carpet when traversed clockwise.\n\n$^{\\text{β}}$The first layer of a carpet of size $n \\times m$ is defined as a closed strip of length $2 \\cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.\n\nInput: The first line of the input contains a single integer $t$ ($1 \\leq t \\leq 100$) β the number of test cases. The following lines describe the test cases.\n\nThe first line of each test case contains a pair of numbers $n$ and $m$ ($2 \\leq n, m \\leq 10^3$, $n, m$ β even integers).\n\nThis is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ β the description of the carpet.\n\nIt is guaranteed that the sum of $n \\cdot m$ across all test cases does not exceed $10^6$.\n\nOutput: For each test case, output a single number β the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.\n\nExamples:\ninput:\n8\n2 4\n1543\n7777\n2 4\n7154\n8903\n2 4\n3451\n8888\n2 2\n54\n13\n2 2\n51\n43\n2 6\n432015\n512034\n4 4\n5431\n1435\n5518\n7634\n6 4\n5432\n1152\n4542\n2432\n2302\n5942\noutput:\n1\n1\n0\n1\n0\n2\n2\n2\n\nNote:  Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.\n\nBuggy Code:\nimport java.util.*;\npublic class Epic{\n\tpublic static void main(String [] args){\n\t\tScanner sc = new Scanner(System.in);\n\t\tint t = sc.nextInt();\n\t\twhile(t-->0){\n\t\t\tint n = sc.nextInt();\n\t\t\tint m = sc.nextInt();\n\t\t\tchar [][] arr = new char[n][m];\n\t\t\tfor(int i=0;i<n;i++){\n\t\t\t\tarr[i] = sc.next().toCharArray();\n\t\t\t}\n\t\t\tint times = n/2;\n\t\t\tint count = 0;\n\t\t\tint x = 0;\n\t\t\twhile(times-- > 0){\n\t\t\t\tStringBuilder st = new StringBuilder();\n\t\t\t\tfor(int j=x;j<m-x;j++){\n\t\t\t\t\tst.append(arr[x][j]);\n\t\t\t\t}\n\t\t\t\tfor(int row=x+1;row<n-x;row++){\n\t\t\t\t\tst.append(arr[row][m-1-x]);\n\t\t\t\t}\n\t\t\t\tfor(int col=m-2-x;col>=x;col--){\n\t\t\t\t\tst.append(arr[n-1-x][col]);\n\t\t\t\t}\n\t\t\t\tfor(int row = n-2-x;row >x; row--){\n\t\t\t\t\tst.append(arr[row][x]);\n\t\t\t\t}\n\t\t\t\tint size = st.length();\n\t\t\t\tst.append(st.toString());\n\t\t\t\tfor(int i=0;i<size;i++){\n\t\t\t\t\tif(st.charAt(i) == '1' && st.charAt(i+1) == '5' && st.charAt(i+2) == '4' && st.charAt(i+3) == '3'){\n\t\t\t\t\t\tcount++;\n\t\t\t\t\t\ti+=3;\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tx++;\n\t\t\t}\n\t\t\tSystem.out.println(count);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2036D",
"submission_id": 289945599,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nconst int p=998244353;\nint t,n,m,f[3011],dp[3011],a[3011],rt;\nint fa[6011],id[6011],sz,lc[6011],rc[6011],siz[6011],dep[6011],cl[3011][21],cr[3011][21],lca[3011][3011],pw[3011];\nvoid build(int L,int R,int &x,int d)\n{\n\tif(L==R)\n\t{\n\t\tx=L;//printf(\"build([%d,%d],%d) dep:%d\\n\",L,R,x,d);\n\t\tdep[x]=d;siz[x]=0;\n\t\treturn;\n\t}\n\tif(!x)x=++sz;\n\tdep[x]=d;//printf(\"build([%d,%d],%d) dep:%d\\n\",L,R,x,d);\n\tid[x]=L+R>>1;\n\tfor(int i=L;i<=(L+R>>1);++i)for(int j=(L+R>>1)+1;j<=R;++j)lca[i][j]=x;\n\tbuild(L,L+R>>1,lc[x]=0,d+1);build((L+R>>1)+1,R,rc[x]=0,d+1);\n\tfa[lc[x]]=fa[rc[x]]=x;siz[x]=siz[lc[x]]+siz[rc[x]]+1;\n}\nvoid proccl(int x)\n{\n\t// printf(\">>>>>>>>proccl(%d)\\n\",x);\n\tint ans=1,u=x;\n\twhile(u!=rt)\n\t{\n\t\tcl[x][dep[u]-1]=ans;\n\t\t// printf(\"u:%d\\n\",u);\n\t\tif(u==lc[fa[u]])\n\t\t{\n\t\t\tif(id[fa[u]]!=x)ans=1ll*ans*(n-a[x]+1)%p;\n\t\t\tans=1ll*ans*pw[siz[rc[fa[u]]]]%p;\n\t\t\t// printf(\"node (%d) subtree pw^%d\\n\",n-a[x]+1,siz[rc[fa[u]]]);\n\t\t}\n\t\tu=fa[u];\n\t}\n\tcl[x][0]=ans;\n}\nvoid proccr(int x)\n{\n\tint ans=1,u=x;\n\twhile(u!=rt)\n\t{\n\t\tcr[x][dep[u]-1]=ans;\n\t\tif(u==rc[fa[u]])\n\t\t{\n\t\t\tif(id[fa[u]]!=x)ans=1ll*ans*(a[x]-1)%p;\n\t\t\tans=1ll*ans*pw[siz[lc[fa[u]]]]%p;\n\t\t}\n\t\tu=fa[u];\n\t}\n\tcr[x][0]=ans;\n}\nint main()\n{\n\tscanf(\"%d\",&t);while(t--)\n\t{\n\t\tscanf(\"%d%d\",&n,&m);\n\t\tpw[0]=1;for(int i=1;i<=n;++i)pw[i]=1ll*pw[i-1]*n%p;\n\t\tfor(int i=1;i<=n;++i)a[i]=-1;\n\t\tfor(int i=1;i<=m;++i)\n\t\t{\n\t\t\tint x,k;scanf(\"%d%d\",&x,&k);\n\t\t\tif(k==1&&x!=1)k=-1;\n\t\t\ta[x]=k;\n\t\t}\n\t\tint cnt=0;\n\t\tfor(int i=1;i<=n;++i)\n\t\t{\n\t\t\tcnt+=(a[i]!=-1);\n\t\t}\n\t\tif(cnt==0)\n\t\t{\n\t\t\tprintf(\"%d\\n\",pw[n]);\n\t\t\tcontinue;\n\t\t}\n\t\tfor(int i=1;i<=n;++i)\n\t\t{\n\t\t\tif(a[i]==-1)dp[i]=-1e9;\n\t\t\telse\n\t\t\t{\n\t\t\t\tdp[i]=1;\n\t\t\t\tfor(int j=1;j<i;++j)if(a[j]<a[i])dp[i]=max(dp[i],dp[j]+1);\n\t\t\t}\n\t\t}\n\t\tsz=n;\n\t\trt=0;\n\t\tbuild(1,n,rt,1);\n\t\tfor(int i=1;i<=n;++i)if(~a[i])proccl(i),proccr(i);\n\t\tfor(int i=1;i<=n;++i)f[i]=0;\n\t\tfor(int i=1;i<=n;++i)if(~a[i])\n\t\t{\n\t\t\tif(dp[i]==1)f[i]=cr[i][0];\n\t\t\telse\n\t\t\t{\n\t\t\t\tfor(int j=1;j<i;++j)if(dp[j]+1==dp[i]&&a[j]<a[i])\n\t\t\t\t{\n\t\t\t\t\tint L=lca[j][i],coe=1;\n\t\t\t\t\tif(id[L]!=j)\n\t\t\t\t\t{\n\t\t\t\t\t\tcoe=a[i]-a[j];\n\t\t\t\t\t}\n\t\t\t\t\t// printf(\"j:%d i:%d L:%d cl:%d cr:%d coe:%d\\n\",j,i,L,cl[j][dep[L]],cr[i][dep[L]],coe);\n\t\t\t\t\tf[i]=(f[i]+1ll*f[j]*cl[j][dep[L]]%p*cr[i][dep[L]]%p*coe)%p;\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t\t// printf(\"a:\");for(int i=1;i<=n;++i)printf(\"%d \",a[i]);putchar(10);\n\t\t// printf(\"dp:\");for(int i=1;i<=n;++i)printf(\"%d \",dp[i]);putchar(10);\n\t\t// printf(\"f:\");for(int i=1;i<=n;++i)printf(\"%d \",f[i]);putchar(10);\n\t\tint ans=0,mx=0;\n\t\tfor(int i=1;i<=n;++i)mx=max(mx,dp[i]);\n\t\tfor(int i=1;i<=n;++i)if(dp[i]==mx)\n\t\t{\n\t\t\tans=(ans+1ll*f[i]*cl[i][0]%p*(i!=n?n:1))%p;\n\t\t}\n\t\tif(dp[n]!=mx)ans=1ll*ans*n%p;\n\t\tprintf(\"%d %d\\n\",m-mx,(ans%p+p)%p);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288426253,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define Loop(x,l,r) for (ll x = (l); x < (ll)(r); ++x)\n#define LoopR(x,l,r) for (ll x = (r)-1; x >= (ll)(l); --x)\ntypedef long long ll;\ntypedef std::pair<int, int> pii;\ntypedef std::pair<ll , ll > pll;\nusing namespace std;\n \nconst int N = 3*1024;\nconst int mod = 998244353;\nint inv[N];\n \nll pw(ll x, ll y)\n{\n\tll a = 1;\n\twhile (y) {\n\t\tif (y&1)\n\t\t\ta = a*x % mod;\n\t\tx = x*x % mod;\n\t\ty /= 2;\n\t}\n\treturn a;\n}\n \nint par[N], height[N];\n \nvoid make_tree(int l, int r, int p, int h)\n{\n\tif (l == r)\n\t\treturn;\n\tint m = (l + r)/2;\n\tpar[m] = p;\n\theight[m] = h;\n\tmake_tree(l, m, m, h+1);\n\tmake_tree(m+1, r, m, h+1);\n}\n \nint get(int l, int r, int i)\n{\n\tif (l == r)\n\t\treturn -1;\n\tint m = (l + r)/2;\n\tint x;\n\tif (i <= m) {\n\t\tx = get(l, m, i);\n\t} else {\n\t\tx = get(m+1, r, i);\n\t}\n\tif (x == -1)\n\t\treturn m;\n\treturn x;\n}\n \nint n;\n \nll cnt1(int p, int i, int k)\n{\n\tll ans = inv[n];\n\tfor (int v = i; v != -1; v = par[v]) {\n\t\tans %= mod;\n\t\tif (v < p) {\n\t\t\tans *= inv[n];\n\t\t\tans %= mod;\n\t\t\tans *= k;\n\t\t} else if (v > p) {\n\t\t\tans *= inv[n];\n\t\t\tans %= mod;\n\t\t\tans *= n-k;\n\t\t}\n\t\tans %= mod;\n\t}\n\treturn ans;\n}\n \nll cnt1r(int x, int k)\n{\n\tll ans = cnt1(x, get(0, n-1, x), k);\n\t//cerr << \"cnt1 \" << x << ' ' << k << \" = \" << ans << '\\n';\n\treturn ans;\n}\n \nbool anc(int v, int a)\n{\n\twhile (v != -1) {\n\t\tif (v == a)\n\t\t\treturn 1;\n\t\tv = par[v];\n\t}\n\treturn 0;\n}\n \nll cnt2(int p1, int i1, int k1, int p2, int i2, int k2)\n{\n\tll ans = anc(i1, p2)? inv[n-k1]: inv[n];\n\twhile (height[i1] > height[i2])\n\t\ti1 = par[i1];\n\t//while (height[i1] < height[i2])\n\t//\ti1 = par[i1];\n\tfor (int v = i2; v != -1; v = par[v]) {\n\t\twhile (height[i1] > height[v])\n\t\t\ti1 = par[i1];\n\t\tif (i1 == v) {\n\t\t\tif (v > p2) {\n\t\t\t\tans *= inv[n];\n\t\t\t\tans %= mod;\n\t\t\t\tans *= n-k2;\n\t\t\t} else if (p1 < v && v < p2) {\n\t\t\t\tans *= inv[n-k1];\n\t\t\t\tans %= mod;\n\t\t\t\t//ans *= k1..k2;\n\t\t\t\tans *= (k2-k1);\n\t\t\t}\n\t\t} else {\n\t\t\tif (v < p2) {\n\t\t\t\tans *= inv[n];\n\t\t\t\tans %= mod;\n\t\t\t\tans *= k2;\n\t\t\t} else if (v > p2) {\n\t\t\t\tans *= inv[n];\n\t\t\t\tans %= mod;\n\t\t\t\tans *= n-k2;\n\t\t\t}\n\t\t}\n\t\tans %= mod;\n\t}\n\treturn ans;\n}\n \nll cnt2r(int x1, int k1, int x2, int k2)\n{\n\tll ans = cnt2(x1, get(0, n-1, x1), k1, x2, get(0, n-1, x2), k2);\n\t//cerr << \"cnt2 \" << x1 << ' ' << k1 << ' ' << x2 << ' ' << k2 << \" = \" << ans << '\\n';\n\treturn ans;\n}\n \nvoid solve()\n{\n\tint m;\n\tcin >> n >> m;\n\tmake_tree(0, n-1, -1, 0);\n\tvector<pii> vec;\n\tint bad = 0;\n\tLoop (i,0,m) {\n\t\tint x, k;\n\t\tcin >> x >> k;\n\t\tx--; k--;\n\t\tif (x != 0 && k == 0) {\n\t\t\tbad++;\n\t\t\tcontinue;\n\t\t}\n\t\tvec.push_back({x, k});\n\t}\n\tif (n == 1) {\n\t\tcout << \"0 1\\n\";\n\t\treturn;\n\t}\n\tm = vec.size();\n\tsort(vec.begin(), vec.end());\n\tvector<ll> sz(m), cnt(m);\n\tint mx = 0;\n\tLoop (i,0,m) {\n\t\tsz[i] = 1;\n\t\tLoop (j,0,i) {\n\t\t\tif (vec[j].second < vec[i].second)\n\t\t\t\tsz[i] = max(sz[i], 1 + sz[j]);\n\t\t}\n\t\tmx = max<int>(mx, sz[i]);\n\t}\n\tLoop (i,0,m) {\n\t\tif (sz[i] == 1) {\n\t\t\tcnt[i] = cnt1r(vec[i].first, vec[i].second);\n\t\t\tcontinue;\n\t\t}\n\t\tcnt[i] = 0;\n\t\tLoop (j,0,i) {\n\t\t\tif (vec[j].second >= vec[i].second)\n\t\t\t\tcontinue;\n\t\t\tif (sz[j] + 1 != sz[i])\n\t\t\t\tcontinue;\n\t\t\tll x = cnt2r(vec[j].first, vec[j].second, vec[i].first, vec[i].second);\n\t\t\tcnt[i] += x * cnt[j];\n\t\t\tcnt[i] %= mod;\n\t\t}\n\t}\n\tll ans = 0;\n\tLoop (i,0,m) {\n\t\tif (sz[i] == mx) {\n\t\t\tans += cnt[i];\n\t\t\tans %= mod;\n\t\t}\n\t}\n\tans *= pw(n, n);\n\tans %= mod;\n\tcout << (m-mx) + bad << ' ' << ans << '\\n';\n}\n \nsigned main()\n{\n\tLoop (i,0,N)\n\t\tinv[i] = pw(i, mod-2);\n\tcin.tie(0) -> sync_with_stdio(false);\n\tint t;\n\tcin >> t;\n\twhile (t--)\n\t\tsolve();\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288372357,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define ll long long\n#define lll __int128\n#define uint unsigned int\n#define ull unsigned long long\n#define db double\n#define pii pair<int,int>\n#define pli pair<ll,int>\n#define vi vector<int>\n#define eb emplace_back\n#define po pop_back\n#define par __builtin_parity\n#define pct __builtin_popcount\n#define ctz __builtin_ctz\n#define clz __builtin_clz\n#define gcd __gcd\n#define fi first\n#define se second\n#define mid ((l+r)/2)\nconst int N=1e6+5,INF=1e9,MOD=998244353;\nint T,n,m,rt,invn,f[N],g[N],fa[N],d[N],bt[N],ch[N][2];\npii a[N];\nint qpow(int x,int y)\n{\n\tint res=1;\n\tfor(;y;y/=2,x=1ll*x*x%MOD) if(y&1)\n\t\tres=1ll*res*x%MOD;\n\treturn res;\n}\nvoid upd(int x,int w)\n{\n\tfor(;x<=n;x+=x&-x)\n\t\tbt[x]=max(bt[x],w);\n}\nint qry(int x)\n{\n\tint res=0;\n\tfor(;x;x-=x&-x)\n\t\tres=max(res,bt[x]);\n\treturn res;\n}\nint build(int l,int r)\n{\n\tif(l==r) return l;\n\tint p=n+mid;\n\tch[p][0]=build(l,mid);\n\tch[p][1]=build(mid+1,r);\n\tfa[ch[p][0]]=fa[ch[p][1]]=p;\n\treturn p;\n}\nvoid dfs(int u,int f)\n{\n\tif(!u) return;d[u]=d[f]+1;\n\tdfs(ch[u][0],u);dfs(ch[u][1],u);\n}\nint lca(int u,int v)\n{\n\tif(d[u]<d[v]) swap(u,v);\n\twhile(d[u]>d[v]) u=fa[u];\n\twhile(u!=v) u=fa[u],v=fa[v];\n\treturn u;\n}\nint get(int u) {return ch[fa[u]][1]==u;}\nint calc(int u,int v,int l,int r)\n{\n\tint res=1;\n\tif(!u)\n\t{\n\t\tfor(int i=v;i!=rt;i=fa[i])\n\t\t\tif(get(i) && fa[i]!=n+v)\n\t\t\t\tres=1ll*res*(r-1)%MOD*invn%MOD;\n\t\treturn res;\n\t}\n\tif(!v)\n\t{\n\t\tfor(int i=u;i!=rt;i=fa[i])\n\t\t\tif(!get(i) && fa[i]!=n+u)\n\t\t\t\tres=1ll*res*(n-l+1)%MOD*invn%MOD;\n\t\treturn res;\n\t}\n\tint t=lca(u,v);\n\tif(t!=n+u && t!=n+v) res=1ll*res*(r-l)%MOD*invn%MOD;\n\tfor(int i=v;fa[i]!=t;i=fa[i])\n\t\tif(get(i) && fa[i]!=n+u && fa[i]!=n+v)\n\t\t\tres=1ll*res*(r-1)%MOD*invn%MOD;\n\tfor(int i=u;fa[i]!=t;i=fa[i])\n\t\tif(!get(i) && fa[i]!=n+u && fa[i]!=n+v)\n\t\t\tres=1ll*res*(n-l+1)%MOD*invn%MOD;\n\treturn res;\n}\nvoid slv()\n{\n\tscanf(\"%d %d\",&n,&m);\n\tif(n==1) {puts(\"1\");return;}\n\tinvn=qpow(n,MOD-2);\n\tfor(int i=1;i<=n*2;++i) fa[i]=ch[i][0]=ch[i][1];\n\trt=build(1,n);\n\tdfs(rt,0);\n\tfor(int i=1;i<=m;++i)\n\t\tscanf(\"%d %d\",&a[i].fi,&a[i].se);\n\tsort(a+1,a+m+1);\n\ta[m+1].se=n;\n\tfill(bt+1,bt+n+1,0);\n\tfor(int i=1;i<=m;++i)\n\t{\n\t\tf[i]=qry(a[i].se)+1;\n\t\tif(a[i].fi>1 && a[i].se==1) f[i]=0;\n\t\tif(a[i].fi<n && a[i].se==n) f[i]=0;\n\t\tif(f[i]>0) upd(a[i].se,f[i]);\n\t}\n\tf[m+1]=qry(n)+1;\n\tg[0]=1;\n\tfor(int i=1;i<=m+1;++i)\n\t{\n\t\tg[i]=0;\n\t\tfor(int j=0;j<i;++j) if(a[i].se>=a[j].se && f[i]==f[j]+1)\n\t\t\tg[i]=(g[i]+1ll*g[j]*calc(a[j].fi,a[i].fi,a[j].se,a[i].se))%MOD;\n\t\tif(i<=m) g[i]=1ll*g[i]*invn%MOD;\n\t}\n\tprintf(\"%d %lld\\n\",m+1-f[m+1],1ll*g[m+1]*qpow(n,n)%MOD);\n}\nint main()\n{\n\tscanf(\"%d\",&T);\n\twhile(T--) slv();\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288371155,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\ndef binary_search(n, k, arr):\n l = 1\n h = n\n while l < h:\n m = (l + h) // 2\n if arr[m-1] < k:\n l = m + 1\n else:\n h = m\n return l\n \ndef check_val(n, tests, removed_tests, arr):\n for i, (x, k) in enumerate(tests):\n if i in removed_tests:\n continue\n \n return False\n \n if binary_search(n, k, arr) != x:\n return False\n return True\n \ndef solvec(n, m, tests):\n MOD = 998244353\n \n \n for r in range(m + 1):\n valid_arrays = set()\n \n \n from itertools import combinations\n for removed in combinations(range(m), r):\n \n def gen_arr(pos=0, current_arr=None):\n if current_arr is None:\n current_arr = [0] * n\n \n if pos == n:\n \n arr_tuple = tuple(current_arr)\n if check_val(n, tests, removed, current_arr):\n valid_arrays.add(arr_tuple)\n return\n \n \n fixed = False\n fixed_value = None\n for i, (x, k) in enumerate(tests):\n if i not in removed and x-1 == pos:\n fixed = True\n fixed_value = k\n break\n \n if fixed:\n current_arr[pos] = fixed_value\n gen_arr(pos+1, current_arr)\n else:\n \n for val in range(1, n+1):\n current_arr[pos] = val\n gen_arr(pos+1, current_arr)\n \n gen_arr()\n \n if valid_arrays:\n return r, len(valid_arrays) % MOD\n \n return m, 0 \ndef main():\n t = int(input())\n for _ in range(t):\n n, m = map(int, input().split())\n tests = []\n for _ in range(m):\n x, k = map(int, input().split())\n tests.append((x, k))\n \n r, count = solvec(n, m, tests)\n print(f\"{r} {count}\")\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288319596,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\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 = 3005;\nint mn[maxn];\nint mx[maxn];\nvoid build(int l, int r) {\n if (l == r) return;\n int mid = (l + r) >> 1;\n chkmin(mn[mid], l);\n chkmax(mx[mid], r);\n build(mid + 1, r);\n build(l, mid);\n}\nint val[maxn];\nint mxlen[maxn];\nint dp[maxn];\nint cnt[2][maxn];\nint main() {\n int t;\n cin >> t;\n while (t--) {\n int m, n;\n cin >> n >> m;\n memset(val, -1, sizeof(val));\n for (int i = 1; i <= n; i++)\n mx[i] = i, mn[i] = i;\n build(1, n);\n for (int i = 1; i <= m; i++) {\n int pos, v;\n cin >> pos >> v;\n val[pos] = v;\n }\n memset(dp, 0, sizeof(dp));\n memset(mxlen, 0, sizeof(mxlen));\n val[0] = 0;\n val[n + 1] = n + 1;\n mxlen[n + 1] = 0;\n dp[n + 1] = 1;\n for (int i = n; i >= 0; i--) {\n if (val[i] == -1) {\n mxlen[i] = 0;\n dp[i] = 0;\n continue;\n }\n for (int j = i + 1; j <= n + 1; j++) \n if (val[j] != -1 && val[j] > val[i]) \n chkmax(mxlen[i], mxlen[j] + 1);\n memset(cnt, 0, sizeof(cnt));\n int h[2][2]; memset(h, 0, sizeof(h));\n for (int j = i + 1; j <= n + 1; j++) {\n for (int s = 0; s < 2; s++) {\n h[s][1] -= cnt[s][j - 1];\n h[s][0] += cnt[s][j - 1];\n }\n if (val[j] != -1 && val[j] > val[i] && mxlen[i] == mxlen[j] + 1) {\n ll mult = 1;\n for (int u = 0; u < 2; u++)\n for (int v = 0; v < 2; v++) {\n int L = 1, R = n;\n if (u == 1) L = max(L, val[i]);\n if (v == 1) R = min(R, val[j] - 1);\n mult = mult * ksm(R - L + 1, h[u][v]) % mod;\n }\n dp[i] = (dp[i] + dp[j] * mult % mod) % mod;\n }\n int tu = 0; \n if (mn[j] <= i) tu = 1;\n cnt[tu][mx[j]] += 1;\n h[tu][1] += 1;\n }\n }\n cout << m + 1 - mxlen[0] << ' ' << dp[0] << 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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288350787,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nconst int N = 3e5;\nconst int mod = 998244353;\nint n, m;\nstruct dpnode{\n int nr, nr2;\n}dp[N];\nvoid add(dpnode &a, dpnode b){\n if(b.nr2 == 0)return;\n if(a.nr < b.nr){\n a = b;\n }else if(a.nr == b.nr){\n a.nr2 = (a.nr2 + b.nr2)%mod;\n }\n}\nint fastexp(int a, int b){\n assert(b >= 0);\n int r = 1;\n while(b){\n if(b&1)r = 1ll*r*a%mod;\n a = 1ll*a*a%mod;\n b/=2;\n }\n return r;\n}\nvector <int> touch(int x){\n vector <int> p;\n int l = 0, r = n - 1;\n while(l != r){\n int mij = (l + r)/2;\n p.push_back(mij);\n if(x <= mij){\n r = mij;\n }else{\n l = mij + 1;\n }\n }\n return p;\n}\nstruct ev{\n int pos, t;\n}v[N];\ndpnode dp2[N], dp3[N];\nint v2[N];\nvoid solve(){\n cin>>n>>m;\n for(int i = 0;i < n;i++){\n dp[i] = {-mod, 0};\n dp2[i] = {-mod, 0};\n dp3[i] = {-mod, 0};\n v2[i] = 0;\n }\n for(int i = 0;i < m;i++){\n int x, p;\n cin>>p>>x;\n p--;\n v[i] = {p, x};\n v2[p] = x;\n }\n sort(v, v + m,[&](ev a, ev b){\n return a.t < b.t;\n });\n for(int i = 0;i < m;i++){\n auto x = touch(v[i].pos);\n int nr = fastexp(n, v[i].pos);\n for(auto j:x){\n if(j < v[i].pos){\n nr = 1ll*nr*fastexp(n, mod - 2)%mod*(v[i].t - 1)%mod;\n }\n }\n add(dp[v[i].pos], {1, nr});\n /// find the before\n nr = 0;\n int nr2 = 0;\n for(auto j:x){\n if(j == v[i].pos){\n break;\n }\n if(j < v[i].pos){\n ///targeting this point\n assert(dp2[j].nr == dp3[j].nr);\n nr2 = 0;\n for(auto k:x){\n if(j < k && k < v[i].pos)nr2++;\n }\n add(dp[v[i].pos], {dp2[j].nr + 1, 1ll*dp2[j].nr2*fastexp(v[i].t - 1, nr2)%mod*v[i].t%mod*fastexp(n, v[i].pos - j - 1 - nr2)%mod});\n add(dp[v[i].pos], {dp3[j].nr + 1, mod - 1ll*dp3[j].nr2*fastexp(v[i].t - 1, nr2)%mod*fastexp(n, v[i].pos - j - 1 - nr2)%mod});\n add(dp[v[i].pos], {dp[j].nr + 1, 1ll*dp[j].nr2*fastexp(v[i].t - 1, nr2)%mod*fastexp(n, v[i].pos - j - 1 - nr2)%mod});\n }\n ///exactly this point!\n }\n /// what occurs when lca is i or j?\n /// bruteforce dp\n /// find the after\n for(auto j:x){\n if(j == v[i].pos){\n break;\n }\n if(v[i].pos < j){\n ///v[i].pos -> j\n nr2 = 0;\n for(auto k:x){\n if(v[i].pos < k && k < j)nr2++;\n }\n add(dp2[j], {dp[v[i].pos].nr, 1ll*dp[v[i].pos].nr2*fastexp(n - v[i].t + 1, nr2)%mod*fastexp(n, j - v[i].pos - nr2 - 1)%mod});\n add(dp3[j], {dp[v[i].pos].nr, 1ll*dp[v[i].pos].nr2*fastexp(n - v[i].t + 1, nr2)%mod*v[i].t%mod*fastexp(n, j - v[i].pos - nr2 - 1)%mod});\n if(nr2 != 0 && v2[j] > v[i].t){\n add(dp[j], {dp[v[i].pos].nr + 1, 1ll*dp[v[i].pos].nr2*fastexp(n - v[i].t + 1, nr2 - 1)%mod*fastexp(n, j - v[i].pos - nr2 - 1)%mod*(v2[j] - v[i].t)%mod});\n }else if(v2[j] > v[i].t){\n add(dp[j], {dp[v[i].pos].nr + 1, 1ll*dp[v[i].pos].nr2*fastexp(n, j - v[i].pos - nr2 - 1)%mod});\n }\n }\n }\n }\n dpnode ans = {0, fastexp(n, n)};\n for(int i = 0;i < n;i++){\n if(!v2[i])continue;\n auto x = touch(i);\n int cnt = 0;\n for(auto j:x){\n if(i < j)cnt++;\n }\n add(ans, {dp[i].nr, 1ll*dp[i].nr2*fastexp(n - v2[i] + 1, cnt)%mod*fastexp(n, n - i - 1 - cnt)%mod});\n }\n cout<<m - ans.nr<<' '<<ans.nr2<<'\\n';\n}\nint main(){\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n int t;\n cin>>t;\n while(t--)solve();\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 290745893,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \ntypedef long long ll;\nconst ll MOD = 998244353;\n \nint t;\nint n, k;\nint arr[3002];\nint DPans[3002];\nbool valid[3002];\nll DPcnt[3002];\nint S[3002], E[3002];\n \nvoid binary(int l, int r){\n if(l==r) return;\n int m = (l+r)/2;\n S[m] = l, E[m] = r;\n binary(l, m);\n binary(m+1, r);\n}\n \nint cnt1[3002][3002], cnt2[3002][3002], cnt3[3002][3002], cnt4[3002][3002];\nll powtable[3002][3002];\n \ninline ll calc(int l, int r, ll vl, ll vr){\n return powtable[vr-vl+1][cnt1[l][r]] * powtable[n-vl+1][cnt2[l][r]] % MOD\n * powtable[vr][cnt3[l][r]] % MOD * powtable[n][cnt4[l][r]] % MOD;\n}\n \nint main(){\n for(int i=0; i<=3001; i++){\n powtable[i][0] = 1;\n for(int j=1; j<=3001; j++) powtable[i][j] = powtable[i][j-1]*i%MOD;\n }\n \n scanf(\"%d\", &t);\n while(t--){\n scanf(\"%d %d\", &n, &k);\n for(int i=1; i<=n; i++) arr[i] = -1, valid[i] = 0;\n for(int i=1; i<=k; i++){\n int x, v;\n scanf(\"%d %d\", &x, &v);\n arr[x] = v;\n if(!(x>1&&v==1)) valid[x] = 1;\n }\n \n binary(1, n);\n S[n] = E[n] = n;\n \n fill(DPans+1, DPans+n+1, 0);\n fill(DPcnt+1, DPcnt+n+1, 0);\n \n for(int i=1; i<=n; i++){\n if(!valid[i]) continue;\n DPans[i] = 1;\n for(int j=1; j<i; j++) if(valid[j] && arr[j] < arr[i]) DPans[i] = max(DPans[i], DPans[j] + 1);\n }\n \n int longest = *max_element(DPans+1, DPans+n+1);\n \n for(int i=0; i<=n+1; i++) for(int j=0; j<=n+1; j++) cnt1[i][j] = cnt2[i][j] = cnt3[i][j] = cnt4[i][j] = 0;\n \n /// board μ±μ°κΈ°\n /// 1. B1 (s<=l, r<=e)\n for(int i=1; i<=n; i++) cnt1[S[i]][E[i]]++, cnt1[i][E[i]]--, cnt1[S[i]][i]--, cnt1[i][i]++;\n for(int i=1; i<=n+1; i++) for(int j=0; j<=n+1; j++) cnt1[i][j] += cnt1[i-1][j];\n for(int j=n; j>=0; j--) for(int i=0; i<=n+1; i++) cnt1[i][j] += cnt1[i][j+1];\n \n /// 2. B2 (s<=l, e<r)\n for(int i=1; i<=n; i++) cnt2[S[i]][E[i]+1]++, cnt2[i][E[i]+1]--;\n for(int i=1; i<=n+1; i++) for(int j=0; j<=n+1; j++) cnt2[i][j] += cnt2[i-1][j];\n for(int j=1; j<=n+1; j++) for(int i=0; i<=n+1; i++) cnt2[i][j] += cnt2[i][j-1];\n \n /// 3. B3 (l<s, r<=e)\n for(int i=1; i<=n; i++) cnt3[S[i]-1][E[i]]++, cnt3[S[i]-1][i]--;\n for(int i=n; i>=0; i--) for(int j=0; j<=n+1; j++) cnt3[i][j] += cnt3[i+1][j];\n for(int j=n; j>=0; j--) for(int i=0; i<=n+1; i++) cnt3[i][j] += cnt3[i][j+1];\n \n /// 4. B4\n for(int i=1; i<=n; i++) cnt4[S[i]-1][E[i]+1]++;\n for(int i=n; i>=0; i--) for(int j=0; j<=n+1; j++) cnt4[i][j] += cnt4[i+1][j];\n for(int j=1; j<=n+1; j++) for(int i=0; i<=n+1; i++) cnt4[i][j] += cnt4[i][j-1];\n \n// printf(\"DPans: \"); for(int i=1; i<=n; i++) printf(\"%d \", DPans[i]); puts(\"\");\n// printf(\"cnt1: \\n\");\n// for(int i=0; i<=n+1; i++){\n// for(int j=0; j<=n+1; j++) printf(\"%d \", cnt1[i][j]);\n// puts(\"\");\n// }\n \n ll ans = 0;\n for(int i=1; i<=n; i++){\n if(arr[i] == -1) continue;\n if(DPans[i] == 1){\n DPcnt[i] = calc(0, i, 1, arr[i]-1);\n continue;\n }\n for(int j=1; j<i; j++){\n if(valid[j] && arr[j] < arr[i] && DPans[j] + 1 == DPans[i]){\n DPcnt[i] = (DPcnt[i] + DPcnt[j] * calc(j, i, arr[j], arr[i]-1)) % MOD;\n }\n }\n if(DPans[i] == longest){\n ans = (ans + DPcnt[i] * calc(i, n+1, arr[i], n)) % MOD;\n }\n }\n printf(\"%d %lld\\n\", k-longest, ans);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288377777,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nusing ll=long long;\nconst int N=3005,mod=998244353;\nint n,m,rt,tot,lca[N][N],fa[N*2],ls[N*2],rs[N*2],fd[N*2],to[N*2],a[N],g[N];\nll fc[N],inv[N],pw[N],f[N];\nll qp(ll a,int b){ll ans=1;for(;b;b>>=1,a=a*a%mod)if(b&1)ans=ans*a%mod;return ans;}\nll C(int n,int m){return n<m||m<0?0:fc[n]*inv[m]%mod*inv[n-m]%mod;}\nvoid init(int n) {\n fc[0]=pw[0]=1;\n for(int i=1;i<=n;i++)fc[i]=fc[i-1]*i%mod,pw[i]=pw[i-1]*2%mod;\n inv[n]=qp(fc[n],mod-2);\n for(int i=n-1;~i;i--)inv[i]=inv[i+1]*(i+1)%mod;\n}\nint bd(int l,int r) {\n if(l==r)return to[l]=l;\n int mid=l+r>>1,p=++tot;to[p]=mid;\n for(int i=l;i<=mid;i++)for(int j=mid+1;j<=r;j++)lca[i][j]=p;\n ls[p]=bd(l,mid),rs[p]=bd(mid+1,r);\n return fd[ls[p]]=0,fd[rs[p]]=1,fa[ls[p]]=fa[rs[p]]=p;\n}\nvoid solve() {\n cin>>n>>m,tot=n,rt=bd(1,n),fa[rt]=0;\n ll iv=qp(n,mod-2);\n for(int i=1;i<=n;i++)a[i]=0;\n for(int i=1,x,k;i<=m;i++)cin>>x>>k,a[x]=k;\n int as=0;\n for(int i=1;i<=n;i++)if(a[i]) {\n g[i]=(a[i]!=1||i==1?1:-n);\n for(int j=1;j<i;j++)if(a[j]&&a[j]<a[i])g[i]=max(g[i],g[j]+1);\n as=max(as,g[i]);\n }\n ll way=!as;\n for(int i=1;i<=n;i++)if(a[i]) {\n f[i]=0;\n if(g[i]==1) {\n f[i]=iv;\n for(int x=fa[i];fa[x];x=fa[x])if(fd[x])f[i]=f[i]*(a[i]-1)%mod*iv%mod;\n }\n for(int j=1;j<i;j++)if(a[j]&&a[j]<a[i]&&g[j]+1==g[i]) {\n int lc=lca[j][i];\n ll v=f[j];\n for(int x=j;fa[x]!=lc;x=fa[x])if(to[fa[x]]!=j&&!fd[x])v=v*(n-a[j]+1)%mod*iv%mod;\n for(int x=i;fa[x]!=lc;x=fa[x])if(fd[x])v=v*(a[i]-1)%mod*iv%mod;\n if(to[lc]!=j)v=v*(a[i]-a[j])%mod*iv%mod;\n f[i]=(f[i]+v*iv)%mod;\n }\n if(g[i]==as) {\n ll v=f[i];\n for(int x=i;fa[x];x=fa[x])if(to[fa[x]]!=i&&!fd[x])v=v*(n-a[i]+1)*iv%mod;\n way=(way+v)%mod;\n }\n }cout<<m-as<<' '<<way*qp(n,n)%mod<<'\\n';\n}\nint main() {\n #ifndef ONLINE_JUDGE\n freopen(\".in\",\"r\",stdin);\n freopen(\".out\",\"w\",stdout);\n #endif\n ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);\n int tt;cin>>tt;\n while(tt--)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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288466504,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \nnamespace Combo {\n const int mod = 998244353;\n vector<int> f = {1}, invf = {1};\n int mul(int x, int y) {\n return (1LL * x * y) % mod;\n }\n int exp(int x, int y) {\n int ret = 1;\n while(y) {\n if(y & 1) ret = mul(ret, x);\n x = mul(x, x), y >>= 1;\n }\n return ret;\n }\n int fact(int x) {\n if(x < 0) return 0;\n while(x >= f.size()) {\n f.push_back(mul(f.size(), f.back()));\n }\n return f[x];\n }\n int ifact(int x) {\n if(x < 0) return 0;\n while(x >= invf.size()) {\n invf.push_back(mul(exp(invf.size(), mod - 2), invf.back()));\n }\n return invf[x];\n }\n int inv(int x) {\n return mul(fact(x - 1), ifact(x));\n }\n int ncr(int x, int y) {\n if(x < y || x < 0) return 0;\n return mul(fact(x), mul(ifact(x - y), ifact(y)));\n }\n}\n \nusing namespace Combo;\n \nint main() {\n ios_base::sync_with_stdio(false); cin.tie(0);\n int t; cin >> t;\n while(t--) {\n int n, m; cin >> n >> m;\n vector<int> fix(n, -1), pn(n + 1, 1);\n for(int i = 0, x, k; i < m; i++) {\n cin >> x >> k; --x, --k; fix[x] = k;\n }\n for(int i = 1; i <= n; i++) {\n pn[i] = mul(pn[i - 1], n);\n }\n vector<int> par(n, -1), st(n + 2, -1); int inv_n = exp(n, mod - 2);\n auto gen = [&](auto rec, int l, int r, int p) -> void {\n int mid = (l + r) / 2;\n if(l == r) {\n st[l + 1] = p;\n return;\n }\n par[mid] = p;\n rec(rec, l, mid, mid);\n rec(rec, mid + 1, r, mid);\n };\n gen(gen, 0, n - 1, -1); map<int, int> mp;\n auto get_btw = [&](int x, int y) {\n int res = 1, tx = st[x + 1], ty = st[y + 1];\n while(true) {\n if(tx == -1 && ty == -1) break;\n if(x < tx && tx < y) mp[tx] |= 2;\n if(x < ty && ty < y) mp[ty] |= 1;\n if(tx == ty) break;\n if(tx != -1) tx = par[tx];\n if(ty != -1) ty = par[ty];\n }\n for(auto [_, msk]: mp) {\n if(msk == 3) res = mul(res, mul(inv_n, fix[y] - fix[x]));\n if(msk == 2) res = mul(res, mul(inv_n, n - fix[x]));\n if(msk == 1) res = mul(res, mul(inv_n, fix[y]));\n }\n mp.clear();\n return res;\n };\n vector<int> dp1(n, 0), dp2(n); int res1 = 0, res2 = 0;\n for(int i = 0; i < n; i++) {\n if(fix[i] == -1) continue;\n dp1[i] = 1;\n dp2[i] = mul(pn[i], get_btw(-1, i));\n if(dp2[i] == 0) dp1[i] = 0;\n for(int j = 0; j < i; j++) {\n if(fix[j] == -1 || fix[j] > fix[i]) continue;\n if(dp1[j] + 1 > dp1[i]) {\n int btw = get_btw(j, i);\n dp1[i] = dp1[j] + 1;\n dp2[i] = mul(dp2[j], mul(pn[i - j - 1], btw));\n } else if(dp1[j] + 1 == dp1[i]) {\n int btw = get_btw(j, i);\n dp2[i] = (dp2[i] + mul(dp2[j], mul(pn[i - j - 1], btw))) % mod;\n }\n }\n res1 = max(res1, dp1[i]);\n }\n for(int i = 0; i < n; i++) {\n if(dp1[i] != res1) continue;\n res2 = (res2 + mul(dp2[i], mul(pn[n - i - 1], get_btw(i, n)))) % mod;\n }\n if(res1 == m) res2 = 1;\n cout << m - res1 << \" \" << res2 << '\\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 290014247,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define int long long\n#define pr pair<int,int>\nconst int N=3005,mods=998244353,inf=1e9;\nint n,m,t,a[N],b[N],dy[N],pw[N],bh[N][N],idx;\nmap<pr,pr>q[2][N];\npr hb(pr a,pr b){\n\tif(a.first==b.first)return {a.first,(a.second+b.second)%mods};\n\tif(a.first>b.first)return a;\n\treturn b;\n}\npr mul(pr a,pr b){\n\treturn {a.first+b.first,a.second*b.second%mods};\n}\npr solve(int l,int r,int a,int b,int op){\n\tif(l>r)return {0,1};\n\tif(a>b)return {0,pw[r-l]};\n\tif(l==r){\n\t\tif(a<=dy[l]&&b>=dy[l])return {op,1};\n\t return {0,1};\n\t}\n\tif(!bh[l][r])bh[l][r]=++idx;\n\tif(q[op][bh[l][r]].count({a,b}))return q[op][bh[l][r]][{a,b}];\n\tint mid=l+r>>1;\n\tpr res={-inf,0};\n\tfor(int i=1;i<=n;i++){\n\t\tif(i!=dy[mid])res=hb(res,mul(solve(l,mid,a,min(b,i),0),solve(mid+1,r,max(a,i+1),b,op)));\n\t\telse res=hb(res,mul(solve(l,mid,a,min(b,i),1),solve(mid+1,r,max(a,i+1),b,op)));\n\t}\n\treturn q[op][bh[l][r]][{a,b}]=res;\n}\nsigned main(){\n\t//freopen(\"in.txt\",\"r\",stdin);\n\tios::sync_with_stdio(0);\n\tcin.tie(0);cout.tie(0);\n\tcin>>t;\n\twhile(t--){\n\t\tcin>>n>>m;\n\t\tpw[0]=1;\n\t\tfor(int i=1;i<=n;i++){\n\t\t\tfor(int j=1;j<=n;j++){\n\t\t\t\tbh[i][j]=0;\n\t\t\t}\n\t\t\tq[0][i].clear();\n\t\t\tq[1][i].clear();\n\t\t}\n\t\tfor(int i=1;i<=n;i++)dy[i]=0,pw[i]=pw[i-1]*n%mods;\n\t\tfor(int i=1;i<=m;i++)cin>>a[i]>>b[i],dy[a[i]]=b[i];\n\t\tpr res;\n\t\tif(dy[n]){\n\t\t\tauto r1=solve(1,n,1,n,1),r2=solve(1,n,1,n,0);\n\t\t\tr2.second=r2.second*(n-1)%mods;\n\t\t\tres=hb(r1,r2);\n\t\t}else{\n\t\t\tres=solve(1,n,1,n,0);\n\t\t\tres.second=res.second*n%mods;\n\t\t}\n\t\tcout<<m-res.first<<\" \"<<res.second<<\"\\n\";\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 296251646,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ll long long\n#define pb push_back\nusing namespace std;\nconst int N = 3050;\nconst int mod = 998244353;\nvector<int> path[N];\nint oper[N],dp[N],cnt[N],used[N],it;\nvoid build(vector<int> a, int l, int r){\n if(l > r)return;\n if(l == r){\n path[l] = a;\n return;\n }\n int mid = (l + r) / 2;\n a.pb(mid);\n build(a, mid + 1, r);\n build(a, l, mid);\n}\nvoid add(int &x, int y){\n x += y;\n if(x >= mod)x -= mod;\n}\nint binpow(int a,int n){\n int rt = 1;\n while(n){\n if(n & 1){\n rt = (rt * 1ll * a) % mod;\n }\n a = (a * 1ll * a) % mod;\n n >>= 1;\n }\n return rt;\n}\nint main() {\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n int qq;\n cin >> qq;\n while(qq--){\n int n,m;\n cin >> n >> m;\n for(int i = 0;i <= n + 1;i++){\n oper[i] = -1;\n dp[i] = cnt[i] = 0;\n }\n for(int i = 1;i <= m;i++){\n int j,x;\n cin >> j >> x;\n oper[j] = x;\n }\n vector<int> tmp;\n build(tmp, 1, n);\n oper[0] = 0;\n assert(path[n + 1].empty());\n oper[n + 1] = n + 1;\n int mx = 0;\n cnt[0] = 1;\n for(int i = 1;i <= n + 1;i++){\n if(oper[i] == -1)continue;\n for(int j = i - 1;j >= 0;j--){\n if(oper[j] == -1)continue;\n if(oper[j] >= oper[i])continue;\n if(dp[j] + (i <= n) < dp[i])continue;\n int kol = 0;\n ++it;\n int cntA = 0,cntB = 0,cntb = 0;\n for(int el: path[j]){\n if((el <= j) || (el >= i))continue;\n used[el] = it;\n cntA++;\n }\n for(int el: path[i]){\n \n if((el <= j) || (el >= i))continue;\n if(used[el] == it){\n cntA--;\n cntb++;\n }else{\n cntB++;\n }\n }\n int vA = (n - oper[j] + 1);\n int vB = oper[i] - 1;\n int vb = oper[i] - oper[j];\n \n \n kol = (binpow(vA, cntA) * 1ll * binpow(vB, cntB)) % mod;\n kol = (kol * 1ll * binpow(vb, cntb)) % mod;\n int ost = i - j - 1 - cntA - cntB - cntb;\n assert(ost >= 0);\n kol = (binpow(n, ost) * 1ll * kol) % mod;\n kol = (kol * 1ll * cnt[j]) % mod;\n if(dp[j] + (i <= n) == dp[i]){\n add(cnt[i], kol);\n }else if(kol > 0){\n cnt[i] = kol;\n dp[i] = dp[j] + (i <= n);\n }\n }\n mx = max(mx, dp[i]);\n }\n assert(mx > 0);\n cout << m - dp[n + 1] << ' ' << cnt[n + 1];\n cout << '\\n';\n int res = 0;\n for(int i = 1;i <= n;i++){\n if(oper[i] == -1)continue;\n if(dp[i] == mx){\n \n }\n }\n for(int i = 1;i <= n;i++){\n path[i].clear();\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 289067675,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n// Cao Quang Hung\n#include<bits/stdc++.h>\n \nusing namespace std;\n#define pb push_back\n#define eb emplace_back\n#define mp make_pair\n#define pii pair<int,int>\n#define pll pair<long long , long long>\n#define vi vector<int>\n#define vpii vector<pii>\n#define SZ(x) ((int)(x.size()))\n#define fi first\n#define se second\n#define IN(x,y) ((y).find((x))!=(y).end())\n#define ALL(t) t.begin(),t.end()\n#define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++)\n#define REP(i,a,b) for(int (i)=(a);(i)<=(b);++i)\n#define REPD(i,a,b) for(int (i)=(a); (i)>=(b);--i)\n#define FOR(i, n) for (int (i) = 0; (i) < (n); ++(i))\n#define dem(x) __builtin_popcount(x)\n#define Mask(x) (1LL << (x))\n#define BIT(x, i) ((x) >> (i) & 1)\n#define ln '\\n'\n#define io_faster ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);\n///mt19937 rnd(time(0));\n \nconst int INF = 1e9 , mod = 998244353;\n \ntemplate <class T1, class T2>\ninline T1 mul(T1& x, const T2 &k){ return x = (1LL * x * k) % mod; }\n \ntemplate <class T1 , class T2>\ninline T1 pw(T1 x, T2 k){T1 res = 1; for (; k ; k >>= 1){ if (k & 1) mul(res, x); mul(x, x); } return res;}\n \ntemplate <class T>\ninline bool minimize(T &x, const T &y){ if (x > y){x = y; return 1;} return 0; }\n \ntemplate <class T>\ninline bool maximize(T &x, const T &y){ if (x < y){x = y; return 1;} return 0; }\n \ntemplate <class T>\ninline void add(T &x , const T &y){ if ((x += y) >= mod) x -= mod; }\n \ntemplate <class T>\ninline T product (const T &x , const T &y) { return 1LL * x * y % mod; }\n \n#define PROB \"a\"\nvoid file(){\n if(fopen(PROB\".inp\", \"r\")){\n freopen(PROB\".inp\",\"r\",stdin);\n freopen(PROB\".out\",\"w\",stdout);\n }\n}\nvoid sinh_(){\n// srand(time(0));\n// freopen(PROB\".inp\" , \"w\" , stdout);\n// int n;\n}\n \ntypedef long long ll;\ntypedef double db;\nconst int N = 3e3 + 5;\n \nint n, m, a[N];\npii test[N], dp[N];\nvector<int> touch[N];\nint p[N][N];\n \nvoid readip(){\n cin >> n >> m;\n REP(i, 1, n) a[i] = 0, touch[i].clear();\n \n REP(i, 1, m) cin >> test[i].first >> test[i].second;\n sort(test + 1, test + m + 1); \n \n \n REP(i, 1, m) {\n int val = test[i].first;\n a[val] = test[i].second; \n \n int l = 1, r = n;\n \n while(l < r) {\n int mid = (l + r) >> 1;\n touch[val].eb(mid);\n \n if (val <= mid) \n r = mid;\n else \n l = mid + 1;\n }\n }\n \n \n p[0][0] = 1;\n REP(i, 1, n+1) {\n p[i][0] = 1;\n REP(j, 1, n+1) p[i][j] = product(p[i][j - 1], i);\n }\n} \n \nbool marki[N], markboth[N];\nvoid solve(){ \n REP(i, 1, n) dp[i] = make_pair(0, 0); \n dp[0] = make_pair(0, 1);\n \n REP(i, 1, n) {\n if (a[i] == 0 ||(i > 1 && a[i] == 1)) continue;\n \n REPD(j, i - 1, 0) if (a[j] < a[i] && dp[j].first >= dp[i].first) {\n vector<int> both, I, J;\n for (auto &pos : touch[i]) if (j < pos && pos < i) marki[pos] = true;\n \n for (auto &pos : touch[j]) if (j < pos && pos < i) {\n if (marki[pos]) both.eb(pos), markboth[pos] = true;\n else J.eb(pos);\n }\n \n for (auto &pos : touch[i]) if (j < pos && pos < i) {\n marki[pos] = false;\n if (markboth[pos]) markboth[pos] = false;\n else I.eb(pos);\n }\n \n int numBoth = both.size();\n int numI = I.size();\n int numJ = J.size();\n \n int numway = 1LL * dp[j].second \n * 1LL * p[n][i - j - 1 - numBoth - numI - numJ] % mod \n * 1LL * p[a[i] - a[j]][numBoth] % mod \n * 1LL * p[n - a[j] + 1][numJ] % mod\n * 1LL * p[a[i] - 1][numI] % mod; \n \n \n if (dp[j].first > dp[i].first) {\n dp[i].first = dp[j].first;\n dp[i].second = numway;\n }\n else {\n add(dp[i].second, numway);\n }\n }\n ++dp[i].first;\n } \n \n \n int mx = -1, res;\n REP(i, 1, n) {\n int cnt = 0;\n for (auto it : touch[i]) cnt += it > i;\n mul(dp[i].second, product(p[n][n - i - cnt], p[n - a[i] + 1][cnt]));\n if (dp[i].first > mx) {\n mx = dp[i].first;\n res = dp[i].second;\n }\n else if (dp[i].first == mx) {\n add(res, dp[i].second);\n }\n }\n cout << m - mx << ' ' << res << ln;\n} \n \nint main(){\n sinh_();\n io_faster\n file();\n int t = 1;\n cin >> t;\n while (t--){\n readip();\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 291168134,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nusing ll=long long;\nconst int N=3005,mod=998244353;\nint n,m,rt,tot,lca[N][N],fa[N*2],ls[N*2],rs[N*2],fd[N*2],to[N*2],a[N],g[N];\nll fc[N],inv[N],pw[N],f[N];\nll qp(ll a,int b){ll ans=1;for(;b;b>>=1,a=a*a%mod)if(b&1)ans=ans*a%mod;return ans;}\nll C(int n,int m){return n<m||m<0?0:fc[n]*inv[m]%mod*inv[n-m]%mod;}\nvoid init(int n) {\n fc[0]=pw[0]=1;\n for(int i=1;i<=n;i++)fc[i]=fc[i-1]*i%mod,pw[i]=pw[i-1]*2%mod;\n inv[n]=qp(fc[n],mod-2);\n for(int i=n-1;~i;i--)inv[i]=inv[i+1]*(i+1)%mod;\n}\nint bd(int l,int r) {\n if(l==r)return to[l]=l;\n int mid=l+r>>1,p=++tot;to[p]=mid;\n for(int i=l;i<=mid;i++)for(int j=mid+1;j<=r;j++)lca[i][j]=p;\n ls[p]=bd(l,mid),rs[p]=bd(mid+1,r);\n return fd[ls[p]]=0,fd[rs[p]]=1,fa[ls[p]]=fa[rs[p]]=p;\n}\nvoid solve() {\n cin>>n>>m,tot=n,rt=bd(1,n),fa[rt]=0;\n ll iv=qp(n,mod-2);\n for(int i=1;i<=n;i++)a[i]=0;\n for(int i=1,x,k;i<=m;i++)cin>>x>>k,a[x]=k;\n int as=0;\n for(int i=1;i<=n;i++)if(a[i]) {\n g[i]=(a[i]!=1||i==1?1:-n);\n for(int j=1;j<i;j++)if(a[j]&&a[j]<a[i])g[i]=max(g[i],g[j]+1);\n as=max(as,g[i]);\n }\n ll way=0;\n for(int i=1;i<=n;i++)if(a[i]) {\n f[i]=0;\n if(g[i]==1) {\n f[i]=iv;\n for(int x=fa[i];fa[x];x=fa[x])if(fd[x])f[i]=f[i]*(a[i]-1)%mod*iv%mod;\n }\n for(int j=1;j<i;j++)if(a[j]&&a[j]<a[i]&&g[j]+1==g[i]) {\n int lc=lca[j][i];\n ll v=f[j];\n for(int x=j;fa[x]!=lc;x=fa[x])if(to[fa[x]]!=j&&!fd[x])v=v*(n-a[j]+1)%mod*iv%mod;\n for(int x=i;fa[x]!=lc;x=fa[x])if(fd[x])v=v*(a[i]-1)%mod*iv%mod;\n if(to[lc]!=j)v=v*(a[i]-a[j])%mod*iv%mod;\n f[i]=(f[i]+v*iv)%mod;\n }\n if(g[i]==as) {\n ll v=f[i];\n for(int x=i;fa[x];x=fa[x])if(to[fa[x]]!=i&&!fd[x])v=v*(n-a[i]+1)*iv%mod;\n way=(way+v)%mod;\n }\n }cout<<m-as<<' '<<way*qp(n,n)%mod<<'\\n';\n}\nint main() {\n #ifndef ONLINE_JUDGE\n freopen(\".in\",\"r\",stdin);\n freopen(\".out\",\"w\",stdout);\n #endif\n ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);\n int tt;cin>>tt;\n while(tt--)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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288465849,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#pragma GCC optimize(\"O3,unroll-loops\")\n// #pragma GCC target(\"avx,popcnt,sse4,abm\")\n#include <bits/stdc++.h>\nusing namespace std;\n \n#ifdef WAIMAI\n#define debug(HEHE...) cout << \"[\" << #HEHE << \"] : \", dout(HEHE)\nvoid dout() {cout << '\\n';}\ntemplate<typename T, typename...U>\nvoid dout(T t, U...u) {cout << t << (sizeof...(u) ? \", \" : \"\"), dout(u...);}\n#else\n#define debug(...) 7122\n#endif\n \n#define ll long long\n#define Waimai ios::sync_with_stdio(false), cin.tie(0)\n#define FOR(x,a,b) for (int x = a, I = b; x <= I; x++)\n#define pb emplace_back\n#define F first\n#define S second\n \nconst int MOD = 998244353;\nconst int SIZE = 3005;\n \nint n, m;\nint mx, ans;\nint pw[SIZE];\nvector<int> p[SIZE];\npair<int, int> a[SIZE], dp[SIZE];\n \nint power(int d, int up) {\n int re = 1;\n while (up) {\n if (up & 1) re = 1ll * re * d % MOD;\n d = 1ll * d * d % MOD;\n up >>= 1;\n }\n return re;\n}\n \nvoid solve() {\n cin >> n >> m;\n FOR (i, 1, m) cin >> a[i].F >> a[i].S;\n sort(a + 1, a + m + 1);\n pw[0] = 1;\n FOR (i, 1, n) {\n int l = 1, r = n;\n p[i].clear();\n pw[i] = 1ll * pw[i - 1] * n % MOD;\n while (l < r) {\n int mid = (l + r) / 2;\n p[i].pb(mid);\n if (i <= mid) r = mid;\n else l = mid + 1;\n }\n sort(p[i].begin(), p[i].end());\n }\n mx = ans = 0;\n FOR (i, 1, m) {\n dp[i] = {-1, 0};\n if (a[i].F > 1 && a[i].S == 1) continue;\n FOR (j, 1, i - 1) if (a[j].S < a[i].S && dp[j].F + 1 >= dp[i].F) {\n if (dp[j].F + 1 > dp[i].F) dp[i] = {dp[j].F + 1, 0};\n int len = a[i].F - a[j].F - 1, pro = 1;\n for (int pj = 0, pi = 0;;) {\n while (pj < p[a[j].F].size() && (p[a[j].F][pj] <= a[j].F || p[a[j].F][pj] >= a[i].F)) pj++;\n while (pi < p[a[i].F].size() && (p[a[i].F][pi] <= a[j].F || p[a[i].F][pi] >= a[i].F)) pi++;\n if (pj >= p[a[j].F].size() && pi >= p[a[i].F].size()) break;\n if (pj < p[a[j].F].size() && pi < p[a[i].F].size() && p[a[j].F][pj] == p[a[i].F][pi]) {\n pro = 1ll * pro * (a[i].S - a[j].S) % MOD;\n pj++, pi++, len--;\n continue;\n }\n if (pj < p[a[j].F].size() && (pi == p[a[i].F].size() || p[a[j].F][pj] < p[a[i].F][pi])) {\n pro = 1ll * pro * (n - a[j].S + 1) % MOD;\n pj++, len--;\n } else {\n pro = 1ll * pro * (a[i].S - 1) % MOD;\n pi++, len--;\n }\n }\n pro = 1ll * pro * pw[len] % MOD;\n dp[i].S = (dp[i].S + 1ll * pro * dp[j].S) % MOD;\n }\n if (dp[i].F <= 0) {\n int c = lower_bound(p[a[i].F].begin(), p[a[i].F].end(), a[i].F) - p[a[i].F].begin();\n dp[i] = {1, 1ll * power(a[i].S - 1, c) * pw[a[i].F - 1 - c] % MOD};\n }\n if (dp[i].F > mx) {\n mx = dp[i].F;\n ans = 0;\n }\n if (dp[i].F == mx) {\n int c = upper_bound(p[a[i].F].begin(), p[a[i].F].end(), a[i].F) - p[a[i].F].begin();\n c = p[a[i].F].size() - c;\n ans = (ans + 1ll * dp[i].S * power(n - a[i].S + 1, c) % MOD * pw[n - a[i].F - c]) % MOD;\n }\n }\n cout << m - mx << ' ' << ans << '\\n';\n}\n \nint main() {\n Waimai;\n int tt;\n cin >> tt;\n while (tt--) 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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 290167952,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nusing ll = long long;\nconst int N = 3e3;\nconst int mod = 998244353;\nint add(int x, int y) { return (x += y) >= mod ? x - mod : x; }\nint sub(int x, int y) { return (x -= y) < 0 ? x + mod : x; }\nint ksm(ll x, int tp, int s = 1)\n{\n for (; tp; x = x * x % mod, tp >>= 1)\n if (tp & 1)\n s = x * s % mod;\n return s;\n}\nint n, m;\nint v[N + 5];\nvector<int> pos[N + 5];\nvoid build(int l, int r, int x)\n{\n if (l == r)\n return;\n int mid = l + r >> 1;\n pos[x].push_back(mid);\n if (x <= mid)\n build(l, mid, x);\n else\n build(mid + 1, r, x);\n}\nstruct F\n{\n int c, s;\n F(int c = 0, int s = 0) : c(c), s(s) {}\n} f[N + 5];\nint cal(int x, int y)\n{\n static bool inx[N + 5];\n int a = 0, b = 0, c = 0;\n for (auto it : pos[x])\n if (it > x && it < y)\n {\n inx[it] = 1;\n a++;\n }\n for (auto it : pos[y])\n if (it > x && it < y)\n {\n b++;\n if (inx[it])\n c++;\n }\n a -= c;\n b -= c;\n return ksm(n, y - x - 1 - a - b - c, ksm(n - v[x] + 1, a, ksm(v[y] - 1, b, ksm(v[y] - v[x], c))));\n}\nvoid solve()\n{\n scanf(\"%d %d\", &n, &m);\n memset(v, -1, sizeof v);\n for (int i = 1, x; i <= m; i++)\n {\n scanf(\"%d\", &x);\n scanf(\"%d\", &v[x]);\n }\n for (int i = 1; i <= n; i++)\n {\n pos[i].clear();\n build(1, n, i);\n }\n int r = 0;\n for (int i = 1; i <= n; i++)\n {\n f[i] = {0, 0};\n if (v[i] == -1)\n continue;\n int a = 0;\n for (auto it : pos[i])\n a += it < i;\n int c = ksm(n, i - a - 1, ksm(v[i] - 1, a));\n if (c != 0)\n f[i] = {1, c};\n for (int j = 1; j < i; j++)\n if (v[j] != -1 && v[j] < v[i] && f[j].c + 1 >= f[i].c)\n {\n int coe = cal(j, i);\n if (f[j].c + 1 > f[i].c)\n f[i] = F(f[j].c + 1, 1ll * f[j].s * coe % mod);\n else\n f[i].s = (f[i].s + 1ll * f[j].s * coe % mod);\n }\n r = max(r, f[i].c);\n }\n int ans = 0;\n for (int i = 1; i <= n; i++)\n {\n if (v[i] == -1)\n continue;\n if (f[i].c == r)\n {\n int a = 0;\n for (auto it : pos[i])\n a += it > i;\n int c = ksm(n, n - i - a, ksm(n - v[i] + 1, a));\n ans = (ans + 1ll * f[i].s * c) % mod;\n }\n }\n printf(\"%d %d\\n\", m - r, ans);\n}\nint main()\n{\n int t = 100;\n scanf(\"%d\", &t);\n while (t--)\n 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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 291446115,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define fi first\n#define se second\ntypedef long long ll;\nconst ll mod=998244353;\nconst int N=3005;\nint n,m;\nll guy[N]; \nint d[N];\nll dp[N];\nll ways[N];\n \nll pn[N][N];\nint filled[N];\nvoid dfs(int l,int r,int frog){\n\tif(l==r){\n\t\td[l]+=frog;//d[l] = number of elements that is forced < guy[l] in [1..l-1]\n\t\treturn;\n\t}\n\tint mid=(l+r)/2;\n\tdfs(l,mid,frog);\n\tdfs(mid+1,r,frog+1);\n}\nvoid aespa(int l,int r){\n\tif(l==r) return;\n\tint mid=(l+r)/2;\n\taespa(l,mid);\n\tfor(int i=l; i<=mid ;i++){\n\t\tfor(int j=mid+1; j<=r ;j++){\n\t\t\tif(guy[i]==0 || guy[j]==0 || guy[i]>guy[j] || dp[i]==0) continue;\n\t\t\t//cout << \"trans \" << i << ' ' << j << ' ' << ways[i] << ' ' << guy[i] << endl;\n\t\t\tint newi=(j-filled[i]-2+(i==mid));\n\t\t\t//cout << mid << ' ' << filled[i] << endl;\n\t\t\tint force=d[j]-d[l]-1;\n\t\t\tll middle=guy[j]-guy[i];\n\t\t\tif(i==mid) middle=1;\n\t\t\t//cout << \"haha \" << newi << ' ' << force << ' ' << middle << endl;\n\t\t\t//and multiply number of ways to pick [mid]\n\t\t\t\n\t\t\tif(dp[j]<dp[i]+1){\n\t\t\t\tdp[j]=dp[i]+1;\n\t\t\t\tways[j]=0;\n\t\t\t}\n\t\t\tif(dp[j]==dp[i]+1){\n\t\t\t\tways[j]=(ways[j]+ways[i]*pn[n][newi-force]%mod*pn[guy[j]-1][force]%mod*middle)%mod;\n\t\t\t}\n\t\t\t//cout << \"fin \" << ways[j] << endl;\n\t\t}\n\t}\n\taespa(mid+1,r);\n\tfor(int i=l; i<mid ;i++){\n\t\tfilled[i]++;\n\t\tif(guy[i]!=0){\n\t\t\tways[i]=ways[i]*(n-guy[i]+1)%mod;\n\t\t}\n\t}\n}\nvoid solve(){\n\tcin >> n >> m;\n\tpn[0][0]=1;\n\tfor(int i=1; i<=n ;i++){\n\t\tguy[i]=0;\n\t\td[i]=0;\n\t\tdp[i]=0;\n\t\tways[i]=0;\n\t\tfilled[i]=i;\n\t\tpn[i][0]=1;\n\t\tfor(int j=1; j<=n ;j++){\n\t\t\tpn[i][j]=pn[i][j-1]*i%mod;\n\t\t}\n\t}\n\tways[0]=1;\n\tfor(int i=1; i<=m ;i++){\n\t\tint x,k;cin >> x >> k;\n\t\tguy[x]=k;\n\t}\n\tdfs(1,n,0);\n\tfor(int i=1; i<=n ;i++){\n\t\tif(guy[i]==0) continue;\n\t\tif(guy[i]==1 && d[i]!=0){\n\t\t\tdp[i]=0;\n\t\t\tways[i]=0;\n\t\t\tcontinue;\n\t\t}\n\t\tdp[i]=1;\n\t\tways[i]=pn[guy[i]-1][d[i]]*pn[n][i-d[i]-1]%mod;\n\t}\n\taespa(1,n);\n\tll mx=1,w=0;\n\tfor(int i=1; i<=n ;i++){\n\t\tif(dp[i]>mx){\n\t\t\tmx=dp[i];w=0;\n\t\t}\n\t\tif(dp[i]==mx){\n\t\t\tw=(w+ways[i]*pn[n][n-filled[i]])%mod;\n\t\t}\n\t\t//cout << \"haha \" << dp[i] << ' ' << ways[i] << endl;\n\t}\n\tcout << m-mx << ' ' << w << '\\n';\n}\nint main(){\n\tios::sync_with_stdio(false);cin.tie(0);\n\tint t;cin >> 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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288345518,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\nfrom collections import defaultdict\nfrom math import prod\nP = 998244353\ndef solve():\n n, m = map(int, input().split())\n a = [list(map(int, input().split())) for _ in range(m)]\n def bsearch(x):\n l, h = 1, n\n while l < h:\n m = l + h >> 1\n yield m\n if m < x: l = m + 1\n else: h = m\n \n a.append([0, 0])\n a.append([n+1, n+1])\n a.sort()\n L = [1] * len(a)\n W = [0] * len(a)\n W[0] = 1\n for i in range(1, len(a)):\n x2, k2 = a[i]\n l = 0\n for j in range(i):\n x1, k1 = a[j]\n if k1 < k2 and W[j]:\n l = max(l, L[j])\n L[i] = l + 1\n for j in range(i):\n x1, k1 = a[j]\n q = defaultdict(int)\n if k1 < k2 and L[j] == l:\n for t in bsearch(x1):\n if x1 < t < x2:\n q[t] += max(0, k1 - 1)\n for t in bsearch(x2):\n if x1 < t < x2:\n q[t] += n - k2 + 1\n z = W[j] * pow(n, x2 - x1 - len(q) - 1, P)\n for v in q.values():\n z = z * (n - v) % P\n W[i] = (W[i] + z) % P\n print(m - L[-1] + 2, W[-1])\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 296009555,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define rep(i,n) for(int i=0,del##i##verme=int(n);i<del##i##verme;++i)\n#define rep1(i,n) for(int i=1,parano##i##a=int(n);i<=parano##i##a;++i)\n#define per(i,n) for(int i=int(n)-1;i>=0;--i)\n#define per1(i,n) for(int i=int(n);i>=1;--i)\n#define pb push_back\n#define mp make_pair\n#define fi first\n#define se second\n#define y0 LingLuo\n#define y1 VividCycle\ntypedef long long ll;\ntypedef unsigned long long ull;\ntypedef long double ldb;\nusing namespace std;\nconst ll mod1=998244353;\nunsigned time_related_rand()\n{\n\treturn unsigned(std::chrono::steady_clock::now().time_since_epoch().count());\n}\nstruct query{\n\tint x,k;\n};\nbool operator<(const query&u,const query&v)\n{\n\treturn u.x<v.x;\n}\nint l[6005],r[6005],p[6005];\nint t,n,m;\nquery q[3005];\nint lis[3005];\nint dep[3005];\nll dp[3005];\nll pw[3005];\nint root;\nint dfs(int L,int R,int d)\n{\n\tif(L==R)\n\t{\n\t\tdep[L+n]=d;\n\t\treturn L+n;\n\t}\n\tint mid=(L+R)>>1;\n\tdep[mid]=d;\n\tl[mid]=dfs(L,mid,d+1);\n\tr[mid]=dfs(mid+1,R,d+1);\n\tp[l[mid]]=p[r[mid]]=mid;\n\treturn mid;\n}\nll val(int x,int y,int f,int g)// f~g\n{\n\tll ret=1;int cnt=y-x-1;\n\tint cx=x+n,cy=y+n;\n\twhile(cx!=root&&cy!=root)\n\t{\n\t\tif(p[cx]==p[cy])\n\t\t{\n\t\t\tcx=p[cx];cy=p[cy];\n\t\t\tif(x<cx&&cy<y)\n\t\t\t{\n\t\t\t\tret=ret*(g-f+1)%mod1;\n\t\t\t\t--cnt;\n\t\t\t}\n\t\t}\n\t\telse\n\t\t{\n\t\t\tif(dep[cx]>dep[cy])\n\t\t\t{\n\t\t\t\tcx=p[cx];\n\t\t\t\tif(x<cx&&cx<y)\n\t\t\t\t{\n\t\t\t\t\tret=ret*(n-f+1)%mod1;\n\t\t\t\t\t--cnt;\n\t\t\t\t}\n\t\t\t}\n\t\t\telse\n\t\t\t{\n\t\t\t\tcy=p[cy];\n\t\t\t\tif(x<cy&&cy<y)\n\t\t\t\t{\n\t\t\t\t\tret=ret*g%mod1;\n\t\t\t\t\t--cnt;\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t}\n\treturn ret*pw[cnt]%mod1;\n}\nll bef(int y,int g)\n{\n\tif(y==1) return 1;\n\tll ret=1;int cnt=y-1;\n\tint cur=y+n;\n\twhile(cur!=root)\n\t{\n\t\tcur=p[cur];\n\t\tif(cur<y)\n\t\t{\n\t\t\tret=ret*g%mod1;\n\t\t\t--cnt;\n\t\t}\n\t}\n\treturn ret*pw[cnt]%mod1;\n}\nll aft(int x,int f)\n{\n\tif(x==n) return 1;\n\tll ret=1;int cnt=n-x;\n\tint cur=x+n;\n\twhile(cur!=root)\n\t{\n\t\tcur=p[cur];\n\t\tif(cur>x)\n\t\t{\n\t\t\tret=ret*(n-f+1)%mod1;\n\t\t\t--cnt;\n\t\t}\n\t}\n\treturn ret*pw[cnt]%mod1;\n}\nll x[3005],y[3005];\nll qaq[3005];\nvoid Q()\n{\n\tcin>>n>>m;\n\trep1(i,m)\n\t{\n\t\tcin>>q[i].x>>q[i].k;\n\t}\n\tif(n==1)\n\t{\n\t\tcout<<\"0 1\\n\";\n\t\treturn ;\n\t}\n\tsort(q+1,q+m+1);\n\troot=dfs(1,n,0);\n//\treturn ;\n\tpw[0]=1;rep1(i,n)pw[i]=pw[i-1]*n%mod1;\n\trep1(i,m)\n\t{\n\t\tx[i]=bef(q[i].x,q[i].k-1);\n\t\ty[i]=aft(q[i].x,q[i].k);\n\t}\n\tint qq=0;\n\trep1(i,m)\n\t{\n\t\tif(x[i])\n\t\tlis[i]=1;\n\t\telse\n\t\tlis[i]=-1145141919;\n\t\tfor(int j=1;j<i;++j) if(lis[j]!=-1145141919&&q[j].k<q[i].k&&(qaq[j]=val(q[j].x,q[i].x,q[j].k,q[i].k-1)))\n\t\t{\n\t\t\tlis[i]=max(lis[i],lis[j]+1);\n\t\t}\n\t\tif(lis[i]==1)\n\t\t{\n\t\t\tdp[i]=x[i];\n\t\t}\n\t\telse\n\t\t{\n\t\t\tdp[i]=0;\n\t\t\tfor(int j=1;j<i;++j) if(q[j].k<q[i].k&&lis[i]==lis[j]+1&&qaq[j])\n\t\t\t{\n\t\t\t\tdp[i]=(dp[i]+dp[j]*qaq[j])%mod1;\n\t\t\t}\n\t\t}\n\t\tif(y[i]) qq=max(qq,lis[i]);\n\t//\tcout<<i<<\";\"<<lis[i]<<\" \"<<dp[i]<<endl;\n\t}\n\tcout<<m-qq<<\" \";\n\tll answer=0;\n\trep1(i,m) if(lis[i]==qq)\n\t{\n\t\tanswer=(answer+dp[i]*y[i])%mod1;\n\t//\tcout<<i<<\" \"<<aft(i,q[i].k)<<\"!!\"<<endl;\n\t}\n\tif(qq==0)\n\t{\n\t\tanswer=1;rep1(i,n)answer=answer*n%mod1;\n\t}\n\tcout<<answer<<\"\\n\";\n}\nint main()\n{\n\t#ifndef DEBUG \n\tios_base::sync_with_stdio(false);cin.tie(0);\n\t#endif\n\tcin>>t;while(t--)Q();return 0;\n} \n/* things to check\n1. int overflow or long long memory need\n2. recursion/array/binary search/dp/loop bounds\n3. precision\n4. special cases(n=1,bounds)\n5. delete debug statements\n6. initialize(especially multi-tests)\n7. = or == , n or m ,++ or -- , i or j , > or >= , < or <=\n8. keep it simple and stupid\n9. do not delete, use // instead\n10. operator priority\n11. is there anything extra to output?\n12. ...\n*/\n \n/* something to think about\n1. greedy? dp? searching? dp with matrix/ segment tree? binary search?\n2. If contains \"not\", why not ?????? or few affect?\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288359686,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include <vector>\n#include <list>\n#include <map>\n#include <set>\n#include <deque>\n#include <vector>\n#include <list>\n#include <map>\n#include <set>\n#include <deque>\n#include <queue>\n#include <stack>\n#include <bitset>\n#include <algorithm>\n#include <functional>\n#include <numeric>\n#include <utility>\n#include <sstream>\n#include <iostream>\n#include <iomanip>\n#include <cstdio>\n#include <cmath>\n#include <cstdlib>\n#include <cctype>\n#include <string>\n#include <cstring>\n#include <ctime>\n#include <random>\n#include <chrono>\n \nusing namespace std;\n \n#define _int64 long long\n#define mo 998244353\n \nmap<pair<int,int>,int> ind;\nint nn;\nmap<pair<int,int>,pair<int,_int64> > d[10000][2];\nint n,m;\n_int64 pown[3100];\n \nvoid upd(pair<int,_int64>& ans,pair<int,_int64>& tmp)\n{\n\tif (tmp.first>ans.first) ans=tmp;\n\telse if (tmp.first==ans.first)\n\t{\n\t\tans.second+=tmp.second;\n\t\tans.second%=mo;\n\t}\n}\n \npair<int,_int64> doit(int ll,int rr,int bit,int vl,int vr,vector<pair<int,int> >& aa)\n{\n\tint id,i,j,prev,cnt,mid;\n\tvector<pair<int,int> > v;\n\tvector<pair<int,int> > a1,a2;\n\tpair<int,_int64> v1,v0;\n\tpair<int,_int64> v2;\n\tif (aa.size()==0)\n\t{\n\t\t//cerr<<\"ll,rr,vl,vr0:\"<<ll<<\" \"<<rr<<\" \"<<vl<<\" \"<<vr<<endl;\n\t\tif (rr==n) return make_pair(0,pown[rr-ll+1]);\n\t\telse return make_pair(0,pown[rr-ll]);\n\t}\n\tif (ll==rr)\n\t{\n\t\t//cerr<<\"ll,rr1:\"<<ll<<\" \"<<rr<<\" \"<<aa.size()<<endl;\n\t\t//for (i=0;i<aa.size();i++)\n\t\t//\tcerr<<\"aa:\"<<aa[i].first<<\" \"<<aa[i].second<<endl;\n\t\tif (bit==1) return make_pair(1,1);\n\t\telse return make_pair(0,1);\n\t}\n\tif (ind.find(make_pair(ll,rr))==ind.end())\n\t{\n\t\tind[make_pair(ll,rr)]=nn;\n\t\td[nn][0].clear();\n\t\td[nn][1].clear();\n\t\tnn++;\n\t}\n\tid=ind[make_pair(ll,rr)];\n\tif (d[id][bit].find(make_pair(vl,vr))!=d[id][bit].end()) return d[id][bit][make_pair(vl,vr)];\n\tmid=(ll+rr)/2;\n\tv.clear();\n\tfor (i=0;i<aa.size();i++)\n\t\tv.push_back(make_pair(aa[i].second,aa[i].first));\n\tsort(v.begin(),v.end());\n\tv.push_back(make_pair(n+1,1));\n\tprev=1;\n\tpair<int,_int64> ans,tmp,tmp1;\n\tans=make_pair(-1,0);\n\tfor (i=0;i<v.size();i++)\n\t{\n\t\t//cerr<<\"i,v,prev:\"<<i<<\" \"<<v[i]<<\" \"<<prev<<endl;\n\t\tcnt=v[i].first-prev;\n\t\tif (cnt==0) continue;\n\t\ta1.clear();\n\t\ta2.clear();\n\t\tfor (j=0;j<aa.size();j++)\n\t\t{\n\t\t\tif ((aa[j].first<=mid)&&(aa[j].second<=v[i].first-1)) a1.push_back(aa[j]);\n\t\t\tif ((aa[j].first>mid)&&(aa[j].second>v[i].first-1)) a2.push_back(aa[j]);\n\t\t}\n\t\tv1=doit(ll,mid,0,vl,min(vr,v[i].first-1),a1);\n\t\tif ((i-1>=0)&&(v[i-1].second==mid))\n\t\t{\n\t\t\tv0=doit(ll,mid,1,vl,min(vr,v[i].first-1),a1);\n\t\t\tif (cnt-1==0) v1=v0;\n\t\t\telse\n\t\t\t{\n\t\t\t\tv1.second*=(cnt-1);\n\t\t\t\tv1.second%=mo;\n\t\t\t\tupd(v1,v0);\n\t\t\t}\n\t\t}\n\t\telse\n\t\t{\n\t\t\tv1.second*=cnt;\n\t\t\tv1.second%=mo;\n\t\t}\n\t\t//cerr<<\"sz:\"<<a1.size()<<\" \"<<a2.size()<<endl;\n\t\tv2=doit(mid+1,rr,bit,max(vl,v[i].first),vr,a2);\n\t\ttmp=make_pair(v1.first+v2.first,v1.second*v2.second%mo);\n\t\tupd(ans,tmp);\n\t\t//cerr<<\"ll,rr,tmp.second:\"<<ll<<\" \"<<rr<<\" \"<<vl<<\" \"<<vr<<\" \"<<tmp.first<<\" \"<<tmp.second<<\" \"<<cnt<<\" \"<<v[i].first<<\" \"<<prev<<endl;\n\t\t//cerr<<\"tmp.second:\"<<tmp.second<<endl;\n\t\tprev=v[i].first;\n\t}\n\td[id][bit][make_pair(vl,vr)]=ans;\n\t//cerr<<\"d:\"<<ll<<\" \"<<rr<<\" \"<<vl<<\" \"<<vr<<\" \"<<ans.first<<\" \"<<ans.second<<\" \"<<aa.size()<<endl;\n\treturn ans;\n}\n \nint main()\n{\n\tint i,j,k,l,t,x,y,o,b1;\n\tscanf(\"%d\",&t);\n\tvector<pair<int,int> > aa;\n\tfor (l=0;l<t;l++)\n\t{\n\t\tscanf(\"%d%d\",&n,&m);\n\t\tpown[0]=1;\n\t\tfor (i=1;i<=n;i++)\n\t\t\tpown[i]=pown[i-1]*n%mo;\n\t\taa.clear();\n\t\tfor (i=0;i<m;i++)\n\t\t{\n\t\t\tscanf(\"%d%d\",&x,&y);\n\t\t\taa.push_back(make_pair(x,y));\n\t\t}\n\t\tind.clear();\n\t\tnn=0;\n\t\tpair<int,_int64> ans=doit(1,n,1,1,n,aa);\n\t\tprintf(\"%d %lld\\n\",m-ans.first,ans.second);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 288383189,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Go Learn! (Easy Version)\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \\leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.\n\nWell, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.\n\nBessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\\ldots,a_n]$ with $n$ elements.\n\n let l = 1 let h = n while l < h: let m = floor((l + h) / 2) if a[m] < k: l = m + 1 else: h = m return l \n\nBessie submitted her code to Farmer John's problem with $m$ ($1 \\leq m \\leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \\leq x, k \\leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.\n\nTest $i$ is correct if the following hold:\n\n 1. The $x_i$-th element in the array is $k_i$. 2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$. \n\nIt might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ so that all remaining tests are correct.\n\nIn addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\\ldots,a_n]$ with $1 \\leq a_i \\leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) β the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\leq m \\leq n \\leq 3000$) denoting the number of the array and the number of tests.\n\nThe following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \\leq x_i, k_i \\leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.\n\nIt is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.\n\nOutput: For each test case, output two integers, $r$ β the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n2\n5 4\n1 1\n2 2\n4 3\n5 4\n5 4\n5 4\n2 5\n1 2\n3 3\noutput:\n0 1\n1 3\n\ninput:\n2\n6 6\n1 3\n2 5\n3 1\n4 2\n5 4\n6 6\n30 8\n19 22\n6 12\n12 1\n28 27\n3 4\n14 25\n29 14\n11 15\noutput:\n3 78\n3 839271911\n\nNote: Consider the first example.\n\nIn the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.\n\nIn the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\ntypedef long long ll;\nconst ll N = 3e5;\nconst ll mod = 998244353;\nll n, m;\nstruct dpnode{\n ll nr, nr2;\n}dp[N];\nvoid add(dpnode &a, dpnode b){\n if(b.nr2 == 0)return;\n if(a.nr < b.nr){\n a = b;\n }else if(a.nr == b.nr){\n a.nr2 = (a.nr2 + b.nr2)%mod;\n }\n}\nll fastexp(ll a, ll b){\n assert(b >= 0);\n ll r = 1;\n while(b){\n if(b&1)r = 1ll*r*a%mod;\n a = 1ll*a*a%mod;\n b/=2;\n }\n return r;\n}\nvector <ll> touch(ll x){\n vector <ll> p;\n ll l = 0, r = n - 1;\n while(l != r){\n ll mij = (l + r)/2;\n p.push_back(mij);\n if(x <= mij){\n r = mij;\n }else{\n l = mij + 1;\n }\n }\n return p;\n}\nstruct ev{\n ll pos, t;\n}v[N];\ndpnode dp2[N], dp3[N];\nll v2[N];\nvoid solve(){\n cin>>n>>m;\n for(ll i = 0;i < n;i++){\n dp[i] = {-mod, 0};\n dp2[i] = {-mod, 0};\n dp3[i] = {-mod, 0};\n v2[i] = 0;\n }\n for(ll i = 0;i < m;i++){\n ll x, p;\n cin>>p>>x;\n p--;\n v[i] = {p, x};\n v2[p] = x;\n }\n sort(v, v + m,[&](ev a, ev b){\n return a.t < b.t;\n });\n for(ll i = 0;i < m;i++){\n auto x = touch(v[i].pos);\n ll nr = fastexp(n, v[i].pos);\n for(auto j:x){\n if(j < v[i].pos){\n nr = 1ll*nr*fastexp(n, mod - 2)%mod*(v[i].t - 1)%mod;\n }\n }\n add(dp[v[i].pos], {1, nr});\n /// find the before\n nr = 0;\n ll nr2 = 0;\n for(auto j:x){\n if(j == v[i].pos){\n break;\n }\n if(j < v[i].pos){\n ///targeting this poll\n assert(dp2[j].nr == dp3[j].nr);\n nr2 = 0;\n for(auto k:x){\n if(j < k && k < v[i].pos)nr2++;\n }\n add(dp[v[i].pos], {dp2[j].nr + 1, 1ll*dp2[j].nr2*fastexp(v[i].t - 1, nr2)%mod*v[i].t%mod*fastexp(n, v[i].pos - j - 1 - nr2)%mod});\n add(dp[v[i].pos], {dp3[j].nr + 1, mod - 1ll*dp3[j].nr2*fastexp(v[i].t - 1, nr2)%mod*fastexp(n, v[i].pos - j - 1 - nr2)%mod});\n if(v[i].t > v2[j]){\n add(dp[v[i].pos], {dp[j].nr + 1, 1ll*dp[j].nr2*fastexp(v[i].t - 1, nr2)%mod*fastexp(n, v[i].pos - j - 1 - nr2)%mod});\n }\n }\n ///exactly this poll!\n }\n /// what occurs when lca is i or j?\n /// bruteforce dp\n /// find the after\n for(auto j:x){\n if(j == v[i].pos){\n break;\n }\n if(v[i].pos < j){\n ///v[i].pos -> j\n nr2 = 0;\n for(auto k:x){\n if(v[i].pos < k && k < j)nr2++;\n }\n add(dp2[j], {dp[v[i].pos].nr, 1ll*dp[v[i].pos].nr2*fastexp(n - v[i].t + 1, nr2)%mod*fastexp(n, j - v[i].pos - nr2 - 1)%mod});\n add(dp3[j], {dp[v[i].pos].nr, 1ll*dp[v[i].pos].nr2*fastexp(n - v[i].t + 1, nr2)%mod*v[i].t%mod*fastexp(n, j - v[i].pos - nr2 - 1)%mod});\n if(nr2 != 0 && v2[j] > v[i].t){\n add(dp[j], {dp[v[i].pos].nr + 1, 1ll*dp[v[i].pos].nr2*fastexp(n - v[i].t + 1, nr2 - 1)%mod*fastexp(n, j - v[i].pos - nr2 - 1)%mod*(v2[j] - v[i].t)%mod});\n }else if(v2[j] > v[i].t){\n add(dp[j], {dp[v[i].pos].nr + 1, 1ll*dp[v[i].pos].nr2*fastexp(n, j - v[i].pos - nr2 - 1)%mod});\n }\n }\n }\n }\n dpnode ans = {0, fastexp(n, n)};\n for(ll i = 0;i < n;i++){\n if(!v2[i])continue;\n //cout<<dp[i].nr<<' '<<dp[i].nr2<<'\\n';\n auto x = touch(i);\n ll cnt = 0;\n for(auto j:x){\n if(i < j)cnt++;\n }\n //cout<<1ll*dp[i].nr2*fastexp(n - v2[i] + 1, cnt)%mod*fastexp(n, n - i - 1 - cnt)%mod<<'\\n';\n add(ans, {dp[i].nr, 1ll*dp[i].nr2*fastexp(n - v2[i] + 1, cnt)%mod*fastexp(n, n - i - 1 - cnt)%mod});\n }\n cout<<m - ans.nr<<' '<<ans.nr2<<'\\n';\n}\nint main(){\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n ll t;\n cin>>t;\n while(t--)solve();\n return 0;\n}\n/**\n1\n8 8\n1 2\n2 1\n3 4\n4 3\n5 6\n6 5\n7 8\n8 7\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2035G1",
"submission_id": 290755435,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \n#define int long long\n#define rep(i, x, y) for(int i = x; i <= y; ++ i)\n#define mid ((l + r) >> 1)\n#define pb push_back\n#define i64 long long\n \nconst int N = 1e6 + 20;\nconst long long inf = 2e18;\nconst int M = 1e9 + 7;\n \nvector <int> pr;\nint n, vis[N * 20];\nvoid sieve()\n{\n\tn = 2e7;\n\tfor(int i = 2; i <= n; ++ i)\n\t{\n\t\tif(vis[i]) continue;\n\t\tpr.pb(i);\n\t\tfor(int j = i + i; j <= n; j += i)\n\t\t\tvis[j] = 1;\n\t}\n//\tint Las = 0, s = 0;\n//\tfor(auto i: pr)\n//\t\ts = max(s, i - Las), Las = i;\n//\tcout << s << \"\\n\";\n}\n \nint f[500][500];\nint m, tsx, tsy;\n \nint Solve(int x, int y, int mx, int my)\n{\n\trep(i, 0, mx) rep(j, 0, my) f[i][j] = 0;\n\t\n\tf[0][0] = 1;\n\trep(i, 0, mx)\n\t\trep(j, 0, my)\n\t\t{\n\t\t\tif(!i && !j) continue;\n\t\t\tif(__gcd(i + x, j + y) > 1) continue;\n\t\t\tif(!i) f[i][j] = f[i][j - 1];\n\t\t\telse if(!j) f[i][j] = f[i - 1][j];\n\t\t\telse f[i][j] = (f[i][j - 1] | f[i - 1][j]);\n\t\t}\n\tint ans = tsx * n + tsy;\n\tans = max(ans, tsx + tsy * m);\n\trep(i, 0, mx)\n\t\trep(j, 0, my)\n\t\t\tif(f[i][j]) ans = max(ans, (x + i) * tsx + (y + j) * tsy);\n\treturn ans;\n}\nvoid Main()\n{\n\tcin >> n >> m >> tsx >> tsy;\n\t\n\tif(tsx < tsy) swap(tsx, tsy), swap(n, m);\n\t\n\tif(n == 2)\n\t{\n\t\tint ans = tsx + tsy * m;\n\t\tif(m % 2) ans = tsx * 2 + tsy * m;\n\t\telse ans = max(ans, tsx * 2 + tsy * (m - 1));\n\t\tcout << ans << \"\\n\";\n\t\treturn;\n\t}\n\tif(m == 2)\n\t{\n\t\tint ans = tsx * n + tsy;\n\t\tif(n % 2) ans = tsx * n + tsy * 2;\n\t\telse ans = max(ans, tsx * (n - 1) + tsy * 2);\n\t\tcout << ans << \"\\n\";\n\t\treturn;\n\t}\n\tint l = 0, r = pr.size() - 1, ans = 0;\n\twhile(l <= r)\n\t\tif(pr[mid] <= n) ans = mid, l = mid + 1;\n\t\telse r = mid - 1;\n\t\n\t\n\tint x = pr[ans];\n\t\n\tl = 0, r = pr.size() - 1, ans = 0;\n\twhile(l <= r)\n\t\tif(pr[mid] <= m) ans = mid, l = mid + 1;\n\t\telse r = mid - 1;\n\tint y = pr[ans], yy = y;\n\t\n\tans = n * tsx + tsy;\n\tans = max(ans, m * tsy + tsx);\n\t\n\tx = max(n - 100, 1LL);\n\ty = max(m - 100, 1LL);\n\tans = max(ans, Solve(x, y, n - x, m - y));\n\tcout << ans << \"\\n\";\n}\nsigned main()\n{\n\tstd::ios::sync_with_stdio(false);\n\tcin.tie(0);\n\tcout.tie(0);\n\tsieve();\n\t\n\tint T = 1;\n\tcin >> T;\n\twhile(T --) Main();\n\treturn 0;\n} \n/*\n1\n4 5 5 4\n \n1\n9 9 9 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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 275960266,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\nbool M1;\n#include<bits/stdc++.h>\n#define look_memory cerr<<abs(&M2-&M1)/1024.0/1024<<'\\n'\n#define look_time cerr<<(clock()-Time)*1.0/CLOCKS_PER_SEC<<'\\n'\nusing namespace std;\n#define rep(i,a,b) for(int i=(a);i<=(b);i++)\n#define per(i,a,b) for(int i=(a);i>=(b);i--)\n#define pb push_back\n#define mk make_pair\n#define fi first\n#define se second\n#define ll long long\n#define pii pair<int,int>\nconst int M=2e7+5;\nll ans;\nvector<int> pr;\nbool prime[M];\nint n,m,l,f;\nvoid init(){\n\trep(i,2,2e7){\n\t\tif(!prime[i]) pr.pb(i);\n\t\tfor(auto j:pr){\n\t\t\tif(i*j>2e7) break;\n\t\t\tprime[i*j]=true;\n\t\t\tif(i%j==0) break;\n\t\t}\n\t}\n\tpr.pb(1);\n}\nint gcd(int x,int y){\n\tif(!y) return x;\n\treturn gcd(y,x%y);\n}\nvoid work(int x,int y){\n\tans=max(ans,1ll*x*l+1ll*y*f);\n\tif(x<m){\n\t\tif(gcd(x+1,y)==1) work(x+1,y);\n\t}\n\tif(y<n){\n\t\tif(gcd(x,y+1)==1) work(x,y+1);\n\t}\n}\nvoid solve(){\n\tcin>>n>>m>>l>>f;\n\tans=0;\n\tif(n<m) swap(n,m);\n\tif(n<=10&&m<=10){\n\t\twork(1,1);\n\t\tcout<<ans<<'\\n';\n\t\treturn;\n\t}\n\tint p=pr[upper_bound(pr.begin(),pr.end(),n)-pr.begin()-1];\n\tint id=upper_bound(pr.begin(),pr.end(),m)-pr.begin()-1;\n\tint q=pr[upper_bound(pr.begin(),pr.end(),m)-pr.begin()-1];\n\tif(q==p) q=pr[--id];\n\tif(2*q>n){\n\t\twork(q,p);\n\t}\n\telse{\n\t\twhile(1){\n\t\t\tbool fl=false;\n\t\t\trep(i,p+1,n){\n\t\t\t\tif(gcd(q,i)>1){\n\t\t\t\t\tfl=true;\n\t\t\t\t\tbreak;\n\t\t\t\t}\n\t\t\t}\n\t\t\tif(!fl) break;\n\t\t\tq=pr[--id];\n\t\t}\n\t\twork(q,p);\n\t}\n\tcout<<ans<<'\\n';\n}\nbool M2;\nint main(){\n\tint Time=clock();\n\tlook_memory;\n\tios::sync_with_stdio(false);\n\tcin.tie(0),cout.tie(0);\n\tinit();\n\tint t;\n\tcin>>t;\n\twhile(t--) solve();\n\tlook_time;\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 300223880,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\n//cpp\n//#define __filed_\n//#define __fasIO_\n//#define __subit_\n#define __moret_\n \n#ifndef __wht_pre_\n#define __wht_pre_\n \n#include <bits/stdc++.h>\n \nusing namespace std;\n \n#ifdef ONLINE_JUDGE\n#define __subit_\n#undef __filed_\n#endif\n \n#ifndef __w_IO__\n#define __w_IO__\n \nnamespace IO {\n #ifdef __fasIO_\n const int maxn = 1 << 20;\n char in[maxn + 1], out[maxn], *p1 = in, *p2 = in, *p3 = out;\n #define getchar() (p1 == p2 && (p2 = (p1 = in) + fread(in + 1, 1, maxn, stdin), p1 == p2) ? EOF : *( ++ p1))\n #define flush() (fwrite(out + 1, 1, p3 - out, stdout))\n #define putchar(x) (p3 == out + maxn && (flush() , p3 = out), *( ++ p3) = x)\n class Flush{public: ~Flush(){flush();}}_;\n #endif\n #define isdigit(x) ('0' <= x && x <= '9')\n \n inline void rd(char *x) {\n do *x = getchar();\n while (*x <= 32 || *x == 127);\n while (*x > 32 && *x != 127)\n * ++ x = getchar();\n *x = 0;\n }\n \n template<typename _Tp>\n inline void rd(_Tp &x) {\n x = 0;\n bool fl(false);\n char ch;\n do ch = getchar(), fl |= ch == '-'; while (!isdigit(ch));\n do x = x * 10 + (ch ^ 48), ch = getchar(); while (isdigit(ch));\n if (fl) x = -x;\n }\n \n template<>\n inline void rd(char &x) {\n do x = getchar();\n while (x <= 32 || x == 127);\n }\n \n template<>\n inline void rd(std::string &x) {\n char c;\n x.clear();\n do c = getchar();\n while (c <= 32 || c == 127);\n while (c > 32 && c != 127)\n x.push_back(c), c = getchar();\n }\n \n template<typename _Tp>\n inline void wr(_Tp x) {\n x < 0 ? x = -x, putchar('-') : 0;\n static short st[50], top(0);\n do st[ ++ top] = x % 10, x /= 10; while (x);\n while (top) putchar(st[top -- ] | 48);\n }\n \n template<>\n inline void wr(char c) {\n putchar(c);\n }\n \n template<>\n inline void wr(char* s) {\n while (*s)\n putchar(*s ++ );\n }\n \n template<>\n inline void wr(const char* s) {\n while (*s)\n putchar(*s ++ );\n }\n \n template<typename _Tp, typename ...Args>\n void rd(_Tp &&x, Args &&...args) {\n rd(x); rd(args...);\n }\n \n template<typename _Tp, typename ...Args>\n void wr(_Tp x, Args ...args) {\n wr(x); wr(args...);\n }\n \n #undef isdigit\n}\n//using namespace IO::us;\n \n#endif\n \n \n#define frer(x) freopen(x, \"r\", stdin) \n#define frew(x) freopen(x, \"w\", stdout) \n#ifdef __filed_\n#define frei(x) frer(x \".in\")\n#define freo(x) frew(x \".out\")\n#define frea(x) frew(x \".ans\")\n#define fre(x) frei(x), freo(x)\n#else\n#define frei(x) void()\n#define freo(x) void()\n#define frea(x) void()\n#define fre(x) void()\n#endif\n \n#ifdef __subit_\n#define erd(...) void()\n#define ewr(...) void()\n#else\n#define erd(...) rd(__VA_ARGS__)\n#define ewr(...) wr(__VA_ARGS__)\n#endif\n \nint Turn;\n#ifndef __moret_\n#define rdT() Turn = 1\n#else\n#define rdT() rd(Turn)\n#endif\nusing IO::rd;\nusing IO::wr;\n \n#define EXPAND5(...) EXPAND4(EXPAND4(EXPAND4(EXPAND4(__VA_ARGS__))))\n#define EXPAND4(...) EXPAND3(EXPAND3(EXPAND3(EXPAND3(__VA_ARGS__))))\n#define EXPAND3(...) EXPAND2(EXPAND2(EXPAND2(EXPAND2(__VA_ARGS__))))\n#define EXPAND2(...) EXPAND1(EXPAND1(EXPAND1(EXPAND1(__VA_ARGS__))))\n#define EXPAND1(...) __VA_ARGS__\n#define PARENS ()\n \n#endif\n \n \n#define int long long\n \nconst int N = 2e7 + 7;\n \nint pri[N];\nint vis[N];\nint low[N];\nint pnt;\n \ninline void INIT() {\n pri[0] = 1;\n for (int i = 2; i < N; ++ i) {\n if (!vis[i]) {\n ++ pnt;\n pri[pnt] = i;\n low[i] = i;\n }\n for (int j = 1; j <= pnt && i * pri[j] < N; ++ j) {\n vis[i * pri[j]] = 1;\n low[i * pri[j]] = pri[j];\n if (i % pri[j] == 0) {\n break;\n }\n }\n }\n}\n \ninline void WORK() {\n int n, m;\n int a, b;\n rd(n, m, a, b);\n if (a < b) swap(a, b), swap(n, m);\n int idx = upper_bound(pri, pri + pnt + 1, n) - pri - 1;\n int idy = upper_bound(pri, pri + pnt + 1, m) - pri - 1;\n int x, y;\n deque<pair<int, int>> q;\n set<int> se;\n int ans = 0;\n auto add = [&](int x, int y) -> void {\n ans = max(ans, x * a + y * b);\n if (!se.count(x * N + y)) {\n se.insert(x * N + y);\n q.push_back({x, y});\n }\n };\n if (idx == idy) {\n add(pri[idx - 1], pri[idy]);\n add(pri[idx], pri[idy - 1]);\n } else {\n add(pri[idx], pri[idy]);\n }\n add(n - 1, 1);\n add(1, m - 1);\n \n while (!q.empty()) {\n auto t = q.back();\n q.pop_back();\n x = t.first;\n y = t.second;\n if (x < n && __gcd(x + 1, y) == 1 && !se.count((x + 1) * N + y)) {\n add(x + 1, y);\n }\n if (y < m && __gcd(x, y + 1) == 1 && !se.count(x * N + y + 1)) {\n add(x, y + 1);\n }\n }\n wr(ans, '\\n');\n}\n \n#undef int\n \nint main() {\n fre(\"\");\n INIT();\n for (rdT(); Turn; -- Turn, ewr('\\n'))\n WORK();\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 277742343,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\nimport math\n \ndef gcd(a, b):\n while b:\n a, b = b, a % b\n return a\n \ndef initialize_dp_table(n, m, l, f):\n dp = [[0] * (m + 1) for _ in range(n + 1)]\n dp[1][0] = l\n dp[0][1] = f\n return dp\n \ndef update_dp_value(dp, i, j, l, f):\n if i > 0:\n dp[i][j] = max(dp[i][j], dp[i-1][j] + l)\n if j > 0:\n dp[i][j] = max(dp[i][j], dp[i][j-1] + f)\n \ndef fill_dp_table(dp, n, m, l, f):\n for i in range(n + 1):\n for j in range(m + 1):\n if gcd(i, j) <= 1:\n update_dp_value(dp, i, j, l, f)\n \ndef find_max_score(dp):\n return max(max(row) for row in dp)\n \ndef solve_test_case(n, m, l, f):\n dp = initialize_dp_table(n, m, l, f)\n fill_dp_table(dp, n, m, l, f)\n return find_max_score(dp)\n \nif __name__ == \"__main__\":\n t = int(input())\n for _ in range(t):\n n, m, l, f = map(int, input().split())\n result = solve_test_case(n, m, l, f)\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 296936953,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\nimport math\ngcd=math.gcd\nx=int(input())\ndef score(wl,wf,l,f):\n return (l*wl+f*wf)\nfor i in range(x):\n s=input()\n p=s.split()\n n=int(p[0])\n m=int(p[1])\n l=int(p[2])\n f=int(p[3])\n c=[[0,0],[0,1],[1,0]]\n d=[]\n for i in range(1,n+1):\n for j in range(1,m+1):\n if gcd(i,j)==0 or gcd(i,j)==1:\n if [i-1,j] in c or [i,j-1] in c:\n z=[i,j]\n c.append(z)\n for i in c:\n d.append(score(int(i[0]),int(i[-1]),l,f))\n print(max(d))\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 275830833,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\ndef gcd(num1, num2):\n if num1 == 0 or num2 == 0:\n return num1 + num2\n if num1 == num2:\n return num1\n if num1 > num2:\n return gcd(num1 - num2, num2)\n else:\n return gcd(num1, num2 - num1)\nx=int(input())\ndef score(wl,wf,l,f):\n return (l*wl+f*wf)\nfor i in range(x):\n s=input()\n p=s.split()\n n=int(p[0])\n m=int(p[1])\n l=int(p[2])\n f=int(p[3])\n c=[[0,0],[0,1],[1,0]]\n d=[]\n for i in range(1,n+1):\n for j in range(1,m+1):\n if gcd(i,j)==0 or gcd(i,j)==1:\n if [i-1,j] in c or [i,j-1] in c:\n z=[i,j]\n c.append(z)\n for i in c:\n d.append(score(int(i[0]),int(i[-1]),l,f))\n print(max(d))\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 275828917,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n private static BufferedReader br;\n private static StringTokenizer tokenizer;\n private static String next() throws IOException{\n if (br == null) br = new BufferedReader(new InputStreamReader(System.in));\n if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());\n while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());\n return tokenizer.nextToken();\n }\n private static int nextInt() throws IOException{\n return Integer.parseInt(next());\n }\n private static long nextLong() throws IOException{\n return Long.parseLong(next());\n }\n \n static StringBuilder sb = new StringBuilder();\n \n static List<Integer> primes;\n static int sup = 20_000_002;\n static int[] prePrimes = new int[sup];\n static int[] minPrimes = new int[sup];\n static int[][] direction = {{1,0},{0,1}};\n static long[][] visited;\n static int n;\n static int m;\n static long l;\n static long f;\n static Map<Long, Long> cached;\n \n static void getPrimes() {\n boolean[] visited = new boolean[sup];\n primes = new ArrayList<>();\n prePrimes[1] = prePrimes[2] = 1;\n minPrimes[1] = 1;\n for (int i = 2; i < sup; ++i) {\n if (i > 2) prePrimes[i] = primes.get(primes.size() - 1);\n if (!visited[i]) {\n minPrimes[i] = i;\n primes.add(i);\n }\n for (int p : primes) {\n if (i * p >= sup) break;\n visited[i * p] = true;\n minPrimes[i * p] = p;\n if (i % p == 0) {\n break;\n }\n }\n }\n }\n \n static long gcd(int x, int y) {\n if (x < y) {\n int t = x; x = y; y = t;\n }\n if (x % y == 0) return y;\n return gcd(y, x % y);\n }\n \n static long getKey(long x, long y) {\n return x * sup + y;\n }\n \n static long dfs(int p, int q, long l, long f, int n, int m, int flag) {\n// System.out.println(\"p:\"+p+\",q:\"+q);\n if (p > n || q > m || gcd(p, q) > 1 || p < q) return 0L;\n long key = getKey(p, q);\n if (cached.containsKey(key)) return cached.get(key);\n long ans = flag == 0 ? Math.max(l * p + f * q, l * q + f * p) : l * p + f * q;\n for (int[] d : direction) {\n ans = Math.max(ans, dfs(p + d[0], q + d[1], l, f, n, m, flag));\n }\n cached.put(key, ans);\n return ans;\n }\n \n static void slove() throws IOException{\n n = nextInt(); m = nextInt(); l = nextInt(); f = nextInt();\n // System.out.println(n + \",\" + m + \",\" + l + \",\" + f);\n if (n < m) {\n int t = n; n = m; m = t; long t2 = l; l = f; f = t2;\n }\n \n if (n == 1) {\n sb.append(l + f).append(\"\\n\");\n return;\n }\n \n int lp = prePrimes[m + 1];\n int lq = prePrimes[lp];\n if (m < 200) {\n lp = lq = 1;\n }\n cached = new HashMap<>();\n long ans = 0L;\n for (int i = lq; i <= m; ++i) ans = Math.max(ans, dfs(lp, i, l, f, m, m, 0));\n for (int i = lp; i <= m; ++i) ans = Math.max(ans, dfs(i, lq, l, f, m, m, 0));\n \n int p = prePrimes[n + 1];\n int q = m;\n while (q > 1) {\n boolean flag = true;\n int tq = q;\n while (tq > 1) {\n // System.out.println(tq);\n int mq = minPrimes[tq];\n if ((p % mq) + (n - p) >= mq) {\n flag = false;\n break;\n }\n while (tq % mq == 0) tq /= mq;\n }\n if (flag) break;;\n --q;\n }\n if (n < 200) {\n p = q = 1;\n }\n // System.out.println(\"q:\" + q + \", p:\" + p);\n cached = new HashMap<>();\n \n for (int i = q; i <= m; ++i) ans = Math.max(ans, dfs(p, i, l, f, n, m, 1));\n for (int i = p; i <= n; ++i) ans = Math.max(ans, dfs(i, q, l, f, n, m, 1));\n sb.append(ans).append(\"\\n\");\n \n }\n \n \n \n public static void main(String[] args) throws IOException {\n getPrimes();\n int tt = nextInt();\n sb = new StringBuilder();\n for (int i = 0; i < tt; ++i) slove();\n System.out.println(sb);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 277504204,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n private static BufferedReader br;\n private static StringTokenizer tokenizer;\n private static String next() throws IOException{\n if (br == null) br = new BufferedReader(new InputStreamReader(System.in));\n if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());\n while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());\n return tokenizer.nextToken();\n }\n private static int nextInt() throws IOException{\n return Integer.parseInt(next());\n }\n private static long nextLong() throws IOException{\n return Long.parseLong(next());\n }\n \n static StringBuilder sb = new StringBuilder();\n \n static List<Integer> primes;\n static int sup = 20_000_002;\n static int[] prePrimes = new int[sup];\n static int[] minPrimes = new int[sup];\n static int[][] direction = {{1,0},{0,1}};\n static long[][] visited;\n static int n;\n static int m;\n static long l;\n static long f;\n static Map<Long, Long> cached;\n \n static void getPrimes() {\n boolean[] visited = new boolean[sup];\n primes = new ArrayList<>();\n prePrimes[1] = prePrimes[2] = 1;\n minPrimes[1] = 1;\n for (int i = 2; i < sup; ++i) {\n if (i > 2) prePrimes[i] = primes.get(primes.size() - 1);\n if (!visited[i]) {\n minPrimes[i] = i;\n primes.add(i);\n }\n for (int p : primes) {\n if (i * p >= sup) break;\n visited[i * p] = true;\n minPrimes[i * p] = p;\n if (i % p == 0) {\n break;\n }\n }\n }\n }\n \n static long gcd(int x, int y) {\n if (x < y) {\n int t = x; x = y; y = t;\n }\n if (x % y == 0) return y;\n return gcd(y, x % y);\n }\n \n static long getKey(long x, long y) {\n return x * sup + y;\n }\n \n static long dfs(int p, int q, long l, long f, int n, int m, int flag) {\n if (p > n || q > m || gcd(p, q) > 1 || p < q) return 0L;\n long key = getKey(p, q);\n if (cached.containsKey(key)) return cached.get(key);\n long ans = flag == 0 ? Math.max(l * p + f * q, l * q + f * p) : l * p + f * q;\n for (int[] d : direction) {\n ans = Math.max(ans, dfs(p + d[0], q + d[1], l, f, n, m, flag));\n }\n cached.put(key, ans);\n return ans;\n }\n \n static void slove() throws IOException{\n n = nextInt(); m = nextInt(); l = nextInt(); f = nextInt();\n // System.out.println(n + \",\" + m + \",\" + l + \",\" + f);\n if (n < m) {\n int t = n; n = m; m = t; long t2 = l; l = f; f = t2;\n }\n \n if (n == 1) {\n sb.append(l + f).append(\"\\n\");\n return;\n }\n \n \n int lp = prePrimes[m + 1];\n int lq = prePrimes[lp];\n if (m < 200) {\n lp = lq = 1;\n }\n cached = new HashMap<>();\n long ans = 0L;\n for (int i = lq; i <= m; ++i) ans = Math.max(ans, dfs(lp, i, l, f, m, m, 0));\n for (int i = lp; i <= m; ++i) ans = Math.max(ans, dfs(i, lq, l, f, m, m, 0));\n \n int p = prePrimes[n + 1];\n int q = m;\n while (q > 1) {\n boolean flag = true;\n int tq = q;\n while (tq > 1) {\n // System.out.println(tq);\n int mq = minPrimes[tq];\n if ((p % mq) + (n - p) >= mq) {\n flag = false;\n break;\n }\n if (!flag) break;\n while (tq % mq == 0) tq /= mq;\n }\n \n --q;\n }\n if (n < 200) {\n p = q = 1;\n }\n // System.out.println(\"q:\" + q + \", p:\" + p);\n cached = new HashMap<>();\n \n for (int i = q; i <= m; ++i) ans = Math.max(ans, dfs(p, i, l, f, n, m, 1));\n for (int i = p; i <= n; ++i) ans = Math.max(ans, dfs(i, q, l, f, n, m, 1));\n sb.append(ans).append(\"\\n\");\n \n }\n \n \n \n public static void main(String[] args) throws IOException {\n getPrimes();\n int tt = nextInt();\n sb = new StringBuilder();\n for (int i = 0; i < tt; ++i) slove();\n System.out.println(sb);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 277502443,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nusing ll=long long;\nconst int N=2e7+5;\nbool isp[N];\nvector<int> p={1};\nauto init=[](){\n\tfor(int i=2;i<N;i++){\n\t\tif(!isp[i]) p.push_back(i);\n\t\tfor(auto v:p){\n\t\t\tif(v==1) continue;\n\t\t\tif(i*v>=N) break;\n\t\t\tisp[i*v]=true;\n\t\t\tif(i%v==0) break;\n\t\t}\n\t}\n\treturn 0;\n}();\nauto solve(int n,int m,int l,int f){\n\tauto find=[&](int n){\n\t\tauto it=lower_bound(begin(p),end(p),n);\n\t\tif(it!=p.begin()) it=prev(it);\n\t\treturn *it;\n\t};\n\tint x=find(n);\n\tm=min(m,n);\n\tint y=find(min(x,m));\n\tif(n<50) x=1;\n\tif(m<50) y=1;\n\tint p=n-x,q=m-y;\n\tvector dp(p+1,vector<int>(q+1));\n\tdp[0][0]=1;\n\tll ans=0;\n\tfor(int i=0;i<=p;i++){\n\t\tfor(int j=0;j<=q;j++){\n\t\t\tif(!dp[i][j]) continue;\n\t\t\tint s=i+x,t=j+y;\n\t\t\tif(s+1<=n&&__gcd(s+1,t)==1) dp[i+1][j]=1;\n\t\t\tif(t+1<=m&&__gcd(s,t+1)==1) dp[i][j+1]=1;\n\t\t\tans=max(ans,1ll*l*s+1ll*f*t);\n\t\t}\n\t}\n\treturn ans;\n}\nauto solve2(int n,int m,int l,int f){\n\tif(m>=n) return 0ll;\n\tauto find=[&](int n){\n\t\tauto it=lower_bound(begin(p),end(p),n);\n\t\tif(it!=p.begin()) it=prev(it);\n\t\treturn *it;\n\t};\n\tint y=find(m);\n\tn=min(n,2*y);\n\tint x=find(n);\n\ty=find(min(x,m));\n\tif(n<50) x=1;\n\tif(m<50) y=1;\n\tif(x<=y) return 0ll;\n\tint p=n-x,q=m-y;\n\tvector dp(p+1,vector<int>(q+1));\n\tdp[0][0]=1;\n\tll ans=0;\n\tfor(int i=0;i<=p;i++){\n\t\tfor(int j=0;j<=q;j++){\n\t\t\tif(!dp[i][j]) continue;\n\t\t\tint s=i+x,t=j+y;\n\t\t\tif(s+1<=n&&__gcd(s+1,t)==1) dp[i+1][j]=1;\n\t\t\tif(t+1<=m&&__gcd(s,t+1)==1) dp[i][j+1]=1;\n\t\t\tans=max(ans,1ll*l*s+1ll*f*t);\n\t\t}\n\t}\n\treturn ans;\n}\nauto solve(){\n\tint n,m,l,f;\n\tcin>>n>>m>>l>>f;\n\treturn max({\n\t\tsolve(n,m,l,f),solve(m,n,f,l),\n\t\tsolve2(n,m,l,f),solve2(m,n,f,l)});\n}\nint main(){\n ios::sync_with_stdio(false);\n cin.tie(0);\n\tint t=1;\n\tcin>>t;\n\twhile(t--){\n\t\t// solve();\n\t\tcout<<solve()<<'\\n';\n\t\t// cout<<(solve()?\"YES\":\"NO\")<<'\\n';\t\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 275827478,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nconst int N = 2e7 + 1000;\nint fl[N + 10], p[N], tot;\nlong long ans;\nint G[1100][1100], FL, L, F;\nvoid calc(int x, int y, int Lx, int Rx, int Ly, int Ry) {\n\t// if(Ly > R)return;\n\tif(__gcd(x, y) > 1)return;\n\tif(G[x - Lx][y - Ly] == FL)return;\n\tG[x - Lx][y - Ly] = FL;\n\tans = max(ans, 1ll * x * L + 1ll * y * F);\n\tif(x < y && x < Rx) {\n\t\tcalc(x + 1, y, Lx, Rx, Ly, Ry);\n\t}\n\tif(y < Ry) {\n\t\tcalc(x, y + 1, Lx, Rx, Ly, Ry);\n\t}\n}\nvoid work(int n, int m, int l, int f) {\n\tL = l;\n\tF = f;\n\tif(m <= 200) {\n\t\tFL++;\n\t\tcalc(1, 1, 1, n, 1, m);\n\t\treturn;\n\t}\n\tint t = upper_bound(p + 1, p + tot + 1, m) - p - 1;\n\tint q = upper_bound(p + 1, p + tot + 1, min(p[t] - 1, n)) - p - 1;\n\tfor(int i = q; i >= 1 && i >= q - 10; i--) {\n\t\tFL++;\n\t\tcalc(p[i], p[t], p[i], n, p[t], m);\n\t}\n}\nint main() {\n\tint T;\n\tfor(int i = 2; i <= N; i++) {\n\t\tif(!fl[i])p[++tot] = i;\n\t\telse continue;\n\t\tfor(int j = i; 1ll * i * j <= N; j++)\n\t\t\tfl[i * j] = 1;\n\t}\n\t// cout<<p[m]<<endl;\n\tscanf(\"%d\", &T);\n\twhile(T--) {\n\t\tint n, m, l, f;\n\t\tscanf(\"%d%d%d%d\", &n, &m, &l, &f);\n\t\tans = 0;\n\t\twork(n, m, l, f);\n\t\twork(m, n, f, l);\n\t\tprintf(\"%lld\\n\", ans);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 279781422,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n private static BufferedReader br;\n private static StringTokenizer tokenizer;\n private static String next() throws IOException{\n if (br == null) br = new BufferedReader(new InputStreamReader(System.in));\n if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());\n while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());\n return tokenizer.nextToken();\n }\n private static int nextInt() throws IOException{\n return Integer.parseInt(next());\n }\n private static long nextLong() throws IOException{\n return Long.parseLong(next());\n }\n \n static StringBuilder sb = new StringBuilder();\n \n static List<Integer> primes;\n static int sup = 20_000_002;\n static int[] prePrimes = new int[sup];\n static int[] minPrimes = new int[sup];\n static int[][] direction = {{1,0},{0,1}};\n static long[][] visited;\n static int n;\n static int m;\n static long l;\n static long f;\n static Map<Long, Long> cached;\n \n static void getPrimes() {\n boolean[] visited = new boolean[sup];\n primes = new ArrayList<>();\n prePrimes[1] = prePrimes[2] = 1;\n minPrimes[1] = 1;\n for (int i = 2; i < sup; ++i) {\n if (i > 2) prePrimes[i] = primes.get(primes.size() - 1);\n if (!visited[i]) {\n minPrimes[i] = i;\n primes.add(i);\n }\n for (int p : primes) {\n if (i * p >= sup) break;\n visited[i * p] = true;\n minPrimes[i * p] = p;\n if (i % p == 0) {\n break;\n }\n }\n }\n }\n \n static long gcd(int x, int y) {\n if (x < y) {\n int t = x; x = y; y = t;\n }\n if (x % y == 0) return y;\n return gcd(y, x % y);\n }\n \n static long getKey(long x, long y) {\n return x * sup + y;\n }\n \n static long dfs(int p, int q, long l, long f, int n, int m, int flag) {\n if (p > n || q > m || gcd(p, q) > 1 || p < q) return 0L;\n long key = getKey(p, q);\n if (cached.containsKey(key)) return cached.get(key);\n long ans = flag == 0 ? Math.max(l * p + f * q, l * q + f * p) : l * p + f * q;\n for (int[] d : direction) {\n ans = Math.max(ans, dfs(p + d[0], q + d[1], l, f, n, m, flag));\n }\n cached.put(key, ans);\n return ans;\n }\n \n static void slove() throws IOException{\n n = nextInt(); m = nextInt(); l = nextInt(); f = nextInt();\n // System.out.println(n + \",\" + m + \",\" + l + \",\" + f);\n if (n < m) {\n int t = n; n = m; m = t; long t2 = l; l = f; f = t2;\n }\n \n if (n == 1) {\n sb.append(l + f).append(\"\\n\");\n return;\n }\n \n \n int lp = prePrimes[m + 1];\n int lq = prePrimes[lp];\n if (m < 100) {\n lp = lq = 1;\n }\n cached = new HashMap<>();\n long ans = 0L;\n for (int i = lq; i <= m; ++i) ans = Math.max(ans, dfs(lp, i, l, f, m, m, 0));\n for (int i = lp; i <= m; ++i) ans = Math.max(ans, dfs(i, lq, l, f, m, m, 0));\n \n int p = prePrimes[n + 1];\n int q = m;\n while (q > 1) {\n boolean flag = true;\n int tq = q;\n while (tq > 1) {\n // System.out.println(tq);\n int mq = minPrimes[tq];\n if ((p % mq) + (n - p) >= mq) {\n flag = false;\n break;\n }\n while (tq % mq == 0) tq /= mq;\n }\n if (!flag) break;\n --q;\n }\n if (n < 100) {\n p = q = 1;\n }\n // System.out.println(\"q:\" + q);\n cached = new HashMap<>();\n \n for (int i = q; i <= m; ++i) ans = Math.max(ans, dfs(p, i, l, f, n, m, 1));\n for (int i = p; i <= n; ++i) ans = Math.max(ans, dfs(i, q, l, f, n, m, 1));\n sb.append(ans).append(\"\\n\");\n \n }\n \n \n \n public static void main(String[] args) throws IOException {\n getPrimes();\n int tt = nextInt();\n sb = new StringBuilder();\n for (int i = 0; i < tt; ++i) slove();\n System.out.println(sb);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 277500075,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class F {\n\t\n\tstatic IOHandler sc = new IOHandler();\n\tstatic PrintWriter out = new PrintWriter(System.out);\n\t\n\tstatic TreeSet<Integer> set;\n\t\n\tstatic List<Integer> [] factors;\n\t\n\tstatic long [][] memo;\n\tstatic boolean [][] visited;\n\t\n\tpublic static void main(String[] args) throws IOException {\n\t\t\n new Thread(null, new Solution(), \"solution\", 1 << 26 ).start();\n }\n\t\n\tprivate static class Solution implements Runnable {\n \n\t\tpublic void run() {\n\t\t\t// TODO Auto-generated method stub\n\t\t\tinit();\n\t\t\tint testCases = sc.nextInt();\n\t\t\t\n\t\t\tfor (int i = 1; i <= testCases; ++i) {\n\t\t\t\tsolve(i);\n\t\t\t}\n\t\t\t\n\t\t\tout.flush();\n\t\t}\n\t\t\n\t\t\n\t\tprivate void advice() {\n\t\t\t// Read question and Test cases thoroughly\n\t\t\t// Check for overflow , Avoid Integer == \n\t\t\t// Handle invalid cases\n\t\t\t// Think before solving\n\t\t\t// Search for concepts\n\t\t\t// expected alg like seg tree, trie, binSearch, sqrt decom ... \n\t\t\t\n\t\t\t// Source of Error\n\t\t\t// Int overflow, Not returning after special case\n\t\t\t// Not taking mod properly\n\t\t\t// sorting input groups after reading certain values\n\t\t\t// Not using the right vairable eg idx in method for i in loop\n\t\t\t// Accounting for special case in general logic. eg special in dp\n\t\t\t// for BIT. Do not use tree maps with update query\n\t\t\t// Assuming Constraints on an unbound variable\n\t\t\t// Not properly calculating time complexity\n\t\t}\n\t\t\n\t\tprivate void init() {\n\t\t\t\n\t\t\tboolean [] primes = new boolean [20_000_001];\n\t\t\tArrays.fill(primes, true);\n\t\t\t\n\t\t\tmemo = new long [400][400];\n\t\t\tvisited = new boolean [400][400];\n\t\t\t\n\t\t\tset = new TreeSet<>();\n\t\t\t\n\t\t\tset.add(1);\n\t\t\t\n\t\t\tfor (int i = 2; i < primes.length; ++i) {\n\t\t\t\tif (!primes[i])\n\t\t\t\t\tcontinue;\n\t\t\t\t\n\t\t\t\tfor (int j = 2*i; j < primes.length; j += i) {\n\t\t\t\t\tprimes[j] = false;\n\t\t\t\t}\n\t\t\t}\n\t\t\t\n\t\t\t\n\t\t\tfor (int i = 2; i < primes.length; ++i) {\n\t\t\t\tif (primes[i]) {\n\t\t\t\t\tset.add(i);\n\t\t\t\t}\n\t\t\t}\n\t\t\t\n\t\t}\n\t\t\n\t\t\n\t\tprivate void solve(int t) {\n\t\t\t\n\t\t\t\n\t\t\tint n = sc.nextInt();\n\t\t\tint m = sc.nextInt();\n\t\t\t\n\t\t\t\n\t\t\tlong l = sc.nextInt();\n\t\t\tlong f = sc.nextInt();\n\t\t\t\n \n\t\t\tint back = 200;\n\t\t\t\n\t\t\tint base1 = set.floor(n);\n\t\t\tint start1 = set.floor(base1 - 1);\n\t\t\t\n\t\t\tstart1 = Math.min(start1, base1 - back);\n\t\t\tstart1 = Math.max(1, start1);\n\t\t\t\n\t\t\tint base2 = set.floor(m);\n\t\t\tint start2 = set.floor(base2 - 1);\n\t\t\t\n\t\t\tstart2 = Math.min(start2, base2 - back);\n\t\t\tstart2 = Math.max(1, start2);\n\t\t\t\n\t\t\t// int start = 1;\n\t\t\t\n\t\t\t\n\t\t\tint max1 = n - start1 + 2;\n\t\t\tint max2 = m - start2 + 2;\n\t\t\t\n\t\t\t\n\t\t\t\n\t\t\tfor (int i = 0; i < max1; ++i) {\n\t\t\t\tArrays.fill(visited[i], false);\n\t\t\t}\n\t\t\t\n\t\t\t\n\t\t\tlong result1 = solve(start1, base2, n, m, l, f , 0, base2 - start2);\n\t\t\tlong result2 = solve(base1, start2, n, m, l, f , base1 - start1, 0);\n\t\t\t\n\t\t\tout.println(Math.max(result1, result2));\n\t\t}\n\t\t\n\t\tprivate int gcd(int a, int b) {\n\t\t\tif (a % b == 0)\n\t\t\t\treturn b;\n\t\t\t\n\t\t\treturn gcd(b, a % b);\n\t\t}\n\t\t\n\t\tprivate long solve(int wL, int wF, int n, int m,\n\t\t\t\tlong l, long f, int la, int fa) {\n\t\t\t\n\t\t\tif (wL > n || wF > m)\n\t\t\t\treturn 0;\n\t\t\telse if (visited[la][fa])\n\t\t\t\treturn memo[la][fa];\n\t\t\t\n\t\t\tint diff = Math.abs(la - fa);\n\t\t\t\n\t\t\tif (diff == 0) {\n\t\t\t\tvisited[la][fa] = true;\n\t\t\t\tmemo[la][fa] = 0;\n\t\t\t\treturn 0;\n\t\t\t} else if (gcd(wL, wF) > 1) {\n\t\t\t\tvisited[la][fa] = true;\n\t\t\t\tmemo[la][fa] = 0;\n\t\t\t\treturn 0;\n\t\t\t}\n\t\t\t\n\t\t\tlong v1 = l;\n\t\t\tv1 *= wL;\n\t\t\t\n\t\t\tlong v2 = f;\n\t\t\tv2 *= wF;\n\t\t\t\n\t\t\tlong res = v1 + v2;\n\t\t\t\n\t\t\tres = Math.max(res, solve(wL + 1, wF, n, m, l, f, la + 1, fa));\n\t\t\tres = Math.max(res, solve(wL, wF + 1, n, m, l, f, la, fa + 1));\n\t\t\t\n\t\t\t\n\t\t\t\n\t\t\tvisited[la][fa] = true;\n\t\t\tmemo[la][fa] = res;\n\t\t\treturn res;\n\t\t}\n\t\t\n\t}\n\t\n\tprivate static class IOHandler {\n BufferedReader br;\n StringTokenizer st;\n \n public IOHandler() {\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 int [] readArray(int n) {\n \t int [] res = new int [n];\n \t \n \t for (int i = 0; i < n; ++i)\n \t\t res[i] = nextInt();\n \t \n \t return res;\n }\n \n int [] readArray2(int n) {\n \t int [] res = new int [n + 1];\n \t \n \t for (int i = 1; i <= n; ++i)\n \t\t res[i] = nextInt();\n \t \n \t return res;\n }\n \n int [][] readGraph(int n, int m, int c) {\n \t int [][] graph = new int [n + 1][];\n \t \n \t int [] count = new int [n + 1];\n \t int [][] arr = new int [m][1 + c];\n \t \n \t int a, b;\n \t \n \t for (int i = 0; i < m; ++i) {\n \t\t a = sc.nextInt();\n \t\t b = sc.nextInt();\n \t\t \n \t\t arr[i][0] = a;\n \t\t arr[i][1] = b;\n \t\t \n \t\t ++count[a];\n \t\t ++count[b];\n \t\t \n \t\t for (int j = 2; j <= c; ++j) {\n \t\t\t arr[i][j] = sc.nextInt();\n \t\t }\n \t }\n \t \n \t for (int i = 0; i <= n; ++i) {\n \t\t graph[i] = new int [count[i] * c];\n \t\t count[i] = 0;\n \t }\n \t int swap;\n \t \n \t for (int [] tArr : arr) {\n \t\t for (int j = 1; j < tArr.length; ++j) {\n \t\t\t graph[tArr[0]][count[tArr[0]]++] = tArr[j]; \n \t\t }\n \t\t \n \t\t swap = tArr[0];\n \t\t tArr[0] = tArr[1];\n \t\t tArr[1] = swap;\n \t\t \n \t\t for (int j = 1; j < tArr.length; ++j) {\n \t\t\t graph[tArr[0]][count[tArr[0]]++] = tArr[j]; \n \t\t }\n \t }\n \t \n \t return graph;\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine(){\n String str = \"\";\n\t\t try {\n\t\t str = br.readLine();\n\t\t } catch (IOException e) {\n\t\t e.printStackTrace();\n\t\t }\n\t\t return str;\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 275863887,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define endl '\\n'\n \nusing namespace std;\n \nbool zadr[20000007]={1, 1};\nint dp[105][105];\nvector<int> primes;\nint queries[4][3005];\nint bst[20000007];\n \nqueue<pair<int, int>> pos;\n \nbool slzh[3005][3005];\n \nvoid dfs(int a, int b, int c, int d)\n{\n int kr=a, kr2=b;\n pos.push({a, b});\n for(int i=b; i<=d; i++) bst[i]=0;\n while(!pos.empty())\n {\n int a=pos.front().first, b=pos.front().second;\n slzh[a-kr][b-kr2]=0;\n pos.pop();\n if(b>d || a>c) continue;\n if(bst[b]<a) bst[b]=a;\n if(__gcd(a, b+1)==1 && !slzh[a-kr][b+1-kr2]) \n {\n pos.push({a, b+1});\n slzh[a-kr][b+1-kr2]=1;\n }\n if(__gcd(a+1, b)==1 && !slzh[a+1-kr][b-kr2])\n {\n pos.push({a+1, b});\n slzh[a+1-kr][b-kr2]=1;\n }\n }\n ///if(br>500) cout<<kr<<\" \"<<kr2<<\" \"<<br<<endl;\n}\n \nvoid precalc()\n{\n for(int i=1; i<=100; i++) {dp[1][i]=dp[i][1]=1; bst[i]=1;}\n for(int j=2; j<=100; j++)\n {\n for(int i=2; i<=100; i++)\n {\n if(__gcd(i, j)>1) continue;\n dp[i][j]=bool(dp[i-1][j]||dp[i][j-1]);\n if(j<i && dp[i][j]) bst[i]=max(bst[i], j);\n }\n }\n for(int i=2; i*i<=20000000; i++)\n {\n if(zadr[i]) continue;\n for(int j=i*i; j<=20000000; j+=i) zadr[j]=1;\n }\n for(int i=2; i<=20000000; i++)\n {\n if(!zadr[i]) primes.push_back(i);\n }\n primes.push_back(20000002);\n}\n \n \n \nvoid solve()\n{\n long long n, m, l, f;\n cin>>n>>m>>l>>f;\n if(n<m) {swap(n, m); swap(l, f);}\n for(int i=max(0, (int)n-7000); i<=n; i++) bst[i]=0;\n for(int i=max(0, (int)m-7000); i<=m; i++) bst[i]=0;\n long long ans=0;\n if(n>100)\n {\n int id1, id2;\n for(id1=2; id1<primes.size(); id1++)\n {\n if(primes[id1]>n)\n {\n id1--;\n break;\n }\n }\n for(id2=2; id2<id1; id2++)\n {\n if(primes[id2]>m)\n {\n id2--;\n break;\n }\n }\n if(id1==id2) id2--;\n dfs(primes[id2], primes[id1], m, n);\n if(primes[id2]<=m) primes[id2]++;\n dfs(primes[id2-1], primes[id2], m, m);\n }\n for(int i=1; i<=min((int)n, 100); i++)\n {\n bst[i]=0;\n for(int j=1; j<min(i, (int)m+1); j++)\n {\n if(dp[i][j]) bst[i]=j;\n } \n }\n for(int i=0; n>=0 && i<7000; i++, n--)\n {\n ///cout<<bst[i]<<endl;\n ///cout<<n<<\" \"<<bst[n]<<endl;\n ans=max(ans, n*l+bst[n]*f);\n }\n for(int i=0; m>=0 && i<7000; i++, m--)\n {\n ///cout<<bst[i]<<endl;\n ///cout<<n<<\" \"<<bst[n]<<endl;\n ans=max(ans, m*l+bst[m]*f);\n ans=max(ans, m*f+bst[m]*l);\n }\n cout<<ans<<endl;\n}\n \nint main()\n{\n ios::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n \n precalc();\n long long tt;\n cin>>tt;\n while(tt--) solve();\n \n return 0;\n}\n/**\n1\n5 4\n1 1 3 4\n2 3 4 5 1\n5 1\n4 5\n3 4\n2 3\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 275901758,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define ll long long\n#define pii pair<int,int>\n#define lowbit(x) (x&-x)\n#define log(x) (31^__builtin_clz(x))\nusing namespace std;\nconst int maxn=2e7+5;\nbool notpri[maxn];\nvector<int>pri;\nvoid init(){\n\tfor(int i=2;i<=2e7;i++){\n\t\tif(!notpri[i]){\n\t\t\tpri.push_back(i);\n\t\t}\n\t\tfor(int num:pri){\n\t\t\tif((ll)i*num>2e7)break;\n\t\t\tnotpri[i*num]=1;\n\t\t\tif(i%num==0){\n\t\t\t\tbreak;\n\t\t\t}\n\t\t}\n\t}\n}\nll get(ll n){\n\tif(n<=2)return 0;\n\tint l=0,r=pri.size()-1;\n\twhile(l<r){\n\t\tint mid=l+r+1>>1;\n\t\tif(pri[mid]<=n)l=mid;\n\t\telse r=mid-1;\n\t}\n\treturn pri[l];\n}\nll n,m,x,y;\nll ans;\nvoid solve1(){\n\tvector<vector<int>>dp(n+5,vector<int>(m+5,0));\n\tdp[0][0]=1;\n\tfor(ll i=0;i<=n;i++){\n\t\tfor(ll j=0;j<=m;j++){\n\t\t\tif(i==0&&j==0||__gcd(i,j)>1)continue;\n\t\t\tif(i&&dp[i-1][j])dp[i][j]=1;\n\t\t\tif(j&&dp[i][j-1])dp[i][j]=1;\n\t\t\tif(dp[i][j])ans=max(ans,i*x+j*y);\n\t\t}\n\t}\n\tcout<<ans<<\"\\n\";\n}\nvoid solve2(ll a,ll b){\n\tll rn=min(n,a+250),rm=min(m,b+250);\n\tvector<vector<int>>dp(rn-a+5,vector<int>(rm-b+5,0));\n\tdp[0][0]=1;\n\tfor(ll i=a;i<=rn;i++){\n\t\tfor(ll j=b;j<=rm;j++){\n\t\t\tif((i==a&&j==b)||__gcd(i,j)>1)continue;\n\t\t\tif(i>a&&dp[i-a-1][j-b])dp[i-a][j-b]=1;\n\t\t\tif(j>b&&dp[i-a][j-b-1])dp[i-a][j-b]=1;\n\t\t\tif(dp[i-a][j-b])ans=max(ans,i*x+j*y);\n\t\t}\n\t}\n}\nvoid solve(){\n\tans=0;\n\tcin>>n>>m>>x>>y;\n\tif(n<=200&&m<=200){\n\t\tsolve1();\n\t\treturn;\n\t}\n\tsolve2(0,get(m));\n\tsolve2(get(n),0);\n\tsolve2(0,0);\n\tll a=get(n),b=get(m);\n\tif(a!=b)solve2(a,b);\n\telse{\n\t\tsolve2(a,get(a-1));\n\t\tsolve2(get(a-1),a);\n\t}\n\tcout<<ans<<\"\\n\";\n}\nsigned main(){\n ios::sync_with_stdio(0), cin.tie(0);\n int t=1;\n\tinit();\n cin>>t;\n while(t--)solve();\n return 0;\n}\n/*\n17051887 17051899\n1\n17051898\n17051898\n1\n1\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 275833098,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define rep(i,j,k) for(int i=j;i<=(k);++i)\n#define drp(i,j,k) for(int i=j;i>=(k);--i)\n#define isdigit(ch) (ch>=48&&ch<=57)\n#define mp std::make_pair\n#define mod 1000000007\n#define MAXN 20000005\ntypedef long long ll;\ntypedef long double ld;\ntypedef unsigned long long ull;\ntypedef std::pair<int,int> pii;\ntypedef std::pair<ll,ll> pll;\ninline int read(){\n\tint x=0;\n\tbool sgn=true;\n\tchar ch=getchar();\n\twhile(!isdigit(ch)) sgn^=ch=='-',ch=getchar();\n\twhile(isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=getchar();\n\treturn sgn?x:-x;\n}\ninline ll readll(){\n\tll x=0;\n\tbool sgn=true;\n\tchar ch=getchar();\n\twhile(!isdigit(ch)) sgn^=ch=='-',ch=getchar();\n\twhile(isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=getchar();\n\treturn sgn?x:-x;\n}\ntemplate <typename tp> inline tp min(tp x,tp y){return x<y?x:y;}\ntemplate <typename tp> inline tp max(tp x,tp y){return x>y?x:y;}\ntemplate <typename tp> inline bool chkmin(tp &x,tp y){return x>y&&(x=y,true);}\ntemplate <typename tp> inline bool chkmax(tp &x,tp y){return x<y&&(x=y,true);}\nint n,m,x,y,rk,pri[MAXN];\nbool vis[MAXN],state[305][305];\ninline void preprocess(){\n\trep(i,2,20000000){\n\t\tif(vis[i]) continue;\n\t\tpri[++rk]=i;\n\t\tfor(int j=i<<1;j<=20000000;j+=i) vis[j]=true;\n\t}\n}\ninline ll solve(int u,int v,int x,int y){\n\tif(state[n-u][m-v]) return 0;\n\tstate[n-u][m-v]=true;\n\tll ret=1ll*u*x+1ll*v*y;\n\tif(u<n&&std::__gcd(u+1,v)==1) chkmax(ret,solve(u+1,v,x,y));\n\tif(v<m&&std::__gcd(u,v+1)==1) chkmax(ret,solve(u,v+1,x,y));\n\treturn ret;\n}\nint main(){\n\tpreprocess();\n\tdrp(task,read(),1){\n\t\tn=read(),m=read(),x=read(),y=read();\n\t\tif(n>m) std::swap(n,m),std::swap(x,y);\n\t\tmemset(state,0,sizeof(state));\n\t\tif(n<=300&&m<=300) printf(\"%lld\\n\",max(solve(1,2,x,y),solve(2,1,x,y)));\n\t\telse if(n<=300){\n\t\t\tint p=std::upper_bound(pri+1,pri+rk+1,m)-pri-1;\n\t\t\tprintf(\"%lld\\n\",solve(1,pri[p],x,y));\n\t\t}\n\t\telse{\n\t\t\tint p1=std::upper_bound(pri+1,pri+rk+1,n)-pri-1;\n\t\t\tint p2=std::upper_bound(pri+1,pri+rk+1,m)-pri-1;\n\t\t\tif(p1==p2){\n\t\t\t\tll ans1=solve(pri[p1-1],pri[p2],x,y);\n\t\t memset(state,0,sizeof(state));\n\t\t\t\tll ans2=solve(pri[p1],pri[p2-1],x,y);\n\t\t\t\tprintf(\"%lld\\n\",max(ans1,ans2));\n\t\t\t}\n\t\t\telse{\n\t\t\t\tll ans1=solve(pri[p1],pri[p2],x,y);\n\t\t memset(state,0,sizeof(state));\n\t\t\t\tll ans2=solve(pri[p1-1],pri[p2],x,y);\n\t\t memset(state,0,sizeof(state));\n\t\t\t\tll ans3=solve(pri[p1-2],pri[p2],x,y);\n\t\t memset(state,0,sizeof(state));\n\t\t\t\tll ans4=solve(pri[p1-3],pri[p2],x,y);\n\t\t\t\tprintf(\"%lld\\n\",max(max(ans1,ans2),max(ans3,ans4)));\n\t\t\t}\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 275841361,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n//#define int long long\ntypedef long long ll;\ntypedef long double ld;\n#define pb push_back\n#define ar(x) array<int, x>\n#define all(x) begin(x),end(x)\n#define rall(x) rbegin(x),rend(x)\nconst ll MAGIC[7] = {2ll, 325ll, 9375ll, 28178ll, 450775ll, 9780504ll, 1795265022ll};\n \nll bpow(__int128 a, ll x, ll mod) {\n a %= mod;\n __int128 ans = 1;\n while (x) {\n if (x % 2) {\n ans *= a;\n ans %= mod;\n }\n a *= a;\n a %= mod;\n x /= 2;\n }\n return (ll) ans;\n}\n \nbool is_prime(int n) {\n if (n == 1) return false;\n if (n <= 3) return true;\n if (n % 2 == 0 || n % 3 == 0) return false;\n ll s = __builtin_ctzll(n - 1), d = n >> s; // $n-1 = 2^s \\cdot d$\n for (auto a: MAGIC) {\n if (a % n == 0) {\n continue;\n }\n ll x = bpow(a, d, n);\n for (int _ = 0; _ < s; _++) {\n ll y = bpow(x, 2, n);\n if (y == 1 && x != 1 && x != n - 1) {\n return false;\n }\n x = y;\n }\n if (x != 1) {\n return false;\n }\n }\n return true;\n}\n \nconst int D = 180;\nint used[D + D][D + D];\nint tmm = 1;\n \nvoid solve() {\n tmm++;\n int n, m, l, f;\n cin >> n >> m >> l >> f;\n if (m - n > D) {\n int p = m;\n while (!is_prime(p))p--;\n int q = max(1, n - D);\n ll ans = 0;\n vector<ar(2) > s = {{q, p}};\n used[0][0] = tmm;\n while (!s.empty()) {\n auto [a,b] = s.back();\n s.pop_back();\n ans = max(ans, (ll) l * (ll) a + (ll) f * (ll) b);\n if (a + 1 <= n && __gcd(a + 1, b) == 1 && used[a + 1 - q][b - p] < tmm) {\n s.pb({a + 1, b});\n used[a + 1 - q][b - p] = tmm;\n }\n if (b + 1 <= m && __gcd(a, b + 1) == 1 && used[a - q][b + 1 - p] < tmm) {\n s.pb({a, b + 1});\n used[a - q][b + 1 - p] = tmm;\n }\n }\n cout << ans << '\\n';\n return;\n }\n int p = n;\n while (!is_prime(p))p--;\n int q = max(1, n - D);\n ll ans = 0;\n vector<ar(2) > s = {{q, p}, {p, q}};\n used[0][p - q] = used[p - q][0] = tmm;\n while (!s.empty()) {\n auto [a,b] = s.back();\n s.pop_back();\n ans = max(ans, (ll) l * (ll) a + (ll) f * (ll) b);\n if (a + 1 <= n && __gcd(a + 1, b) == 1 && used[a + 1 - q][b - q] < tmm) {\n s.pb({a + 1, b});\n used[a + 1 - q][b - q] = tmm;\n }\n if (b + 1 <= m && __gcd(a, b + 1) == 1 && used[a - q][b + 1 - q] < tmm) {\n s.pb({a, b + 1});\n used[a - q][b + 1 - q] = tmm;\n }\n }\n cout << ans << '\\n';\n}\n \nsigned main() {\n ios_base::sync_with_stdio(false);\n cin.tie(nullptr);\n int t;\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 275823928,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\n#https://codeforces.com/contest/2002/problem/F2?locale=en\n \nprimes = [1]\ndef sieve_of_eratosthenes(n):\n global primes\n a = [True] * (n + 1)\n \n i = 2\n while i * i <= n:\n if a[i] == True:\n primes.append(i)\n for j in range(i, n+1, i):\n a[j] = False\n i += 1\n \n while i <= n:\n if (a[i] == True):\n primes.append(i)\n i += 1\n \ndef find_the_nearest_prime_index(n):\n global primes\n l = 0\n r = len(primes) - 1\n while True:\n if (primes[r] == n):\n return r\n if (primes[l] == n):\n return l\n if r - l == 1:\n return l\n mid = (r + l) // 2\n if (primes[mid] < n):\n l = mid\n else:\n r = mid\n \ndef gcd(a, b):\n if b == 0:\n return a\n return gcd(b, a % b)\n \ndef depth_search(a, b, n, m, l, f):\n ###a < b <= n\n ans = a * l + b * f\n \n if a + 1 <= n and gcd(a + 1, b) == 1:\n ans = max(ans, depth_search(a+1, b, n, m, l, f))\n if b + 1 <= m and gcd(a, b + 1) == 1:\n ans = max(ans, depth_search(a, b + 1, n, m, l, f))\n \n return ans\n \ndef find_less_prime(n, m, big_prime):\n p_n = find_the_nearest_prime_index(n)\n while primes[p_n] != 1:\n flag = True\n for i in range(big_prime, m + 1):\n if gcd(primes[p_n], i) != 1:\n flag = False\n break\n \n if flag:\n return primes[p_n]\n p_n -= 1\n return 1\n \ndef dp(n, m, l, f):\n p_m = find_the_nearest_prime_index(m)\n big_prime = primes[p_m]\n less_prime = find_less_prime(n, m, big_prime)\n arr = []\n ans = 0\n for i in range(m - big_prime + 1):\n arr.append([])\n for j in range(n - less_prime + 1):\n arr[i].append(False)\n \n for i in range(0, m - big_prime + 1):\n arr[i][0] = True\n if (less_prime) * l + (i + big_prime) * f > ans:\n ans = (less_prime) * l + (i + big_prime) * f\n if (i + big_prime <= n) and (less_prime <= m):\n ans = max(ans, (less_prime) * f + (i + big_prime) * l)\n \n for j in range(0, n - less_prime + 1):\n if j + less_prime < big_prime:\n arr[0][j] = True\n if (j + less_prime) * l + (big_prime) * f > ans:\n ans = (j + less_prime) * l + (big_prime) * f\n if (j + less_prime <= m) and (big_prime <= n):\n ans = max(ans, (j + less_prime) * f + (big_prime) * l)\n \n for i in range(1, m - big_prime + 1):\n for j in range(1, n - less_prime + 1):\n if arr[i - 1][j] == True or arr[i][j - 1] == True:\n if gcd(i + big_prime, j + less_prime) == 1:\n arr[i][j] = True\n if (j + less_prime) * l + (i + big_prime) * f > ans:\n ans = (j + less_prime) * l + (i + big_prime) * f\n \n if (i + big_prime <= n) and (j + less_prime <= m):\n ans = max(ans, (j + less_prime) * f + (i + big_prime) * l)\n \n return ans\n \nsieve_of_eratosthenes(20000000)\n \nt = int(input())\n \nfor k in range(t):\n x = input().split(\" \")\n n,m,l,f = int(x[0]), int(x[1]), int(x[2]), int(x[3])\n \n ans = dp(n, m, l, f)\n \n # if n <= 10 and m <= 10:\n # ans = depth_search(1, 1, n, m, l, f)\n # else:\n # ans = dp(n, m, l, f)\n \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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 277610137,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n private static BufferedReader br;\n private static StringTokenizer tokenizer;\n private static String next() throws IOException{\n if (br == null) br = new BufferedReader(new InputStreamReader(System.in));\n if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());\n while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());\n return tokenizer.nextToken();\n }\n private static int nextInt() throws IOException{\n return Integer.parseInt(next());\n }\n private static long nextLong() throws IOException{\n return Long.parseLong(next());\n }\n \n static StringBuilder sb = new StringBuilder();\n \n static List<Integer> primes;\n static int sup = 20_000_002;\n static int[] prePrimes = new int[sup];\n static int[][] direction = {{1,0},{0,1}};\n static long[][] visited;\n static int n;\n static int m;\n static long l;\n static long f;\n static Map<Long, Long> cached;\n \n static void getPrimes() {\n boolean[] visited = new boolean[sup];\n primes = new ArrayList<>();\n prePrimes[1] = prePrimes[2] = 1;\n for (int i = 2; i < sup; ++i) {\n if (i > 2) prePrimes[i] = primes.get(primes.size() - 1);\n if (!visited[i]) primes.add(i);\n for (int p : primes) {\n if (i * p >= sup) break;\n visited[i * p] = true;\n if (i % p == 0) break;\n }\n }\n }\n \n static long gcd(int x, int y) {\n if (x < y) {\n int t = x; x = y; y = t;\n }\n if (x % y == 0) return y;\n return gcd(y, x % y);\n }\n \n static long getKey(long x, long y) {\n return x * sup + y;\n }\n \n static long dfs(int p, int q, long l, long f, int n, int m, int flag) {\n if (p > n || q > m || gcd(p, q) > 1 || p < q) return 0L;\n long key = getKey(p, q);\n if (cached.containsKey(key)) return cached.get(key);\n long ans = flag == 0 ? Math.max(l * p + f * q, l * q + f * p) : l * p + f * q;\n for (int[] d : direction) {\n ans = Math.max(ans, dfs(p + d[0], q + d[1], l, f, n, m, flag));\n }\n cached.put(key, ans);\n return ans;\n }\n \n static void slove() throws IOException{\n n = nextInt(); m = nextInt(); l = nextInt(); f = nextInt();\n if (n < m) {\n int t = n; n = m; m = t; long t2 = l; l = f; f = t2;\n }\n \n if (n == 1) {\n sb.append(l + f).append(\"\\n\");\n return;\n }\n \n \n int lp = prePrimes[m + 1];\n int lq = prePrimes[lp];\n if (m < 100) {\n lp = lq = 1;\n }\n cached = new HashMap<>();\n long ans = 0L;\n for (int i = lq; i <= m; ++i) ans = Math.max(ans, dfs(lp, i, l, f, m, m, 0));\n for (int i = lp; i <= m; ++i) ans = Math.max(ans, dfs(i, lq, l, f, m, m, 0));\n \n int p = prePrimes[n + 1];\n int q = prePrimes[Math.min(p - 1, m)];\n while (q > 1 && (p % q) + (n - p) >= q) q = prePrimes[q];\n int tq = Math.min(p - 1, m) / q * q;\n while (tq > q) {\n boolean flag = true;\n for (int i = p; i <= n; ++i) {\n if (gcd(i, tq) > 1) {\n flag = false;\n break;\n }\n }\n if (flag) break;\n tq -= q;\n }\n q = tq;\n if (n < 100) {\n p = q = 1;\n }\n cached = new HashMap<>();\n \n for (int i = q; i <= m; ++i) ans = Math.max(ans, dfs(p, i, l, f, n, m, 1));\n for (int i = p; i <= n; ++i) ans = Math.max(ans, dfs(i, q, l, f, n, m, 1));\n sb.append(ans).append(\"\\n\");\n \n }\n \n \n \n public static void main(String[] args) throws IOException {\n getPrimes();\n int tt = nextInt();\n sb = new StringBuilder();\n for (int i = 0; i < tt; ++i) slove();\n System.out.println(sb);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 277497564,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n private static BufferedReader br;\n private static StringTokenizer tokenizer;\n private static String next() throws IOException{\n if (br == null) br = new BufferedReader(new InputStreamReader(System.in));\n if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());\n while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());\n return tokenizer.nextToken();\n }\n private static int nextInt() throws IOException{\n return Integer.parseInt(next());\n }\n private static long nextLong() throws IOException{\n return Long.parseLong(next());\n }\n \n static StringBuilder sb = new StringBuilder();\n \n static List<Integer> primes;\n static int sup = 20_000_002;\n static int[] prePrimes = new int[sup];\n static int[][] direction = {{1,0},{0,1}};\n static long[][] visited;\n static int n;\n static int m;\n static long l;\n static long f;\n static Map<Long, Long> cached;\n \n static void getPrimes() {\n boolean[] visited = new boolean[sup];\n primes = new ArrayList<>();\n prePrimes[1] = prePrimes[2] = 1;\n for (int i = 2; i < sup; ++i) {\n if (i > 2) prePrimes[i] = primes.get(primes.size() - 1);\n if (!visited[i]) primes.add(i);\n for (int p : primes) {\n if (i * p >= sup) break;\n visited[i * p] = true;\n if (i % p == 0) break;\n }\n }\n }\n \n static long gcd(int x, int y) {\n if (x < y) {\n int t = x; x = y; y = t;\n }\n if (x % y == 0) return y;\n return gcd(y, x % y);\n }\n \n static long getKey(long x, long y) {\n return x * sup + y;\n }\n \n static long dfs(int p, int q, long l, long f, int n, int m, int flag) {\n if (p > n || q > m || gcd(p, q) > 1 || p < q) return 0L;\n long key = getKey(p, q);\n if (cached.containsKey(key)) return cached.get(key);\n long ans = flag == 0 ? Math.max(l * p + f * q, l * q + f * p) : l * p + f * q;\n for (int[] d : direction) {\n ans = Math.max(ans, dfs(p + d[0], q + d[1], l, f, n, m, flag));\n }\n cached.put(key, ans);\n return ans;\n }\n \n static void slove() throws IOException{\n n = nextInt(); m = nextInt(); l = nextInt(); f = nextInt();\n if (n < m) {\n int t = n; n = m; m = t; long t2 = l; l = f; f = t2;\n }\n \n if (n == 1) {\n sb.append(l + f).append(\"\\n\");\n return;\n }\n \n \n int lp = prePrimes[m + 1];\n int lq = prePrimes[lp];\n if (m < 200) {\n lp = lq = 1;\n }\n cached = new HashMap<>();\n long ans = 0L;\n for (int i = lq; i <= m; ++i) ans = Math.max(ans, dfs(lp, i, l, f, m, m, 0));\n for (int i = lp; i <= m; ++i) ans = Math.max(ans, dfs(i, lq, l, f, m, m, 0));\n \n int p = prePrimes[n + 1];\n int q = prePrimes[Math.min(p - 1, m)];\n while (q > 1 && (p % q) + (n - p) >= q) q = prePrimes[q];\n if (n < 200) {\n p = q = 1;\n }\n cached = new HashMap<>();\n \n for (int i = q; i <= m; ++i) ans = Math.max(ans, dfs(p, i, l, f, n, m, 1));\n for (int i = p; i <= n; ++i) ans = Math.max(ans, dfs(i, q, l, f, n, m, 1));\n sb.append(ans).append(\"\\n\");\n \n }\n \n \n \n public static void main(String[] args) throws IOException {\n getPrimes();\n int tt = nextInt();\n sb = new StringBuilder();\n for (int i = 0; i < tt; ++i) slove();\n System.out.println(sb);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 277493280,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Court Blue (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: This is the hard version of the problem. In this version, it is not guaranteed that $n=m$, and the time limit is higher. You can make hacks only if both versions of the problem are solved.\n\nIn the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.\n\nLet $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:\n\n * after every round, $\\gcd(W_L,W_F)\\le 1$; * at the end of the match, $W_L\\le n, W_F\\le m$. \n\nNote that $\\gcd(0,x)=\\gcd(x,0)=x$ for every non-negative integer $x$.\n\nLelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \\cdot W_L + f \\cdot W_F$.\n\nPlease help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.\n\nInput: The first line contains an integer $t$ ($1\\leq t \\leq 10^3$) β the number of test cases.\n\nThe only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\\leq n\\leq m \\leq 2\\cdot 10^7$, $1\\leq l,f \\leq 10^9$): $n$, $m$ give the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.\n\nUnusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.\n\nOutput: For each test case, output a single integer β the maximum total score of a successful performance.\n\nExamples:\ninput:\n8\n3 4 2 5\n4 4 1 4\n6 6 2 2\n7 9 2 3\n8 9 9 1\n2 7 1 4\n5 9 1 4\n5 6 6 7\noutput:\n22\n17\n18\n37\n77\n30\n41\n59\n\ninput:\n2\n3082823 20000000 1341 331\n20000000 20000000 3 5\noutput:\n10754065643\n159999991\n\ninput:\n1\n139 1293 193 412\noutput:\n559543\n\nNote: In the first test case, a possible performance is as follows:\n\n * Flamm wins, $\\gcd(0,1)=1$. * Lelle wins, $\\gcd(1,1)=1$. * Flamm wins, $\\gcd(1,2)=1$. * Flamm wins, $\\gcd(1,3)=1$. * Flamm wins, $\\gcd(1,4)=1$. * Lelle and Flamm agree to stop the match. \n\nThe final score is $1\\cdot2+4\\cdot5=22$.\n\nBuggy Code:\n#https://codeforces.com/contest/2002/problem/F2?locale=en\n \nprimes = [1]\ndef sieve_of_eratosthenes(n):\n global primes\n a = [True] * (n + 1)\n \n i = 2\n while i * i <= n:\n if a[i] == True:\n primes.append(i)\n for j in range(i, n+1, i):\n a[j] = False\n i += 1\n \n while i <= n:\n if (a[i] == True):\n primes.append(i)\n i += 1\n \ndef find_the_nearest_prime_index(n):\n global primes\n l = 0\n r = len(primes) - 1\n while True:\n if (primes[r] == n):\n return r\n if (primes[l] == n):\n return l\n if r - l == 1:\n return l\n mid = (r + l) // 2\n if (primes[mid] < n):\n l = mid\n else:\n r = mid\n \ndef gcd(a, b):\n if b == 0:\n return a\n return gcd(b, a % b)\n \ndef depth_search(a, b, n, m, l, f):\n ###a < b <= n\n ans = a * l + b * f\n \n if a + 1 <= n and gcd(a + 1, b) == 1:\n ans = max(ans, depth_search(a+1, b, n, m, l, f))\n if b + 1 <= m and gcd(a, b + 1) == 1:\n ans = max(ans, depth_search(a, b + 1, n, m, l, f))\n \n return ans\n \ndef find_less_prime(n, m, big_prime):\n for i in range(n, 0, -1):\n flag = True\n for j in range(big_prime, m + 1):\n if gcd(i, j) != 1:\n flag = False\n break\n \n if flag:\n return i\n \ndef dp(n, m, l, f):\n p_n = find_the_nearest_prime_index(n)\n p_m = find_the_nearest_prime_index(m)\n big_prime = primes[p_m]\n \n less_prime = find_less_prime(n, m, big_prime)\n arr = []\n ans = 0\n ##j < i <= n\n for i in range(m - big_prime + 1):\n arr.append([])\n for j in range(n - less_prime + 1):\n arr[i].append(False)\n \n # for i in range(0, m - big_prime + 1):\n # if gcd(i + big_prime, less_prime) == 1:\n # arr[i][0] = True\n # if (less_prime) * l + (i + big_prime) * f > ans:\n # ans = (less_prime) * l + (i + big_prime) * f\n # if (i + big_prime <= n) and (less_prime <= m):\n # ans = max(ans, (less_prime) * f + (i + big_prime) * l)\n \n # for j in range(0, n - less_prime + 1):\n # if gcd(big_prime, j + less_prime) == 1:\n # if j + less_prime < big_prime:\n # arr[0][j] = True\n # if (j + less_prime) * l + (big_prime) * f > ans:\n # ans = (j + less_prime) * l + (big_prime) * f\n # if (j + less_prime <= m) and (big_prime <= n):\n # ans = max(ans, (j + less_prime) * f + (big_prime) * l)\n \n for i in range(0, m - big_prime + 1):\n for j in range(0, n - less_prime + 1):\n if (i == 0 and j == 0) or (i != 0 and arr[i-1][j] == True) or (j != 0 and arr[i][j-1] == True):\n if gcd(i + big_prime, j + less_prime) == 1:\n arr[i][j] = True\n if (j + less_prime) * l + (i + big_prime) * f > ans:\n ans = (j + less_prime) * l + (i + big_prime) * f\n \n if (i + big_prime <= n) and (j + less_prime <= m):\n ans = max(ans, (j + less_prime) * f + (i + big_prime) * l)\n \n return ans\n \nsieve_of_eratosthenes(20000000)\n \nt = int(input())\n \nfor k in range(t):\n x = input().split(\" \")\n n,m,l,f = int(x[0]), int(x[1]), int(x[2]), int(x[3])\n \n ans = dp(n, m, l, f)\n \n # if n <= 10 and m <= 10:\n # ans = depth_search(1, 1, n, m, l, f)\n # else:\n # ans = dp(n, m, l, f)\n \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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2002F2",
"submission_id": 277529112,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n \n#define ll long long\n \n \n \nvoid solve(ll m, ll n,vector<vector<ll>>&vec, ll x1, ll y1, ll x2, ll y2){\n // cout<<\"---\"<<x1<<x2<<y1<<y2;\n \n vector<vector<ll>>dp(x2,vector<ll>(y2,0)); \n vector<vector<ll>>mulpi(x2,vector<ll>(y2,0)); \n \n //set mulpi 1,2,3 ... n multiply kro\n \n ll val=1;\n for (ll i = x1-1; i <=x2-1; i++)\n {\n for (ll j = y1-1; j <=y2-1; j++)\n {\n vec[i][j]*=val;\n val++;\n }\n \n }\n \n//dp matrix -> each elem sum of elem starting from(0,0)\n \n dp[0][0]=vec[0][0];\n \n //1st row\n for (ll j = 1; j < y2; j++)\n {\n dp[0][j]=dp[0][j-1]+vec[0][j];\n }\n \n //1st column\n for (ll i = 1; i<x2; i++)\n {\n dp[i][0]=dp[i-1][0]+vec[i][0];\n }\n \n //rest values\n for (ll i = 1; i < x2; i++)\n {\n for (ll j = 1; j < y2; j++)\n {\n dp[i][j]=dp[i-1][j]+dp[i][j-1]+vec[i][j] -dp[i-1][j-1];\n }\n \n }\n \n \n \n \n \n \n///////////////--------------------------/////////////////////////\n \n// for submatrix sum i need to chk if gaps before it above it \n//req sum= total_sum- left_block_sum- top_block_sum+ common_block_sum;\n// 1<=x1<=x2<=n; 1<=y1<=y2<=n;\nx1--;x2--;y1--;y2--;\n \nll total_sum=dp[x2][y2];\nll left_block_sum=0;\nll upper_block_sum=0;\nll common_block_sum=0;\nif(y1>0){\n left_block_sum=dp[x2][y1-1];\n}\nif (x1>0)\n{\n upper_block_sum=dp[x1-1][y2];\n}\nif(x1>0 && y1>0){\n common_block_sum=dp[x1-1][y1-1];\n}\n \nll req_sum=total_sum-left_block_sum-upper_block_sum+common_block_sum;\ncout<<req_sum<<endl;\n \n \nval=1;\n for (ll i = x1; i <=x2; i++)\n {\n for (ll j = y1; j <=y2; j++)\n {\n vec[i][j]/=val;\n val++;\n }\n \n }\n \n}\n//////////////////////////////////////////////////////\n \nint main(){\n ll test;\n cin>>test;\n ll n; ll x1; ll y1; ll x2; ll y2;\n ll q;\n \n \n while (test--)\n { cin>>n>>q;\n vector<vector<ll>>vec2(n,vector<ll>(n,0));\n for (ll i = 0; i < n; i++)\n {\n for (ll j = 0; j < n; j++)\n {\n cin>>vec2[i][j];\n }\n \n }\n while(q--){\n cin>>x1>>y1>>x2>>y2;\n solve(n,n,vec2,x1,y1,x2,y2);\n }\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 298075683,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define FAST ios::sync_with_stdio(0), cin.tie(0) //, cout.tie(0)\n#define ll long long\n#define ld long double\n#define pb push_back\n#define pi pair<ll,ll>\n#define F first\n#define S second\n#define all(x) (x).begin(), (x).end()\n#define alll(x) ((x).begin()+1), (x).end()\n#define yes cout<<\"Yes\"<<endl;\n#define no cout<<\"No\"<<endl;\n#define endl \"\\n\"\nusing namespace std;\nconst ll MOD = 1e9+7 ;\n// const int MOD = 998244353 ;\nconst int N = 505 ;\nconst ll INF = 1e18 ;\nconst ll MIN = -1e18 ;\nconst ll prime = 1000003;\n//const ll prime2 = 101 ;\n#include<iostream>\nusing namespace std;\nvoid __print(int x) {cout << x;}\nvoid __print(long x) {cout << x;}\nvoid __print(long long x) {cout << x;}\nvoid __print(unsigned x) {cout << x;}\nvoid __print(unsigned long x) {cout << x;}\nvoid __print(unsigned long long x) {cout << x;}\nvoid __print(float x) {cout << x;}\nvoid __print(double x) {cout << x;}\nvoid __print(long double x) {cout << x;}\nvoid __print(char x) {cout << '\\'' << x << '\\'';}\nvoid __print(const char *x) {cout << '\\\"' << x << '\\\"';}\nvoid __print(const string &x) {cout << '\\\"' << x << '\\\"';}\nvoid __print(bool x) {cout << (x ? \"true\" : \"false\");}\n \ntemplate<typename T, typename V>\nvoid __print(const pair<T, V> &x) {cout << '{'; __print(x.first); cout << ','; __print(x.second); cout << '}';}\ntemplate<typename T>\nvoid __print(const T &x) {int f = 0; cout << '{'; for (auto &i: x) cout << (f++ ? \",\" : \"\"), __print(i); cout << \"}\";}\nvoid _print() {cout << \"]\"<<endl;}\ntemplate <typename T, typename... V>\nvoid _print(T t, V... v) {__print(t); if (sizeof...(v)) cout << \", \"; _print(v...);}\n#define deb(x...) cout << #x << \" = [\"; _print(x)\ntemplate<typename T,typename T1>T amax(T &a,T1 b){if(b>a)a=b;return a;}\ntemplate<typename T,typename T1>T amin(T &a,T1 b){if(b<a)a=b;return a;}\n \n \nint querySubarraySum(const vector<vector<int>>& prefix, int r1, int c1, int r2, int c2) {\n int total = prefix[r2][c2];\n if (r1 > 0) total -= prefix[r1-1][c2];\n if (c1 > 0) total -= prefix[r2][c1-1];\n if (r1 > 0 && c1 > 0) total += prefix[r1-1][c1-1];\n return total;\n}\n \nvector<vector<int>> computePrefixSum(const vector<vector<int>>& arr) {\n int n = arr.size();\n int m = arr[0].size();\n vector<vector<int>> prefix(n, vector<int>(m, 0));\n \n for (int i = 0; i < n; i++) {\n for (int j = 0; j < m; j++) {\n prefix[i][j] = arr[i][j]\n + (i > 0 ? prefix[i-1][j] : 0)\n + (j > 0 ? prefix[i][j-1] : 0)\n - (i > 0 && j > 0 ? prefix[i-1][j-1] : 0);\n }\n }\n return prefix;\n}\n \nvector<vector<int>> computePrefixSum2(const vector<vector<int>>& arr) {\n int n = arr.size();\n int m = arr[0].size();\n vector<vector<int>> prefix(n, vector<int>(m, 0));\n \n for (int i = 0; i < n; i++) {\n for (int j = 0; j < m; j++) {\n prefix[i][j] = (i*m+j +1 ) * arr[i][j]\n + (i > 0 ? prefix[i-1][j] : 0)\n + (j > 0 ? prefix[i][j-1] : 0)\n - (i > 0 && j > 0 ? prefix[i-1][j-1] : 0);\n }\n }\n return prefix;\n}\n \nvoid solve(){\n \n int n,q;\n cin>>n>>q;\n vector<vector<int>>a(n,vector<int>(n));\n for(int i=0;i<n;i++){\n for(int j=0;j<n;j++){\n cin>>a[i][j];\n \n }\n }\n \n \n vector<vector<int>> prefix = computePrefixSum(a);\n vector<vector<int>> ans = computePrefixSum2(a);\n \n \n while (q--) {\n int x1,y1,x2,y2;\n cin>>x1>>y1>>x2>>y2;\n \n \n x1--; y1--; x2--; y2--;\n \n if (x1==0 && y1==0 && x2==n-1 && y2==n-1) {\n cout<<querySubarraySum(ans,x1,y1,x2,y2)<<\" \";\n continue;\n }\n \n ll ans2=0;\n for(int i=x1;i<=x2;i++) {\n ans2+=querySubarraySum(ans,i,y1,i,y2);\n ans2-=querySubarraySum(prefix,i,y1,i,y2)*((x1-i)*(n-1-y2));\n ans2-=querySubarraySum(prefix,i,y1,i,y2)*(i*n+y1);\n \n ans2+=querySubarraySum(prefix,i,y1,i,y2)*(i-x1);\n \n \n }\n \n \n cout<<ans2<<\" \";\n }\n cout<<endl;\n \n \n \n}\n \nint main()\n{\n \n \n FAST ;\n int t = 1 ;\n cin>> t;\n while( t--) {\n solve();\n }\n}\n \n/* stuff you should look for\n* int overflow, array bounds\n* special cases (n=1?)\n* do smth instead of nothing and stay organized\n* WRITE STUFF DOWN\n* DON'T GET STUCK ON ONE APPROACH\n* think backwards\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296735133,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\n#include <iostream>\n#include <algorithm>\n#include <vector>\n#include <array>\n#include <set>\n#include <map>\n#include <queue>\n#include <stack>\n#include <list>\n#include <chrono>\n#include <random>\n#include <cstdlib>\n#include <cmath>\n#include <ctime>\n#include <cstring>\n#include <iomanip>\n#include <bitset>\n#include <cassert>\ntypedef long long ll;\nusing namespace std;\n \nstruct segtree {\n \n\tvector<ll> tree, tree_sum;\n\tint size;\n \n\tvoid init(int n) {\n\t\tsize = 1;\n\t\twhile (size < n)\n\t\t\tsize *= 2;\n\t\ttree.assign(2 * size - 1, 0);\n\t\ttree_sum.assign(2 * size - 1, 0);\n\t}\n \n\tvoid build(vector<long long>& a, int x, int lx, int rx) {\n\t\tif (rx - lx == 1) {\n\t\t\tif (lx < a.size()) {\n\t\t\t\ttree_sum[x] = a[lx];\n\t\t\t\ttree[x] = a[lx];\n\t\t\t}\n\t\t\t\t\n\t\t}\n\t\telse {\n\t\t\tint m = (lx + rx) / 2;\n\t\t\tbuild(a, 2 * x + 1, lx, m);\n\t\t\tbuild(a, 2 * x + 2, m, rx);\n\t\t\ttree_sum[x] = tree_sum[2 * x + 1] + tree_sum[2 * x + 2];\n\t\t\ttree[x] = tree[2 * x + 1] + tree_sum[2 * x + 2] * (long long)(m - lx) + tree[2 * x + 2];\n\t\t}\n\t}\n \n\tvoid build(vector<long long>& a) {\n\t\tinit(a.size());\n\t\tbuild(a, 0, 0, size);\n\t}\n \n\tll sum(int l, int r, int x, int lx, int rx) {\n\t\tif (l >= rx || r <= lx) {\n\t\t\treturn 0;\n\t\t}\n\t\tif (lx >= l && rx <= r) {\n\t\t\treturn tree_sum[x];\n\t\t}\n\t\tint m = (lx + rx) / 2;\n\t\tlong long mi1 = sum(l, r, 2 * x + 1, lx, m);\n\t\tlong long mi2 = sum(l, r, 2 * x + 2, m, rx);\n\t\treturn mi1 + mi2;\n\t}\n \n\tlong long query(int l, int r, int x, int lx, int rx) {\n\t\tif (l >= rx || r <= lx) {\n\t\t\treturn 0;\n\t\t}\n\t\tif (lx >= l && rx <= r) {\n\t\t\treturn tree[x];\n\t\t}\n\t\tint m = (lx + rx) / 2;\n\t\tlong long mi1 = query(l, r, 2 * x + 1, lx, m);\n\t\tlong long mi2 = query(l, r, 2 * x + 2, m, rx);\n\t\tlong long s = sum(l, r, 2 * x + 2, m, rx);\n\t\treturn mi1 + mi2 + s * (long long)(m - l);\n\t}\n \n\tlong long query(int l, int r, long long mult) {\n\t\tlong long rr = query(l, r, 0, 0, size);\n\t\tlong long s = sum(l, r, 0, 0, size);\n\t\treturn (query(l, r, 0, 0, size) - sum(l, r, 0, 0, size)) * mult;\n\t}\n\tlong long query(int l, int r) {\n\t\treturn query(l, r, 0, 0, size);\n\t}\n};\n \nint main()\n{\n\tint t;\n\tcin >> t;\n\twhile (t--) {\n\t\tint n, q;\n\t\tcin >> n >> q;\n\t\tvector<vector<long long>> ma(n, vector<long long>(n));\n\t\tvector<vector<long long>> pref_row(n + 1, vector<long long>(n));\n\t\tvector<vector<long long>> pref_column(n + 1, vector<long long>(n));\n\t\tfor (int i = 0; i < n; i++) {\n\t\t\tfor (int j = 0; j < n; j++) {\n\t\t\t\tcin >> ma[i][j];\n\t\t\t}\n\t\t}\n\t\tfor (int i = 0; i < n; i++) {\n\t\t\tfor (int j = 0; j < n; j++) {\n\t\t\t\tpref_row[j + 1][i] = pref_row[j][i] + ma[i][j];\n\t\t\t\tpref_column[j + 1][i] = pref_column[j][i] + ma[j][i];\n\t\t\t}\n\t\t}\n\t\tvector<segtree> seg1(n + 1), seg2(n + 1);\n\t\tfor (int i = 0; i <= n; i++) {\n\t\t\tseg1[i].build(pref_column[i]);\n\t\t\tseg2[i].build(pref_row[i]);\n\t\t}\n\t\twhile (q--) {\n\t\t\tint x1, y1, x2, y2;\n\t\t\tcin >> y1 >> x1 >> y2 >> x2;\n\t\t\tlong long mult = x2 - x1 + 1;\n\t\t\tlong long res = 0;\n\t\t\tres += seg2[x2].query(y1 - 1, y2, mult) - seg2[x1 - 1].query(y1 - 1, y2, mult);\n\t\t\tres += seg1[y2].query(x1 - 1, x2) - seg1[y1 - 1].query(x1 - 1, x2);\n\t\t\tcout << res << '\\n';\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296747035,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\ndef process_queries(t, test_cases):\n results = []\n \n for n, q, matrix, queries in test_cases:\n for x1, y1, x2, y2 in queries:\n # Convert to 0-based indexing\n x1 -= 1\n y1 -= 1\n x2 -= 1\n y2 -= 1\n \n # Flatten the submatrix\n A = []\n for i in range(x1, x2 + 1):\n for j in range(y1, y2 + 1):\n A.append(matrix[i][j])\n \n # Calculate the weighted sum\n weighted_sum = sum(A[i] * (i + 1) for i in range(len(A)))\n results.append(weighted_sum)\n \n return results\n \n# Input and output handling\nt = int(input())\ntest_cases = []\nfor _ in range(t):\n n, q = map(int, input().split())\n matrix = [list(map(int, input().split())) for _ in range(n)]\n queries = [tuple(map(int, input().split())) for _ in range(q)]\n test_cases.append((n, q, matrix, queries))\n \nresults = process_queries(t, test_cases)\nfor res in results:\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296697131,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\nMAX_N = 2000\n \nprefix_sum = [[0] * (MAX_N + 1) for _ in range(MAX_N + 1)]\n \ndef compute_prefix(n, grid):\n for row in range(1, n + 1):\n for col in range(1, n + 1):\n prefix_sum[row][col] = (\n grid[row][col]\n + (prefix_sum[row - 1][col] if row > 1 else 0)\n + (prefix_sum[row][col - 1] if col > 1 else 0)\n - (prefix_sum[row - 1][col - 1] if row > 1 and col > 1 else 0)\n )\n \ndef query_sum(x1, y1, x2, y2):\n return (\n prefix_sum[x2][y2]\n - (prefix_sum[x1 - 1][y2] if x1 > 1 else 0)\n - (prefix_sum[x2][y1 - 1] if y1 > 1 else 0)\n + (prefix_sum[x1 - 1][y1 - 1] if x1 > 1 and y1 > 1 else 0)\n )\n \ndef main():\n test_cases = int(input())\n for _ in range(test_cases):\n grid_size, queries = map(int, input().split())\n grid = [[0] * (MAX_N + 1) for _ in range(grid_size + 1)]\n \n for row in range(1, grid_size + 1):\n values = list(map(int, input().split()))\n for col in range(1, grid_size + 1):\n grid[row][col] = values[col - 1]\n \n compute_prefix(grid_size, grid)\n \n for _ in range(queries):\n x1,y1, x2, y2 = map(int, input().split())\n total_sum = 0\n multiplier = 1\n \n for row in range(x1, x2 + 1):\n for col in range(y1, y2 + 1):\n total_sum += grid[row][col] * multiplier\n multiplier += 1\n \n print(total_sum, end=\" \")\n print()\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296743509,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\ndef process_pancake_factory(t, test_cases): \n results = [] \n \n for n, q, matrix, queries in test_cases: \n for (x1, y1, x2, y2) in queries: \n flatten_list = [] \n for i in range(x1 - 1, x2): # converting to 0-indexed for Python access \n for j in range(y1 - 1, y2): \n flatten_list.append(matrix[i][j]) \n \n weighted_sum = sum(value * (idx + 1) for idx, value in enumerate(flatten_list)) \n results.append(str(weighted_sum)) \n \n return ' '.join(results) \n \n# Example of input processing and calling the function \nif __name__ == \"__main__\": \n import sys \n input = sys.stdin.read \n data = input().splitlines() \n \n index = 0 \n t = int(data[index]) # number of test cases \n index += 1 \n test_cases = [] \n \n for _ in range(t): \n n, q = map(int, data[index].split()) \n index += 1 \n matrix = [] \n for _ in range(n): \n row = list(map(int, data[index].split())) \n matrix.append(row) \n index += 1 \n \n queries = [] \n for _ in range(q): \n x1, y1, x2, y2 = map(int, data[index].split()) \n queries.append((x1, y1, x2, y2)) \n index += 1 \n \n test_cases.append((n, q, matrix, queries)) \n \n results = process_pancake_factory(t, test_cases) \n print(results)\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296715231,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class Main {\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 q = sc.nextInt();\n long[][] pA = new long[n + 1][n + 1];\n long[][] pB = new long[n + 1][n + 1];\n long[][] pC = new long[n + 1][n + 1];\n for (int i = 1; i <= n; i++){\n for (int j = 1; j <= n; j++){\n int v = sc.nextInt();\n pA[i][j] = pA[i - 1][j] + pA[i][j - 1] - pA[i - 1][j - 1] + v;\n pB[i][j] = pB[i - 1][j] + pB[i][j - 1] - pB[i- 1][j - 1] + (long)i * v;\n pC[i][j] = pC[i - 1][j] + pC[i][j - 1] - pC[i - 1][j - 1] + (long)j * v;\n }\n }\n for (int k = 0; k < q; k++){\n int x1 = sc.nextInt();\n int y1 = sc.nextInt();\n int x2 = sc.nextInt();\n int y2 = sc.nextInt();\n long c = (long)(y2 - y1 + 1);\n long sS = pA[x2][y2] - pA[x1 - 1][y2] - pA[x2][y1 - 1] + pA[x1 - 1][y1 - 1];\n long sI = pB[x2][y2] - pB[x1 - 1][y2] - pB[x2][y1 - 1] + pB[x1 - 1][y1 - 1];\n long sJ = pC[x2][y2] - pC[x1 - 1][y2] - pC[x2][y1 - 1] + pC[x1 - 1][y1 - 1];\n long ans = c * sI + sJ + (1L - c * x1 - y1) * sS;\n System.out.println(ans);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 306267370,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Main{\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n \n int t = scanner.nextInt();\n \n while (t-- > 0) {\n int n = scanner.nextInt();\n int q = scanner.nextInt();\n \n long[][] matrix = new long[n][n];\n for (int i = 0; i < n; i++) {\n for (int j = 0; j < n; j++) {\n matrix[i][j] = scanner.nextLong();\n }\n }\n \n // Compute prefix sums\n long[][] prefixSum = new long[n + 1][n + 1];\n for (int i = 1; i <= n; i++) {\n for (int j = 1; j <= n; j++) {\n prefixSum[i][j] = prefixSum[i - 1][j] + prefixSum[i][j - 1] - prefixSum[i - 1][j - 1] + matrix[i - 1][j - 1];\n }\n }\n \n // Process each query\n for (int qi = 0; qi < q; qi++) {\n int x1 = scanner.nextInt() - 1;\n int y1 = scanner.nextInt() - 1;\n int x2 = scanner.nextInt() - 1;\n int y2 = scanner.nextInt() - 1;\n \n // Calculate the sum of the submatrix\n long submatrixSum = prefixSum[x2 + 1][y2 + 1] - prefixSum[x1][y2 + 1] - prefixSum[x2 + 1][y1] + prefixSum[x1][y1];\n \n // Flatten the submatrix and compute the required sum\n long result = 0;\n for (int i = 0; i <= x2 - x1; i++) {\n for (int j = 0; j <= y2 - y1; j++) {\n result += (i * (y2 - y1 + 1) + j + 1) * matrix[x1 + i][y1 + j];\n }\n }\n \n System.out.print(result + \" \");\n }\n System.out.println();\n }\n \n scanner.close();\n }\n}\n \n \n// XBUMhTsEK45kLbJuReS\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 302351652,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define ll long long\n#define vvi vector<vector<ll>>\n#define vi vector<ll>\n#define get(n) int n; cin >> n;\nusing namespace std;\n \n \nvvi M(2001, vi(2001));\nvvi Sum(2001, vi(2001));\nvvi x_Sum(2001, vi(2001));\nvvi y_Sum(2001, vi(2001));\nll n, x1, y_start, x2, y2;\nint q;\nvi dp_x(2001);\nvi dp_y(2001);\nvi dp_sum(2001);\n \n \nll return_sum_x(){\n ll ans = 0;\n ans += x_Sum[x2][y2];\n ans -= x_Sum[x2][y_start-1];\n ans -= x_Sum[x1-1][y2];\n ans += x_Sum[x1-1][y_start-1];\n return ans;\n}\n \nll return_sum_y(){\n ll ans = 0;\n ans += y_Sum[x2][y2];\n ans -= y_Sum[x2][y_start-1];\n ans -= y_Sum[x1-1][y2];\n ans += y_Sum[x1-1][y_start-1];\n return ans;\n}\n \nll return_sum_normal(){\n ll ans = 0;\n ans += Sum[x2][y2];\n ans -= Sum[x2][y_start-1];\n ans -= Sum[x1-1][y2];\n ans += Sum[x1-1][y_start-1];\n return ans;\n}\n \nvoid cout_vec(vvi &v){\n for(int i = 1; i<= n ; i++){\n for(int j = 1; j <= n ; j++){\n cout << v[i][j] << \" \";\n }\n cout << \"\\n\";\n }\n cout <<\"\\nCOMP\\n\\n\";\n}\n \n \n \nvoid solve(){\n ll n;\n int q;\n \n cin >> n >> q;\n for(int i = 1; i <= n ; i++){\n for(int j =1 ; j <= n ; j++){\n cin >> M[i][j];\n }\n }\n for(ll i = 1 ; i<= n ; i++){\n for(ll j = 1 ; j <=n ; j++){\n Sum[i][j] = Sum[i-1][j] + dp_sum[j-1] + M[i][j];\n dp_sum[j] = dp_sum[j-1] + M[i][j];\n x_Sum[i][j] = x_Sum[i-1][j] + dp_x[j-1] + i*M[i][j];\n dp_x[j] = dp_x[j-1] + (i*M[i][j]);\n y_Sum[i][j] = y_Sum[i-1][j] + dp_y[j-1] + j*M[i][j];\n dp_y[j] = dp_y[j-1] + (j*M[i][j]);\n }\n }\n while(q--){\n cin >> x1 >> y_start >> x2 >> y2;\n printf(\"%d \", (return_sum_x())*(y2-y_start+1) + return_sum_y() - (return_sum_normal())*((y2-y_start+1)*x1 + y_start - 1));\n }\n \n printf(\"\\n\");\n}\n \nint main(){\n // freopen(\"input.txt\",\"r\",stdin);\n // freopen(\"output.txt\",\"w\",stdout);\n get(testcases);\n while(testcases--){\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 297224929,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nint main()\n{\n int tt;\n cin >> tt;\n while (tt > 0) {\n int n, q;\n cin >> n >> q;\n vector<vector<int>> v(n, vector<int>(n));\n for (int i = 0; i < n; i++) {\n for (int j = 0; j < n; j++) {\n cin >> v[i][j];\n }\n }\n for (int z = 1; z <= q; z++) {\n int x1, y1, x2, y2;\n cin >> x1 >> y1 >> x2 >> y2;\n int cnt = 1; // Reset cnt for every query\n int sum = 0;\n for (int k = x1 - 1; k < x2; k++) {\n for (int l = y1 - 1; l < y2; l++) {\n sum += cnt * v[k][l];\n cnt++; // Increment weight\n }\n }\n cout << sum << \" \";\n }\n cout << endl;\n tt--;\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 297727816,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\nt = int(input()) \nresults = []\nfor _ in range(t):\n n, q = map(int, input().split()) \n M = [list(map(int, input().split())) for _ in range(n)] \n prefix_sum = [[0] * (n + 1) for _ in range(n + 1)]\n for i in range(1, n + 1):\n for j in range(1, n + 1):\n prefix_sum[i][j] = M[i-1][j-1] + prefix_sum[i-1][j] + prefix_sum[i][j-1] - prefix_sum[i-1][j-1] \n case_results = []\n for __ in range(q):\n x1, y1, x2, y2 = map(int, input().split()) \n x1 -= 1; y1 -= 1; x2 -= 1; y2 -= 1\n weighted_sum = 0\n index = 1 \n for i in range(x1, x2 + 1):\n for j in range(y1, y2 + 1):\n weighted_sum += M[i][j] * index\n index += 1 \n case_results.append(weighted_sum) \n results.append(\" \".join(map(str, case_results))) \nprint(\"\\n\".join(results))\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296683375,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\nt = int(input())\nfor _ in range(t):\n n, q = map(int, input().split())\n \n M = [[0] * (n+1) for _ in range(n+1)]\n for r in range(1, n+1):\n row = list(map(int, input().split()))\n for c in range(1, n+1):\n M[r][c] = row[c-1]\n \n S = [[0] * (n+1) for _ in range(n+1)]\n Sr = [[0] * (n+1) for _ in range(n+1)]\n Sc = [[0] * (n+1) for _ in range(n+1)]\n \n for r in range(1, n+1):\n for c in range(1, n+1):\n S[r][c] = S[r-1][c] + S[r][c-1] - S[r-1][c-1] + M[r][c]\n Sr[r][c] = Sr[r-1][c] + Sr[r][c-1] - Sr[r-1][c-1] + r * M[r][c]\n Sc[r][c] = Sc[r-1][c] + Sc[r][c-1] - Sc[r-1][c-1] + c * M[r][c]\n \n output = []\n for _ in range(q):\n x1, y1, x2, y2 = map(int, input().split())\n width = y2 - y1 + 1\n \n sum_S = S[x2][y2] - S[x1-1][y2] - S[x2][y1-1] + S[x1-1][y1-1]\n sum_Sr = Sr[x2][y2] - Sr[x1-1][y2] - Sr[x2][y1-1] + Sr[x1-1][y1-1]\n sum_Sc = Sc[x2][y2] - Sc[x1-1][y2] - Sc[x2][y1-1] + Sc[x1-1][y1-1]\n \n term1 = width * (sum_Sr - x1 * sum_S)\n term2 = sum_Sc - (y1 - 1) * sum_S\n total = term1 + term2\n output.append(str(total))\n \n print(\" \".join(output))\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296743366,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main{\n \n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n PrintWriter pw = new PrintWriter(System.out);\n \n int t = Integer.parseInt(br.readLine());\n while (t-- > 0) {\n StringTokenizer st = new StringTokenizer(br.readLine());\n int n = Integer.parseInt(st.nextToken());\n int q = Integer.parseInt(st.nextToken());\n \n int[][] matrix = new int[n][n];\n for (int i = 0; i < n; i++) {\n st = new StringTokenizer(br.readLine());\n for (int j = 0; j < n; j++) {\n matrix[i][j] = Integer.parseInt(st.nextToken());\n }\n }\n \n for (int i = 0; i < q; i++) {\n st = new StringTokenizer(br.readLine());\n int x1 = Integer.parseInt(st.nextToken()) - 1; // 0-based indexing\n int y1 = Integer.parseInt(st.nextToken()) - 1;\n int x2 = Integer.parseInt(st.nextToken()) - 1;\n int y2 = Integer.parseInt(st.nextToken()) - 1;\n \n long sum = 0;\n int pos = 1;\n for (int x = x1; x <= x2; x++) {\n for (int y = y1; y <= y2; y++) {\n sum += matrix[x][y] * pos++;\n }\n }\n pw.print(sum + \" \");\n }\n pw.println();\n }\n pw.flush();\n }\n}\n \n \n// UwGzOeZJ9QrtWVXS1g7\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 304986223,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.StringTokenizer;\n \npublic class Main{\n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n StringBuilder sb = new StringBuilder();\n StringTokenizer st = new StringTokenizer(br.readLine());\n int t = Integer.parseInt(st.nextToken());\n for (int test = 0; test < t; test++) {\n st = new StringTokenizer(br.readLine());\n int n = Integer.parseInt(st.nextToken());\n int q = Integer.parseInt(st.nextToken());\n int[][] M = new int[n + 1][n + 1];\n for (int i = 1; i <= n; i++) {\n st = new StringTokenizer(br.readLine());\n for (int j = 1; j <= n; j++) {\n M[i][j] = Integer.parseInt(st.nextToken());\n }\n }\n int[][] prefix_sum = new int[n + 1][n + 1];\n int[][] prefix_sum_j = new int[n + 1][n + 1];\n for (int i = 1; i <= n; i++) {\n for (int j = 1; j <= n; j++) {\n prefix_sum[i][j] = prefix_sum[i][j - 1] + M[i][j];\n prefix_sum_j[i][j] = prefix_sum_j[i][j - 1] + M[i][j] * j;\n }\n }\n for (int query = 0; query < q; query++) {\n st = new StringTokenizer(br.readLine());\n int x1 = Integer.parseInt(st.nextToken());\n int y1 = Integer.parseInt(st.nextToken());\n int x2 = Integer.parseInt(st.nextToken());\n int y2 = Integer.parseInt(st.nextToken());\n int sum1 = 0;\n int sum2 = 0;\n for (int i = x1; i <= x2; i++) {\n int row_sum = prefix_sum[i][y2] - prefix_sum[i][y1 - 1];\n sum1 += (i - x1) * row_sum;\n int sum_j = prefix_sum_j[i][y2] - prefix_sum_j[i][y1 - 1];\n sum2 += sum_j - (y1 - 1) * row_sum;\n }\n int total_sum = (y2 - y1 + 1) * sum1 + sum2;\n sb.append(total_sum).append(\" \");\n }\n sb.append(\"\\n\");\n }\n System.out.print(sb.toString());\n }\n}\n \n \n// somlLeDvfT6GMIAXzWU\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 305079113,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define ll long long\n#define pb push_back\n#define po pop_back\n#define vl vector<long long>\n#define vi vector<int>\n#define vt vector<int64_t>\n#define vs vector<string>\n#define vii vector<pair<int, int>>\n#define vll vector<pair<long long, long long>>\n#define si set<int>\n#define mii map<int,int>\n#define umii unordered_map<int, int>\n#define umll unordered_map<long long, long long>\n#define all(v) sort(v.begin(),v.end())\n#define rall(v) sort(v.rbegin(),v.rend())\n#define sortall(a) sort(a,a+n)\n#define string(s) sort(s.begin(), s.end())\n#define rstring(s) reverse(s.begin(), s.end())\n#define repa(a,n) for (ll i = a; i < (ll)(n); i++)\n#define repb(b,n) for (ll i = b; i <= (ll)(n); i++)\n#define PI 3.1415926535897932384626433832795\n#define print(x) cout<<(x)<<endl;\n \n \n \n \n \n \nint main() {\n#ifndef ONLINE_JUDGE\nfreopen(\"inputf.in\", \"r\", stdin);\nfreopen(\"outputf.in\", \"w\", stdout);\n#endif \n \n \nint t;\ncin>>t;\nwhile(t--) {\n ll n, q;\n cin >> n >> q;\n \n vector<vector<ll>> M(n, vector<ll>(n, 0));\n for(ll r = 0; r < n; r++){\n for(ll c = 0; c < n; c++){\n cin >> M[r][c];\n }\n }\n \n vector<vector<ll>> S(n+1, vector<ll>(n+1, 0));\n vector<vector<ll>> Sr(n+1, vector<ll>(n+1, 0));\n vector<vector<ll>> Sc(n+1, vector<ll>(n+1, 0));\n \n for(ll r = 1; r <= n; r++){\n ll current_S = 0;\n ll current_Sr = 0;\n ll current_Sc = 0;\n for(ll c = 1; c <= n; c++){\n ll val = M[r-1][c-1];\n current_S += val;\n current_Sr += val * r;\n current_Sc += val * c;\n \n S[r][c] = S[r-1][c] + current_S;\n Sr[r][c] = Sr[r-1][c] + current_Sr;\n Sc[r][c] = Sc[r-1][c] + current_Sc;\n }\n }\n \n string s;\n s.reserve(q * 20);\n for(ll i = 0; i < q; i++){\n ll x1, y1, x2, y2;\n cin >> x1 >> y1 >> x2 >> y2;\n \n ll S_sub = S[x2][y2] - S[x1-1][y2] - S[x2][y1-1] + S[x1-1][y1-1];\n ll Sr_sub = Sr[x2][y2] - Sr[x1-1][y2] - Sr[x2][y1-1] + Sr[x1-1][y1-1];\n ll Sc_sub = Sc[x2][y2] - Sc[x1-1][y2] - Sc[x2][y1-1] + Sc[x1-1][y1-1];\n \n ll W = y2 - y1 + 1;\n ll sum_val = (ll)W * (Sr_sub - (ll)x1 * S_sub) + (Sc_sub - (ll)(y1-1) * S_sub);\n \n s += to_string(sum_val) + ' ';\n }\n \n cout << s << '\\n';\n}\n \n \n \n \n \n \n \n \n \n \n \n \n \nreturn 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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296731220,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\n#include \"bits/stdc++.h\"\n#define int long long\n#define uint unsigned long long\n#define vi vector<int>\n#define vvi vector<vi > \n#define vb vector<bool>\n#define vvb vector<vb >\n#define fr_3(i,n) for(int i=0; i<(n); i++)\n#define fb_3(i,a,n) for(int i=(a); i<=(n); i++)\n#define nl cout<<\"\\n\"\n#define dbg(var) cout<<#var<<\"=\"<<var<<\" \"\n#define all_3(v) v.begin(),v.end()\n#define srt_3(v) sort(v.begin(),v.end()) // sort \n#define mxe_3(v) *max_element(v.begin(),v.end()) // find max element in vector\n#define mne_3(v) *min_element(v.begin(),v.end()) // find min element in vector\n#define unq_3(v) v.resize(distance(v.begin(), unique(v.begin(), v.end())));\n// make sure to sort before applying unique // else only consecutive duplicates would be removed \n#define bin(x,y) bitset<y>(x) \nusing namespace std;\n // Hardcoded, directly change from here for functions!\n \nint MOD=1e9+7; \n \nvoid mod_add(int &a , int b) {a=((a%MOD)+(b%MOD))%MOD;}\nvoid mo_dsub(int &a , int b) {a=((a%MOD)-(b%MOD)+MOD)%MOD;}\nvoid mo_dmul(int &a , int b) {a=((a%MOD)*(b%MOD))%MOD;}\n// ================================== take ip/op like vector,pairs directly!==================================\ntemplate<typename typC,typename typD> istream &operator>>(istream &cin,pair<typC,typD> &a) { return cin>>a.first>>a.second; }\ntemplate<typename typC> istream &operator>>(istream &cin,vector<typC> &a) { for (auto &x:a) cin>>x; return cin; }\ntemplate<typename typC,typename typD> ostream &operator<<(ostream &cout,const pair<typC,typD> &a) { return cout<<a.first<<' '<<a.second; }\ntemplate<typename typC,typename typD> ostream &operator<<(ostream &cout,const vector<pair<typC,typD>> &a) { for (auto &x:a) cout<<x<<'\\n'; return cout; }\ntemplate<typename typC> ostream &operator<<(ostream &cout,const vector<typC> &a) { int n=a.size(); if (!n) return cout; cout<<a[0]; for (int i=1; i<n; i++) cout<<' '<<a[i]; return cout; }\n// ===================================END Of the input module ==========================================\n \n \n \nvoid funny_3(){\n // int n=1,m=0;\n // string s_3;\n int n_08o, q_08o;\n cin >> n_08o >> q_08o;\n \n vector<vector<int>> Munni(n_08o, vector<int>(n_08o));\n vector<vector<long long>> p_o_x(n_08o + 1, vector<long long>(n_08o + 1, 0));\n \n for (int i = 0; i < n_08o; ++i) {\n for (int j = 0; j < n_08o; ++j) {\n cin >> Munni[i][j];\n p_o_x[i + 1][j + 1] = Munni[i][j] + p_o_x[i + 1][j] + p_o_x[i][j + 1] - p_o_x[i][j];\n }\n }\n \n \n while (q_08o--) {\n int x10_, y10_, x20_, y20_;\n cin >> x10_ >> y10_ >> x20_ >> y20_;\n \n \n long long t_l_sum = 0;\n int index = 1;\n \n \n for (int i = x10_ - 1; i < x20_; ++i) {\n for (int j = y10_ - 1; j < y20_; ++j) {\n t_l_sum += Munni[i][j] * index;\n index++;\n }\n }\n \n cout << t_l_sum << \" \";\n }\n cout << endl;\n // cin>>n;\n // vi v(n);\n // cin>>v;\n \n}\n \nint32_t main()\n{\n \n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n \n int t_c_3= 1;\n cin >> t_c_3;\n while (t_c_3--)\n {\n funny_3();\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296668765,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\nt = int(input())\nfor _ in range(t):\n n, q = map(int, input().split())\n matrix = []\n for _ in range(n):\n row = list(map(int, input().split()))\n matrix.append(row)\n results = []\n for _ in range(q):\n x1, y1, x2, y2 = map(int, input().split())\n # θ°ζ΄δΈΊ 0 η΄’εΌ\n x1 -= 1\n y1 -= 1\n x2 -= 1\n y2 -= 1\n A = []\n for i in range(x1, x2 + 1):\n for j in range(y1, y2 + 1):\n A.append(matrix[i][j])\n total = 0\n for i, num in enumerate(A, start=1):\n total += num * i\n results.append(total)\n print(\" \".join(map(str, results)))\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 304236858,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\nt=int(input())\nfor i in range(t):\n (n,q)=map(int,input().split())\n mi=[]\n for j in range(n):\n \n rdl=input()\n mir=list(map(int,rdl.split()))\n mi.append(mir)\n qryinp=[]\n for f in range(q):\n rdll=input()\n lqry=list(map(int,rdll.split()))\n qryinp.append(lqry)\n \n \n smml=[]\n \n for g in range(len(qryinp)):\n x1=qryinp[g][0]\n y1=qryinp[g][1]\n x2=qryinp[g][2]\n y2=qryinp[g][3]\n lansqry=[]\n \n smm=0\n k=1\n for rw in range(x1-1,x2):\n \n for cl in range(y1-1,y2):\n # print(cl)\n lansqry.append(mi[rw][cl])\n smm+=(mi[rw][cl])*k\n k+=1\n smml.append(smm)\n \n \n # smm=0\n # for b in range(len(lansqry)):\n # smm+=lansqry[b]*(b+1)\n \n # smml.append(smm)\n \n print(*smml)\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296832311,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\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\t\n\t\tfor(int i = 1; i <= T; i++){\n\t\t\t\n\t\t\tString[] s = br.readLine().split(\" \");\n\t\t\tint N = Integer.parseInt(s[0]);\n\t\t\tint Q = Integer.parseInt(s[1]);\n\t\t\tlong[][] p = new long[N][N];\n\t\t\tlong[][] p2 = new long[N][N];\n\t\t\tlong[][] p3 = new long[N][N];\n\t\t\tfor(int j = 0; j < N; j++){\n\t\t\t\ts = br.readLine().split(\" \");\n\t\t\t\tfor(int k = 0; k < N; k++){\n\t\t\t\t\tlong val = Long.parseLong(s[k]);\n\t\t\t\t\tp[j][k] = (j >= 1 ? p[j-1][k] : 0) + (k >= 1 ? p[j][k-1] : 0) - \n\t\t\t\t\t((j >= 1 && k >= 1) ? p[j-1][k-1] : 0) + val;\n\t\t\t\t\tp2[j][k] = (j >= 1 ? p2[j-1][k] : 0) + (k >= 1 ? p2[j][k-1] : 0) - \n\t\t\t\t\t((j >= 1 && k >= 1) ? p2[j-1][k-1] : 0) + (j * val);\n\t\t\t\t\tp3[j][k] = (j >= 1 ? p3[j-1][k] : 0) + (k >= 1 ? p3[j][k-1] : 0) - \n\t\t\t\t\t((j >= 1 && k >= 1) ? p3[j-1][k-1] : 0) + (k * val);\n \n\t\t\t\t}\n\t\t\t}\n\t\t\tfor(int l = 1; l <= Q; l++){\n\t\t\t\ts = br.readLine().split(\" \");\n\t\t\t\tint x1 = Integer.parseInt(s[0]);\n\t\t\t\tint y1 = Integer.parseInt(s[1]);\n\t\t\t\tint x2 = Integer.parseInt(s[2]);\n\t\t\t\tint y2 = Integer.parseInt(s[3]);\n\t\t\t\tx1--; y1--; x2--; y2--;\n\t\t\t\t// First take the actual 2D prefix sum within these bounds\n\t\t\t\tlong plainTotal = p[x2][y2] - (y1 >= 1 ? p[x2][y1-1] : 0) - \n\t\t\t\t(x1 >= 1 ? p[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p[x1-1][y1-1] : 0);\n\t\t\t\tlong total2 = p2[x2][y2] - (y1 >= 1 ? p2[x2][y1-1] : 0) - \n\t\t\t\t(x1 >= 1 ? p2[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p2[x1-1][y1-1] : 0);\n\t\t\t\ttotal2 -= x1 * plainTotal;\n\t\t\t\tlong total3 = p3[x2][y2] - (y1 >= 1 ? p3[x2][y1-1] : 0) - \n\t\t\t\t(x1 >= 1 ? p3[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p3[x1-1][y1-1] : 0);\n\t\t\t\ttotal3 -= y1 * plainTotal;\n\t\t\t\tint numCols = y2 - y1 + 1;\n\t\t\t\tSystem.out.print((plainTotal + total3 + numCols * total2) + \" \");\n \n\t\t\t}\n\t\t\tSystem.out.println();\n\t\t\t\n \n\t\t}\n\t}\n}\n \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 i = 1; i <= T; i++){\n\t\t\tString[] s = br.readLine().split(\" \");\n\t\t\tint N = Integer.parseInt(s[0]);\n\t\t\tint Q = Integer.parseInt(s[1]);\n\t\t\tlong[][] grid = new long[N][N];\n\t\t\tlong[][] grid2 = new long[N][N];\n\t\t\tlong[][] grid3 = new long[N][N];\n\t\t\tlong[][] p = new long[N][N];\n\t\t\tlong[][] p2 = new long[N][N];\n\t\t\tlong[][] p3 = new long[N][N];\n\t\t\tfor(int j = 0; j < N; j++){\n\t\t\t\ts = br.readLine().split(\" \");\n\t\t\t\tfor(int k = 0; k < N; k++){\n\t\t\t\t\tgrid[j][k] = Long.parseLong(s[k]);\n\t\t\t\t\tgrid2[j][k] = j * grid[j][k];\n\t\t\t\t\tgrid3[j][k] = k * grid[j][k];\n\t\t\t\t\tp[j][k] = (j >= 1 ? p[j-1][k] : 0) + (k >= 1 ? p[j][k-1] : 0) - \n\t\t\t\t\t((j >= 1 && k >= 1) ? p[j-1][k-1] : 0) + grid[j][k];\n\t\t\t\t\tp2[j][k] = (j >= 1 ? p2[j-1][k] : 0) + (k >= 1 ? p2[j][k-1] : 0) - \n\t\t\t\t\t((j >= 1 && k >= 1) ? p2[j-1][k-1] : 0) + grid2[j][k];\n\t\t\t\t\tp3[j][k] = (j >= 1 ? p3[j-1][k] : 0) + (k >= 1 ? p3[j][k-1] : 0) - \n\t\t\t\t\t((j >= 1 && k >= 1) ? p3[j-1][k-1] : 0) + grid3[j][k];\n \n\t\t\t\t}\n\t\t\t}\n\t\t\tfor(int l = 1; l <= Q; l++){\n\t\t\t\ts = br.readLine().split(\" \");\n\t\t\t\tint x1 = Integer.parseInt(s[0]);\n\t\t\t\tint y1 = Integer.parseInt(s[1]);\n\t\t\t\tint x2 = Integer.parseInt(s[2]);\n\t\t\t\tint y2 = Integer.parseInt(s[3]);\n\t\t\t\tx1--; y1--; x2--; y2--;\n\t\t\t\t// First take the actual 2D prefix sum within these bounds\n\t\t\t\tlong plainTotal = p[x2][y2] - (y1 >= 1 ? p[x2][y1-1] : 0) - \n\t\t\t\t(x1 >= 1 ? p[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p[x1-1][y1-1] : 0);\n\t\t\t\tlong total2 = p2[x2][y2] - (y1 >= 1 ? p2[x2][y1-1] : 0) - \n\t\t\t\t(x1 >= 1 ? p2[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p2[x1-1][y1-1] : 0);\n\t\t\t\ttotal2 -= x1 * plainTotal;\n\t\t\t\tlong total3 = p3[x2][y2] - (y1 >= 1 ? p3[x2][y1-1] : 0) - \n\t\t\t\t(x1 >= 1 ? p3[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p3[x1-1][y1-1] : 0);\n\t\t\t\ttotal3 -= y1 * plainTotal;\n\t\t\t\tint numCols = y2 - y1 + 1;\n\t\t\t\tSystem.out.print((plainTotal + total3 + numCols * total2) + \" \");\n \n\t\t\t}\n\t\t\tSystem.out.println();\n\t\t\t\n \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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 296914000,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Hard Demon Problem\ntime_limit_ms: 3500\nmemory_limit_mb: 512\nDescription: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.\n\nSwing is given an $$$n \\times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.\n\nFor each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \\ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \\ldots, M_{(x2,y2)}]$$$.\n\nThe following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.\n\nAfterwards, he asks you for the value of $$$\\sum_{i=1}^{|A|} A_i \\cdot i$$$ (sum of $$$A_i \\cdot i$$$ over all $$$i$$$).\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 10^3$$$) β the number of test cases.\n\nThe first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \\leq n \\leq 2000, 1 \\leq q \\leq 10^6$$$) β the length of $$$M$$$ and the number of queries.\n\nThe following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \\ldots, M_{(i,n)}$$$ ($$$1 \\leq M_{(i, j)} \\leq 10^6$$$).\n\nThe following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \\leq x_1 \\leq x_2 \\leq n, 1 \\leq y_1 \\leq y_2 \\leq n$$$) β the bounds of the query.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.\n\nOutput: For each test case, output the results of the $$$q$$$ queries on a new line.\n\nExamples:\ninput:\n2\n4 3\n1 5 2 4\n4 9 5 3\n4 5 2 3\n1 5 5 2\n1 1 4 4\n2 2 3 3\n1 2 4 3\n3 3\n1 2 3\n4 5 6\n7 8 9\n1 1 1 3\n1 3 3 3\n2 2 2 2\noutput:\n500 42 168 \n14 42 5\n\nNote: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \\cdot 9 + 2 \\cdot 5 + 3 \\cdot 5 + 4 \\cdot 2 = 42$$$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n static int n, q;\n static int [][] M;\n public static void main(String [] args) throws Exception {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n StringTokenizer st;\n \n int t = Integer.parseInt(br.readLine());\n while (t --> 0) {\n st = new StringTokenizer(br.readLine());\n n = Integer.parseInt(st.nextToken());\n q = Integer.parseInt(st.nextToken());\n M = new int[n + 1][n + 1];\n for (int i = 1; i <= n; i++) {\n st = new StringTokenizer(br.readLine());\n for (int j = 1; j <= n; j++) {\n M[i][j] = Integer.parseInt(st.nextToken());\n }\n }\n long [][] prefix1 = new long[n + 1][n + 1]; // Regular\n long [][] prefix2 = new long[n + 1][n + 1]; // Multiplied by index\n long [][] prefix3 = new long[n + 1][n + 1]; // Multiplied by row\n for (int i = 1; i <= n; i++) {\n for (int j = 1; j <= n; j++) {\n prefix1[i][j] = prefix1[i - 1][j] + prefix1[i][j - 1] - prefix1[i - 1][j - 1];\n prefix2[i][j] = prefix2[i - 1][j] + prefix2[i][j - 1] - prefix2[i - 1][j - 1];\n prefix3[i][j] = prefix3[i - 1][j] + prefix3[i][j - 1] - prefix3[i - 1][j - 1];\n prefix1[i][j] += M[i][j]; prefix2[i][j] += (long) M[i][j] * (i * n + j); prefix3[i][j] += (long) M[i][j] * i;\n }\n }\n \n while (q --> 0) {\n st = new StringTokenizer(br.readLine());\n int x1 = Integer.parseInt(st.nextToken());\n int y1 = Integer.parseInt(st.nextToken());\n int x2 = Integer.parseInt(st.nextToken());\n int y2 = Integer.parseInt(st.nextToken());\n \n long res = sum(prefix2, x1, y1, x2, y2);\n res -= sum(prefix3, x1, y1, x2, y2) * (n - (y2 - y1 + 1));\n res += sum(prefix1, x1, y1, x2, y2) * (x1 * (n - (y2 - y1 + 1)) - (x1 * n + y1 - 1));\n System.out.print(res + \" \");\n }\n System.out.println();\n }\n \n br.close();\n }\n \n public static long sum(long [][] prefix, int l1, int r1, int l2, int r2) {\n return prefix[l2][r2] - prefix[l1 - 1][r2] - prefix[l2][r1 - 1] + prefix[l1 - 1][r1 - 1];\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044H",
"submission_id": 298474531,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\n#include <iostream>\n#include <algorithm>\nusing namespace std;\n \n#define X 15\n#define Y 2\n \nint main() {\n\tint t; cin >> t;\n\twhile (t--) {\n\t\tint x, y; cin >> x >> y;\n\t\tint cnt = 0;\n\t\tcnt += y / Y;\n\t\tif (y % 2) cnt++;\n\t\tint num = cnt * X - y * 4;\n\t\tif (x <= num) cout << cnt << endl;\n\t\telse {\n\t\t\tx -= num;\n\t\t\tif (x <= X) cout << cnt + 1 << endl;\n\t\t\telse cout << x / X + 1 << endl;\n\t\t}\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261869266,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define ll long long\n#define cy cout << \"YES\" << endl\n#define cn cout << \"NO\" << endl\n#define c1 cout<< -1 <<endl\n#define PI 3.141592653589793238462643383279\nconst ll mod= 1e9 + 7;\n \nll BinExp(ll a,ll b,ll mod){ll ans=1;while(b>0){if(b&1){ans=(ans*1LL*a)%mod;}a=(a*1LL*a)%mod;b>>=1;}return ans;}\nll m_mod(ll x, ll y) { return ((x % mod) * (y % mod)) % mod;}\nll m_divmod(ll x,ll y,ll m){return (x%mod*BinExp(y,m-2,m))%m;}\nint main(){\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n int t;\n cin>>t;\n while(t--){\n int x,y;\n cin>>x>>y;\n int ct,yt;\n if(y==0){\n if(x==0){\n cout<<0<<endl;\n }\n else{\n ct = (int)(x/15);\n if(x%15 != 0){\n ct++;\n }\n cout<<ct<<endl;\n }\n }\n else if(x==0){\n yt = (int)(y/2);\n if(y%2 != 0){\n yt++;\n }\n cout<<yt<<endl;\n }\n else{\n ct = (int)(x/7);\n yt = (int)(y/2);\n if(x%7 != 0){\n ct++;\n }\n if(y%2 != 0){\n yt++;\n }\n if(yt>=ct){\n cout<<yt<<endl;\n }\n else{\n y = y%2;\n x = x%7;\n if(y==1){\n if(x<=4){\n cout<<yt<<endl;\n }\n else{\n cout<<ct<<endl;\n }\n }\n else{\n cout<<ct<<endl;\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 262368618,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n// Macros\n#define fast_io ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);\n#define ll long long\n#define ull unsigned long long\n#define vi vector<int>\n#define vll vector<long long>\n#define pii pair<int, int>\n#define pll pair<long long, long long>\n#define all(v) v.begin(), v.end()\n#define rall(v) v.rbegin(), v.rend()\n#define pb push_back\n#define mp make_pair\n#define F first\n#define S second\n#define sz(x) (int)(x).size()\n#define rep(i,a,b) for(int i = a; i < b; ++i)\n#define rrep(i,a,b) for(int i = a; i >= b; --i)\n \n// Constants\nconst int MOD = 1e9 + 7;\nconst int INF = 1e9;\nconst ll LINF = 1e18;\nconst double EPS = 1e-9;\n \nint ceil_div(int a, int b) {\n return (a + b - 1) / b;\n}\n \nvoid solve() {\n int a, b;\n cin >> a >> b;\n int count = 0;\n if(b==0){\n count=ceil_div(a,15);\n }\n if(b%2==1){\n count=b/2+1;\n int x=9+7*(count);\n if(a>x){\n count+=ceil_div(a-x,15);\n }\n }\n if(b%2==0){\n count=b/2;\n int x=7*(count);\n \n if(a>x){\n count+=ceil_div(a-x,15);\n }\n }\n \n \n \n \n cout << count << endl;\n}\n \nint main() {\n fast_io\n \n int t;\n cin >> t; \n while (t--) {\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261834176,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\nt = int(input())\nif (t>=1) and (t<=10000) :\n A = []\n for i in range(t) :\n a = []\n x,y = map(int, input().split())\n a.append(x)\n a.append(y)\n A.append(a)\n \n \n for i in range(len(A)) :\n if (A[i][0]>=0) and (A[i][0]<=99) and (A[i][1]>=0) and (A[i][1]<=99) :\n count = A[i][0] + A[i][1]*4\n if A[i][1]%2==0 :\n if A[i][1]==0 :\n if A[i][0]%15==0 :\n screens = A[i][0]/15\n else :\n screens = A[i][0]//15 + 1\n else :\n screens = A[i][1]/2\n else :\n screens = A[i][1]//2 + 1\n if (A[i][0]>((screens*15)-(A[i][1]*4))) :\n screens += (A[i][0] - (screens*7))//15 + 1 \n print(int(screens))\n else :\n print(int(screens))\n else :\n print(\"Error\")\nelse :\n print(\"Error\")\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261832169,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\ndef box(x,y):\n if x <= 0 and y <= 0:\n return 0\n if x > 0 and y <= 0:\n return 1 + box(x-15,y)\n elif x <= 0 and y >0:\n return 1 + box(x,y-2)\n else:\n if y <= 2:\n if x + 4*y <= 15: return 1\n else: return 1 + box(x-7,0)\n elif y > 2:\n return 1 + box(x-7,y-2)\nn = int(input())\nfor i in range(n):\n k = input()\n s = []\n for j in k.split():\n s.append(int(j))\n print(box(s[0],s[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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261855536,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\ndef box(x,y):\n if x <= 0 and y <= 0:\n return 0\n if x > 0 and y <= 0:\n return 1 + box(x-15,y)\n elif x <= 0 and y >0:\n return 1 + box(x,y-2)\n else:\n if x + y*4 <= 15: return 1\n else:\n return 1 + box(x-7,y-2)\nn = int(input())\nfor i in range(n):\n k = input()\n s = []\n for j in k.split():\n s.append(int(j))\n print(box(s[0],s[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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261840599,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\n// Online Java Compiler\n// Use this editor to write, compile and run your Java code online\nimport java.util.*;\npublic class HelloWorld {\n public static void main(String[] args) {\n Scanner Sc=new Scanner(System.in);\n int N=Sc.nextInt();\n Sc.nextLine();\n for(int j=0;j<N;j++){\n int XX=0;\n String S=Sc.nextLine();\n String L[]=S.split(\" \");\n int X=Integer.parseInt(L[0]);\n int Y=Integer.parseInt(L[1]);\n \n if(Y%2==0){\n XX+=Y/2;\n }\n else{\n XX+=(Y/2)+1;\n }\n int TB=15*XX;\n int BL=TB-(Y*4);\n if(BL>=X){\n System.out.println(XX);\n }\n else{\n XX+=1;\n System.out.println(XX);\n }\n \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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 269241700,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\nimport java.util.*;\npublic class Cf{\n public static void main(String args[]){\n Scanner s=new Scanner(System.in);\n int t=s.nextInt();\n // int t=1;\n while(t-- >0){\n int x=s.nextInt();\n int y=s.nextInt();\n // int x=18;\n // int y=3;\n int res=0;\n if(x==0 && y==0)\n System.out.println(0);\n else if (x==0)\n System.out.println((y+1)/2);\n else if(y==0)\n System.out.println((x+14)/15);\n else{\n while(x>0 || y>0){\n if(y<=0){\n x=x-15;\n res++;\n }\n else if(y%2==0){\n x=x-7;\n y=y-2;\n res++;\n }\n else{\n x=x-11;\n y=y-2;\n res++;\n }\n }\n System.out.println(res);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 263111374,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \nint main() {\n int t;\n cin >> t;\n \n while(t--) {\n int x, y, ans = 0;\n cin >> x >> y;\n \n int c = y / 2;\n if(y % 2 == 1) c++;\n \n ans = c;\n \n if(c == 0) {\n ans = x / 15;\n if(x % 15 != 0) ans++;\n \n cout << ans << endl;\n continue;\n }\n \n if(y % 2 == 0) {\n if(x > c * 7) {\n ans++;\n x -= 15;\n }\n else x = 0;\n \n while(x > 0) {\n ans++;\n x -= 15;\n }\n }\n else {\n if(x > (c - 1) * 7 + 11) {\n ans++;\n x -= 15;\n }\n else x = 0;\n \n while(x > 0) {\n ans++;\n x -= 15;\n }\n }\n \n cout << ans << 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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 262888992,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define fio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);\n#define ll long long int\n#define nl cout<<endl;\nint main()\n{\n ll t;\n cin>>t;\n while(t--){\n ll x,y;\n cin>>x>>y;\n ll r=ceil(y/2.0);\n if(x==0 && y==0)cout<<\"0\"<<endl;\n else if(x==0)cout<<ceil(y/2.0)<<endl;\n else if(y==0)cout<<ceil(x/15.0)<<endl;\n else if(r*(15-y*4)>=x)cout<<r<<endl;\n else{\n ll d = x-(r*7);\n cout<<r+ceil(d/15.0)<<endl;\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261842139,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\nimport math\nt=int(input())\nfor _ in range(t):\n a,b=map(int,input().split())\n if b<=2:\n x=(a+b*4)\n print(math.ceil(x/15))\n else:\n x=math.ceil(b/2)\n n=x*15\n m=b*4\n o=a//(n-m)\n print(x+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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261837199,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\nfrom collections import Counter\nimport math as mt\n \n \ndef si():\n return input()\n \n \ndef ii():\n return int(input())\n \n \ndef fi():\n return float(input())\n \n \ndef ili(n=None):\n if n:\n return list(map(int, input().split()))[:n]\n return list(map(int, input().split()))\n \n \ndef fli(n=None):\n if n:\n return list(map(float, input().split()))[:n]\n return list(map(float, input().split()))\n \n \ndef sli(n=None):\n if n:\n return list(map(str, input().split()))[:n]\n return list(map(str, input().split()))\n \n \ndef minInd(m: list):\n mn = m[0]\n ind = 0\n l = len(m)\n for i in range(1, l):\n if m[i] < mn:\n mn = m[i]\n ind = i\n return [mn, ind]\n \n \ndef main():\n t = ii()\n for _ in range(t):\n n, k = ili(2)\n if k == 0 and n == 0:\n print(0)\n elif k == 0:\n print(mt.ceil(n / 15))\n elif n == 0:\n print(mt.ceil(k / 2))\n elif k == 1:\n n -= 11\n if n < 0:\n print(1)\n else:\n print(mt.ceil(n / 15) + 1)\n else:\n kdis = mt.ceil(k / 2)\n n -= (kdis * 7)\n if n < 0:\n print(kdis)\n else:\n print(kdis + mt.ceil(n / 15))\n \n \n \nif __name__ == \"__main__\": # 12 4 6\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261840122,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\nimport java.util.*;\n \npublic class Main{\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 x = sc.nextInt();\n int y = sc.nextInt();\n if((x * y) != 0 && ((x + 4*y) > 15) ){\n System.out.println((int)Math.ceil((x + 4*y)/15.0));\n }\n else if(x == 0 && y==0){\n System.out.println(0);\n }\n else if(x != 0 && y == 0){\n System.out.println((int)Math.ceil(x/15.0));\n }\n else if(x == 0 && y != 0){\n System.out.println((int)Math.ceil(y/2.0));\n }\n else{\n System.out.println(1);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261868253,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\nimport java.util.*;\npublic class Main {\n public static void main(String[]args)\n {Scanner sc=new Scanner(System.in);\n// System.out.println(Math.ceil(1.0/7));\n int t=sc.nextInt();\n for(int i=0;i<t;i++)\n {\n int x=sc.nextInt();\n double y=sc.nextDouble();\n int m=(int)Math.ceil(y/2);\n double num=7.0*m;\n double hh=x-num;\n if(m==0)\n {\n System.out.println((int)Math.ceil((double)x/15));\n continue;\n }\n if(y%2==1)\n {\n if((int)num>=x-1)\n {\n System.out.println(m);continue;\n }\n else{\n System.out.println((int)Math.ceil((hh-1)/7)+m);\n continue;\n }\n \n }\n if((int)num>=x)\n {\n System.out.println(m);\n }\n else{\n System.out.println((int)Math.ceil(hh/7)+m);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261827496,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\n#include <iostream>\n#include <cmath>\n#include <string>\n#include <algorithm>\n#include <iomanip>\n#include <vector>\n#include <bitset>\n#include <set>\n#include <unordered_set>\n \nusing namespace std;\n \n#define DestroeD ios::sync_with_stdio(false);cout.tie(NULL);cin.tie(NULL);\n#define ll long long\n \nvoid solve() {\n int a, b; cin >> a >> b;\n int num;\n if(b%2 == 0){\n num = b/2;\n a -= num*7;\n if(a <= 0){\n cout << num << endl;\n }\n else{\n cout << num + ceil((double)a / 15) << endl;\n }\n }\n else{\n num = b/2+1;\n a -= num*7 + 11;\n if(a <= 0){\n cout << num << endl;\n }\n else{\n cout << num + ceil((double)a / 15) << endl;\n }\n }\n}\n \nint main() {\n DestroeD\n \n int t; cin >> t;\n \n while(t--) {\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261813311,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n \nint main()\n{\n int t;\n cin >> t;\n int i = 1;\n while(t--)\n {\n int x, y;\n cin >> x >> y;\n if(i == 1503) cout << \"x = \" << x << \" y = \" << y << \"\\n\";\n i++;\n \n int screens = (y + 1) / 2;\n \n int remain = screens * 15 - 4 * y;\n \n if(remain > 0)\n screens += ((x - 1) / remain);\n \n else\n screens += ((x+14)/15);\n \n cout << screens << \"\\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 263245080,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\nimport math\n \ndef min_screens(x, y):\n count = 0\n a = 15\n count = math.ceil(y/2)\n total_cells = x + (2 * count)\n o = (15*count) -(4*y) - x\n if o < 0 :\n return(math.ceil(x/15) + count)\n else:\n return(count)\n \n \n \nt = int(input())\n \nfor _ in range(t):\n \n x, y = map(int, input().split())\n \n print(min_screens(x, y))\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261884789,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\nn = int(input())\n \nfor i in range(n):\n a, b = list(map(int, input().split()))\n if b>a:\n k = (-(-b//2))\n k += a//15\n print(k)\n else:\n tg = a + (b*4)\n print(-(-tg//15))\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 263543897,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Q1\n{\n public static void main(String[] args)\n {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n \n while(t-- > 0)\n {\n int x = sc.nextInt(); // 1 x 1\n int y = sc.nextInt(); // 2 x 2\n \n // screen size = 5 x 3\n int count = (int) Math.ceil((float) y / 2.0);\n \n if (y % 2 == 0)\n {\n if (x > count * 7)\n {\n // System.out.println(\"Here\" + count);\n count += (int) Math.ceil((x - count * 7) / 15.0);\n }\n }\n else\n {\n if (x > (count - 1) * 7 + 11)\n {\n // System.out.println(\"Here\" + count);\n count += (int) Math.ceil((x - (count - 1) * 7) / 15.0);\n }\n }\n System.out.println(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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 261821324,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Phone Desktop\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \\times 3$, i.e., five rows and three columns.\n\nThere are $x$ applications with an icon size of $1 \\times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \\times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.\n\nRosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.\n\nInput: The first line of the input contains $t$ ($1 \\leq t \\leq 10^4$) β the number of test cases.\n\nThe first and only line of each test case contains two integers $x$ and $y$ ($0 \\leq x, y \\leq 99$) β the number of applications with a $1 \\times 1$ icon and the number of applications with a $2 \\times 2$ icon, respectively.\n\nOutput: For each test case, output the minimal number of required screens on a separate line.\n\nExamples:\ninput:\n11\n1 1\n7 2\n12 4\n0 3\n1 0\n8 1\n0 0\n2 0\n15 0\n8 2\n0 9\noutput:\n1\n1\n2\n2\n1\n1\n0\n1\n1\n2\n5\n\nNote: The solution for the first test case can look as follows:\n\n Blue squares represent empty spaces for icons, green squares represent $1 \\times 1$ icons, red squares represent $2 \\times 2$ icons\n\nThe solution for the third test case can look as follows:\n\n\n\nBuggy Code:\nimport java.util.*;\npublic class Cf{\n public static void main(String args[]){\n Scanner s=new Scanner(System.in);\n int t=s.nextInt();\n // int t=1;\n while(t-- >0){\n int x=s.nextInt();\n int y=s.nextInt();\n // int x=18;\n // int y=3;\n int res=0;\n if(x==0 && y==0)\n System.out.println(0);\n else if (x==0)\n System.out.println((y+1)/2);\n else if(y==0)\n System.out.println((x+14)/15);\n else{\n while(x>0 || y>0){\n if(y==0){\n x=x-15;\n res++;\n }\n else if(y%2==0){\n x=x-7;\n y=y-2;\n res++;\n }\n else{\n x=x-11;\n y=y-2;\n res++;\n }\n }\n System.out.println(res);\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1974A",
"submission_id": 263110569,
"task_type": "TestCase-Eval-Task2"
} |
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