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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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nconst long double PI = acos(-1);\n//#define int long long\ntypedef long long ll;\ntypedef long double ld;\ntypedef unsigned long long ull;\ntypedef pair<int,int> PII;\ntypedef pair<ll, ll> PLL;\ntypedef tuple<ll, ll, ll> TLL;\nint lowbit(int x) {return x & (- x);};\nconst int N = 200010, mod = 998244353;\nconst int Mod = 1e9 + 7, INF = 0x3f3f3f3f;\n \nvector<ll> e[N];\n \nvoid solve(){\n ll n;\n cin >> n;\n ll l = 1, r = n, L = n + 1, R = 0;\n for(ll i = 1; i <= n; i ++){\n ll x;\n cin >> x;\n l = max(l, i - x + 1);\n r = min(r, i + x - 1);\n e[x].push_back(i);\n }\n for(int i = 1; i <= n; i ++){\n if(e[i].size() == 0) continue;\n L = min(L, e[i][0]), R = max(R, e[i][e[i].size() - 1]);\n e[i].clear();\n if(R - L + 1 > i){\n cout << 0 <<'\\n';\n return;\n }\n }\n if(l > r) cout << 0 << '\\n';\n else cout << r - l + 1 << '\\n';\n}\n \nint main(){\n ios::sync_with_stdio(0);\n cin.tie(0);cout.tie(0);\n int t = 1;\n cin >> t;\n while(t --){\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2018B",
"submission_id": 293645085,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\n#include <iostream>\n#include <map>\n#include <vector>\n \nvoid solve() {\n int n;\n std::cin >> n;\n std::vector<int> v(n);\n for (int i = 0; i < n; ++i) std::cin >> v[i];\n \n std::vector<int> first(n + 1);\n int cur = n;\n for (int i = 0; i < n; ++i) {\n while (cur >= v[i]) first[cur] = i, --cur;\n }\n cur = n;\n for (int i = n - 1; i >= 0; --i) {\n while (cur >= v[i]) {\n if (i - first[v[i]] >= v[i]) {\n std::cout << \"0\\n\";\n return;\n }\n --cur;\n }\n }\n \n std::vector<int> diff(n + 1);\n for (int i = 0; i < n; ++i) {\n int l = std::max(0, i - v[i] + 1);\n int r = std::min(n - 1, i + v[i] - 1);\n ++diff[l];\n --diff[r + 1];\n }\n cur = 0;\n int cnt = 0;\n for (auto el : diff) {\n cur += el;\n cnt += cur == n;\n }\n std::cout << cnt << '\\n';\n}\n \nint main() {\n int t;\n std::cin >> t;\n while (t--) solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2018B",
"submission_id": 293476357,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\n/*\n Những mùa thu khác có thể trở về\nNhưng mùa thu của chúng ta đã chết...\n*/\n \n// #pragma GCC optimize(\"Ofast,O3\") \n// #pragma GCC target(\"avx,avx2,fma,sse4\")\n#include <bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n \nusing namespace std;\nusing namespace __gnu_pbds;\n \ntypedef long long ll;\n \n#ifdef Debug\n#include \"./namsee_lib/Debug.h\"\n#else\n#define dbg(...)\n#endif\n \n#define FOR(i,a,b) for(int i=(a),_b=(b); i<=_b; i++)\n#define FORD(i,a,b) for(int i=(a),_b=(b); i>=_b; i--)\n#define REP(i,a) for(int i=0,_a=(a); i<_a; i++)\n \n#define F first\n#define S second\n#define MP make_pair\n#define PB push_back\n \n#define all(x) x.begin(), x.end()\n#define rall(x) x.rbegin(), x.rend()\n#define lb lower_bound\n#define ub upper_bound\n#define ins insert\n \n#define nl \"\\n\"\n \n#define vi vector<long long>\n#define vii vector<pair<long long, long long>>\n#define pi pair<ll, ll>\n \ntemplate <class T>\nusing Tree =\n tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\n \ntemplate<class X, class Y> bool ckmin(X& a, const Y& b) {return a > b ? a = b, 1 : 0;}\ntemplate<class X, class Y> bool ckmax(X& a, const Y& b) {return a < b ? a = b, 1 : 0;}\n \n// const ll M = 1e9 + 9; // (1ll << 61) - 1\n// mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());\n// const ll B = uniform_int_distribution<ll>(311, M - 1)(rng);\n \n \n \nint main()\n{\n ios_base::sync_with_stdio(false);\n cin.tie(0);\n \n #ifdef Debug\n freopen(\"in.txt\", \"r\", stdin);\n freopen(\"out.txt\", \"w\", stdout);\n #endif\n \n int t; cin >> t;\n while (t--) {\n int n; cin >> n;\n vector<vector<int>> pos(n + 1);\n REP(i, n) {\n int val; cin >> val;\n pos[val].PB(i);\n }\n int ans = 1e9, lf = 1e9, rt = 0;\n FOR(i, 1, n) {\n if (pos[i].empty()) continue;\n int l = pos[i][0], r = pos[i].back();\n if (r - l + 1 > i) {\n ans = 0;\n }\n if (ans == 1e9) {\n ans = r - l + 1;\n rt = r, lf = l;\n }\n else {\n ckmax(r, rt);\n ckmin(l, lf);\n if (r - l + 1 > i) ans = 0;\n }\n }\n cout << ans << nl;\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": "2018B",
"submission_id": 283786369,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class CodeForcesSolution {\n static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n static StringTokenizer st;\n /*static int mod = (int) 1e9+7;*/\n public static void main(String[] args) throws IOException {\n int t = readInt();\n while (t-- > 0) {\n int n = readInt();\n int[] a = new int[n + 1], cnt = new int[n + 1], mx = new int[n + 1], mi = new int[n + 1], df = new int[n + 5];\n for (int i = 1; i <= n; i++) {\n a[i] = readInt();\n if (mi[a[i]] == 0) mi[a[i]] = i;\n mx[a[i]] = i;\n cnt[a[i]]++;\n }\n Arrays.sort(a);\n int ans = 0, l = n, r = 1;\n for (int i = 1; i <= n; i += cnt[a[i]]) {\n l = Math.min(l, mi[a[i]]);\n r = Math.max(r, mx[a[i]]);\n if (a[i] < r - l + 1) {\n ans = 0; break;\n }\n int nl = Math.max(l, r - a[i] + 1), nr = Math.min(l + a[i] - 1, r);\n df[nl] += cnt[a[i]]; df[nr + 1] -= cnt[a[i]];\n }\n for (int i = 1; i <= n; i++) {\n df[i] += df[i - 1];\n if (df[i] == n) ans++;\n }\n System.out.println(ans);\n }\n }\n static String next() throws IOException {\n while (st == null || !st.hasMoreTokens())\n st = new StringTokenizer(br.readLine().trim());\n return st.nextToken();\n }\n static long readLong() throws IOException {\n return Long.parseLong(next());\n }\n static int readInt() throws IOException {\n return Integer.parseInt(next());\n }\n static double readDouble() throws IOException {\n return Double.parseDouble(next());\n }\n static char readCharacter() throws IOException {\n return next().charAt(0);\n }\n static String readLine() throws IOException {\n return br.readLine().trim();\n }\n static int readALotInt() throws IOException{\n int x = 0, c;\n while((c = br.read()) != ' ' && c != '\\n')\n x = x * 10 + (c - '0');\n return x;\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": "2018B",
"submission_id": 299705611,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class CodeForcesSolution {\n static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n static StringTokenizer st;\n /*static int mod = (int) 1e9+7;*/\n static int MM = (int) 1e6, tree[], lazy[];\n public static void main(String[] args) throws IOException {\n int t = readInt();\n while (t-- > 0) {\n tree = new int[MM]; lazy = new int[MM];\n int n = readInt();\n int[] a = new int[n + 1];\n for (int i = 1; i <= n; i++) {\n a[i] = readInt();\n updateRange(1, 1, n, i, i, a[i] - i + 1);\n }\n int ans = 0;\n for (int i = 1; i <= n; i++) {\n TreeSet<int[]> ts = new TreeSet<>((f, s) -> f[0] == s[0] ? f[1] - s[1] : f[0] - s[0]);\n for (int j = 1; j <= n; j++) {\n ts.add(new int[] {a[j], j});\n }\n int time = 1, l = i, r = i;\n boolean flg = true;\n while (!ts.isEmpty()) {\n int[] cur = ts.first(); ts.remove(cur);\n if (l <= cur[1] && cur[1] <= r) continue; // conquered\n if (time + Math.min(Math.abs(cur[1] - l), Math.abs(cur[1] - r)) > cur[0]) {\n flg = false; break;\n }\n time += Math.min(Math.abs(cur[1] - l), Math.abs(cur[1] - r));\n l = Math.min(l, cur[1]);\n r = Math.max(r, cur[1]);\n }\n if (flg) ans++;\n }\n System.out.println(ans);\n }\n }\n // l -> update left bound, r -> update right bound\n static void updateRange(int node, int start, int end, int l, int r, int val) {\n if(start > end || l > r) return; // Out of range\n if(lazy[node] != 0) { // Propagate lazy flag (if exists one)\n // This node needs to be updated\n tree[node] += (end - start + 1) * lazy[node]; // Update it\n if(start != end) {\n lazy[node*2] += lazy[node]; // Propagate lazy flag\n lazy[node*2+1] += lazy[node]; // Propagate lazy flag\n }\n lazy[node] = 0; // Reset lazy flag\n }\n if(start >= l && end <= r) { // reached the segment where we want to update\n // Segment is fully within range\n tree[node] += (end - start + 1) * val;\n if(start != end) { // Propagate\n // Not leaf node\n lazy[node*2] += val;\n lazy[node*2+1] += val;\n }\n return;\n }\n int mid = (start + end) / 2;\n if (l > mid) updateRange(node*2 + 1, mid + 1, end, l, r, val); // Updating right child\n else if (r<=mid) updateRange(node*2, start, mid, l, r, val); // Updating left child\n else {\n updateRange(2*node, start, mid, l, mid, val);\n updateRange(2*node+1, mid+1, end, mid+1, r, val);\n }\n tree[node] = Math.min(tree[node*2], tree[node*2+1]); // Updating root with min value\n }\n \n // l -> query left bound, r -> query right bound\n static int queryRange(int node, int start, int end, int l, int r) {\n if(start > end) return 0; // Out of range\n if(lazy[node] != 0) {\n // This node needs to be updated\n tree[node] += (end - start + 1) * lazy[node]; // Update it\n if(start != end)\n {\n lazy[node*2] += lazy[node]; // Mark child as lazy\n lazy[node*2+1] += lazy[node]; // Mark child as lazy\n }\n lazy[node] = 0; // Reset it\n }\n if (start == l && end == r) return tree[node];\n int m = (start+end)/2;\n if (l > m) return queryRange(2*node+1, m+1, end, l, r);\n else if (r<=m) return queryRange(2*node, start, m, l, r);\n else return Math.min(queryRange(2*node, start, m, l, m), queryRange(2*node+1, m+1, end, m+1, r));\n }\n static String next() throws IOException {\n while (st == null || !st.hasMoreTokens())\n st = new StringTokenizer(br.readLine().trim());\n return st.nextToken();\n }\n static long readLong() throws IOException {\n return Long.parseLong(next());\n }\n static int readInt() throws IOException {\n return Integer.parseInt(next());\n }\n static double readDouble() throws IOException {\n return Double.parseDouble(next());\n }\n static char readCharacter() throws IOException {\n return next().charAt(0);\n }\n static String readLine() throws IOException {\n return br.readLine().trim();\n }\n static int readALotInt() throws IOException{\n int x = 0, c;\n while((c = br.read()) != ' ' && c != '\\n')\n x = x * 10 + (c - '0');\n return x;\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": "2018B",
"submission_id": 299698651,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <string>\n#include <algorithm>\n#include <fstream>\n#include <cmath>\n#include <set>\n#include <cassert>\n#include <map>\nusing namespace std;\n \n#define int long long\n \ntypedef pair <int, int> ii;\n \nvector <int> suf, pref;\n \nint Right(int left, int pos, int right) {\n\treturn suf[right + 1] + right - (pos - left) - (right - pos);\n}\n \nint Left(int left, int pos, int right) {\n\treturn pref[left - 1] - left - (right - pos) - (pos - left);\n}\n \nsigned main() {\n \n#ifdef _DEBUG\n\tfstream cin(\"input.txt\");\n\tofstream cout(\"output.txt\");\n#endif\n \n\tios::sync_with_stdio(false);\n\tcin.tie(NULL);\n \n\tint t; cin >> t;\n\twhile (t--) {\n\t\t\n\t\tsuf.clear(); pref.clear();\n \n\t\tint n, ans = 0, success = -1; cin >> n;\n\t\tvector <int> val(n), sufIdx(n), prefIdx(n);\n\t\tsuf.resize(n); pref.resize(n);\n \n\t\tfor (int i = 0; i < n; i++)\n\t\t\tcin >> val[i];\n \n\t\tif (n == 1) {\n\t\t\tcout << 1 << '\\n';\n\t\t\tcontinue;\n\t\t}\n \n\t\tfor (int i = n - 1; i >= 0; i--) {\n\t\t\tsuf[i] = val[i] - i;\n\t\t\tsufIdx[i] = i;\n\t\t}\n\t\tfor (int i = n - 2; i >= 0; i--) {\n\t\t\tif (suf[i] > suf[i + 1])\n\t\t\t\tsufIdx[i] = sufIdx[i + 1];\n\t\t\tsuf[i] = min(suf[i], suf[i + 1]);\n\t\t}\n\t\tfor (int i = 0; i < n; i++) {\n\t\t\tpref[i] = val[i] + i;\n\t\t\tprefIdx[i] = i;\n\t\t\tbool ok = suf[i] + i > 0;\n\t\t\tif (i) {\n\t\t\t\tif (pref[i] > pref[i - 1])\n\t\t\t\t\tprefIdx[i] = prefIdx[i - 1];\n\t\t\t\tpref[i] = min(pref[i], pref[i - 1]);\n\t\t\t\tok = min(ok, pref[i] - i > 0);\n\t\t\t}\n\t\t\tif (ok && success == 1) ans++;\n\t\t\telse if (ok && success == -1) {\n\t\t\t\tint left = i, right = i;\n\t\t\t\twhile (true) {\n\t\t\t\t\tif (left == 0) {\n\t\t\t\t\t\tif (Right(left, i, right) > 0) {\n\t\t\t\t\t\t\tsuccess = 1;\n\t\t\t\t\t\t\tans++;\n\t\t\t\t\t\t}\n\t\t\t\t\t\telse success = 0;\n\t\t\t\t\t\tbreak;\n\t\t\t\t\t}\n\t\t\t\t\telse if (right == n - 1) {\n\t\t\t\t\t\tif (Left(left, i, right) > 0) {\n\t\t\t\t\t\t\tsuccess = 1;\n\t\t\t\t\t\t\tans++;\n\t\t\t\t\t\t}\n\t\t\t\t\t\telse success = 0;\n\t\t\t\t\t\tbreak;\n\t\t\t\t\t}\n\t\t\t\t\tint l = Left(left, i, right), r = Right(left, i, right);\n\t\t\t\t\tif (min(l, r) <= 0) {\n\t\t\t\t\t\tsuccess = 0;\n\t\t\t\t\t\tbreak;\n\t\t\t\t\t}\n\t\t\t\t\telse if (l > r) {\n\t\t\t\t\t\tright = sufIdx[right + 1];\n\t\t\t\t\t}\n\t\t\t\t\telse\n\t\t\t\t\t\tleft = prefIdx[left - 1];\n \n\t\t\t\t}\n\t\t\t}\n\t\t}\n \n\t\tcout << ans << '\\n';\n \n\t}\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": "2018B",
"submission_id": 283431031,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n \nint main(){\n int t;\n cin >> t;\n while(t--) {\n int a;\n cin >> a;\n int arr[a], j;\n vector<int> vec,vec1,vec2;\n for(int i = 0; i < a; i++) {\n cin >> arr[i];\n }\n int minn = *min_element(arr, arr + a);\n for(int i = 0; i < a; i++) {\n if (arr[i] == minn) {\n vec.push_back(i);\n j = i;\n }\n }\n float minz = vec[0];\n float leftend = j, rightend = j, x, y, z = 0, n = 0, k = a;\n float t2 = 1e9, t1 = 1e9;\n while(k--) {\n t1 = 1e9;\n t2 = 1e9;\n for(int i = 0; i < leftend; i++) {\n if(i < leftend) {\n t1 = min((arr[i] - z - 1) / (leftend-i), t1);\n if(t1 < 1) {\n cout << 0 << endl;\n n=1;\n goto b;\n }\n if((arr[i] - z - 1)/ (leftend - i) == t1) {\n vec1.push_back(i);\n x = i;\n }\n }\n }\n for(int i = rightend+1; i < a; i++){\n if(i > rightend) {\n t2 = min((arr[i] - z - 1) / (i - rightend), t2);\n if(t2 < 1) {\n cout << 0 << endl;\n n=1;\n goto b;\n }\n if((arr[i] - z - 1) /(i - rightend) == t2) {\n y = i;\n // cout<<t2 <<\" \"<<y<<endl;\n }\n }\n }\n if(t1 < t2) {\n z = (leftend - x) + z;\n leftend = x;\n // cout<<\"l \"<<leftend<<\" \"<<t2<<\" \"<<t1<<\" \"<<endl;\n } if(t2<t1) {\n z = (y - rightend) + z;\n rightend = y;\n // cout<<\"r \"<<rightend<<\" \"<<t2<<\" \"<<t1<<\" \"<<endl;\n }\n }\n if(n==0){\n int ans = 0 , h;\n h=arr[j]-1;\n if(j==minz){\n if(j-h<0){\n ans=ans+j;\n }\n else{\n ans = h +ans;\n }\n if(j+h>a-1){\n ans = (a-1)-j;\n }\n else{\n ans = ans + h;\n }\n \n cout<<ans+1<<endl;}\n else{\n if(j-h<0){\n ans = j + ans;\n }\n else{ \n ans = ans +h;\n }\n if(minz+h>a-1){\n ans = a - 1 - minz;\n }\n else{\n ans = ans + h; \n }\n ans = ans - (j - minz - 1);\n cout<<ans<<endl;\n }\n }\n b:;\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": "2018B",
"submission_id": 283829234,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#include<bitset>\n#include<nmmintrin.h>\nusing namespace std;\n#define oOo {ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);}\n#define ll long long\n#define vc vector\n#define str string\n#define e \"\\n\"\n#define all(v) (v).begin() , (v).end() \n#define File freopen(\"two2.in\", \"r\", stdin);\n#define ONES(x) _mm_popcnt_u64(x)\n#define ONESs(x) __builtin_popcountll(x)\n#define MEM(x , y) memset(x , y , sizeof(x))\n#define MASE(data , n , m , df) vc<vc<data>>(n + 1, vc<data>(m + 1, df))\nconst int mod = 1e9 + 7;\nconst ll modd = 998244353;\nconst int oo = 0x3f3f3f3f;\nbool ok = false, sat = false, ent = false;\nint dx[8] = { 0, 0, 1, -1, -1, -1, 1, 1 };\nint dy[8] = { -1, 1, 0, 0, -1, 1, -1, 1 };\nll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }\nll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }\nll add(ll a, ll b, ll md) {\n\treturn (((a % md) + (b % md)) % md);\n}\nll mul(ll a, ll b, ll md) {\n\treturn (((a % md) * (b % md)) % md);\n}\nll fix_mod(ll a, ll b, ll md) {\n\treturn ((a % md - b % md) + md) % md;\n}\n//---------------------------- [ We need a learn ] -------------------------</end>\nconst int N = 1e6 + 5, M = 21;\nvc<int>v,v2;\n//class segTree {\n//private:\n//\tint n;\n//\tvc<int>tree , tree2;\n//public:\n//\tsegTree(int _n) {\n//\t\tn = _n;\n//\t\twhile (ONES(n) != 1)\n//\t\t{\n//\t\t\tn++; v.push_back(0); v2.push_back(0);\n//\t\t}\n//\t\ttree=tree2 = vc<int>(n * 2);\n//\t}\n//\tint up_to() {\n//\t\treturn n;\n//\t}\n//\tll buildSegTree(ll idx, ll fm, ll to) {\n//\t\tif (fm == to) {\n//\t\t\treturn tree[idx] = v[idx - n];\n//\t\t}\n//\t\tll mid = (fm + to) / 2;\n//\t\treturn tree[idx] = min(buildSegTree(idx * 2, fm, mid) , buildSegTree(idx * 2 + 1, mid + 1, to));\n//\t\t\n//\t}\n//\t\n//\tll quary(int idx, int fm, int to, int fm_q, int to_q) {\n//\t\tif (fm_q <= fm && to_q >= to) {\n//\t\t\treturn tree[idx];\n//\t\t}\n//\t\tif (to < fm_q || to_q < fm) {\n//\t\t\treturn INT_MAX;\n//\t\t}\n//\t\tll mid = (fm + to) / 2;\n//\t\treturn min(quary(idx * 2, fm, mid, fm_q, to_q) , quary(idx * 2 + 1, mid + 1, to, fm_q, to_q));\n//\t\t\n//\t}\n//\t///\n//\tll buildSegTree2(ll idx, ll fm, ll to) {\n//\t\tif (fm == to) {\n//\t\t\treturn tree2[idx] = v2[idx - n];\n//\t\t}\n//\t\tll mid = (fm + to) / 2;\n//\t\treturn tree2[idx] = min(buildSegTree2(idx * 2, fm, mid), buildSegTree2(idx * 2 + 1, mid + 1, to));\n//\n//\t}\n//\n//\tll quary2(int idx, int fm, int to, int fm_q, int to_q) {\n//\t\tif (fm_q <= fm && to_q >= to) {\n//\t\t\treturn tree2[idx];\n//\t\t}\n//\t\tif (to < fm_q || to_q < fm) {\n//\t\t\treturn INT_MAX;\n//\t\t}\n//\t\tll mid = (fm + to) / 2;\n//\t\treturn min(quary2(idx * 2, fm, mid, fm_q, to_q), quary2(idx * 2 + 1, mid + 1, to, fm_q, to_q));\n//\n//\t}\n//\t///\n//\tvoid update(ll idx, int val) {\n//\t\ttree[(n + idx - 1)] = val;\n//\t\tfor (ll a = (n + idx - 1) / 2; a >= 1; a /= 2) {\n//\t\t\ttree[a] = tree[a * 2] + tree[a * 2 + 1];\n//\t\t}\n//\t}\n//\t/*void quary2(int idx, int fm, int to, int fm_q, int to_q , int g) {\n//\t\tif (geet)return;\n//\t\tif (fm == to && (fm_q <= fm && to_q >= to) && gcd(tree[idx], g) != g) {\n//\t\t\tgeet = fm;\n//\t\t\treturn ;\n//\t\t}\n//\t\tif (fm_q <= fm && to_q >= to && gcd(tree[idx] , g) == g) {\n//\t\t\treturn ;\n//\t\t}\n//\t\tif (to < fm_q || to_q < fm) {\n//\t\t\treturn ;\n//\t\t}\n//\t\tll mid = (fm + to) / 2;\n//\t\tquary2(idx * 2, fm, mid, fm_q, to_q , g); quary2(idx * 2 + 1, mid + 1, to, fm_q, to_q , g);\n//\t}*/\n//\t/*void lazy(int idx, int fm, int to, int fm_upt, int to_upt, ll ad) {\n//\t\tif (fm == fm_upt && to == to_upt) {\n//\t\t\ttree[idx] += ad;\n//\t\t\treturn;\n//\t\t}\n//\t\tint mid = (fm + to) / 2;\n//\t\t\n//\t\tif (fm_upt <= mid) {\n//\t\t\tlazy(idx * 2, fm, mid, fm_upt, min(mid, to_upt), ad);\n//\t\t}\n//\t\t\n//\t\tif (to_upt >= mid + 1) {\n//\t\t\tlazy(idx * 2 + 1, mid + 1, to, max(mid + 1, fm_upt), to_upt, ad);\n//\t\t}\n//\t}*/\n//\t/*void lazy2(int idx, int fm, int to, int fm_upt, int to_upt, ll ad) {\n//\t\tif (fm == fm_upt && to == to_upt) {\n//\t\t\ttree2[idx] = max(tree2[idx], ad);\n//\t\t\treturn;\n//\t\t}\n//\t\tint mid = (fm + to) / 2;\n//\t\t/\n//\t\tif (fm_upt <= mid) {\n//\t\t\tlazy2(idx * 2, fm, mid, fm_upt, min(mid, to_upt), ad);\n//\t\t}\n//\t\t\n//\t\tif (to_upt >= mid + 1) {\n//\t\t\tlazy2(idx * 2 + 1, mid + 1, to, max(mid + 1, fm_upt), to_upt, ad);\n//\t\t}\n//\t}*/\n//};\n \nint main() {\n\toOo;\n\tint t; cin >> t;\n\twhile (t--)\n\t{\n\t\tint n; cin >> n;\n\t\tv=v2 = vc<int>(n);\n\t\tfor (int a = 0; a < n; a++) {\n\t\t\tcin >> v[a];\n\t\t}\n\t\t/*segTree sg(n) , sg2(n);\n\t\tsg.buildSegTree(1, 1, sg.up_to());\n\t\tsg2.buildSegTree(1, 1, sg2.up_to());*/\n\t\tint cnt = 0;\n\t\tpriority_queue<pair<int, int>, vc<pair<int, int>>, greater<pair<int, int>>>pq, pq2;\n\t\tset<int>st;\n\t\tfor (int a = n-1; a >=0; a--) {\n\t\t\tpq2.push({ v[a] + (n - a) , -(a + 1) });\n\t\t}\n\t\tpq.push({ v[0] + 1 , 1 }); st.insert(1);\n\t\tfor (int a = 1; a < n-1; a++) {\n\t\t\tst.insert(a + 1);\n\t\t\twhile (pq2.size())\n\t\t\t{\n\t\t\t\tif (st.count(-pq2.top().second))pq2.pop();\n\t\t\t\telse break;\n\t\t\t}\n\t\t\tint mn = pq.top().first, mn2 = pq2.top().first, idx = pq.top().second, idx2 = -pq2.top().second;\n\t\t\tif (mn > a +1 && mn2 > n - a) {\n\t\t\t\tint len = idx2 - idx + 1;\n\t\t\t\tif (v[idx - 1] >= len || v[idx2 - 1] >= len) {\n\t\t\t\t\tcnt++;\n\t\t\t\t}\n\t\t\t}\n\t\t\tpq.push({ v[a] + a + 1 , a + 1 });\n\t\t}\n\t\tint ad = 0;\n\t\tok = 1;\n\t\tfor (int a = 0; a < n; a++) {\n\t\t\tif (v[a] < a + 1) {\n\t\t\t\tok = 0;\n\t\t\t}\n\t\t}\n\t\tad += ok;\n\t\tok = 1;\n\t\tfor (int a = n - 1; a >= 0; a--) {\n\t\t\tif (v[a] < n - a) {\n\t\t\t\tok = 0;\n\t\t\t}\n\t\t}\n\t\tad += ok;\n\t\tif (n == 1)ad--;\n\t\tcout << cnt+ad << e;\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": "2018B",
"submission_id": 283528155,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nt = int(input())\nfor _ in range(t):\n n = int(input())\n a = list(map(int, input().split())) \n \n mnmtm = float('inf')\n count = 0\n \n \n for i in range(n - 1, -1, -1):\n mnmtm = min(mnmtm, a[i])\n if mnmtm >= i + 1:\n count += 1\n \n print(count-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": "2018B",
"submission_id": 283214477,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nfor _ in range(int(input())):\n n = int(input())\n array = [int(i) for i in input().split()]\n l_a = [float('inf')]\n r_a = [float('inf')]\n result = 0\n for i in range(n):\n l_a.append(min(l_a[-1]-1, array[i]))\n for i in range(n-1,-1,-1):\n r_a.append(min(r_a[-1]-1, array[i]))\n r_a.reverse()\n \n for i in range(n):\n count = 1\n if l_a[i] >0 and r_a[i]>0:\n l,r = i,i+1\n while(l_a[l]>count and r_a[r]>count):\n if l_a[l]<r_a[r]:\n l-=1 \n count+=1 \n else:\n r+=1 \n count +=1\n if l==0 and r==n :\n result+=1\n break\n \n \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": "2018B",
"submission_id": 284304681,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nimport java.io.*;\n \n/**\n * @version 1.0\n * @Author 杨雨彤\n * @Date 2024/10/13 19:57\n * @注释\n */\npublic class Main {\n static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));\n static StreamTokenizer st = new StreamTokenizer(br);\n \n static int nextInt() throws IOException {\n st.nextToken();\n return (int) st.nval;\n }\n \n static long nextLong() throws IOException {\n st.nextToken();\n return (long) st.nval;\n }\n \n static String nextLine() throws IOException {\n return br.readLine();\n }\n \n public static void main(String[] args) throws IOException{\n int T=nextInt();\n while(T-->0) {\n int n = nextInt();//城市数\n int[] arr = new int[n + 1];\n int ll = 1;\n int rr = n;\n for (int i = 1; i <= n; i++) {\n arr[i] = nextInt();\n ll = Math.max(ll, i - arr[i] + 1);\n rr = Math.min(rr, i + arr[i] - 1);\n }\n int ans=rr-ll+1;\n int l=1;\n int r=n;\n a:while(l<=r) {\n if (arr[l] >= r - l + 1 && arr[r] >= r - l + 1) {\n r--;\n l++;\n } else if (arr[l] >= r - l + 1) {\n r--;\n } else if (arr[r] >= r - l + 1) {\n l++;\n } else {\n ans=0;\n break a;\n }\n }\n pw.println(ans);\n pw.flush();\n }\n pw.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": "2018B",
"submission_id": 285646798,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nimport java.io.*;\n \n/**\n * @version 1.0\n * @Author 杨雨彤\n * @Date 2024/10/13 19:57\n * @注释\n */\npublic class Main {\n static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));\n static StreamTokenizer st = new StreamTokenizer(br);\n \n static int nextInt() throws IOException {\n st.nextToken();\n return (int) st.nval;\n }\n \n static long nextLong() throws IOException {\n st.nextToken();\n return (long) st.nval;\n }\n \n static String nextLine() throws IOException {\n return br.readLine();\n }\n \n public static void main(String[] args) throws IOException{\n int T=nextInt();\n while(T-->0){\n int n=nextInt();//城市数\n int []arr=new int[n+1];\n int l=1;\n int r=n;\n for(int i=1;i<=n;i++){\n arr[i]=nextInt();\n l=Math.max(l,i-arr[i]+1);\n r=Math.min(r,i+arr[i]-1);\n }\n \n if(arr[1]<n&&arr[n]<n){\n pw.println(\"0\");\n }else{\n int ans=r-l+1;\n pw.println(ans);\n }\n }\n pw.flush();\n pw.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": "2018B",
"submission_id": 285642442,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class D_Speedbreaker {\n \n public static void main(String[] args) {\n Scanner sc =new Scanner(System.in);\n\t\tint t=sc.nextInt();\n\t\twhile(t-->0){\n\t\t int n = sc.nextInt();\n\t\t\tint[] a = new int[n];\n for(int i=0;i<n;i++) a[i] = sc.nextInt();\n Map<Integer,Integer> first = new HashMap<>();\n Map<Integer,Integer> last = new HashMap<>();\n for(int i=0;i<n;i++){\n first.putIfAbsent(a[i], i);\n last.put(a[i],i);\n }\n int interval = 0;\n int l = -1, r = -1;\n boolean f = true;\n boolean flag = true;\n for(int i=1;i<=n;i++){\n if(first.containsKey(i)){\n if(f){\n f = false;\n interval = i;\n l = first.get(i);\n r = last.get(i);\n }\n l = Math.min(l,first.get(i));\n r = Math.max(r,last.get(i));\n if(r-l+1>i){\n flag = false;\n break;\n }\n }\n }\n if(!flag) System.out.println(0);\n else System.out.println(interval);\n\t\t}\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": "2018B",
"submission_id": 284683338,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class D_Speedbreaker {\n \n public static void main(String[] args) {\n Scanner sc =new Scanner(System.in);\n\t\tint t=sc.nextInt();\n\t\twhile(t-->0){\n\t\t int n = sc.nextInt();\n\t\t\tint[] a = new int[n];\n for(int i=0;i<n;i++) a[i] = sc.nextInt();\n Map<Integer,Integer> first = new HashMap<>();\n Map<Integer,Integer> last = new HashMap<>();\n for(int i=0;i<n;i++){\n first.putIfAbsent(a[i], i);\n last.put(a[i],i);\n }\n int interval = 0;\n int l = -1, r = -1;\n boolean f = true;\n boolean flag = true;\n for(int i=1;i<=n;i++){\n if(first.containsKey(i)){\n if(f){\n f = false;\n interval = last.get(i)-first.get(i)+1;\n l = first.get(i);\n r = last.get(i);\n }\n l = Math.min(l,first.get(i));\n r = Math.max(r,last.get(i));\n if(r-l+1>i){\n flag = false;\n break;\n }\n }\n }\n if(!flag) System.out.println(0);\n else System.out.println(interval);\n\t\t}\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": "2018B",
"submission_id": 284683097,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nclass Solution():\n def __init__(self):\n pass\n \n def getSegem(self):\n self.segs = {}\n for i in range(len(self.xs)):\n if self.xs[i] in self.segs.keys():\n old_seg = self.segs[self.xs[i]]\n self.segs[self.xs[i]] = (old_seg[0], i + 1)\n else:\n self.segs[self.xs[i]] = (i + 1, i + 1)\n \n def getVal(self):\n val = (1, self.n)\n for key in self.keys:\n if key < self.segs[key][1] - self.segs[key][0] + 1:\n return 0\n good_seg = (self.segs[key][1] - key + 1, self.segs[key][0] + key - 1)\n # print(good_seg)\n val = (max(good_seg[0], val[0]), min(good_seg[1], val[1]))\n if val[0] > val[1]:\n return 0\n else:\n pass\n return val[1] - val[0] + 1\n \n def solve(self):\n self.n = int(input())\n self.xs = tuple(map(int, input().split()))\n \n self.getSegem()\n self.keys = self.segs.keys()\n self.keys = sorted(self.keys)\n \n print(self.getVal())\n # print(self.segs)\n \n \ndef main():\n case_size = int(input())\n solution = Solution()\n for _ in range(case_size):\n solution.solve()\n \n \n \n\"\"\"\n4\n3\n5 4 5\n3\n4 5 4\n10\n3 3 3 3 4 1 2 3 4 5\n9\n17 89 92 42 29 92 14 70 45\n\"\"\"\n \nmain()\n# test()\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": "2018B",
"submission_id": 283242946,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nfor _ in range(int(input())):\n n = int(input())\n a = list(map(int,input().split(' ')))\n started = []\n mn = n\n mx = 0\n # m_num = 2*10**5 +1\n # p_sum = [0 for _ in range (m_num)]\n # print('m:',m_num)\n for i in range(n):\n l = max(0,i-a[i]+1)\n r = min(n,i+a[i]-1)\n started.append([l,r])\n mx = max(mx,l)\n mn = min(mn,r)\n # p_sum[a[i]] +=1\n l = mx\n r = mn\n cost = 1\n can =1\n while l>=0 and r<n:\n # print('leftL',l,'right:',r,'cost:',cost)\n cost +=2\n if l >0:\n if cost - a[l-1] >1:\n can = 0\n # print('me:', cost)\n break\n if r<=n-2:\n if cost - a[r+1] >1:\n can = 0\n # print('me:', cost)\n break\n \n r+=1\n l-=1\n # print('can:',can)\n \n # print(started)\n # print(mx,mn)\n print(mn-mx+1 if can else can)\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": "2018B",
"submission_id": 289109919,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\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 __gnu_pbds; \nusing namespace std;\n#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>\n#define ll long long \n#define vl vector<ll>\n#define vs vector<string>\n#define mll map<ll,ll>\n#define pll pair<ll,ll>\n#define sl set<ll>\n#define pql priority_queue<ll>\n#define rep(i,n) for(i=0;i<n;i++)\n#define repr(i,n) for(i=n-1;i>=0;i--)\n#define REP(i,a,b) for(i=a;i<b;i++)\n#define all(a) a.begin(),a.end()\n#define rall(a) a.rbegin(), a.rend()\n#define vsort(a) sort(all(a))\n#define rvsort(a) sort(rall(a))\n#define F first\n#define endl \"\\n\" \n#define S second\n#define pi acos(-1)\n#define inf 10000000000000000\n#define mod 1000000007\n#define MA LLONG_MAX\n#define MI LLONG_MIN\nmt19937 rng(chrono::steady_clock::now().time_since_epoch().count());\n \nlong long power(long long x,ll y)\n{\nlong long res = 1;\nx = x % mod;\nwhile (y > 0) \n{\nif (y & 1)\nres = (res * x) % mod;\ny = y >> 1; \nx = (x * x) % mod;\n}\nreturn res;\n}\n \nlong modInverse(long long n)\n{\n return power(n, mod - 2);\n}\n \nlong long ncr(long long n, ll r)\n{\nif (n < r)\nreturn 0;\nif (r == 0)\nreturn 1;\nlong long fac[n + 1];\nfac[0] = 1;\nfor (int i = 1; i <= n; i++)\nfac[i] = (fac[i - 1] * i) % mod;\n \nreturn (fac[n] * modInverse(fac[r]) % mod*modInverse(fac[n - r])% mod)% mod;\n}\n \n \nstruct custom_hash {\n static uint64_t splitmix64(uint64_t x) {\n // http://xorshift.di.unimi.it/splitmix64.c\n x += 0x9e3779b97f4a7c15;\n x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;\n x = (x ^ (x >> 27)) * 0x94d049bb133111eb;\n return x ^ (x >> 31);\n }\n \n size_t operator()(uint64_t x) const {\n static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();\n return splitmix64(x + FIXED_RANDOM);\n }\n};\n \nvoid solve()\n{\n ll n,i,j,ans=-1,leftans=0,rightans=0,firstleft,firstright,leftextra,rightextra;\n cin>>n;\n vector<ll> a(n),first(n+1,-1),last(n+1,-1);\n rep(i,n)\n {\n cin>>a[i];\n if(first[a[i]]==-1)\n first[a[i]]=i;\n last[a[i]]=i;\n }\n ll curleft=n,curright=0;\n for(i=1;i<=n;i++)\n {\n if(last[i]!=-1)\n {\n curleft=min(curleft,first[i]);\n curright=max(curright,last[i]);\n if(curright-curleft+1>i)\n {\n cout<<0<<endl;\n return;\n }\n if(ans==-1)\n {\n ans=curright-curleft+1;\n firstleft=first[i];\n firstright=last[i];\n leftextra=i-ans;\n rightextra=i-ans;\n }\n \n if(curleft<firstleft)\n {\n leftans=max(leftans,min(leftextra+leftans,firstleft-curleft));\n \n // firstleft=firstleft-temp;\n // leftans+=temp;\n }\n if(curright>firstright)\n {\n rightans=max(rightans,min(rightextra+rightans,curright-firstright));\n // firstright=firstright+temp;\n // rightans+=temp;\n }\n ll temp=max(0ll,i-(curright-curleft+1));\n leftextra=max(0ll,min(leftextra-leftans,temp));\n rightextra=max(0ll,min(rightextra-rightans,temp));\n //cout<<ans<<\" \"<<leftans<<\" \"<<rightans<<\" \"<<leftextra<<\" \"<<rightextra<<endl;\n }\n }\n cout<<ans+leftans+rightans<<endl;\n \n}\n \nint32_t main() {\nios_base::sync_with_stdio(false);\ncin.tie(NULL); cout.tie(NULL);\nint t=1;\ncin>>t;\nwhile (t--) {\nsolve();\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": "2018B",
"submission_id": 296972726,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define FOR(i,a,b) for(int i=(a); i<=(b); ++i)\n#define ROF(i,a,b) for(int i=(a); i>=(b); --i)\n#define sz(x) (int)(x).size()\n#define all(x) (x).begin(),(x).end()\n#define make_unique(x) sort(all(x)), (x).erase(unique(all(x)), (x).end())\n#define x first\n#define y second\n#define pb push_back\n#define eb emplace_back\n \nusing namespace std;\n \ntypedef pair<int,int> pii;\ntypedef long long i64;\ntypedef __int128 i128;\ntypedef tuple<int,int,int> iii;\ntypedef vector<int> vi;\ntypedef vector<vi> vvi;\ntypedef vector<i64> vl;\ntypedef long double ld;\n \nint readInt(){ int a; scanf(\"%d\",&a); return a;}\ni64 readLong(){ i64 a; scanf(\"%lld\",&a); return a;}\nvoid readString(char *s){ scanf(\" %s\",s); }\n \nconst int mod = 1000000007;\nint add(int a, int b){ return ((a+=b)>=mod)?a-mod:a; }\nvoid adding(int &a, int b){ a = add(a,b); }\nint mul(int a, int b){ return a*1ll*b%mod; }\nint modpw(int a, int b){\n if(a == 0) return 0;\n if(b == 0) return 1;\n int ans = 1;\n for(i64 i=1,res=a; i<=b; i<<=1, res=mul(res,res)){\n if(b&i) ans = mul(ans,res);\n }\n return ans;\n}\n \nconst int N = 200005;\nint MiL[N*4],MiR[N*4];\nint num[N];\nvoid init(int l, int r, int now) {\n if(l == r) {\n MiL[now] = MiR[now] = num[l];\n return;\n }\n int mid = (l+r)/2;\n init(l,mid,now*2); init(mid+1,r,now*2+1);\n MiL[now] = min(MiL[now*2], MiL[now*2+1]-(mid-l+1));\n MiR[now] = min(MiR[now*2]-(r-mid), MiR[now*2+1]);\n}\nint queryL(int ll, int rr, int l, int r, int now) {\n if(l > r) return (1<<30);\n if(l > rr || r < ll) return (1<<30);\n if(ll <= l && r <= rr) {\n //printf(\"L (%d, %d) = %d\\n\",l,r,MiR[now]);\n return MiL[now];\n }\n int mid = (l+r)/2;\n if(rr <= mid) return queryL(ll, rr, l, mid, now*2);\n if(mid < ll) return queryL(ll, rr, mid+1, r, now*2+1);\n return min(queryL(ll,rr,l,mid,now*2), queryL(ll,rr,mid+1,r,now*2+1)-(mid-ll+1));\n}\nint queryR(int ll, int rr, int l, int r, int now) {\n if(l > r) return (1<<30);\n if(l > rr || r < ll) return (1<<30);\n if(ll <= l && r <= rr){\n //printf(\"R (%d, %d) = %d\\n\",l,r,MiR[now]);\n return MiR[now];\n }\n int mid = (l+r)/2;\n if(rr <= mid) return queryR(ll, rr, l, mid, now*2);\n if(mid < ll) return queryR(ll, rr, mid+1, r, now*2+1);\n return min(queryR(ll,rr,l,mid,now*2)-(rr-mid), queryR(ll,rr,mid+1,r,now*2+1));\n}\n \nvoid solve(){\n int n = readInt();\n FOR(i,1,n) num[i] = readInt();\n init(1,n,1);\n int l = 1, r = n;\n int ans = 0;\n while(l <= r) {\n //printf(\"%d %d\\n\",l,r);\n if(num[l] >= r-l+1) {\n int x = queryL(l+1, r, 1, n, 1);\n if(x > 1) {\n //printf(\"[%d]\\n\",x);\n //puts(\"F\");\n ++ans;\n }\n ++l;\n } else if(num[r] >= r-l+1) {\n int x = queryR(l, r-1, 1, n, 1);\n if(x > 1) {\n //printf(\"[%d\\n\",x);\n //puts(\"F\");\n ++ans;\n }\n --r;\n } else {\n break;\n }\n }\n printf(\"%d\\n\",ans);\n \n // clean up\n FOR(i,1,4*n) MiL[i] = MiR[i] = 0;\n}\n \nint main(){\n int t = 1;\n t = readInt(); // comment this out is q = 1\n while(t--){\n solve();\n }\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2018B",
"submission_id": 283240800,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class B_Speedbreaker {\n public static void main(String[] args) {\n FastScanner sc = new FastScanner();\n int t = sc.nextInt();\n \n while(t-- > 0) {\n int n = sc.nextInt();\n int[] a = new int[n];\n \n for(int i = 0; i < n; i++) \n a[i] = sc.nextInt();\n \n int count = 0;\n for(int i = 0; i < n; i++) {\n HashSet<Integer> set = new HashSet<>();\n set.add(i);\n if(helper(set, a, 2)) {\n count++;\n } \n set.remove(i); \n }\n \n System.out.println(count);\n }\n \n }\n private static boolean helper(HashSet<Integer> set, int[] a, int time) {\n if(set.size() == a.length) return true;\n \n boolean ans = false;\n for(int i = 0; i < a.length; i++) {\n if(!set.contains(i)) {\n if(a[i] < time) return false;\n \n if(set.contains(i + 1) || set.contains(i - 1)) {\n set.add(i);\n ans |= helper(set, a, time + 1);\n set.remove(i);\n }\n }\n }\n \n return ans;\n }\n static class FastScanner {\n BufferedReader br;\n StringTokenizer st;\n public FastScanner() {\n br = new BufferedReader(new InputStreamReader(System.in));\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}\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": "2018B",
"submission_id": 284359993,
"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: Speedbreaker\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: [Djjaner - Speedbreaker](https://soundcloud.com/luciano- ferrari-151560131/speedbreaker)\n\n⠀\n\nThere are $n$ cities in a row, numbered $1, 2, \\ldots, n$ left to right.\n\n * At time $1$, you conquer exactly one city, called the starting city. * At time $2, 3, \\ldots, n$, you can choose a city adjacent to the ones conquered so far and conquer it. \n\nYou win if, for each $i$, you conquer city $i$ at a time no later than $a_i$. A winning strategy may or may not exist, also depending on the starting city. How many starting cities allow you to win?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 2 \\cdot 10^5$) — the number of cities.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le n$) — the deadlines for conquering the cities.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer: the number of starting cities that allow you to win.\n\nExamples:\ninput:\n3\n6\n6 3 3 3 5 5\n6\n5 6 4 1 4 5\n9\n8 6 4 2 1 3 5 7 9\noutput:\n3\n0\n1\n\nNote: In the first test case, cities $2$, $3$, and $4$ are good starting cities.\n\nIn the second test case, there are no good starting cities.\n\nIn the third test case, the only good starting city is city $5$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\npublic class SpeedBreaker {\n \n static HashSet<Integer> set;\n public static void counter(int arr[],int s,int e,int t){\n if( s > e || t > e - s + 1) return;\n if(t >= 1 && s == e){\n set.add(s);\n }\n \n counter(arr, s + 1, e, Math.min(t, arr[s]) - 1);\n counter(arr, s, e - 1, Math.min(t, arr[e]) - 1);\n }\n public static void main(String[] args) throws Exception{\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n PrintWriter pw = new PrintWriter(System.out);\n int tt = Integer.parseInt(br.readLine());\n while(tt-- > 0){\n int n = Integer.parseInt(br.readLine());\n int arr[] = new int[n];\n StringTokenizer st = new StringTokenizer(br.readLine());\n for(int i = 0; i < n;i++){\n arr[i] = Integer.parseInt(st.nextToken());\n }\n set = new HashSet<>();\n \n counter(arr, 0, n - 1, n);\n pw.println(set.size());\n \n \n \n \n \n }\n pw.flush();\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": "2018B",
"submission_id": 283432721,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\n#include<stdio.h>\n#include<iostream>\n#include<algorithm>\nusing namespace std;\nint n,m,ans,sum;\nint a[1000];\nint main()\n{\n\tcin>>n;\n\tfor(int i=0;i<n;i++){\n\t\tcin>>m;\n\t\tif(m<3){\n\t\t\tfor(int k=0;k<m;k++){\n\t\t\tcin>>a[k];\n\t\t\t}\n\t\t\tcout<<0<<endl;\n\t\t\tcontinue;\n\t\t}\n\t\tif(m>2){\n\t\t\tsum=0;\n\t\t\tans=0;\n\t\t\tfor(int k=0;k<m;k++){\n\t\t\tcin>>a[k];\n\t\t\t}\n\t\t\tsort(a,a+m);\n\t\t\tfor(int s=1;s<m;s++){\n\t\t\t\tif(a[s]==a[s-1]){\n\t\t\t\t\tans+=1;\n\t\t\t\t//\tprintf(\"ans=%d\\n\",ans);\n\t\t\t\t}\n\t\t\t\tif(a[s]!=a[s-1]){\n\t\t\t\t\tif(ans>2){\n\t\t\t\t\t\tsum+=1;\n\t\t\t\t\t//\tprintf(\"sum=%d\\n\",sum);\n\t\t\t\t\t\tans=0;\n\t\t\t\t\t}\n\t\t\t\t\telse ans=0;\n\t\t\t\t}\n\t\t\t\tif(s==m-1&&ans>1) sum+=1;\n\t\t\t}\n\t\t}\n\t\tcout<<sum<<endl;\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": "1957A",
"submission_id": 257627342,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\n#include<iostream>\n#include<algorithm>\nusing namespace std;\n \nint main()\n{\n short int t;\n cin >> t;\n \n while(t--)\n {\n short int n, i, j, k, cot=0;\n cin >> n;\n short int stick[n];\n \n for(i=0; i<n; i++)\n {\n cin >> stick[i];\n }\n \n if(n>2)\n {\n sort(stick, stick+n);\n \n for(i=0; i<n-1; )\n {\n k=1;\n for(j=i+1; j<n; j++)\n {\n if(stick[i]==stick[j])\n {\n k++;\n }\n else\n {\n break;\n }\n }\n if(k>=3)\n {\n cot++;\n }\n i=j;\n }\n \n cout << cot << '\\n';\n \n }\n else\n {\n cout << \"0\" << '\\n';\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": "1957A",
"submission_id": 257625062,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\n#include <iostream>\nusing namespace std;\nint main()\n{\nint n;\ncin>>n;\nint e[n];\nfor(int i=0;i<n;i++)\n{\n int m;\n cin>>m;\n int a[m];\n for(int j=0;j<m;j++)\n {\n cin>>a[j];\n }\n int y[101]={0};\n for(int k=0;k<m;k++)\n {\n y[a[k]]++;\n }\n int u;\n int s=0;\n for(int z=0;z<m;z++)\n {\n u=y[z]/4;\n s+=u;\n }\n e[i]=s;\n}\nfor(int q=0;q<n;q++)\n{\n cout<<e[q]<<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": "1957A",
"submission_id": 266128732,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\nt=int(input())\nfor _ in range(t):\n n=int(input())\n arr=list(map(int,input().split()))\n c=0\n d={}\n for num in arr:\n if num not in d:\n d[num]=1\n else:\n d[num]+=1\n for i in d:\n if d[i]>2:\n c+=1\n print(c)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957A",
"submission_id": 257590068,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\n# Function to count the frequency of stick lengths\ndef count_sticks(sticks):\n freq = {}\n for stick in sticks:\n if stick in freq:\n freq[stick] += 1\n else:\n freq[stick] = 1\n return freq\n \n# Function to calculate the maximum number of polygons\ndef max_polygons(freq):\n max_poly = 0\n for length in freq:\n max_poly += freq[length] // 4\n return max_poly\n \n# Input the number of test cases\nt = int(input())\n \n# Iterate through each test case\nfor _ in range(t):\n # Input the number of sticks and their lengths\n n = int(input())\n sticks = list(map(int, input().split()))\n \n # Count the frequency of stick lengths\n freq = count_sticks(sticks)\n \n # Calculate the maximum number of polygons\n max_poly = max_polygons(freq)\n \n # Output the result\n print(max_poly)\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": "1957A",
"submission_id": 257588341,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\ndef solution():\n n:int = int(input())\n nums = list(map(int,input().split()))\n freq = {}\n ans:int =0\n for i in nums:\n if i in freq:\n freq[i]+=1\n else:\n freq[i]=1 \n for key in freq:\n if freq[key]>2:\n no:int=3\n while freq[key]%no!=0:\n no+=1\n if freq[key]!=no:\n ans+=freq[key]//no\n else:\n ans+=freq[key]//3\n return ans\nn:int = int(input())\nfor i in range(n):\n print(solution())\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": "1957A",
"submission_id": 257756854,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\n/* \n \n ॐ भूर् भुवः स्वः।\n तत् सवितुर्वरेण्यं।\n भर्गो देवस्य धीमहि।\n धियो यो नः प्रचोदयात् ॥\n \n @Author : ANKUSH\n @Country : INDIA \n \n*/\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {// Codechef {\n static class pair {\n long val1;\n long val2;\n \n public pair(long val1, long val2) {\n this.val1 = val1;\n this.val2 = val2;\n }\n \n public boolean equals(Object o) {\n if (this == o)\n return true;\n if (o == null || getClass() != o.getClass())\n return false;\n pair pair = (pair) o;\n return val1 == pair.val1 && val2 == pair.val2;\n }\n \n @Override\n public int hashCode() {\n return 31 * Long.hashCode(val1) + Long.hashCode(val2);\n }\n \n @Override\n public String toString() {\n return \"pair{\" + \"x=\" + val1 + \", y=\" + val2 + '}';\n }\n }\n \n static long MOD = (long) (1e9 + 7);\n static long dp[][];\n \n public static void main(String[] args) {\n int TTT = 0;\n try {\n FastReader in = new FastReader();\n FastWriter out = new FastWriter();\n int testCases = in.nextInt();\n while (TTT++ != testCases) {\n // write code here\n int n=in.nextInt();\n int a[]=new int[n];\n int f[]=new int[n+1];\n for(int i=0;i<n;i++){\n a[i]=in.nextInt();\n f[a[i]]++;\n }int res=0;\n for(int i=0;i<n;i++){\n res+=f[i]/3;\n }\n \n out.println(res);\n }\n out.close();\n } catch (Exception e) {\n System.out.println(e);\n return;\n }\n }\n /*------------------------------------------- MATHS MATHS MATHS ------------------------------------ */\n public static long sqrt(long x) {\n long low = 0, high = 3037000499L;\n while (low <= high) {\n long mid = low + (high - low) / 2;\n long square = mid * mid;\n \n if (square == x) {\n return mid;\n } else if (square < x) {\n low = mid + 1;\n } else {\n high = mid - 1;\n }\n }\n return high;\n }\n public static boolean isPrime(long number) {\n if (number <= 1) {\n return false;\n }\n if (number <= 3) {\n return true;\n }\n if (number % 2 == 0 || number % 3 == 0) {\n return false;\n }\n \n long sqrt = (long) Math.sqrt(number);\n for (long i = 5; i <= sqrt; i += 6) {\n if (number % i == 0 || number % (i + 2) == 0) {\n return false;\n }\n }\n return true;\n }\n \n public static long power(long a, long b) {\n long ans = 1;\n while (b > 0) {\n if ((b & 1) == 1)\n ans *= a;\n a *= a;\n b >>= 1;\n }\n return ans;\n }\n \n public static long power_mod(long a, long b) {\n long ans = 1;\n while (b > 0) {\n if ((b & 1) == 1) {\n ans *= a;\n ans %= MOD;\n }\n a *= a;\n a %= MOD;\n b >>= 1;\n }\n return ans;\n }\n \n static final int SIZE = (int) (2000);\n private static long[] fact = new long[SIZE + 1];\n private static boolean[] isPrime = new boolean[SIZE + 1];\n \n public static long add_mod(long a, long b) {\n return ((a % MOD) + (b % MOD)) % MOD;\n }\n \n public static long sub_mod(long a, long b) {\n return (((a % MOD) - (b % MOD)) + MOD) % MOD;\n }\n \n public static long multiply_mod(long a, long b) {\n return ((a % MOD) * (b % MOD)) % MOD;\n }\n \n public static long inv(long a) {\n return power_mod(a, MOD - 2);\n }\n \n public static long divide_mod(long a, long b) {\n return multiply_mod(a, inv(b));\n }\n \n public static long nCr(long n, long r) {\n if (n < r)\n return 0;\n return divide_mod(fact[(int) n], multiply_mod(fact[(int) r], fact[(int) (n - r)]));\n }\n \n public static long nPr(long n, long r) {\n if (n < r)\n return 0;\n return divide_mod(fact[(int) n], fact[(int) (n - r)]);\n }\n \n public static void preFactorial() {\n fact[0] = 1;\n for (int i = 1; i <= SIZE; i++) {\n fact[i] = multiply_mod(i, fact[i - 1]);\n }\n }\n \n public static void Sieve_Of_Eratosthenes() {\n Arrays.fill(isPrime, true);\n isPrime[0] = false;\n isPrime[1] = false;\n \n for (int i = 2; i * i <= SIZE; i++) {\n if (isPrime[i]) {\n for (int j = i * i; j <= SIZE; j += i) {\n isPrime[j] = false;\n }\n }\n }\n }\n \n public static long gcd(long x, long y) {\n if (x > y)\n return gcd(y, x);\n if (y == 0) {\n return x;\n }\n if (x == 0) {\n return y;\n }\n return gcd(y % x, x);\n }\n \n public static long lcm(long a, long b) {\n return (a * b) / gcd(a, b);\n }\n \n /*---------------------------------------- ADVANCED DATA STRUCTURES ADVANCED DATA STRUCTURES ------------------------------------ */\n public static class DisjointSet {\n private long[] parent;\n private long[] size;\n \n public DisjointSet(int n) {\n parent = new long[n];\n size = new long[n];\n for (int i = 0; i < n; i++) {\n parent[i] = i;\n size[i] = 1;\n }\n }\n \n public long findParent(long node) {\n if (node == parent[(int) node]) {\n return node;\n }\n return parent[(int) node] = findParent(parent[(int) node]);\n }\n \n public void unionSize(long node1, long node2) {\n long p1 = findParent(node1);\n long p2 = findParent(node2);\n if (p1 == p2)\n return;\n if (size[(int) p1] > size[(int) p2]) {\n size[(int) p1] += size[(int) p2];\n parent[(int) p2] = p1;\n } else {\n size[(int) p2] += size[(int) p1];\n parent[(int) p1] = p2;\n }\n }\n \n public long findSize(long node) {\n long p1 = findParent(node);\n return size[(int) p1];\n }\n }\n \n static class Segment_Tree {\n long arr[];\n int n;\n long seg[];\n \n public Segment_Tree(long arr[], int n) {\n this.arr = arr;\n this.n = n;\n this.seg = new long[(4 * n) + 1];\n build(0, 0, n - 1);\n }\n \n public void build(int ind, int low, int high) {\n if (low == high) {\n seg[ind] = arr[low];\n return;\n }\n int mid = (low + high) / 2;\n build((2 * ind) + 1, low, mid);\n build((2 * ind) + 2, mid + 1, high);\n seg[ind] = Math.max(seg[2 * ind + 1], seg[2 * ind + 2]);\n }\n \n public long query2(int ind, int low, int high, int left, int right) {\n if (low >= left && high <= right) {\n return seg[ind];\n }\n if (low > right || high < left) {\n return 0;\n }\n int mid = (low + high) / 2;\n long l = query2(2 * ind + 1, low, mid, left, right);\n long r = query2(2 * ind + 2, mid + 1, high, left, right);\n return Math.max(l, r);\n }\n \n public long query(int l, int r) {\n return query2(0, 0, n - 1, l, r);\n }\n \n public void pointupdate2(int ind, int low, int high, int node, int val) {\n if (low == high) {\n seg[ind] = val;\n } else {\n int mid = (low + high) / 2;\n if (node <= mid && node >= low)\n pointupdate2(2 * ind + 1, low, mid, node, val);\n else\n pointupdate2(2 * ind + 2, mid + 1, high, node, val);\n seg[ind] = Math.max(seg[2 * ind + 1], seg[2 * ind + 2]);\n }\n }\n \n public void point_update(int node, int val) {\n pointupdate2(0, 0, n - 1, node, val);\n }\n }\n \n /*------------------------------------- FAST READER AND WRITER FAST READER AND WRITER ------------------------------------ */\n static class FastReader {\n BufferedReader br;\n StringTokenizer st;\n \n public FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() {\n while (st == null || !st.hasMoreTokens()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine() {\n String str = \"\";\n try {\n str = br.readLine().trim();\n } catch (Exception e) {\n e.printStackTrace();\n }\n return str;\n }\n }\n \n static class FastWriter {\n \n private final BufferedWriter bw;\n \n public FastWriter() {\n this.bw = new BufferedWriter(new OutputStreamWriter(System.out));\n }\n \n public void print(Object object) throws IOException {\n bw.append(\"\" + object);\n }\n \n public void println(Object object) throws IOException {\n print(object);\n bw.append(\"\\n\");\n }\n \n public void close() throws IOException {\n bw.close();\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": "1957A",
"submission_id": 285043530,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\nimport java.util.HashMap;\nimport java.util.Scanner;\n \npublic class Main {\n public static void main(String[] args) {\n \n var sc = new Scanner(System.in);\n \n var t = sc.nextInt();\n \n while (t-- > 0) {\n var n = sc.nextInt();\n int[] nums = new int[n];\n \n for (var i=0; i < n; i++)\n nums[i] = sc.nextInt();\n \n var map = addToMap(nums);\n var cnt = 0;\n \n for (var v : map.values())\n if (v == 3 || v == 4 || v == 6 || v == 8 || v == 5)\n cnt++;\n \n System.out.println(cnt);\n }\n \n }\n \n static HashMap<Integer, Integer> addToMap(int[] nums) {\n var map = new HashMap<Integer, Integer>();\n \n for (var num : nums) {\n if (!map.containsKey(num))\n map.put(num, 1);\n else {\n var index = map.get(num);\n map.put(num, ++index);\n }\n }\n \n return map;\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": "1957A",
"submission_id": 257591137,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define all(x) x.begin(),x.end()\n#define pp push_back\n#define ones(x) __builtin_popcountll(x)\n#define yes cout<<\"YES\\n\";\n#define no cout<<\"NO\\n\";\n// 32bit(1)=2147483647\nusing namespace std;\n#define ll long long\nbool const sorted (const pair<long long,long long>&p1,const pair<long long,long long>&p2){\n if(p1.second==p2.second){return p1.first>p2.first;}\n // else{return p1.first<p2.first;}\n else{return p1.second>p2.second;}\n}\nint fastpower(int x,int n){\n int ans=1;\n while(n>0){\n if(n%2==1){ans*=x;}\n x*=x;\n n/=2;\n }\n return ans;\n}\nvector<ll>v; map<ll,ll>mp;\nvector<ll> piles(ll n,vector<ll> &v,map<ll,ll>&mp){\n if(n%3==0){\n v.pp(n/3);\n v.pp(((n/3)*2));\n if(mp[n/3]<1){ piles((n/3),v,mp);mp[n/3]++;}\n if(mp[((n/3)*2)]<1){ piles(((n/3)*2),v,mp);mp[((n/3)*2)]++;}\n \n }\n return v;\n \n}\n \nvoid solve() {\n#ifndef ONLINE_JUDGE\n freopen(\"input.txt\", \"r\", stdin);\n freopen(\"output.txt\", \"w\", stdout);\n#endif\n ll t;cin>>t;\n while(t--) {\n ll n;cin>>n;\n map<ll,ll>mp;\n for (long long i = 0; i < n; ++i) {\n ll x;cin>>x;mp[x]++;\n }\n ll ans=0;\n for(auto it:mp){\n ll v=it.second;\n if(v<4){ continue;}\n else if(v>=4&&v<7){ans++;}\n else if (v==7){ans+=2;}\n else{ans+=2;v-=7;ans+=(v/2);}\n }\n cout<<ans<<'\\n';\n }\n}\nint main(){\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": "1957A",
"submission_id": 258941214,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nconstexpr auto sp = \" \";\nconstexpr auto en = \"\\n\";\n \nusing ull = unsigned long long;\nusing ll = long long;\nusing ld = long double;\n \n#define isz(x) int((x).size())\n#define all(x) begin(x), end(x)\n#define rall(x) rbegin(x), rend(x)\n#define fin(x) for (auto& it : x) cin >> it;\n#define foutsp(x) for (auto& it : x) cout << it << \" \"; cout << en;\n#define fouten(x) for (auto& it : x) cout << it << en;\n#define fyesno(x) cout << (x ? \"YES\" : \"NO\") << en;\n \nusing namespace std;\n \nvoid solve();\nvoid global();\n \null gcd(ull a, ull b) {\n return b == 0 ? a : gcd(b, a % b);\n}\n \null lcm(ull a, ull b) {\n return a * b / gcd(a, b);\n}\n \nvector<int> to_z(string& s, bool flag = false) {\n vector<int> z(isz(s));\n if (flag) z[0] = isz(z);\n int l = 0, r = 1;\n for (int i = 1; i < isz(z); i++) {\n if (r > i) z[i] = min(z[i - l], r - i);\n while (z[i] + i < isz(z) && s[z[i]] == s[z[i] + i]) z[i]++;\n if (r < z[i] + i) { l = i; r = z[i] + i; }\n }\n return z;\n}\n \null get_c(ull n, ull k) {\n ull r = 1, dif = n - k;\n for (int i = 1; i <= k; i++)\n r += r * dif / i;\n return r;\n}\n \nstring pi_str = \"31415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679\";\n \ndouble get_pi() {\n return acos(-1);\n}\n \nint my_rand(int left, int right) {\n return rand() % (right - left + 1) + left;\n}\n \nvoid test() {\n solve();\n}\n \nvoid tests() {\n int t; cin >> t;\n while (t--) solve();\n}\n \nint main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr); cout.tie(nullptr);\n cout.setf(ios::fixed); cout.precision(12);\n \n srand(time(0));\n global();\n tests();\n \n return 0;\n}\n \n \nvoid global() {\n \n}\n \nvoid solve() {\n int n;\n cin >> n;\n \n map<int, int> d;\n for (int i = 0; i < n; i++) {\n int x;\n cin >> x;\n \n d[x]++;\n }\n \n int answer = 0;\n for (auto& p : d) {\n int x = 3;\n \n while (p.second - x >= 0) {\n p.second -= x++;\n answer++;\n }\n }\n \n cout << answer << en;\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": "1957A",
"submission_id": 257579548,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\nt=int(input())\nwhile t:\n n=int(input())\n side_length=[]\n count={}\n x=input().split()\n side_length=list(x)\n for i in side_length:\n if i in count:\n count[i]=count[i]+1\n else:\n count[i]=1\n sum=0\n for i in count:\n if count[i]>=3:\n sum=sum+1\n \n print(sum)\n \n t=t-1\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957A",
"submission_id": 276393056,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\nfrom sys import stdin\ndef input(): return stdin.readline()\n \ndef solve():\n l=int(input())\n a = list(map(int, input().split()))\n c=[0]*100\n i=0\n while(i<l):\n c[a[i]]=c[a[i]]+1\n i=i+1\n i=0\n count=0\n while(i<l):\n if(c[i]>=4):\n count=count+1\n i=i+1\n print(count)\n \n \n \n \nfor _ in range(int(input())):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957A",
"submission_id": 257582846,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class A940 {\n static class Pair implements Comparable<Pair>{\n int first;\n int second;\n public Pair(int first, int second){\n this.first=first;\n this.second=second;\n }\n @Override\n public int compareTo(Pair other){\n return this.first-other.first;\n }\n }\n public static int gcd(int a, int b) {\n while (b != 0) {\n int temp = b;\n b = a % b;\n a = temp;\n }\n return a;\n }\n public static int gcdOfArray(int[] arr) {\n int result = arr[0];\n for (int i = 1; i < arr.length; i++) {\n result = gcd(result, arr[i]);\n }\n return result;\n }\n public static int findMax(int[] arr) {\n int max = arr[0];\n for (int i = 1; i < arr.length; i++) {\n if (arr[i] > max) {\n max = arr[i];\n }\n }\n return max;\n }\n public static int findMin(int[] arr) {\n int min = arr[0];\n for (int i = 1; i < arr.length; i++) {\n if (arr[i] < min) {\n min = arr[i];\n }\n }\n return min;\n }\n public static int[] update(int[] arr, int tickets, int day){\n for (int i = day; i < arr.length; i++) {\n arr[i]+=tickets;\n }\n return arr;\n }\n public static int Solve(int n, int[] arr){\n if (n==1 || n==2) {\n return 0;\n }\n HashMap<Integer, Integer> map = new HashMap<>();\n for (int i = 0; i < arr.length; i++) {\n map.put(arr[i], map.getOrDefault(arr[i], 0)+1);\n }\n int count = 0;\n for (int value : map.values()) {\n if (value > 2) {\n count++;\n }\n }\n return count;\n }\n public static void main(String[] args) {\n Scanner sc= new Scanner(System.in);\n int t = sc.nextInt();\n for (int i = 0; i < t; i++) {\n int n= sc.nextInt();\n int[] arr= new int[n];\n for (int j = 0; j < n; j++) {\n arr[j]= sc.nextInt();\n }\n System.out.println(Solve(n ,arr));\n }\n sc.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": "1957A",
"submission_id": 257583377,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\nimport java.lang.*;\nimport java.util.*;\npublic class A\n{\n public static void main(String args[])\n {\n Scanner s=new Scanner(System.in);\n int t=s.nextInt();\n while(t-->0)\n {\n int n=s.nextInt();int a[]=new int[n];int b[]=new int[n];\n // if(n<3)\n // System.out.println(0);\n // else{\n for(int i=0;i<n;i++)\n {\n a[i]=s.nextInt();\n }\n if(n<3)\n System.out.println(0);\n else{\n Arrays.sort(a);int c=1,k=0;\n for(int i=0;i<n-1;i++)\n {\n if(a[i]==a[i+1])\n c++;\n else{\n \n b[k]=c;\n k=k+1;\n c=1;\n }\n }\n b[k]=c;\n int no=0;\n for(int i=0;i<n;i++)\n {\n if(b[i]>=3)\n no++;\n }\n System.out.println(no);\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": "1957A",
"submission_id": 257623842,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define\tendl '\\n'\n#define\tint long long\nusing namespace std;\nconst int N=1e5+5;\nconst int mod=998244353;\nconst int INF=0x3f3f3f3f; \t\nint arr[N];\nsigned main()\n{\n\tios::sync_with_stdio(false), cin.tie(0), cout.tie(0);\n\tint t;\n\tcin>>t;\n\twhile(t--)\n\t{\n\t\tint n,cnt=0;\n\t\tcin>>n;\n\t\tpriority_queue<int,vector<int>,greater<int>> q;\n\t\tfor(int i=0;i<n;i++)\n\t\t{\n\t\t\tint a;\n\t\t\tcin>>a;\n\t\t\tq.push(a);\n\t\t}\n\t\tif(n<=2)\n\t\t{\n\t\t\tcout<<0<<endl;\n\t\t\tcontinue;\n\t\t}\n\t\twhile(q.size()>=3)\n\t\t{\n\t\t\tint arr[5];\n\t\t\tfor(int i=1;i<=3;i++)\n\t\t\t{\n\t\t\t\tarr[i]=q.top();\n\t\t\t\tq.pop(); \n\t\t\t}\n\t\t\tif(arr[1]==arr[2]&&arr[2]==arr[3])\n\t\t\t\tcnt++;\t\t\n\t\t}\n\t\tcout<<cnt<<endl;\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": "1957A",
"submission_id": 271930652,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define ll long long\n#define vi vector<long long >\n#define vpi vector<pair<long long, long long >>\n#define vvi vector<vector<int>>\n#define sl set<ll>;\n#define f(a,i) a[i].second\n#define f(i, a, b) for (ll i = a; i < b; i++)\n#define fors(n) for (ll i = 0; i < n; i++)\n#define show(c) cout << c << \"\\n\";\n#define spcshow(c) cout << c << ' ';\n//const ll m1 = 1e12 + 7;\n//const int m2 = 998244353;\nconst int big = 1e6;\n#define min3(a, b, c) min(c, min(a, b)),\n#define min4(a, b, c, d) min(d, min(c, min(a, b))),\n#define newline cout<<endl;\nlong long factorial(int n)\n{\n if(n==0)\n return 1;\n else\n return(n*factorial(n-1));\n}\n \nlong long ncr(long long n,long long r)\n{\n long long value;\n value=factorial(n)/(factorial(r)*factorial(n-r));\n return value;\n}\nlong long len(long long x){\n ll p = int(log10(x) + 1);\n return p;\n}\nll power(ll a, ll b, ll mod = 0){\n ll result = 1;\n if(mod == 0){\n while(b){\n if(b & 1) result *= a;\n a *= a;\n b = b >> 1;\n }\n }\n else {\n while(b){\n if(b & 1){\n result *= a;\n result = result % mod;\n }\n a *= a;\n a = a % mod;\n b = b >> 1;\n }\n }\n return result;\n}\nbool checkprime(ll n)\n{\n for(ll i=2;i*i<=n;i++)\n {\n if(n==4)\n {return false;}\n if(n%i==0)\n {\n return false;\n }\n }\n return true;\n}\n//Code starts Here\n//Dont forget while(t--) for questions which donot involve it.\n//made by kushal.\nvoid solve()\n{\n ll n;\n cin>>n;\n vi a(n);\n map<ll,ll>m;\n fors(n){\n cin>>a[i];\n m[a[i]]++;\n } \n ll ans=0;\n auto it=m.begin();\n while(it!=m.end()){\n if(it->second>2)ans++;\n it++;\n } \n show(ans)\n}\n \nint main()\n{\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n ll t = 1;\n cin >> t;\n \n while(t--){\n solve();\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": "1957A",
"submission_id": 257698641,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\nx = int(input())\nfor _ in range(x):\n y = int(input())\n z = list(map(int, input().split()))\n z.sort()\n z2 = z[:]\n sum = 0\n for i in range(len(z)-1):\n if z[i] == z[i+1]:\n z2.remove(z[i])\n for k in z2:\n if z.count(k)>2:\n sum += 1\n print(sum)\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": "1957A",
"submission_id": 257763898,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\nT = int(input())\nfor t in range(T):\n n = int(input())\n l = [int(i) for i in input().split()]\n # print(l)\n freq = {}\n for i in l:\n freq[i] = freq.get(i, 0)+1\n count = 0\n for k, v in freq.items():\n if v > 2:\n 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": "1957A",
"submission_id": 260198046,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Stickogon {\n private static int solve(int n, int[] sticks) {\n int[] freq = new int[101];\n \n int res = 0;\n \n for (int stick : sticks) {\n freq[stick]++;\n \n if (freq[stick] == 3) {\n res++;\n }\n }\n \n return res;\n }\n \n public static void main(String[] args) {\n Scanner scan = new Scanner(System.in);\n \n int T = scan.nextInt();\n \n for (int t = 0; t < T; t++) {\n int n = scan.nextInt();\n \n int[] sticks = new int[n];\n \n for (int i = 0; i < n; i++) {\n sticks[i] = scan.nextInt();\n }\n \n System.out.println(solve(n, sticks));\n }\n \n scan.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": "1957A",
"submission_id": 268642057,
"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: Stickogon\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \\ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:\n\n- Each side of a polygon is formed by exactly one stick.\n- No stick is used in more than $$$1$$$ polygon.\n\nNote: Sticks cannot be broken.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 100$$$) — the number of sticks available.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 100$$$) — the stick lengths.\n\nOutput: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.\n\nExamples:\ninput:\n4\n1\n1\n2\n1 1\n6\n2 2 3 3 3 3\n9\n4 2 2 2 2 4 2 4 4\noutput:\n0\n0\n1\n2\n\nNote: In the first test case, we only have one stick, hence we can't form any polygon.\n\nIn the second test case, the two sticks aren't enough to form a polygon either.\n\nIn the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.\n\nIn the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.FileReader;\nimport java.io.IOException;\nimport java.io.InputStream;\nimport java.io.InputStreamReader;\nimport java.io.PrintWriter;\nimport java.util.*;\n \npublic class lol{\n static Scanner input = new Scanner(System.in);\n static PrintWriter output = new PrintWriter(System.out);\n \n \n public static void main(String[] args) throws IOException {\n \n int n=input.nextInt();\n while (n-->0) {\n int length=input.nextInt();\n int[] nums=new int[length];\n HashMap<Integer,Integer> hmap=new HashMap<>();\n int[] count=new int[1];\n for(int i=0;i<length;i++) {\n nums[i]=input.nextInt();\n hmap.put(nums[i],hmap.getOrDefault(nums[i],0)+1);\n }\n hmap.forEach((key,value)->{\n count[0]+=value>2? 1:0;\n });\n output.println(count[0]);\n \n }\n \n output.flush();\n }\n \n \n \n //**********************Scanner&helpers**************\\\\ \n static class Scanner {\n StringTokenizer st;\n BufferedReader br;\n \n public Scanner(InputStream s) {\n br = new BufferedReader(new InputStreamReader(s));\n }\n public Scanner(FileReader r) {\n br = new BufferedReader(r);\n }\n public Scanner(String file) throws IOException {\n br = new BufferedReader(new FileReader(file));\n }\n public String next() throws IOException {\n while (st == null || !st.hasMoreTokens())\n st = new StringTokenizer(br.readLine());\n return st.nextToken();\n }\n public int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n public long nextLong() throws IOException {\n return Long.parseLong(next());\n }\n public String nextLine() throws IOException {\n return br.readLine();\n }\n public double nextDouble() throws IOException {\n String x = next();\n StringBuilder sb = new StringBuilder(\"0\");\n double res = 0, f = 1;\n boolean dec = false, neg = false;\n int start = 0;\n if (x.charAt(0) == '-') {\n neg = true;\n start++;\n }\n for (int i = start; i < x.length(); i++)\n if (x.charAt(i) == '.') {\n res = Long.parseLong(sb.toString());\n sb = new StringBuilder(\"0\");\n dec = true;\n } else {\n sb.append(x.charAt(i));\n if (dec)\n f *= 10;\n }\n res += Long.parseLong(sb.toString()) / f;\n return res * (neg ? -1 : 1);\n }\n \n public long[] nextlongArray(int n) throws IOException {\n long[] a = new long[n];\n for (int i = 0; i < n; i++)\n a[i] = nextLong();\n return a;\n }\n \n public Long[] nextLongArray(int n) throws IOException {\n Long[] a = new Long[n];\n for (int i = 0; i < n; i++)\n a[i] = nextLong();\n return a;\n }\n \n public int[] nextIntArray(int n) throws IOException {\n int[] a = new int[n];\n for (int i = 0; i < n; i++)\n a[i] = nextInt();\n return a;\n }\n public Integer[] nextIntegerArray(int n) throws IOException {\n Integer[] a = new Integer[n];\n for (int i = 0; i < n; i++)\n a[i] = nextInt();\n return a;\n }\n public boolean ready() throws IOException {\n return br.ready();\n }\n }\n static class Pair implements Comparable<Pair> {\n long first;\n long second;\n \n public Pair(long first, long second) {\n this.first = first;\n this.second = second;\n }\n \n public int compareTo(Pair p) {\n if (first != p.first)\n return Long.compare(first, p.first);\n else if (second != p.second)\n return Long.compare(second, p.second);\n else\n return 0;\n }\n public String toString() {\n return first + \" \" + second;\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": "1957A",
"submission_id": 258311475,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define lowbit(x) x&(-x)\n#define int long long\ntypedef long long ll;\nint INF = 1e18;\nconst int N = 2e5+10;\n \n//int mod = 1e9+7;\n//int ways[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};\n \nint n,q;\nint oc[N],a[N],nex[N],ac[N];\nvector<int> have[N];\n \n#define se second\n#define fi first\n \ntypedef struct{\n #define ls(x) (x<<1)\n #define rs(x) (x<<1|1)\n const int MAXN = N;\n \n typedef struct info{\n int val;\n }info;\n \n typedef struct{\n info val;\n }Node;\n \n friend info operator + (const info &a,const info &b){\n info c;\n c.val = max(a.val , b.val);\n return c;\n };\n \n Node tree[N*4];\n int input[N];\n \n void update(int p) {\n tree[p].val = tree[ls(p)].val + tree[rs(p)].val;\n }\n \n void build(int p,int l,int r){\n if(l==r) tree[p] = {input[l]};\n else{\n int mid = (l+r)>>1;\n build(ls(p),l,mid);\n build(rs(p),mid+1,r);\n update(p);\n }\n }\n \n info ask(int p, int l,int r,int tl, int tr) {\n if (tl <= l and r <= tr) return tree[p].val;\n int mid = l + r >> 1;\n if (tr <= mid) return ask(ls(p),l,mid,tl, tr);\n else if (tl > mid) return ask(rs(p),mid+1,r,tl, tr);\n else return ask(ls(p),l,mid,tl, tr) + ask(rs(p),mid+1,r,tl, tr);\n }\n \n void change(int p, int val,int l,int r,int tl,int tr) {\n if (l == r) {\n input[tl] = val;\n tree[p].val = {val};\n return;\n }\n int mid = (l+r) >> 1;\n if (tl <= mid) change(ls(p), val,l,mid, tl,tr);\n else change(rs(p), val,mid+1,r, tl,tr);\n update(p);\n }\n}SGT;\n \nSGT tree;\n \nvoid solve() {\n cin>>n>>q;\n \n for(int i = 0;i<=n+1;i++){\n have[i].clear();\n nex[i] = ac[i] = oc[i] = 0;\n }\n \n for(int i = 1;i<=n;i++) cin>>a[i];\n \n for(int i = 1;i<=n;i++) oc[i] = n+1;\n \n for(int i = n;i>=1;i--){\n if(a[i] == a[i+1]) nex[i] = nex[i+1];\n else nex[i] = oc[a[i]];\n \n have[nex[i]].push_back(i);\n oc[a[i]] = i;\n }\n \n for(int i = 1;i<=n;i++) tree.input[i] = nex[i];\n tree.build(1,1,n);\n \n int l = n;\n for(auto i:have[n+1]) tree.change(1,0,1,n,i,i);\n \n for(int r = n;r>=1;r--){\n while(l >= 1 && tree.ask(1,1,n,l,nex[l]).val <= nex[l]){\n l--;\n }\n \n for(auto i:have[r]) tree.change(1,0,1,n,i,i);\n ac[r] = l+1;\n }\n \n int r;\n while(q--){\n cin>>l>>r;\n \n if(ac[r] <= l) cout<<\"YES\"<<endl;\n else cout<<\"NO\"<<endl;\n }\n \n \n}\nsigned main() {\n std::ios::sync_with_stdio(false);\n cin.tie(0);\n cout.tie(0);\n //cout<<fixed<<setprecision(10);\n \n int caset = 1;\n cin >> caset;\n while (caset--) { 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": "2030F",
"submission_id": 290826791,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n \nusing pii = pair<int,int>;\n \n#define int long long\n \nconst int N = 2e5 + 10;\n \nstruct Info {\n int mx;\n \n Info operator+(const Info &o) const {\n Info ret;\n ret.mx = max(mx, o.mx);\n return ret;\n }\n} tr[N << 2];\n \n \nint pre[N], pre1[N];\n \nvoid push_up(int p) {\n tr[p] = tr[p << 1] + tr[p << 1|1];\n}\n \nvoid build(int s, int t, int p) {\n if (s == t) {\n tr[p].mx = pre1[s];\n return;\n }\n int mid = s + t >> 1;\n build(s,mid,p << 1);\n build(mid + 1,t,p << 1|1);\n push_up(p);\n}\n \nvoid modify(int k, int c, int s, int t, int p) {\n if (s == t) { \n tr[p].mx = c;\n return;\n }\n int mid = s + t >> 1;\n if (k <= mid) modify(k,c,s,mid,p << 1);\n else modify(k,c,mid + 1,t,p << 1|1);\n push_up(p);\n}\n \nInfo query(int l, int r, int s, int t, int p) {\n if (l <= s && t <= r) {\n return tr[p];\n }\n int mid = s + t >> 1;\n if (l > mid) return query(l,r,mid + 1,t,p << 1|1);\n if (r <= mid) return query(l,r,s,mid,p << 1);\n return query(l,r,s,mid,p << 1) + query(l,r,mid + 1,t,p << 1|1);\n}\n \n \nvoid solve() {\n int n, q; cin >> n >> q;\n vector<int> a(n + 1);\n for (int i = 1; i <= n; i++) {\n cin >> a[i];\n }\n \n for (int i = 1; i <= n; i++) {\n modify(1,0,1,n,1);\n pre1[i] = pre[i] = 0;\n }\n \n vector<int> f(n + 1);\n vector<pii> b; \n for (int i = 1; i <= n; i++) {\n pre[i] = f[a[i]];\n f[a[i]] = i;\n // cout << pre[i] << \" \\n\"[i == n];\n b.push_back({pre[i], i});\n }\n \n sort(b.begin(),b.end());\n \n for (auto [pr, i] : b) {\n if (pr + 1 > i - 1) {\n pre1[i] = 0;\n modify(i,pr,1,n,1);\n continue;\n }\n Info ret = query(pr + 1, i - 1,1,n,1);\n pre1[i] = ret.mx;\n modify(i,pr,1,n,1);\n }\n \n build(1,n,1);\n while (q--) {\n int l,r; cin >> l >> r;\n if (query(l,r,1,n,1).mx >= l) cout << \"NO\\n\";\n else cout << \"YES\\n\";\n }\n \n}\n \nsigned main() {\n ios::sync_with_stdio(false);\n cin.tie(0);\n int t = 1;\n cin >> t;\n while (t--)solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2030F",
"submission_id": 286824745,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \ntypedef long long ll;\n \nconst int MAXN = 1e6 + 10;\nconst int mod = 998244353;\n \ninline int add(int x, int y) { return x += y, x < mod || (x -= mod), x; }\ninline void cadd(int &x, int y) { x += y, x < mod || (x -= mod); }\n \nint T, n, m, a[MAXN], p[MAXN], pre[MAXN], nxt[MAXN], val[MAXN << 2], res[MAXN];\n \ninline void pushup(int p) { val[p] = max(val[p << 1], val[p << 1 | 1]); }\n \nvoid build(int l = 1, int r = n, int p = 1) {\n\tif (l == r) return val[p] = nxt[l], void(); int mid = l + r >> 1;\n\tbuild(l, mid, p << 1), build(mid + 1, r, p << 1 | 1), pushup(p);\n}\n \nvoid modify(int k, int l = 1, int r = n, int p = 1) {\n\tif (l == r) return val[p] = 0, void(); int mid = l + r >> 1;\n\tif (k <= mid) modify(k, l, mid, p << 1);\n\telse modify(k, mid + 1, r, p << 1 | 1); pushup(p);\n}\n \nint ask(int ql, int qr, int l = 1, int r = n, int p = 1) {\n\tif (l > r || r < ql || l > qr) return 0;\n\tif (ql <= l && r <= qr) return val[p]; int mid = l + r >> 1;\n\treturn max(ask(ql, qr, l, mid, p << 1), ask(ql, qr, mid + 1, r, p << 1 | 1));\n}\n \nint __cnt = 0;\n \nint main() {\n\tfor (scanf(\"%d\", &T); T--; ) {\n\t\tscanf(\"%d%d\", &n, &m);\n\t\tfor (int i = 1; i <= n; i++) scanf(\"%d\", &a[i]), p[i] = 0;\n\t\tfor (int i = n; i; i--) nxt[i] = p[a[i]], pre[nxt[i]] = i, p[a[i]] = i; build();\n\t\tfor (int i = n, j = n; i; i--) {\n\t\t\tfor (; j > 1 && ask(j, nxt[j - 1] - 1) <= nxt[j - 1]; j--);\n\t\t\tres[i] = j; if (pre[i]) modify(pre[i]);\n\t\t}\n\t\tfor (int l, r; m--; ) {\n\t\t\tscanf(\"%d%d\", &l, &r);\n\t\t\tif (++__cnt == 8201) return printf(\"%d %d %d %d %d %d %d\\n\", a[1], a[2], a[3], a[4], a[5], l, r), 0;\n\t\t\tputs(res[r] <= l ? \"YES\" : \"NO\");\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": "2030F",
"submission_id": 296265658,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\nimport sys, os, io\ninput = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline\n \ndef update(i, x):\n i ^= l1\n tree[i] = x\n i >>= 1\n while i:\n tree[i] = max(tree[i << 1], tree[i << 1 ^ 1])\n i >>= 1\n return\n \ndef get_max(s, t):\n s, t = s ^ l1, t ^ l1\n ans = 0\n while s <= t:\n if s & 1:\n ans = max(ans, tree[s])\n s += 1\n s >>= 1\n if not t & 1:\n ans = max(ans, tree[t])\n t -= 1\n t >>= 1\n return ans\n \nt = int(input())\nans = []\nfor _ in range(t):\n n, q = map(int, input().split())\n a = [0] + list(map(int, input().split()))\n nx, u = [0] * (n + 1), [0] * (n + 1)\n for i in range(n, 0, -1):\n j = a[i]\n if u[j]:\n nx[i] = u[j]\n u[j] = i\n l1 = pow(2, (n + 1).bit_length())\n l2 = 2 * l1\n tree = [0] * l2\n ng = [0] * (n + 1)\n for l in range(1, n + 1):\n if not nx[l]:\n continue\n r = nx[l]\n ng[r] = get_max(l, r)\n update(r, l)\n for i in range(1, n + 1):\n ng[i] = max(ng[i], ng[i - 1])\n for _ in range(q):\n l, r = map(int, input().split())\n ans0 = \"YES\" if ng[r] < l else \"NO\"\n ans.append(ans0)\nsys.stdout.write(\"\\n\".join(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": "2030F",
"submission_id": 291479591,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\nimport heapq\n \nclass SegmentTree:\n def __init__(self, data, default=0, func=max):\n \"\"\"initialize the segment tree with data\"\"\"\n self._default = default\n self._func = func\n self._len = len(data)\n self._size = _size = 1 << (self._len - 1).bit_length()\n \n self.data = [default] * (2 * _size)\n self.data[_size:_size + self._len] = data\n for i in reversed(range(_size)):\n self.data[i] = func(self.data[i + i], self.data[i + i + 1])\n \n def __delitem__(self, idx):\n self[idx] = self._default\n \n def __getitem__(self, idx):\n return self.data[idx + self._size]\n \n def __setitem__(self, idx, value):\n idx += self._size\n self.data[idx] = value\n idx >>= 1\n while idx:\n self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])\n idx >>= 1\n \n def __len__(self):\n return self._len\n \n def query(self, start, stop):\n \"\"\"func of data[start, stop)\"\"\"\n start += self._size\n stop += self._size\n \n res_left = res_right = self._default\n while start < stop:\n if start & 1:\n res_left = self._func(res_left, self.data[start])\n start += 1\n if stop & 1:\n stop -= 1\n res_right = self._func(self.data[stop], res_right)\n start >>= 1\n stop >>= 1\n \n return self._func(res_left, res_right)\n \n def __repr__(self):\n return \"SegmentTree({0})\".format(self.data)\n \n \nfor _ in range(int(input())):\n n, q = map(int, input().split())\n a = list(map(int, input().split()))\n \n seg = SegmentTree([0] * n)\n prv = [-1] * (n + 1)\n intervals = []\n for i in range(n):\n if prv[a[i]] != -1:\n intervals.append([prv[a[i]], i])\n seg[i] = prv[a[i]]\n prv[a[i]] = i\n \n pq = []\n max_l = [-1] * n\n for l, r in sorted(intervals):\n while pq and pq[0][0] <= l:\n heapq.heappop(pq)\n if pq and pq[0][0] < r:\n max_l[r] = seg.query(l, r)\n heapq.heappush(pq, [r, l])\n \n for i in range(1, n):\n max_l[i] = max(max_l[i], max_l[i-1])\n \n for i in range(q):\n l, r = map(int, input().split())\n print(\"YES\" if max_l[r-1] < l-1 else \"NO\")\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": "2030F",
"submission_id": 290476883,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n# vai proshno onk difficult, eta dia hoibo na\n# def is_orangutan_approved(subarray):\n# S = set(subarray)\n# while True:\n# found = False\n# l = 0\n# while l < len(subarray):\n# r = l\n# while r + 1 < len(subarray) and subarray[r + 1] == subarray[l]:\n# r += 1\n# if r > l and subarray[l] in S:\n# S.remove(subarray[l])\n# del subarray[l:r + 1]\n# found = True\n# break\n# l += 1\n# if not found:\n# break\n# return len(subarray) == 0\n \n# t = int(input())\n# for _ in range(t):\n# n, q = map(int, input().split())\n# a = list(map(int, input().split()))\n# for __ in range(q):\n# l, r = map(int, input().split())\n# subarray = a[l - 1:r]\n# if is_orangutan_approved(subarray):\n# print(\"YES\")\n# else:\n# print(\"NO\")\n \n# idk if it works\ndef is_orangutan_approved(a: list) -> bool:\n if len(a) == len(set(a)):\n return True\n test1_failed = False\n while True:\n found = False\n l = 0\n while l < len(a):\n r = l\n while r + 1 < len(a) and a[r + 1] == a[l]:\n r += 1\n if r > l:\n del a[l:r + 1]\n found = True\n break\n l += 1\n if not found:\n test1_failed = True\n break\n while test1_failed:\n if len(a) == 0:\n return True\n found = False\n l, r = 0, len(a) - 1\n while l <= r:\n if a[l] == a[r]:\n if l == r:\n a.pop(l)\n else:\n a.pop(r)\n a.pop(l)\n found = True\n break\n else:\n l += 1\n r -= 1\n if not found:\n break\n \n return len(a) == 0\n \nt = int(input())\nfor _ in range(t):\n n, q = map(int, input().split())\n a = list(map(int, input().split()))\n for __ in range(q):\n l, r = map(int, input().split())\n result = is_orangutan_approved(a[l - 1: r])\n print(\"YES\" if result else \"NO\")\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": "2030F",
"submission_id": 287989060,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\nimport java.util.ArrayList;\nimport java.util.Arrays;\nimport java.util.HashSet;\nimport java.util.List;\nimport java.util.Scanner;\n \npublic class F {\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 Integer[] arr = new Integer[n];\n for(int i=0; i<n; i++){\n arr[i] = sc.nextInt();\n }\n \n queries:\n while(q-->0){\n int l = sc.nextInt();\n int r = sc.nextInt();\n \n if(r-l+1==3){\n System.out.println(\"YES\");\n continue queries;\n } \n \n List<Integer> al = new ArrayList<>(Arrays.asList(arr).subList(l-1, r-1));\n HashSet<Integer> hs = new HashSet<>();\n for(int i:al) hs.add(i);\n int siz = al.size();\n for(int i=0; i<siz-1; i++){\n //System.out.println(\"looking at\" + al.get(i));\n if(al.get(i)==al.get(i+1) && hs.contains(al.get(i))){\n al.remove(i+1); al.remove(i);\n siz = al.size();\n hs.remove(i);\n i--;\n }\n else{\n al.remove(i);\n siz = al.size();\n hs.remove(i);\n i--;\n }\n \n if(al.isEmpty()){\n System.out.println(\"YES\");\n continue queries;\n }\n }\n \n int limit = ((r-l)+1)/2;\n for(int i=0; i<limit; i++){\n if(arr[l+i-1]!=arr[r-i-1]){\n System.out.println(\"NO\");\n continue queries;\n }\n }\n \n System.out.println(\"YES\");\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": "2030F",
"submission_id": 286822719,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class F {\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 int[] arr = new int[n];\n for(int i=0; i<n; i++){\n arr[i] = sc.nextInt();\n }\n queries:\n while(q-->0){\n int l = sc.nextInt();\n int r = sc.nextInt();\n int limit = ((r-l)+1)/2;\n if(r-l+1==3){\n System.out.println(\"YES\");\n continue queries;\n } \n for(int i=0; i<limit; i++){\n if(arr[l+i-1]!=arr[r-i-1]){\n System.out.println(\"NO\");\n continue queries;\n }\n }\n \n System.out.println(\"YES\");\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": "2030F",
"submission_id": 286812254,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\ntypedef long long ll;\n \ntemplate<class Info>\nstruct SegmentTree\n{\n\t#define l(p) (p<<1)\n\t#define r(p) (p<<1|1)\n\tint n;\n\tvector<Info> info;\n \n\tSegmentTree() {}\n SegmentTree(int n) {\n this->n = n;\n info.resize(4 * n + 10);\n }\n\tSegmentTree(vector<Info> _init) {\n\t\tn=_init.size()-1;\n\t\tinfo.resize(4*n+10);\n\t\tauto build=[&](auto build,int p,int l,int r)->void {\n\t\t\tif(l==r) info[p]=_init[l];\n\t\t\telse {\n\t\t\t\tint mid=l+r>>1;\n\t\t\t\tbuild(build,l(p),l,mid);\n\t\t\t\tbuild(build,r(p),mid+1,r);\n\t\t\t\tpushup(p);\n\t\t\t}\n\t\t};\n\t\tbuild(build,1,1,n);\n\t}\n \n\tvoid pushup(int p) {\n\t\tinfo[p]=info[l(p)]+info[r(p)];\n\t}\n \n\tvoid modify(int pos,const Info &v) {\n\t\tmodify(1,1,n,pos,v);\n\t}\n \n\tvoid modify(int p,int l,int r,int pos,const Info &v) {\n\t\tif(l==r) info[p]=v;\n\t\telse {\n\t\t\tint mid=l+r>>1;\n\t\t\tif(mid>=pos) modify(l(p),l,mid,pos,v);\n\t\t\telse modify(r(p),mid+1,r,pos,v);\n\t\t\tpushup(p);\n\t\t}\n\t}\n \n\tInfo query(int l,int r) {\n\t\treturn query(1,1,n,l,r);\n\t}\n \n\tInfo query(int p,int pl,int pr,int l,int r) { \n\t\tif(pl>=l&&pr<=r) return info[p];\n\t\telse {\n\t\t\tint pmid=pl+pr>>1;\n\t\t\t\n\t\t\tif(pmid>=l&&pmid+1<=r) {\n\t\t\t\treturn query(l(p),pl,pmid,l,r)+query(r(p),pmid+1,pr,l,r);\n\t\t\t}\n\t\t\telse if(pmid>=l) return query(l(p),pl,pmid,l,r);\n\t\t\telse if(pmid+1<=r) return query(r(p),pmid+1,pr,l,r);\n\t\t}\n\t}\n \n\t#undef l(p)\n\t#undef r(p)\n};\n \nstruct Info {\n\tint x;\n};\n \nInfo operator+(Info a,Info b) {\n\tInfo c;\n c.x = a.x + b.x;\n\treturn c;\n}\n \nvoid solve() {\n int n, q;\n cin >> n >> q;\n \n vector<int> a(n);\n for (int i = 0; i < n; i++) {\n cin >> a[i];\n a[i]--;\n }\n \n vector<int> pre(n), nxt(n), pos(n, -1);\n \n for (int i = n - 1; i >= 0; i--) {\n nxt[i] = pos[a[i]] == -1 ? i : pos[a[i]];\n pos[a[i]] = i;\n }\n \n iota(pre.begin(), pre.end(), 0);\n for (int i = 0; i < n; i++) {\n if (nxt[i] != i) pre[nxt[i]] = i;\n }\n \n // for (int i = 0; i < n; i++) {\n // cerr << pre[i] << '*' << nxt[i] << \" \\n\"[i == n - 1];\n // }\n \n vector<int> f(n);\n int idx = 0;\n for (int i = 0; i < n; i++) {\n f[i] = i == 0 ? 0 : f[i - 1];\n while (f[i] + 1 < n) {\n int nl = f[i] + 1;\n \n SegmentTree<Info> seg(n);\n for (int j = i; j < pre[nl]; j++) {\n if (nxt[j] + 1 <= n) {\n seg.modify(nxt[j] + 1, {1});\n }\n }\n if (seg.query(pre[nl] + 1, nl + 1).x == 0) {\n f[i]++;\n }\n else {\n break;\n }\n }\n \n // seg.modify(nxt[i] + 1, {0});\n }\n \n while (q--) {\n int l, r;\n cin >> l >> r;\n \n l--, r--;\n \n if (f[l] >= r) {\n cout << \"YES\\n\";\n } else {\n cout << \"NO\\n\";\n }\n }\n}\n \nint main() {\n ios::sync_with_stdio(0);\n cin.tie(0);\n \n int T = 1;\n cin >> T;\n \n while (T--) {\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2030F",
"submission_id": 286898276,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n#define _CRT_SECURE_NO_WARNINGS\n#include<bits/stdc++.h>\n#define int long long\n#define YES cout<<\"YES\\n\"\n#define NO cout<<\"NO\\n\"\n#define endl \"\\n\"\n#define debug cout<<'*'<<'\\n'\nconst int N = 2e5 + 500, M = 5e5 + 500;\nconst int mod = 1e9 + 7, inf = 0x3f3f3f3f3f3f3f3f;\nusing namespace std;\nusing pii=pair<int,int>;\n \nstruct Tree {\n\tint l,r,mi;\n}tr[N<<3];\nint n,m;\nint pre[N],a[N],b[N],lf[N];\nqueue<int> q[N];\ninline int ls(int p){return p<<1;}\ninline int rs(int p){return p<<1|1;}\nvoid push_up(int p) {\n\ttr[p].mi=min(tr[ls(p)].mi,tr[rs(p)].mi);\n}\nvoid build(int p,int l,int r) {\n\ttr[p].l=l,tr[p].r=r,tr[p].mi=n+1;\n\tif(l==r)return;\n\tint mid=(l+r)>>1;\n\tbuild(ls(p),l,mid);\n\tbuild(rs(p),mid+1,r);\n\tpush_up(p);\n}\nvoid change(int p,int inv,int k) {\n\tif(inv==tr[p].l&&tr[p].r==inv) {\n\t\ttr[p].mi=k;\n\t\treturn;\n\t}\n\tif(tr[ls(p)].r>=inv)change(ls(p),inv,k);\n\tif(tr[rs(p)].l<=inv)change(rs(p),inv,k);\n\tpush_up(p);\n}\nint query(int p,int l,int r) {\n if(l>r)return n+1;\n\tif(l<=tr[p].l&&tr[p].r<=r) {\n\t\treturn tr[p].mi;\n\t}\n\tint res=n+1;\n\tif(tr[ls(p)].r>=l)res=min(res,query(ls(p),l,r));\n\tif(tr[rs(p)].l<=r)res=min(res,query(rs(p),l,r));\n\treturn res;\n}\nvoid solve()\n{\n\tcin>>n>>m;\n\tfor(int i=1;i<=n;i++) {\n\t\ta[i]=lf[i]=0;\n\t\tb[i]=pre[i]=n+1;\n\t\twhile(!q[i].empty())q[i].pop();\n\t}\n\tfor(int i=1;i<=n;i++)cin>>a[i];\n\tfor(int i=1;i<=n;i++) {\n\t\tpre[i]=b[a[i]];\n\t\tb[a[i]]=i;\n\t}\n\tbuild(1,1,n);\n\t//for(int i=1;i<=n;i++)cout<<query(1,i,i)<<\" \";cout<<endl;\n\t//for(int i=1;i<=n;i++) cout<<q[i].size()<<\" \";cout<<endl;\n\tfor(int l=1,r=1;r<=n;r++) {\n\t\twhile(!q[a[r]].empty()) {\n\t\t\t//cout<<\"? \"<<query(1,q[a[r]].front()+1,r-1)<<\" \"<<q[a[r]].front()<<endl;\n\t\t\tif(query(1,q[a[r]].front()+1,r-1)<q[a[r]].front()) {\n\t\t\t\twhile(!q[a[r]].empty()&&l<q[a[r]].front()) {\n\t\t\t\t\tq[a[l]].pop();\n\t\t\t\t\tif(!q[a[l]].empty())change(1,q[a[l]].front(),n+1);\n\t\t\t\t\tl++;\n\t\t\t\t}\n\t\t\t\tq[a[r]].pop();\n\t\t\t}else break;\n\t\t}\n\t\tq[a[r]].push(r);\n\t\tchange(1,r,pre[r]);\n\t\t//cout<<l<<\" \"<<r<<endl;\n\t\t//for(int i=1;i<=n;i++)cout<<query(1,i,i)<<\" \";cout<<endl;\n\t\tlf[r]=l;\n\t}\n\t//for(int i=1;i<=n;i++)cout<<lf[i]<<' ';cout<<endl;\n\twhile(m--) {\n\t\tint l,r;cin>>l>>r;\n\t\tif(lf[r]<=l)YES;\n\t\telse NO;\n\t}\n}\nsigned main()\n{\n\t//freopen(\"C:\\\\Users\\\\win11\\\\Desktop\\\\test.txt\", \"r\", stdin);\n\tios::sync_with_stdio(false);cin.tie(0);cout.tie(0);\n \n\tint T = 1;\n\tcin >> T;\n\twhile (T--)solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2030F",
"submission_id": 291308983,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n#!/usr/bin/env python\nimport os\nimport sys\nfrom io import BytesIO, IOBase\n \n \nclass RangeQuery:\n def __init__(self, data, func=min):\n self.func = func\n self._data = _data = [list(data)]\n i, n = 1, len(_data[0])\n while 2 * i <= n:\n prev = _data[-1]\n _data.append([func(prev[j], prev[j + i]) for j in range(n - 2 * i + 1)])\n i <<= 1\n \n def query(self, start, stop):\n \"\"\"func of data[start, stop)\"\"\"\n depth = (stop - start).bit_length() - 1\n return self.func(self._data[depth][start], self._data[depth][stop - (1 << depth)])\n \n def __getitem__(self, idx):\n return self._data[0][idx]\n \nclass SegmentTree:\n def __init__(self, data, default=0, func=max):\n \"\"\"initialize the segment tree with data\"\"\"\n self._default = default\n self._func = func\n self._len = len(data)\n self._size = _size = 1 << (self._len - 1).bit_length()\n \n self.data = [default] * (2 * _size)\n self.data[_size:_size + self._len] = data\n for i in reversed(range(_size)):\n self.data[i] = func(self.data[i + i], self.data[i + i + 1])\n \n def __delitem__(self, idx):\n self[idx] = self._default\n \n def __getitem__(self, idx):\n return self.data[idx + self._size]\n \n def __setitem__(self, idx, value):\n idx += self._size\n self.data[idx] = value\n idx >>= 1\n while idx:\n self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])\n idx >>= 1\n \n def __len__(self):\n return self._len\n \n def query(self, start, stop):\n \"\"\"func of data[start, stop)\"\"\"\n start += self._size\n stop += self._size\n \n res_left = res_right = self._default\n while start < stop:\n if start & 1:\n res_left = self._func(res_left, self.data[start])\n start += 1\n if stop & 1:\n stop -= 1\n res_right = self._func(self.data[stop], res_right)\n start >>= 1\n stop >>= 1\n \n return self._func(res_left, res_right)\n \n def __repr__(self):\n return \"SegmentTree({0})\".format(self.data)\n \n \ndef discrete_binary_search(func, lo, hi):\n \"\"\" Locate the first value x s.t. func(x) = True within [lo, hi] \"\"\"\n while lo < hi:\n mi = lo + (hi - lo) // 2\n if func(mi):\n hi = mi\n else:\n lo = mi + 1\n \n return lo\n \n \ndef main():\n t = int(input())\n \n for _ in range(t):\n n, q = map(int, input().split())\n a = [int(ai) - 1 for ai in input().split()]\n \n pre_curr = [-1] * n\n \n pre = [-1] * n\n nex = SegmentTree([-1] * n)\n \n late_start = [-1] * n\n \n for i, ai in enumerate(a):\n if i != 0 and a[i - 1] == ai:\n pre[i] = pre[i - 1]\n else:\n pre[i] = pre_curr[ai]\n if pre[i] != -1:\n nex[pre[i]] = i\n \n if pre[i] > 0 and pre[i] < nex.query(0, pre[i]):\n late_start[i] = discrete_binary_search(lambda x: nex.query(x, pre[i]) <= pre[i], 0, pre[i])\n else:\n late_start[i] = 0\n \n pre_curr[ai] = i\n \n for _ in range(q):\n l, r = map(int, input().split())\n l -= 1\n \n print(\"YES\" if late_start[r - 1] <= l else \"NO\")\n \n \n# region fastio\n \nBUFSIZE = 8192\n \n \nclass FastIO(IOBase):\n newlines = 0\n \n def __init__(self, file):\n self._file = file\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n \n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n \n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n \n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\n \n \nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\n \n \nsys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\ninput = lambda: sys.stdin.readline().rstrip(\"\\r\\n\")\n \n# endregion\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": "2030F",
"submission_id": 286823427,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n# for I/O for local system\nimport sys\nfrom os import path\nif(path.exists('Input.txt')):\n sys.stdin = open('Input.txt','r')\n sys.stdout = open('Output.txt','w')\n \n# For fast I/O\ninput = sys.stdin.buffer.readline\n# input = sys.stdin.readline\nprint = sys.stdout.write\n \n# Import libraries here whenever required\nfrom random import randint\n \n# Use this because normal dict can sometimes give TLE\nclass mydict:\n def __init__(self, func=lambda: 0):\n self.random = randint(0, 1 << 32)\n self.default = func\n self.dict = {}\n def __getitem__(self, key):\n mykey = self.random ^ key\n if mykey not in self.dict:\n self.dict[mykey] = self.default()\n return self.dict[mykey]\n def get(self, key, default):\n mykey = self.random ^ key\n if mykey not in self.dict:\n return default\n return self.dict[mykey]\n def __setitem__(self, key, item):\n mykey = self.random ^ key\n self.dict[mykey] = item\n def getkeys(self):\n return [self.random ^ i for i in self.dict]\n def __str__(self):\n return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'\n \nout = []\n \nfor _ in range(int(input())):\n n, q = [int(x) for x in input().split()]\n a = [int(x) for x in input().split()]\n helper = [[] for i in range(n)]\n for i in range(q):\n l, r = [int(x) - 1 for x in input().split()]\n helper[l].append([r, i])\n for i in range(n):\n helper[i].sort()\n ans = [\"YES\"] * q\n for i in range(n):\n if(len(helper[i]) == 0):\n continue\n L = 0\n U = len(helper[i]) - 1\n take = -1\n while L <= U:\n m = (L + U) >> 1\n r, __ = helper[i][m]\n st = i\n en = r\n qq = []\n d = mydict()\n f = 1\n for ii in range(st, en + 1):\n v = a[ii]\n if(d[v]):\n cc = 0\n while qq:\n vv = qq.pop()\n if(vv == v):\n cc = 1\n break\n if(cc == 0):\n f = 0\n break\n else:\n qq.append(v)\n d[v] = 1\n if(f):\n L = m + 1\n take = m\n else:\n U = m - 1\n for j in range(take + 1, len(helper[i])):\n ans[helper[i][j][1]] = \"NO\"\n for val in ans:\n out.append(val)\n \nprint('\\n'.join(map(str, out)))\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": "2030F",
"submission_id": 286817042,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class OrangutanApprovedSubarrays {\n \n static class Pair {\n int first, last;\n Pair(int f, int l){\n first = f;\n last = l;\n }\n }\n \n // Binary search to find the first index >= target\n static int lowerBound(ArrayList<Integer> list, int target){\n int low = 0, high = list.size();\n while(low < high){\n int mid = (low + high) / 2;\n if(list.get(mid) >= target){\n high = mid;\n }\n else{\n low = mid + 1;\n }\n }\n return low;\n }\n \n // Binary search to find the first index > target\n static int upperBound(ArrayList<Integer> list, int target){\n int low = 0, high = list.size();\n while(low < high){\n int mid = (low + high) / 2;\n if(list.get(mid) > target){\n high = mid;\n }\n else{\n low = mid + 1;\n }\n }\n return low;\n }\n \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 while(t-- > 0){\n st = new StringTokenizer(br.readLine());\n int n = Integer.parseInt(st.nextToken());\n int q = Integer.parseInt(st.nextToken());\n int[] a = new int[n+1];\n st = new StringTokenizer(br.readLine());\n for(int i = 1; i <= n; i++) {\n a[i] = Integer.parseInt(st.nextToken());\n }\n // Preprocess: Map each unique element to its occurrence positions\n HashMap<Integer, ArrayList<Integer>> occMap = new HashMap<>();\n for(int i = 1; i <= n; i++) {\n occMap.computeIfAbsent(a[i], k -> new ArrayList<>()).add(i);\n }\n // Process each query\n for(int i = 0; i < q; i++){\n st = new StringTokenizer(br.readLine());\n int l = Integer.parseInt(st.nextToken());\n int r = Integer.parseInt(st.nextToken());\n List<Pair> list = new ArrayList<>();\n for(Map.Entry<Integer, ArrayList<Integer>> entry : occMap.entrySet()){\n ArrayList<Integer> pos = entry.getValue();\n // Find the first occurrence >= l\n int firstIdx = lowerBound(pos, l);\n if(firstIdx == pos.size()) continue;\n int first = pos.get(firstIdx);\n if(first > r) continue;\n // Find the last occurrence <= r\n int lastIdx = upperBound(pos, r) - 1;\n if(lastIdx < firstIdx) continue;\n int last = pos.get(lastIdx);\n list.add(new Pair(first, last));\n }\n // Sort the intervals by their starting positions\n list.sort(Comparator.comparingInt(p -> p.first));\n // Validate the sorted intervals\n boolean ok = true;\n int last_max = -1;\n for(Pair p : list){\n if(p.first > last_max){\n last_max = p.last;\n }\n else{\n if(p.last <= last_max){\n continue;\n }\n else{\n ok = false;\n break;\n }\n }\n }\n if(ok)\n sb.append(\"YES\\n\");\n else\n sb.append(\"NO\\n\");\n }\n }\n System.out.print(sb.toString());\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": "2030F",
"submission_id": 286804043,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class OrangutanApprovedSubarrays {\n \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 while (t-- > 0) {\n st = new StringTokenizer(br.readLine());\n int n = Integer.parseInt(st.nextToken());\n int q = Integer.parseInt(st.nextToken());\n int[] a = new int[n+1];\n st = new StringTokenizer(br.readLine());\n for(int i=1;i<=n;i++) {\n a[i] = Integer.parseInt(st.nextToken());\n }\n // Preprocess: for each unique value, store sorted list of occurrence indices\n HashMap<Integer, ArrayList<Integer>> occMap = new HashMap<>();\n for(int i=1;i<=n;i++) {\n occMap.computeIfAbsent(a[i], k -> new ArrayList<>()).add(i);\n }\n // Process queries\n for(int i=0;i<q;i++) {\n st = new StringTokenizer(br.readLine());\n int l = Integer.parseInt(st.nextToken());\n int r = Integer.parseInt(st.nextToken());\n // Collect unique values in [l..r]\n // To collect unique values efficiently, iterate through each unique key and check if it exists in [l..r]\n List<Pair> list = new ArrayList<>();\n for(Map.Entry<Integer, ArrayList<Integer>> entry : occMap.entrySet()) {\n ArrayList<Integer> positions = entry.getValue();\n // Find the first occurrence >= l\n int firstIdx = lowerBound(positions, l);\n if(firstIdx == positions.size()) continue;\n int first = positions.get(firstIdx);\n if(first > r) continue;\n // Find the last occurrence <= r\n int lastIdx = upperBound(positions, r) -1;\n if(lastIdx < firstIdx) continue;\n int last = positions.get(lastIdx);\n list.add(new Pair(first, last));\n }\n // Sort the list by first ascending\n Collections.sort(list, (p1, p2) -> {\n if(p1.first != p2.first)\n return Integer.compare(p1.first, p2.first);\n else\n return Integer.compare(p1.last, p2.last);\n });\n // Check the condition\n boolean ok = true;\n int last_max = -1;\n for(Pair p : list) {\n if(p.first > last_max) {\n last_max = p.last;\n }\n else {\n if(p.last > last_max) {\n ok = false;\n break;\n }\n // else, last_max remains the same (since p.last <= last_max)\n }\n }\n if(ok)\n sb.append(\"YES\\n\");\n else\n sb.append(\"NO\\n\");\n }\n }\n System.out.print(sb.toString());\n }\n \n // Find the first index in list where value >= target\n static int lowerBound(ArrayList<Integer> list, int target) {\n int low = 0, high = list.size();\n while(low < high) {\n int mid = (low + high) /2;\n if(list.get(mid) >= target)\n high = mid;\n else\n low = mid +1;\n }\n return low;\n }\n \n // Find the first index in list where value > target\n static int upperBound(ArrayList<Integer> list, int target) {\n int low = 0, high = list.size();\n while(low < high) {\n int mid = (low + high) /2;\n if(list.get(mid) > target)\n high = mid;\n else\n low = mid +1;\n }\n return low;\n }\n \n // Simple Pair class\n static class Pair {\n int first;\n int last;\n Pair(int f, int l){\n first = f;\n last = l;\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": "2030F",
"submission_id": 286801842,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing std::bitset;\nusing std::cout;\nusing std::deque;\nusing std::endl;\nusing std::greater;\nusing std::lower_bound;\nusing std::make_pair;\nusing std::map;\nusing std::max;\nusing std::min;\nusing std::multimap;\nusing std::multiset;\nusing std::nth_element;\nusing std::pair;\nusing std::priority_queue;\nusing std::queue;\nusing std::reverse; \nusing std::set;\nusing std::sort;\nusing std::sqrt;\nusing std::stable_sort;\nusing std::string;\nusing std::swap;\nusing std::unique;\nusing std::upper_bound;\nusing std::vector;\ntypedef long long li;\ntypedef long double lf; \n \ninline li read(){\n\tli ans = 0, f = 1;\n\tchar ch = getchar();\n\twhile(ch < '0' || ch > '9'){\n\t\tf = (ch == '-') ? -1 : 1;\n\t\tch = getchar();\n\t}\n\twhile(ch <= '9' && ch >= '0'){\n\t\tans = ans * 10 + (ch ^ 48);\n\t\tch = getchar();\n\t}\n\treturn ans * f;\n} \n \nli tree[1000010];\n \nvoid modify(li u, li l, li r, li x, li v){\n if(l > x || x > r) return;\n if(l == r){\n tree[u] = v;\n return;\n }\n li mid = (l + r) >> 1;\n modify(u * 2, l, mid, x, v), modify(u * 2 + 1, mid + 1, r, x, v);\n tree[u] = max(tree[u * 2], tree[u * 2 + 1]);\n}\n \nli query(li u, li l, li r, li x, li y){\n if(l > y || x > r) return -1e18;\n if(x <= l && r <= y) return tree[u];\n li mid = (l + r) >> 1;\n return max(query(u * 2, l, mid, x, y), query(u * 2 + 1, mid + 1, r, x, y));\n}\n \nli n, next[200010];\nli ansl[200010], a[200010], cnt[200010], pre[200010];\n \nint main(){\n // freopen(\"wonderful.ans\", \"r\", stdin);\n // freopen(\"www.ww\", \"w\", stdout);\n li T = read();\n while(T--){\n n = read();\n li Q = read();\n for(li i = 1; i <= n; i++){\n cnt[i] = 0, pre[i] = 0, next[i] = 0, a[i] = read();\n }\n for(li i = n; i; i--){\n next[i] = cnt[a[i]];\n cnt[a[i]] = i;\n }\n for(li i = 1; i <= n; i++) cnt[i] = 0;\n for(li i = 1; i <= n; i++){\n pre[i] = cnt[a[i]];\n cnt[a[i]] = i;\n }\n for(li i = 0; i <= 4 * n + 4; i++) tree[i] = -1e18;\n li l = n + 1;\n // cout << \"?\" << endl;\n for(li r = n; r; ){\n l = min(l, r);\n while(l > 1){\n li maxn = query(1, 1, n, l, next[l - 1] - 1);\n // cout << maxn << \" \" << next[l - 1] << endl;\n if(maxn < next[l - 1]){\n l--;\n modify(1, 1, n, l, next[l]);\n } else break;\n }\n ansl[r] = l; \n // cout << l << \" \" << r << endl;\n if(pre[r] >= l){\n modify(1, 1, n, pre[r], 0);\n }\n r--;\n }\n while(Q--){\n li l = read(), r = read();\n if(l >= ansl[r]) printf(\"YES\\n\");\n else printf(\"NO\\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": "2030F",
"submission_id": 286794862,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nnamespace cs{\n\t#define LL long long\n\tconst int N=2e5;\n\tconst int INF=2e9;\n\tint T,n,Q,a[N+5],pt,ma[N+5],cnt[N+5],lst[N+5];\n\tbool End[N+5];\n\tdeque<int>dq;\n\tvector<int>v[N+5];\n\tvoid Init(){\n\t\tpt=0;\n\t\tfor(int i=1;i<=n;i++){\n\t\t\tma[i]=0;cnt[i]=0;\n\t\t\tEnd[i]=0;\n\t\t\tv[i].clear();\n\t\t}\n\t\tdq.clear();\n\t}\t\n\tint main(){\n\t\tios::sync_with_stdio(false);cin.tie(0);cout.tie(0);\n//\t\tfreopen(\"input.in\",\"r\",stdin);\n//\t\tfreopen(\"output.out\",\"w\",stdout);\n\t\tcin>>T;\n//\t\tT=min(625,T);\n\t\tint l,r;\n\t\twhile(T--){\n\t\t\tInit();\n\t\t\tcin>>n>>Q;\n\t\t\tfor(int i=1;i<=n;i++){\n\t\t\t\tcin>>a[i];\n\t\t\t}\n\t\t\tfor(int i=1;i<=n;i++){\n\t\t\t\tif(i!=1){\n\t\t\t\t\tdq.pop_front();\n\t\t\t\t\treverse(v[i-1].begin(),v[i-1].end());\n\t\t\t\t\tfor(auto u:v[i-1]){\n\t\t\t\t\t\tdq.push_front(u);\n\t\t\t\t\t\tEnd[u]=0;\n\t\t\t\t\t}\n\t\t\t\t\tcnt[a[i-1]]--;\n\t\t\t\t}\n//\t\t\t\tprintf(\"pass\\n\");\n\t\t\t\twhile(pt<n){\n\t\t\t\t\tif(End[a[pt+1]])break;\n\t\t\t\t\tpt++;\n\t\t\t\t\tif(cnt[a[pt]]!=0){\n\t\t\t\t\t\twhile(!dq.empty()&&dq.back()!=a[pt]){\n\t\t\t\t\t\t\tv[lst[a[pt]]].push_back(dq.back());\n\t\t\t\t\t\t\tEnd[dq.back()]=1;\n\t\t\t\t\t\t\tdq.pop_back();\n\t\t\t\t\t\t}\n\t\t\t\t\t}\n\t\t\t\t\tdq.push_back(a[pt]);\n\t\t\t\t\tcnt[a[pt]]++;\n\t\t\t\t\tlst[a[pt]]=pt;\n\t\t\t\t}\n\t\t\t\tma[i]=pt;\n//\t\t\t\tprintf(\"ma[%d]:%d\\n\",i,pt);\n\t\t\t}\n\t\t\tfor(int i=2;i<=n;i++){\n\t\t\t\tma[i]=max(ma[i],ma[i-1]);\n\t\t\t}\n\t\t\twhile(Q--){\n\t\t\t\tcin>>l>>r;\n\t\t\t\tif(ma[l]>=r)cout<<\"YES\\n\";\n\t\t\t\telse cout<<\"NO\\n\";\n\t\t\t}\n\t\t}\n\t\treturn 0;\n\t}\n}\nint main(){\n\tcs::main();\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": "2030F",
"submission_id": 296368984,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\nimport sys\nimport heapq\ninput = sys.stdin.readline\ndef fg():\n return int(input())\ndef fgh():\n return [int(xx) for xx in input().split()]\ndef tt():\n sys.stdout.flush()\ndef sd():\n print('Yes')\ndef df():\n print('No')\nfor __ in range(fg()):\n n, q = fgh()\n a = fgh()\n ind = {}\n for i in range(n):\n ind[a[i]] = []\n for i in range(n):\n ind[a[i]].append(i)\n z = []\n for i in ind:\n for j in range(len(ind[i]) - 1):\n z.append((ind[i][j], ind[i][j + 1]))\n \n k = len(z)\n zz = []\n k = len(z)\n for i in range(k):\n zz.append((z[i][1], z[i][0]))\n z.sort()\n zz.sort()\n h = []\n t = []\n T = []\n l = 0\n for i in range(len(zz)):\n while l < k and z[l][0] < zz[i][0]:\n heapq.heappush(h, (z[l][1], z[l][0]))\n l += 1\n while h and h[0][0] <= zz[i][0]:\n heapq.heappop(h)\n if h:\n tt = []\n for _______ in range(len(h)):\n if h[0][1] >= zz[i][1]:\n t.append((zz[i][1], h[0][0]))\n tt.append(h[0])\n heapq.heappop(h)\n for ii in tt:\n heapq.heappush(h, ii)\n pr0 = [0] * (n + 1)\n pr1 = [0] * (n + 1)\n for i in range(len(t)):\n pr0[t[i][0] + 1] += 1\n pr1[t[i][1] + 1] += 1\n for i in range(1, n + 1):\n pr0[i] += pr0[i - 1]\n pr1[i] += pr1[i - 1]\n for _ in range(q):\n l, r = fgh()\n if pr0[l - 1] >= pr1[r]:\n sd()\n else:\n df()\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": "2030F",
"submission_id": 286831492,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\nimport sys\nimport heapq\ninput = sys.stdin.readline\ndef fg():\n return int(input())\ndef fgh():\n return [int(xx) for xx in input().split()]\ndef tt():\n sys.stdout.flush()\ndef sd():\n print('Yes')\ndef df():\n print('No')\nfor __ in range(fg()):\n n, q = fgh()\n a = fgh()\n ind = {}\n for i in range(n):\n ind[a[i]] = []\n for i in range(n):\n ind[a[i]].append(i)\n z = []\n for i in ind:\n for j in range(len(ind[i]) - 1):\n z.append((ind[i][j], ind[i][j + 1]))\n k = len(z)\n zz = []\n k = len(z)\n for i in range(k):\n zz.append((z[i][1], z[i][0]))\n z.sort()\n zz.sort()\n h = []\n t = []\n l = 0\n for i in range(len(zz)):\n while l < k and z[l][0] < zz[i][0]:\n heapq.heappush(h, (z[l][0], z[l][1]))\n l += 1\n #print(*h)\n while h and h[0][0] <= zz[i][1]:\n heapq.heappop(h)\n if h and h[0][1] > zz[i][0]:\n t.append((zz[i][1], h[0][1]))\n pr0 = [0] * (n + 1)\n pr1 = [0] * (n + 1)\n for i in range(len(t)):\n pr0[t[i][0] + 1] += 1\n pr1[t[i][1] + 1] += 1\n for i in range(1, n + 1):\n pr0[i] += pr0[i - 1]\n pr1[i] += pr1[i - 1]\n for _ in range(q):\n l, r = fgh()\n if pr0[l - 1] >= pr1[r]:\n sd()\n else:\n df()\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": "2030F",
"submission_id": 286810931,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.util.HashMap;\nimport java.util.Map;\nimport java.util.StringTokenizer;\nimport java.util.function.BiFunction;\n \npublic class F_PrimeArray {\n public static void main(String[] args) throws IOException {\n BufferedScanner input = new BufferedScanner();\n BufferedOutput output = new BufferedOutput();\n \n int tasks = input.nextInt();\n while (tasks-- > 0) {\n int length = input.nextInt();\n int questionsCount = input.nextInt();\n int[] numbers = input.nextIntArray(length);\n \n int[] prevIdx = new int[length];\n Map<Integer, Integer> cache = new HashMap<>();\n for (int i = 0; i < length; i++) {\n prevIdx[i] = cache.getOrDefault(numbers[i], -1);\n cache.put(numbers[i], i);\n }\n \n int[] nextIdx = new int[length];\n cache = new HashMap<>();\n for (int i = length - 1; i >= 0; i--) {\n nextIdx[i] = cache.getOrDefault(numbers[i], -1);\n cache.put(numbers[i], i);\n }\n \n Tree maxNextTree = new Tree(0, length - 1, nextIdx, Math::max);\n int[] intervalStarts = new int[length];\n intervalStarts[0] = 0;\n for (int i = 1; i < length; i++) {\n intervalStarts[i] = intervalStarts[i - 1];\n while (prevIdx[i] != -1 && intervalStarts[i] <= prevIdx[i] - 1\n && maxNextTree.getValue(intervalStarts[i], prevIdx[i] - 1) > prevIdx[i]\n && maxNextTree.getValue(intervalStarts[i], prevIdx[i] - 1) <= i) {\n \n intervalStarts[i]++;\n }\n }\n \n for (int i = 0; i < questionsCount; i++) {\n int left = input.nextInt() - 1;\n int right = input.nextInt() - 1;\n output.println(intervalStarts[right] <= left ? \"Yes\" : \"No\");\n }\n }\n \n output.flush();\n }\n \n private static class Tree {\n private final int leftBorder;\n private final int rightBorder;\n private final int value;\n private final Tree leftChildren;\n private final Tree rightChildren;\n private final BiFunction<Integer, Integer, Integer> operation;\n \n public Tree(int leftBorder, int rightBorder, int[] numbers, BiFunction<Integer, Integer, Integer> operation) {\n this.leftBorder = leftBorder;\n this.rightBorder = rightBorder;\n this.operation = operation;\n \n if (leftBorder == rightBorder) {\n this.value = numbers[leftBorder];\n this.leftChildren = null;\n this.rightChildren = null;\n } else {\n int mid = (leftBorder + rightBorder) / 2;\n \n this.leftChildren = new Tree(leftBorder, mid, numbers, operation);\n this.rightChildren = new Tree(mid + 1, rightBorder, numbers, operation);\n this.value = operation.apply(this.leftChildren.value, this.rightChildren.value);\n }\n }\n \n private int getValue(int left, int right) {\n if (leftBorder == left && rightBorder == right) {\n return value;\n }\n \n if (right <= leftChildren.rightBorder) {\n return leftChildren.getValue(left, right);\n } else if (rightChildren.leftBorder <= left) {\n return rightChildren.getValue(left, right);\n } else {\n return operation.apply(\n leftChildren.getValue(left, leftChildren.rightBorder),\n rightChildren.getValue(rightChildren.leftBorder, right)\n );\n }\n }\n }\n \n private static class BufferedScanner {\n private final BufferedReader reader;\n private StringTokenizer tokenizer;\n \n public BufferedScanner() {\n reader = new BufferedReader(new InputStreamReader(System.in));\n }\n \n public int nextInt() throws IOException {\n return Integer.parseInt(nextToken());\n }\n \n public long nextLong() throws IOException {\n return Long.parseLong(nextToken());\n }\n \n public int[] nextIntArray(int length) throws IOException {\n int[] numbers = new int[length];\n for (int i = 0; i < length; i++) {\n numbers[i] = nextInt();\n }\n \n return numbers;\n }\n \n public long[] nextLongArray(int length) throws IOException {\n long[] numbers = new long[length];\n for (int i = 0; i < length; i++) {\n numbers[i] = nextLong();\n }\n \n return numbers;\n }\n \n public String nextLine() throws IOException {\n tokenizer = null;\n return reader.readLine();\n }\n \n private String nextToken() throws IOException {\n while (tokenizer == null || !tokenizer.hasMoreTokens()) {\n tokenizer = new StringTokenizer(reader.readLine());\n }\n \n return tokenizer.nextToken();\n }\n }\n \n private static class BufferedOutput {\n private final StringBuilder result;\n \n public BufferedOutput() {\n result = new StringBuilder();\n }\n \n public void print(Object object) {\n result.append(object);\n }\n \n public void println(Object object) {\n result.append(object).append(\"\\n\");\n }\n \n public void flush() {\n System.out.println(result);\n System.out.flush();\n result.setLength(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": "2030F",
"submission_id": 290458678,
"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: Orangutan Approved Subarrays\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:\n\n * In one operation, select indices $l$ and $r$ ($1 \\leq l \\leq r \\leq |b|$) such that $v = b_l = b_{l+1} = \\ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \\leq i \\leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \\ldots$ as $b_l, b_{l+1}, \\ldots$ accordingly. \n\nYou are given an array $a$ of length $n$ and $q$ queries.\n\nEach query consists of two indices $l$ and $r$ ($1 \\le l \\le r \\le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \\ldots, a_r$ is orangutan-approved.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains integers $n$ and $q$ ($1 \\leq n,q \\leq 2 \\cdot 10^5$) — the size of $a$ and the number of queries, respectively.\n\nThe following line contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq n$) — the elements of the array $a$.\n\nThe following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \\leq l \\leq r \\leq n$).\n\nIt is guaranteed that the sum of $n$ and $q$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each query, output \"YES\" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and \"NO\" (without quotes) otherwise.\n\nYou can output \"YES\" and \"NO\" in any case (for example, strings \"yES\", \"yes\" and \"Yes\" will be recognized as a positive response).\n\nExamples:\ninput:\n3\n4 2\n1 2 2 1\n1 4\n1 3\n5 3\n1 2 1 2 1\n2 5\n3 5\n1 3\n8 4\n1 2 3 2 1 3 2 3\n1 5\n2 8\n3 5\n6 8\noutput:\nYES\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\n\nNote: In the first query of the first testcase, the answer is YES.\n\n * Initially, $S=\\\\{1,2\\\\}$ and $b=[1,2,2,1]$ * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\\\{1\\\\}$ and the array becomes $[1,1]$. * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\\\{\\\\}$ and the array becomes $[]$. * Since the array is now empty, we can say the original array is orangutan-approved. \n\nIn the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nconst int N=2e5+10;\nconst int inf=0x3f3f3f3f;\n \nint T,n,q;\nint a[N]; int lst[N],ps[N],nxt[N];\n \nstruct segtree{\n #define maxx N*4\n #define ls p<<1\n #define rs p<<1|1\n int t[maxx];\n void push_up(int p) {t[p]=min(t[ls],t[rs]);}\n void build(int p,int l,int r)\n {\n if(l==r) {t[p]=inf; return;}\n int mid=(l+r)>>1;\n build(ls,l,mid); build(rs,mid+1,r);\n push_up(p);\n }\n void update(int p,int l,int r,int a,int w)\n {\n if(l==r) {t[p]=w; return;}\n int mid=(l+r)>>1;\n if(mid>=a) update(ls,l,mid,a,w);\n else update(rs,mid+1,r,a,w);\n push_up(p);\n }\n int query(int p,int l,int r,int a,int b)\n {\n if(a>b) return inf;\n if(l>=a&&r<=b) return t[p];\n int mid=(l+r)>>1;\n if(mid>=b) return query(ls,l,mid,a,b);\n else if(mid+1<=a) return query(rs,mid+1,r,a,b);\n else return min(query(ls,l,mid,a,b),query(rs,mid+1,r,a,b));\n }\n}t;\n \nvoid append(int l)\n{\n if(nxt[l]==-1) return;\n t.update(1,1,n,nxt[l],inf);\n}\n \nbool check(int r)\n{\n if(lst[r]==-1) return 1;\n int pos=t.query(1,1,n,lst[r]+1,r-1);\n if(pos<lst[r]) return 0;\n return 1;\n}\nint ver[N];\nvoid solve()\n{\n //freopen(\"T.out\",\"w\",stdout);\n t.build(1,1,n);\n for(int i=1;i<=n;i++) lst[i]=ps[i]=0,nxt[i]=-1,ver[i]=i;\n for(int i=1;i<=n;i++)\n {\n if(ps[a[i]]) lst[i]=ps[a[i]];\n else lst[i]=-1;\n ps[a[i]]=i;\n }\n for(int i=1;i<=n;i++)\n if(lst[i]!=-1) nxt[lst[i]]=i;\n for(int l=1,r=0;l<=n;append(l),l++)\n {\n while(r<n&&check(r+1))\n {\n r++;\n if(lst[r]!=-1) t.update(1,1,n,r,lst[r]);\n }\n ver[l]=r;\n if(ver[l]<l) assert(0);\n }\n while(q--)\n {\n int l,r;\n scanf(\"%d%d\",&l,&r);\n if(ver[l]>=r) puts(\"YES\");\n else puts(\"NO\");\n }\n} \n \nint main()\n{\n scanf(\"%d\",&T);\n while(T--)\n {\n scanf(\"%d%d\",&n,&q);\n for(int i=1;i<=n;i++) scanf(\"%d\",&a[i]);\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": "2030F",
"submission_id": 286984514,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <queue>\n#include <tuple>\n \nusing namespace std;\n \nconst int MAX_N = 10000;\nconst int INF = 2e9;\nconst int directions[4][2] = { {1, 0}, {-1, 0}, {0, 1}, {0, -1} }; // Down, Up, Right, Left\n \nstruct State {\n int x, y, dir, steps, dist;\n};\n \nint bfs(vector<string>& maze, int n, int m, int startX, int startY, int endX, int endY) {\n queue<State> q;\n vector<vector<vector<int>>> visited(n, vector<vector<int>>(m, vector<int>(4, INF)));\n \n // Initialize the queue with starting positions in all four directions\n for (int i = 0; i < 4; i++) {\n q.push({startX, startY, i, 0, 0});\n visited[startX][startY][i] = 0;\n }\n \n // Perform BFS\n while (!q.empty()) {\n State s = q.front(); q.pop();\n int x = s.x, y = s.y, dir = s.dir, steps = s.steps, dist = s.dist;\n \n // Check if we have reached the end position\n if (x == endX && y == endY) return dist;\n \n // Try to move in all four directions\n for (int i = 0; i < 4; i++) {\n int nx = x + directions[i][0];\n int ny = y + directions[i][1];\n \n // Check if the new position is within bounds and not a wall\n if (nx >= 0 && nx < n && ny >= 0 && ny < m && maze[nx][ny] != '#') {\n int newSteps = (dir == i) ? steps + 1 : 1;\n \n // Check if we can move in this direction (no more than three consecutive steps)\n if (newSteps <= 3 && dist + 1 < visited[nx][ny][i]) {\n visited[nx][ny][i] = dist + 1;\n q.push({nx, ny, i, newSteps, dist + 1});\n }\n }\n }\n }\n \n return -1; // Return -1 if there is no valid path\n}\n \nint main() {\n int n, m;\n cin >> n >> m;\n vector<string> maze(n);\n int startX, startY, endX, endY;\n \n for (int i = 0; i < n; i++) {\n cin >> maze[i];\n for (int j = 0; j < m; j++) {\n if (maze[i][j] == 'S') {\n startX = i;\n startY = j;\n }\n if (maze[i][j] == 'T') {\n endX = i;\n endY = j;\n }\n }\n }\n \n cout << bfs(maze, n, m, startX, startY, endX, endY) << endl;\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": "2041D",
"submission_id": 298735098,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <queue>\n#include <vector>\n#include <tuple>\n \nusing namespace std;\n \nstruct State {\n int row, col, dir, step_count, total_steps;\n};\n \nint drow[] = {1, 0, -1, 0}; // D, R, U, L\nint dcol[] = {0, 1, 0, -1}; \n \nint main() {\n int n, m;\n cin >> n >> m;\n \n vector<vector<char>> board(n, vector<char>(m));\n vector<vector<vector<bool>>> visited(n, vector<vector<bool>>(m, vector<bool>(4, false)));\n \n int start_x, start_y, end_x, end_y;\n \n for (int i = 0; i < n; i++) {\n for (int j = 0; j < m; j++) {\n cin >> board[i][j];\n if (board[i][j] == 'S') {\n start_x = i;\n start_y = j;\n } else if (board[i][j] == 'T') {\n end_x = i;\n end_y = j;\n }\n }\n }\n \n queue<State> q;\n \n for (int i = 0; i < 4; i++) {\n int nrow = start_x + drow[i], ncol = start_y + dcol[i];\n if (nrow >= 0 && nrow < n && ncol >= 0 && ncol < m && board[nrow][ncol] != '#') {\n q.push({nrow, ncol, i, 1, 1});\n visited[nrow][ncol][i] = true;\n }\n }\n \n while (!q.empty()) {\n auto [row, col, dir, step_count, total_steps] = q.front();\n q.pop();\n \n if (row == end_x && col == end_y) {\n cout << total_steps << endl;\n return 0;\n }\n \n for (int i = 0; i < 4; i++) {\n int nrow = row + drow[i], ncol = col + dcol[i];\n if (nrow >= 0 && nrow < n && ncol >= 0 && ncol < m && board[nrow][ncol] != '#') {\n int new_step_count = (i == dir) ? step_count + 1 : 1;\n \n if (new_step_count <= 3 && !visited[nrow][ncol][i]) {\n visited[nrow][ncol][i] = true;\n q.push({nrow, ncol, i, new_step_count, total_steps + 1});\n }\n }\n }\n }\n \n cout << -1 << endl;\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": "2041D",
"submission_id": 303962947,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n \n#define vcoistnt ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); \n#define int long long\n#define ull unsigned long long\n#define bit __builtin_popcountll\n#define lowbit(x) ((x)&-(x))\n#define vi vector<int>\n#define vb vector<bool>\ntypedef pair<int,int> pll;\n \nconst int N=2e5+10;\nconst int INF=0x3f3f3f3f3f3f3f3f;\nconst int inf=INT_MIN;\nconst int mod=998244353;\nconst int base=283;\n \nconst int dx[4]={0,0,1,-1};\nconst int dy[4]={1,-1,0,0};\n \n \nint n,m;\nstring s[10005];\nint sx,sy,ex,ey;\nint ans=INF;\n \n \n \nvoid solve(){\n cin>>n>>m;\n for(int i=1;i<=n;i++){\n cin>>s[i];\n s[i]=\" \"+s[i];\n }\n for(int i=1;i<=m;i++){\n for(int j=1;j<=m;j++){\n if(s[i][j]=='S'){\n sx=i,sy=j;\n }\n if(s[i][j]=='T'){\n ex=i,ey=j;\n }\n }\n }\n \n vector<vector<vector<vector<bool>>>> vis(n+10, vector<vector<vector<bool>>>(m+10, vector<vector<bool>>(5, vector<bool>(5, false))));\n \n auto dfs=[&](auto self,int x,int y,int d,int cnt,int step){\n \n if(x==ex&&y==ey){\n ans=min(ans,step);\n return;\n }\n \n for(int i=0;i<4;i++){\n int xx=x+dx[i];\n int yy=y+dy[i];\n if(xx>=1 && xx<=n && yy>=1 && yy<=m && s[xx][yy]!='#'){\n \n if(d==i){\n if(cnt<3){\n if(vis[xx][yy][i][cnt+1]==false){\n vis[xx][yy][i][cnt+1]=true;\n self(self,xx,yy,i,cnt+1,step+1);\n // vis[xx][yy][i][cnt+1]=false;\n } \n }\n \n }else{\n if(vis[xx][yy][i][1]==false){\n vis[xx][yy][i][1]=true;\n self(self,xx,yy,i,1,step+1);\n // vis[xx][yy][i][1]=false;\n }\n }\n \n }\n }\n };\n \n for(int i=0;i<4;i++){\n vis[sx][sy][i][0]=true;\n dfs(dfs,sx,sy,i,0,0);\n vis[sx][sy][i][0]=false;\n }\n \n if(ans==INF){\n cout<<\"-1\\n\";\n }else\n cout<<ans<<\"\\n\";\n \n}\nsigned main() {\n vcoistnt\n int _=1;\n // cin>>_;\n while(_--) {\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": "2041D",
"submission_id": 293228948,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nfrom collections import deque\n \ndef bfs(maze, start, end):\n n = len(maze)\n m = len(maze[0])\n directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]\n \n # Queue for BFS: (x, y, steps, consecutive_steps_in_each_direction)\n queue = deque([(start[0], start[1], 0, [0, 0, 0, 0])])\n visited = set((start[0], start[1], 0, 0, 0, 0))\n \n while queue:\n x, y, steps, cons_steps = queue.popleft()\n \n if (x, y) == end:\n return steps\n \n for i, (dx, dy) in enumerate(directions):\n nx, ny = x + dx, y + dy\n new_cons_steps = cons_steps[:]\n \n if cons_steps[i] < 3:\n new_cons_steps[i] += 1\n else:\n continue\n \n for j in range(4):\n if j != i:\n new_cons_steps[j] = 0\n \n if 0 <= nx < n and 0 <= ny < m and maze[nx][ny] != '#':\n if (nx, ny, new_cons_steps[0], new_cons_steps[1], new_cons_steps[2], new_cons_steps[3]) not in visited:\n visited.add((nx, ny, new_cons_steps[0], new_cons_steps[1], new_cons_steps[2], new_cons_steps[3]))\n queue.append((nx, ny, steps + 1, new_cons_steps))\n \n return -1\n \ndef solve_drunken_maze(n, m, maze_str):\n maze = [list(row) for row in maze_str.strip().split('\\n')]\n start = end = None\n \n for i in range(n):\n for j in range(m):\n if maze[i][j] == 'S':\n start = (i, j)\n elif maze[i][j] == 'T':\n end = (i, j)\n \n if start is None or end is None:\n return -1\n \n return bfs(maze, start, end)\n \n# Example usage:\nn, m = 7, 12\nmaze_str = \"\"\"\\\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\"\"\"\nprint(solve_drunken_maze(n, m, maze_str)) # Output: 15\n \n \n# H7NgOIhuC5iFmEJZzA6\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": "2041D",
"submission_id": 304392350,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nfrom collections import deque\n \ndef drunken_maze(maze):\n # Directions for moving up, down, left, and right\n directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]\n \n # Find the start and end positions\n for i in range(len(maze)):\n for j in range(len(maze[0])):\n if maze[i][j] == 'S':\n start = (i, j)\n elif maze[i][j] == 'T':\n end = (i, j)\n \n # Create a visited set and a queue for BFS\n visited = set()\n queue = deque([(start, 0, (0, 0, 0, 0))]) # (position, steps, consecutive steps)\n \n while queue:\n (x, y), steps, (up, down, left, right) = queue.popleft()\n \n # Check if the end position is reached\n if (x, y) == end:\n return steps\n \n # Mark the current position as visited\n visited.add((x, y))\n \n # Explore all possible directions\n for dx, dy in directions:\n nx, ny = x + dx, y + dy\n \n # Check if the new position is within the maze boundaries and is not a wall\n if (0 <= nx < len(maze)) and (0 <= ny < len(maze[0])) and maze[nx][ny] != '#':\n # Calculate the new number of consecutive steps in the same direction\n if dx == 0 and dy == 1: # right\n new_right = right + 1\n new_up, new_down, new_left = 0, 0, 0\n elif dx == 0 and dy == -1: # left\n new_left = left + 1\n new_up, new_down, new_right = 0, 0, 0\n elif dx == 1 and dy == 0: # down\n new_down = down + 1\n new_up, new_right, new_left = 0, 0, 0\n elif dx == -1 and dy == 0: # up\n new_up = up + 1\n new_down, new_right, new_left = 0, 0, 0\n \n # Check if the new number of consecutive steps exceeds 3\n if max(new_up, new_down, new_left, new_right) <= 3 and (nx, ny) not in visited:\n queue.append(((nx, ny), steps + 1, (new_up, new_down, new_left, new_right)))\n \n # Handle the case where the end position is not reachable\n return -1\n \n \n# Read the input\nn, m = map(int, input().split())\nmaze = [input() for _ in range(n)]\n \n# Solve the problem\nresult = drunken_maze(maze)\n \n# Print the result\nprint(result)\n \n \n# FSJb7H5L3xiKU9qtvI4\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": "2041D",
"submission_id": 304594167,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nfrom collections import deque\n \ndef drunken_maze(maze):\n # Define the possible directions\n directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]\n # Define the opposite directions\n opposites = {(0, 1): (0, -1), (0, -1): (0, 1), (1, 0): (-1, 0), (-1, 0): (1, 0)}\n \n # Read the maze and find the start position\n rows, cols = len(maze), len(maze[0])\n start = None\n for i in range(rows):\n for j in range(cols):\n if maze[i][j] == 'S':\n start = (i, j)\n break\n if start:\n break\n \n # Perform BFS traversal\n queue = deque([(start, 0, None, 0)])\n visited = set([start])\n while queue:\n (x, y), steps, last_dir, last_dir_count = queue.popleft()\n if maze[x][y] == 'T':\n return steps\n \n for dx, dy in directions:\n nx, ny = x + dx, y + dy\n if (0 <= nx < rows and 0 <= ny < cols and maze[nx][ny] != '#' and\n (nx, ny) not in visited):\n if last_dir == (dx, dy):\n if last_dir_count < 3:\n queue.append(((nx, ny), steps + 1, last_dir, last_dir_count + 1))\n else:\n queue.append(((nx, ny), steps + 1, (dx, dy), 1))\n visited.add((nx, ny))\n \n # If we cannot reach the end position, return -1\n return -1\n \n# Read the input\nrows, cols = map(int, input().split())\nmaze = [input() for _ in range(rows)]\n \n# Print the result\nprint(drunken_maze(maze))\n \n \n# apnsCuSqE9DxcYvULVP\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": "2041D",
"submission_id": 304460734,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.util.LinkedList;\nimport java.util.Queue;\n \npublic class Main{\n static int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};\n static char[] dirChars = {'U', 'D', 'L', 'R'};\n \n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n String[] input = br.readLine().split(\" \");\n int n = Integer.parseInt(input[0]);\n int m = Integer.parseInt(input[1]);\n \n char[][] maze = new char[n][m];\n int startX = -1, startY = -1;\n for (int i = 0; i < n; i++) {\n maze[i] = br.readLine().toCharArray();\n for (int j = 0; j < m; j++) {\n if (maze[i][j] == 'S') {\n startX = i;\n startY = j;\n }\n }\n }\n \n System.out.println(bfs(maze, startX, startY));\n }\n \n static int bfs(char[][] maze, int startX, int startY) {\n int n = maze.length;\n int m = maze[0].length;\n boolean[][][] visited = new boolean[n][m][4];\n Queue<Position> queue = new LinkedList<>();\n queue.offer(new Position(startX, startY, 0, -1, -1, -1));\n \n while (!queue.isEmpty()) {\n Position curr = queue.poll();\n if (maze[curr.x][curr.y] == 'T') {\n return curr.steps;\n }\n \n for (int i = 0; i < 4; i++) {\n int newX = curr.x + directions[i][0];\n int newY = curr.y + directions[i][1];\n \n if (newX < 0 || newX >= n || newY < 0 || newY >= m || maze[newX][newY] == '#') {\n continue;\n }\n \n if (curr.lastMove == i && curr.secondLastMove == i && curr.thirdLastMove == i) {\n continue;\n }\n \n if (visited[newX][newY][i]) {\n continue;\n }\n \n visited[newX][newY][i] = true;\n queue.offer(new Position(newX, newY, curr.steps + 1, i, curr.lastMove, curr.secondLastMove));\n }\n }\n \n return -1;\n }\n \n static class Position {\n int x, y, steps, lastMove, secondLastMove, thirdLastMove;\n \n public Position(int x, int y, int steps, int lastMove, int secondLastMove, int thirdLastMove) {\n this.x = x;\n this.y = y;\n this.steps = steps;\n this.lastMove = lastMove;\n this.secondLastMove = secondLastMove;\n this.thirdLastMove = thirdLastMove;\n }\n }\n}\n \n \n// CdN18SEMHetPDxWiVXZ\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": "2041D",
"submission_id": 304625645,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class Main{\n private static final int[][] DIRECTIONS = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};\n private static final int MAX_STEPS = 3;\n \n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int n = scanner.nextInt();\n int m = scanner.nextInt();\n scanner.nextLine();\n \n char[][] maze = new char[n][m];\n int startX = -1;\n int startY = -1;\n \n for (int i = 0; i < n; i++) {\n String row = scanner.nextLine();\n for (int j = 0; j < m; j++) {\n maze[i][j] = row.charAt(j);\n if (maze[i][j] == 'S') {\n startX = i;\n startY = j;\n }\n }\n }\n \n int result = bfs(maze, startX, startY);\n System.out.println(result);\n }\n \n private static int bfs(char[][] maze, int startX, int startY) {\n int n = maze.length;\n int m = maze[0].length;\n boolean[][][] visited = new boolean[n][m][4];\n Queue<Node> queue = new LinkedList<>();\n queue.offer(new Node(startX, startY, -1, 0));\n \n int steps = 0;\n \n while (!queue.isEmpty()) {\n int size = queue.size();\n for (int i = 0; i < size; i++) {\n Node node = queue.poll();\n int x = node.x;\n int y = node.y;\n int direction = node.direction;\n int count = node.count;\n \n if (maze[x][y] == 'T') {\n return steps;\n }\n \n for (int k = 0; k < 4; k++) {\n int newX = x + DIRECTIONS[k][0];\n int newY = y + DIRECTIONS[k][1];\n \n if (newX >= 0 && newX < n && newY >= 0 && newY < m && maze[newX][newY] != '#') {\n int newCount = (k == direction) ? count + 1 : 1;\n if (newCount <= MAX_STEPS && !visited[newX][newY][k]) {\n visited[newX][newY][k] = true;\n queue.offer(new Node(newX, newY, k, newCount));\n }\n }\n }\n }\n steps++;\n }\n \n return -1;\n }\n \n private static class Node {\n int x;\n int y;\n int direction;\n int count;\n \n Node(int x, int y, int direction, int count) {\n this.x = x;\n this.y = y;\n this.direction = direction;\n this.count = count;\n }\n }\n}\n \n \n// AdowOpNyrDFetu64LgR\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": "2041D",
"submission_id": 304464864,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\n#include<format>\n \n#define ll long long\n#define ull unsigned long long\n \nusing namespace std;\n \nchar mp[10010][10010];\nbool visited[10010][10010];\n \nstruct pos\n{\n\tint x, y, t;\n\tint lian; // 连续向一个方向数\n\tint lpos; // 连续的朝向\n \n\tpos(int a, int b, int c, int l = 0, int d = 0) { x = a, y = b, t = c, lian = l, lpos = d; }\n\tbool operator< (const pos& u) const { return t > u.t; }\n};\n \nint dis[][2] = { {0,0}, { -1,0 },{1,0},{0,1},{0,-1} };\n \nint main()\n{\n\tios::sync_with_stdio(false);\n\tcin.tie(0); cout.tie(0);\n \n\tint n, m;\n\tcin >> n >> m;\n\tpriority_queue<pos> q;\n\tfor (int i = 1; i <= n; i++)\n\t\tfor (int j = 1; j <= m; j++)\n\t\t{\n\t\t\tcin >> mp[i][j];\n\t\t\tif (mp[i][j] == 'S')\n\t\t\t\tq.push({ i,j,0 });\n\t\t}\n\tbool fl = 0;\n\twhile (!q.empty())\n\t{\n\t\tauto& cur = q.top();\n\t\tq.pop();\n\t\tvisited[cur.x][cur.y] = 1;\n\t\tif (mp[cur.x][cur.y] == 'T')\n\t\t{\n\t\t\tfl = 1;\n\t\t\tcout << cur.t << endl;\n\t\t\tbreak;\n\t\t}\n\t\tfor (int i = 1; i <= 4; i++)\n\t\t{\n\t\t\tint newx = cur.x + dis[i][0];\n\t\t\tint newy = cur.y + dis[i][1];\n \n\t\t\tif (newx > 0 && newx <= n && newy > 0 && newy <= m && !visited[newx][newy])\n\t\t\t{\n\t\t\t\tif (mp[newx][newy] != '#')\n\t\t\t\t{\n\t\t\t\t\tif (cur.lpos == i && cur.lian >= 3)\n\t\t\t\t\t{\n\t\t\t\t\t\tbool li = 0;\n\t\t\t\t\t\tif (i <= 2)\n\t\t\t\t\t\t{\n\t\t\t\t\t\t\tif (mp[cur.x + dis[3][0]][cur.y + dis[3][1]] || mp[cur.x + dis[4][0]][cur.y + dis[4][1]])\n\t\t\t\t\t\t\t\tli = 1;\n\t\t\t\t\t\t}\n\t\t\t\t\t\telse\n\t\t\t\t\t\t{\n\t\t\t\t\t\t\tif (mp[cur.x + dis[1][0]][cur.y + dis[1][1]] || mp[cur.x + dis[2][0]][cur.y + dis[2][1]])\n\t\t\t\t\t\t\t\tli = 1;\n\t\t\t\t\t\t}\n\t\t\t\t\t\tq.push({ newx,newy,cur.t + 3, 2 - li,i });\n\t\t\t\t\t\t// 如果左右有可走的点，则cur.t+3，且lian变1\n\t\t\t\t\t\t// 反之cur.t+3，且lian变2\n\t\t\t\t\t\tcontinue; // 连续同向过多\n\t\t\t\t\t}\n\t\t\t\t\tif (cur.lpos == i)\n\t\t\t\t\t\tq.push({ newx,newy,cur.t + 1,cur.lian + 1,i });\n\t\t\t\t\telse\n\t\t\t\t\t\tq.push({ newx,newy,cur.t + 1,1,i });\n \n\t\t\t\t}\n\t\t\t}\n\t\t}\n \n \n\t}\n\tif (!fl)\n\t\tcout << -1 << endl;\n \n \n \n \n\t\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": "2041D",
"submission_id": 293457772,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ff first\n#define ss second\n#define ll long long int\n#define mod 1e9+7\n#define pb push_back \n#define eb emplace_back\n#define mpp make_pair \n#define vi vector<int>\n#define vc vector<char>\n#define mpf map<int,int>\n#define SI set<int>\n#define rr(i,a,b) for(int i=a;i>=b;i--)\n#define r(i,a,b) for(int i=a;i<b;i++)\n#define rv(i,mpp) for(auto i:mpp)\n#define debug(x) cerr << (#x) << \" is \" << (x) << endl\n#define minele(x) *min_element(all(x))\n#define maxele(x) *max_element(all(x))\n#define all(x) x.begin(), x.end()\n#define set_bits(x) __builtin_popcountll(x)\n#define leading_zero(x) __builtin_clz(x)\n#define trailing_zero(x) __builtin_ctz(x)\nusing namespace std;\n \nvoid speed()\n{\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n #ifndef ONLINE_JUDGE\n freopen(\"input.txt\",\"r\",stdin);\n freopen(\"output.txt\",\"w\",stdout);\n #endif\n}\n// ll dfs(vector<string>& a,vector<vi>& vis,int i,int j,int left,int right,int top,int bot)\n// {\n// int n=a.size(),m=a[0].size();\n// priority_queue<\n// if(i>=n || j>=m || i<0 || j<0 || vis[i][j]>=2 || a[i][j]=='#')\n// return INT_MAX;\n// if(a[i][j]=='T')\n// {\n// return 0;\n// }\n// vis[i][j]++;\n// ll L=INT_MAX,R=INT_MAX,B=INT_MAX,T=INT_MAX;\n// if(left>0)\n// L=1+dfs(a,vis,i,j-1,left-1,3,3,3);\n// if(right>0)\n// R=1+dfs(a,vis,i,j+1,3,right-1,3,3);\n// if(top>0)\n// T=1+dfs(a,vis,i+1,j,3,3,top,3);\n// if(bot>0)\n// B=1+dfs(a,vis,i-1,j,3,3,3,bot-1);\n \n// return min({L,R,T,B});\n// }\nll bfs(vector<string>& a,vector<vi>& vis,int i,int j)\n{\n int n=a.size(),m=a[0].size();\n priority_queue<pair<int,vi>,vector<pair<int,vi>>,greater<pair<int,vi>>> pq;\n pq.push({0,{i,j,3,3,3,3}});\n while(!pq.empty())\n {\n auto it=pq.top();\n pq.pop();\n int curd=it.ff;\n vi t=it.ss;\n int x=t[0],y=t[1],left=t[2],right=t[3],top=t[4],bot=t[5];\n if(x>=n || y>=m || x<0 || y<0 || vis[x][y]>10 || a[x][y]=='#')\n continue;\n // cout<<x<<\" \"<<y<<\" \";\n if(a[x][y]=='T')\n return curd;\n vis[x][y]++;\n if(left>0)\n pq.push({curd+1,{x,y-1,left-1,3,3,3}});\n if(right>0)\n pq.push({curd+1,{x,y+1,3,right-1,3,3}});\n if(top>0)\n pq.push({curd+1,{x-1,y,3,3,top-1,3}});\n if(bot>0)\n pq.push({curd+1,{x+1,y,3,3,3,bot-1}});\n // vis[x][y]=0;\n }\n return -1;\n}\nvoid phan()\n{\n int n,m,I,J;\n ll ans=0;\n cin>>n>>m;\n vector<string> a(n);\n r(i,0,n)\n cin>>a[i];\n r(i,0,n)\n {\n r(j,0,m)\n {\n if(a[i][j]=='S')\n {\n I=i;\n J=j;\n }\n }\n }\n vector<vi> vis(n,vi(m,0));\n cout<<bfs(a,vis,I,J);\n cout<<\"\\n\";\n}\n \n \nint main() { \n speed(); \n phan(); \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": "2041D",
"submission_id": 293596742,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nfrom collections import deque\n \ndef find_start(maze_x, maze_y, maze):\n for x in range(maze_x):\n for y in range(maze_y):\n if maze[x][y] == \"S\":\n return x, y\n return -1, -1\n \ndef check_cell(new_x, new_y, maze_x, maze_y, maze):\n if (not (0 <= new_x < maze_x and 0 <= new_y < maze_y)):\n return False\n if maze[new_x][new_y] == \"#\":\n return False\n return True\n \ndef find_drunk_direction(x, y, dx, dy, maze_x, maze_y, maze):\n if not dx:\n return 1 if check_cell(x + 1, y, maze_x, maze_y, maze) or check_cell(x - 1, y, maze_x, maze_y, maze) else 2\n if not dy:\n return 1 if check_cell(x, y + 1, maze_x, maze_y, maze) or check_cell(x, y - 1, maze_x, maze_y, maze) else 2\n \ndef flood(maze_x, maze_y, maze):\n \n start_x, start_y = find_start(maze_x, maze_y, maze)\n search = deque()\n search.append((start_x, start_y, 0, 0, 0, 0))\n \n while search:\n \n x, y, dir_x, dir_y, dir_times, time = search.popleft()\n if maze[x][y] == \"T\":\n return time\n elif maze[x][y] == \"s\":\n continue\n maze[x][y] = \"s\"\n \n for dx, dy in ((-1, 0), (1, 0), (0, -1), (0, 1)):\n new_x = x + dx\n new_y = y + dy\n if not check_cell(new_x, new_y, maze_x, maze_y, maze):\n continue\n if maze[new_x][new_y] == \"s\":\n continue\n new_dir_times = 1\n if (dx, dy) == (dir_x, dir_y):\n new_dir_times = dir_times + 1\n if new_dir_times != 4 or maze[new_x][new_y] == \"T\":\n search.append((new_x, new_y, dx, dy, new_dir_times, time + 1))\n else:\n search.append((new_x, new_y, dx, dy, find_drunk_direction(x, y, dx, dy, maze_x, maze_y, maze), time + 3))\n return -1\n \nx, y = [int(i) for i in input().split()]\nmaze = [[j for j in input()] for i in range(x)]\nprint(flood(x, y, maze))\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": "2041D",
"submission_id": 293934003,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nfrom collections import deque\n \ndef bfs(grid, n, m):\n # find x and y\n \n direction = {\n 0: \"U\",\n 1: \"R\",\n 2: \"D\",\n 3: \"L\",\n }\n \n opp = {\n 'R' : 'L',\n 'L' : 'R',\n 'U' : 'D',\n 'D' : 'U',\n }\n \n for row in range(n):\n for col in range(m):\n if 'S' == grid[row][col]:\n sx, sy = row, col\n if 'T' == grid[row][col]:\n dx, dy = row, col\n \n visited = set()\n queue = deque()\n queue.append((sx, sy, \"\"))\n visited.add((sx, sy))\n \n X = [-1, 0, 1, 0]\n Y = [0, 1, 0, -1]\n \n while queue:\n x, y, steps = queue.popleft()\n \n # check whether last four steps are same or not?\n if len(steps) >= 5 and steps[-1] == steps[-2] == steps[-3] == steps[-4] == steps[-5]:\n steps += steps[-1]\n steps += steps[-1]\n \n steps_list = list(steps)\n steps_list[-4] = opp[steps_list[-1]]\n steps = ''.join(steps_list)\n \n if x == dx and y == dy:\n print(len(steps))\n return\n \n for i in range(4):\n newX = x + X[i]\n newY = y + Y[i]\n if 0 <= newX < n and 0 <= newY < len(grid[0]):\n if grid[newX][newY] != '#' and (newX, newY) not in visited:\n visited.add((newX, newY))\n queue.append((newX, newY, steps + direction[i]))\n \n print(-1)\n \nn, m= map(int, input().split(' '))\ngrid = []\n \nfor _ in range(n):\n grid.append(input().strip())\n \nbfs(grid, n, m)\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": "2041D",
"submission_id": 298745823,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nimport java.io.BufferedOutputStream;\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.io.PrintWriter;\nimport java.util.*;\n \n \npublic class p2041D{\n public static void main(String[] args) {\n MyScanner sc = new MyScanner();\n out = new PrintWriter(new BufferedOutputStream(System.out));\n \n int n = sc.nextInt();\n int m = sc.nextInt();\n \n char[][] grid = new char[n][m];\n \n int[] start = new int[4];\n \n int[][] steps = new int[n][m];\n \n for (int[] step : steps) {\n Arrays.fill(step, Integer.MAX_VALUE);\n }\n \n \n for(int i=0;i<n;i++){\n String temp = sc.next();\n for(int j=0;j<m;j++){\n grid[i][j] = temp.charAt(j);\n if(grid[i][j] == 'S') start = new int[]{i,j,0,0,-1};\n }\n }\n int res = -1;\n \n //bfs with int[] we need to keep track of coordinates, steps taken until this point, cur number of steps\n //don't need to keep track of bounds because it is padded by #\n Deque<int[]> dq = new ArrayDeque<>();\n steps[start[0]][start[1]] = 0;\n dq.add(start);\n \n while(!dq.isEmpty()){\n int[] data = dq.removeFirst();\n int row = data[0], col = data[1], total = data[2] ,step = data[3], dir = data[4];\n if(res > -1 && total >= res) continue;\n if(grid[row][col] == 'T'){\n if(res < 0) res = total;\n else res = Math.min(res,total);\n continue;\n }\n \n \n //set total and step for future\n total++;\n step++; \n \n \n \n if(grid[row+1][col] != '#'){\n int value = steps[row+1][col];\n int tempTotal = total;\n if(step ==4 && dir == 1 && grid[row+1][col] != 'T') tempTotal+=2;\n int tempStep = step;\n if(1 != dir) tempStep=1;\n if(tempStep == 4) tempStep=2;\n if(value > tempTotal){\n steps[row+1][col] = tempTotal;\n dq.addLast(new int[]{row+1,col,tempTotal,tempStep,1});\n }\n \n }\n \n if(grid[row-1][col] != '#'){\n int value = steps[row-1][col];\n int tempTotal = total;\n int tempStep = step;\n if(2 != dir) tempStep=1;\n if(step ==4 && dir == 2&& grid[row-1][col] != 'T') tempTotal+=2;\n if(tempStep == 4) tempStep=2;\n if(value > tempTotal){\n steps[row-1][col] = tempTotal;\n dq.addLast(new int[]{row-1,col,tempTotal,tempStep,2});\n }\n \n } \n \n if(grid[row][col+1] != '#'){\n int value = steps[row][col+1];\n int tempTotal = total;\n int tempStep = step;\n if(3 != dir) tempStep=1;\n if(step ==4 && dir == 3&& grid[row][col+1] != 'T') tempTotal+=2;\n if(tempStep == 4) tempStep=2;\n if(value > tempTotal){\n steps[row][col+1] = tempTotal;\n \n dq.addLast(new int[]{row,col+1,tempTotal,tempStep,3});\n }\n \n }\n \n if(grid[row][col-1] != '#'){\n int value = steps[row][col-1];\n int tempTotal = total;\n if(step ==4 && dir == 4&& grid[row][col-1] != 'T') tempTotal+=2;\n int tempStep = step;\n if(4 != dir) tempStep=1;\n if(tempStep == 4) tempStep=2;\n if(value > tempTotal){\n steps[row][col-1] = tempTotal;\n dq.addLast(new int[]{row,col-1,tempTotal,tempStep,4});\n }\n \n }\n \n }\n \n \n \n \n out.println(res);\n \n out.close();\n \n }\n \n public static PrintWriter out;\n \n public static class MyScanner {\n BufferedReader br;\n StringTokenizer st;\n \n public MyScanner() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() {\n while (st == null || !st.hasMoreElements()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n 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": "2041D",
"submission_id": 293382069,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class D_Drunken_Maze {\n \n public static void main(String[] args) {\n int n = sc.nextInt();\n int m = sc.nextInt();\n char[][] a = new char[n][m];\n for (int i = 0; i < n; i++) {\n String s = sc.next();\n for (int j = 0; j < m; j++)\n a[i][j] = s.charAt(j);\n }\n \n int startX = -1, startY = -1, targetX = -1, targetY = -1;\n for (int i = 0; i < n; i++) {\n for (int j = 0; j < m; j++) {\n if (a[i][j] == 'S') {\n startX = i;\n startY = j;\n }\n if (a[i][j] == 'T') {\n targetX = i;\n targetY = j;\n }\n }\n }\n \n PriorityQueue<int[]> pq = new PriorityQueue<>((p, q) -> Integer.compare(p[2], q[2]));\n pq.add(new int[] { startX, startY, 0, 0, 0, 0, 0 });\n boolean[][][][][] vis = new boolean[n][m][4][4][4]; // Tracks visited states\n vis[startX][startY][0][0][0] = true;\n \n while (!pq.isEmpty()) {\n int[] t = pq.poll();\n int x = t[0], y = t[1], cost = t[2], up = t[3], down = t[4], left = t[5], right = t[6];\n \n if (x == targetX && y == targetY) {\n System.out.println(cost);\n return;\n }\n \n // Move Up\n if (x > 0 && a[x - 1][y] != '#') {\n int newUp = Math.min(up + 1, 3);\n if (!vis[x - 1][y][newUp][0][0]) {\n vis[x - 1][y][newUp][0][0] = true;\n pq.add(new int[] { x - 1, y, cost + 1 + (up == 3 ? 2 : 0), newUp, 0, 0, 0 });\n }\n }\n \n // Move Down\n if (x < n - 1 && a[x + 1][y] != '#') {\n int newDown = Math.min(down + 1, 3);\n if (!vis[x + 1][y][0][newDown][0]) {\n vis[x + 1][y][0][newDown][0] = true;\n pq.add(new int[] { x + 1, y, cost + 1 + (down == 3 ? 2 : 0), 0, newDown, 0, 0 });\n }\n }\n \n // Move Left\n if (y > 0 && a[x][y - 1] != '#') {\n int newLeft = Math.min(left + 1, 3);\n if (!vis[x][y - 1][0][0][newLeft]) {\n vis[x][y - 1][0][0][newLeft] = true;\n pq.add(new int[] { x, y - 1, cost + 1 + (left == 3 ? 2 : 0), 0, 0, newLeft, 0 });\n }\n }\n \n // Move Right\n if (y < m - 1 && a[x][y + 1] != '#') {\n int newRight = Math.min(right + 1, 3);\n if (!vis[x][y + 1][0][0][newRight]) {\n vis[x][y + 1][0][0][newRight] = true;\n pq.add(new int[] { x, y + 1, cost + 1 + (right == 3 ? 2 : 0), 0, 0, 0, newRight });\n }\n }\n }\n \n System.out.println(-1);\n }\n \n static FastReader sc = new FastReader();\n \n static class FastReader {\n BufferedReader br;\n StringTokenizer st;\n \n public FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() {\n while (st == null || !st.hasMoreTokens()) {\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 String nextLine() {\n String str = \"\";\n try {\n str = br.readLine().trim();\n } catch (Exception 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": "2041D",
"submission_id": 294980620,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ll long long int\n#define F(i, a, b) for (ll i = a; i < b; i++)\n#define RF(i, a, b) for (ll i = a - 1; i >= b; i--)\n#define FF(a) for (auto it : a)\n#define pb push_back\n#define ff first\n#define ss second\n#define yes cout << \"YES\\n\"\n#define no cout << \"NO\\n\"\n#define all(v) v.begin(), v.end()\n#define rall(v) v.rbegin(), v.rend()\n#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);\nconst ll inf = 1e18 + 5;\nconst ll modd = 998244353;\nconst ll mod = 1000000007;\nusing namespace std;\nvoid solve() {\n ll n,m;\n cin>>n>>m;\n vector<string>a(n);\n ll ii,jj;\n F(i,0,n){\n string s;\n cin>>s;\n a[i]=s;\n F(j,0,m){\n if(a[i][j]=='S'){\n // cout<<i<<\" \"<<j<<endl;\n ii=i;\n jj=j;\n }\n // if(a[i][j]=='T'){\n // cout<<i<<\" \"<<j<<endl;\n // // ii=i;\n // // jj=j;\n // }\n }\n \n }\n// cout<<ii<<\" \"<<jj<<endl;\n priority_queue<\n pair<pair<ll, pair<ll, ll>>, vector<ll>>, \n vector<pair<pair<ll, pair<ll, ll>>, vector<ll>>>, \n greater<pair<pair<ll, pair<ll, ll>>, vector<ll>>>\n > q;\n q.push({{0,{ii,jj}},{3,3,3,3}});\n //0-->D \n //1-->U\n //2-->R\n //3-->L\n map<pair<ll,ll>,ll>mp;\n // mp[{ii,jj}]=1;\n while(!q.empty()){\n pair< pair<ll,pair<ll,ll>>,vector<ll>>p=q.top();\n q.pop();\n // cout<<p.ss[0]<<\" \"<<p.ss[1]<<\" \"<<p.ss[2]<<\" \"<<p.ss[3]<<\" \"<<p.ff.ff<<\" \"<<p.ff.ss.ff<<\" \"<<p.ff.ss.ss<<endl;\n if(mp[{p.ff.ss.ff,p.ff.ss.ss}]){\n continue;\n }\n mp[{p.ff.ss.ff,p.ff.ss.ss}]=1;\n if(a[p.ff.ss.ff][p.ff.ss.ss]=='T'){\n cout<<p.ff.ff<<endl;\n return;\n }\n if( p.ff.ss.ff<n-1 && a[p.ff.ss.ff+1][p.ff.ss.ss]!='#' ){\n vector<ll>tmp;\n if(p.ss[0]==0){\n tmp={1,3,3,3};\n p.ff.ff+=2;\n }\n else tmp={p.ss[0]-1,3,3,3};\n q.push({{p.ff.ff+1,{p.ff.ss.ff+1,p.ff.ss.ss}},tmp});\n // mp[{p.ff.ss.ff+1,p.ff.ss.ss}]++;\n }\n if( p.ff.ss.ff>0 && a[p.ff.ss.ff-1][p.ff.ss.ss]!='#' ){\n vector<ll>tmp;\n if(p.ss[1]==0){\n tmp={3,1,3,3};\n p.ff.ff+=2;\n }\n else tmp={3,p.ss[1]-1,3,3};\n q.push({{p.ff.ff+1,{p.ff.ss.ff-1,p.ff.ss.ss}},tmp});\n // mp[{p.ff.ss.ff-1,p.ff.ss.ss}]++;\n }\n if( p.ff.ss.ss<m-1 && a[p.ff.ss.ff][p.ff.ss.ss+1]!='#' ){\n vector<ll>tmp;\n // if(p.ff.ss.ff==3)\n // cout<<p.ff.ss.ss<<\" \"<<p.ss[2]<<endl; \n if(p.ss[2]==0){\n tmp={3,3,1,3};\n p.ff.ff+=2;\n }\n else tmp={3,3,p.ss[2]-1,3};\n q.push({{p.ff.ff+1,{p.ff.ss.ff,p.ff.ss.ss+1}},tmp});\n // mp[{p.ff.ss.ff,p.ff.ss.ss+1}]++;\n }\n if( p.ff.ss.ss>0 && a[p.ff.ss.ff][p.ff.ss.ss-1]!='#'){\n vector<ll>tmp;\n if(p.ss[3]==0){\n tmp={3,3,3,1};\n p.ff.ff+=2;\n }\n else tmp={3,3,3,p.ss[3]-1};\n q.push({{p.ff.ff+1,{p.ff.ss.ff,p.ff.ss.ss-1}},tmp});\n // mp[{p.ff.ss.ff,p.ff.ss.ss-1}]++;\n }\n \n }\n cout<<-1<<endl;\n}\nint main() {\n IOS;\n ll test = 1;\n // cin >> test;\n while (test--) {\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": "2041D",
"submission_id": 295749476,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n \n \n#pragma GCC optimize(\"Ofast,fast-math,unroll-loops\")\n#pragma GCC target(\"avx2,fma\")\nusing namespace std;\n#define ll long long\n#define int ll\n#define all(a) a.begin(),a.end()\n#define allr(a) a.rbegin(),a.rend()\n#define pb push_back\n#define yes cout<<\"YES\"\n#define no cout<<\"NO\"\n#define endl '\\n'\n#define endll cout<<endl\n#define Fast ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);\n#define F first\n#define S second\n#define im cout<<\"IMPOSSIBLE\"\nconst int N = 6e5 + 100;\nconst int NN = 106;\nconst int mo = 1e9 + 123;\nconst int mod = 2e6 + 3;\nconst int Mod = 998244353;\nconst ll inf = 1e17;\nconst int LOG = 19;\n#define PI 3.14159265\n \nint Lcm(int x, int y) {\n return x / __gcd(x, y) * y;\n}\n \nint n, m;\nstring s[N];\nint dx[] = {-1, 1, 0, 0};\nint dy[] = {0, 0, -1, 1};\n \nbool insd(int x, int y) {\n return x >= 0 && x < n && y >= 0 && y < m;\n}\n \nvoid solve() {\n cin >> n >> m;\n pair<int, int> st, e;\n vector<vector<int>> dist(n, vector<int>(m, inf));\n for (int i = 0; i < n; ++i) {\n cin >> s[i];\n for (int j = 0; j < m; ++j) {\n if (s[i][j] == 'S')\n st.F = i, st.S = j;\n if (s[i][j] == 'T')\n e.F = i, e.S = j;\n }\n }\n priority_queue<pair<pair<int, pair<int, int>>, pair<int, int>>> q;\n q.push({{0, st},\n {-1, 0}});\n dist[st.F][st.S] = 0;\n while (q.size()) {\n auto it = q.top();\n q.pop();\n int x = it.F.S.F;\n int y = it.F.S.S;\n int dst = it.F.F;\n int pre_dir = it.S.F;\n int cnt = it.S.S;\n \n for (int i = 0; i < 4; ++i) {\n int nx = x + dx[i];\n int ny = y + dy[i];\n if (insd(nx, ny) && s[nx][ny] != '#') {\n if (pre_dir != i || pre_dir == i && cnt < 3) {\n if (dist[nx][ny] > dst + 1)\n dist[nx][ny] = dst + 1, q.push({{dist[nx][ny], {nx, ny}},\n {i, (pre_dir == i) * cnt + 1}});\n } else if (pre_dir == i && cnt >= 3) {\n if (dist[nx][ny] > dst + 3)\n dist[nx][ny] = dst + 3, q.push({{dist[nx][ny], {nx, ny}},\n {i, 2}});\n }\n }\n }\n }\n if (dist[e.F][e.S] == inf)\n cout << -1;\n else cout << dist[e.F][e.S];\n endll;\n}\n \nint32_t main() {\n Fast;\n int t = 1;\n#ifndef ONLINE_JUDGE\n freopen(\"input.txt\", \"r\", stdin);\n freopen(\"output.txt\", \"w\", stdout);\n#endif\n //cin >> t;\n for (int i = 1; i <= t; ++i) {\n // cout << \"Case #\" << i << \": \";\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": "2041D",
"submission_id": 293417347,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nfrom collections import deque\nfrom heapq import heappush, heappop, heapify\nfrom math import ceil\nn, m = map(int, input().split())\ngrid = [input() for i in range(n)]\nMAX_N = 200000 + 1\nAL = [[] for _ in range(MAX_N)]\ns = 0\nt = 0\ncc = 1\nmapper = {}\nmapper_index = {}\nfor i in range(n):\n for j in range(m):\n mapper[(i, j)] = cc\n mapper_index[cc] = (i, j)\n if grid[i][j] == \"S\":\n s = cc\n if grid[i][j] == \"T\":\n t = cc\n cc+=1\n#print(cc)\nfor i in range(n):\n for j in range(m):\n if i - 1 >= 0:\n up = grid[i - 1][j]\n id_up = mapper[(i-1, j)]\n if grid[i][j] != \"#\" and up != \"#\":\n AL[mapper[(i, j)]].append(id_up)\n if i + 1 < n:\n down = grid[i + 1][j]\n id_down = mapper[(i + 1, j)]\n if grid[i][j] != \"#\" and down != \"#\":\n AL[mapper[(i, j)]].append(id_down)\n if j - 1 >= 0:\n left = grid[i][j - 1]\n id_left = mapper[(i, j - 1)]\n if grid[i][j] != \"#\" and left != \"#\":\n AL[mapper[(i, j)]].append(id_left)\n if j + 1 < m:\n right = grid[i][j + 1]\n id_right = mapper[(i, j + 1)]\n if grid[i][j] != \"#\" and right != \"#\":\n AL[mapper[(i, j)]].append(id_right)\nINF = int(1e9)\ndist = [INF for i in range(MAX_N)]\n#q = deque()\npq = []\n# 1- w, 2- u, 3- steps, 4- dir\nheappush(pq, [0, t, 0, -1])\n# -1: S, 1 izq, 2 der, 3 up, 4, down\n#q.append([s, 0, -1])\ndist[t] = 0\nwhile pq:\n #u, step, dir = q.popleft()\n w, u, step_, dir = heappop(pq)\n #print(w, u, step_, dir)\n #print(u, step, dir)\n if w > dist[u]: continue\n for v in AL[u]:\n v_i, v_j = mapper_index[v]\n u_i, u_j = mapper_index[u]\n cur = dir\n step = step_\n if dir == 1 and v_i == u_i and v_j - u_j == 1:\n step += 1\n elif dir == 2 and v_i == u_i and u_j - v_j == 1:\n step += 1\n elif dir == 3 and u_j == v_j and u_i - v_i == 1:\n step += 1\n elif dir == 4 and u_j == v_j and v_i - u_i == 1:\n step += 1\n else:\n step = 1\n dif_row = u_i - v_i\n dif_col = u_j - v_j\n if dif_col == 0 and dif_row == 1:\n cur = 3\n if dif_col == 0 and dif_row == -1:\n cur = 4\n if dif_col == 1 and dif_row == 0:\n cur = 2\n if dif_col == -1 and dif_row == 0:\n cur = 1\n new_d = dist[u] + 1 if step <= 3 else dist[u] + 3\n step = step if step <= 3 else 2\n if new_d >= dist[v]: continue\n dist[v] = new_d\n #step_= step\n heappush(pq, [new_d, v, step, cur])\n # q.append([v, step, cur])\n \nprint(dist[s] if dist[s] != INF else -1)\n#print(AL[:41], s, t)\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": "2041D",
"submission_id": 296500257,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\ndef dfs(u):\n dl.append(u)\n bs[u[0]][u[1]] = 0\n dl.append(u)\n sd = [[[[0, 0], 0]] * m for i in range(n)]\n while len(dl):\n u = dl.pop()\n if u == zzz:\n bs[zzz[0]][zzz[1]] = min(bs[u[0]][u[1]], bs[zzz[0]][zzz[1]])\n for i in fx:\n bbb = 1\n v = [u[0] + i[0], u[1] + i[1]]\n if sd[u[0]][u[1]][1] == 3 and sd[u[0]][u[1]][0] == i:\n bbb = 3\n if ma[v[0]][v[1]] != '#':\n if bs[v[0]][v[1]] == -1 or bs[v[0]][v[1]] > bs[u[0]][u[1]] + bbb:\n if i == sd[u[0]][u[1]][0]:\n sd[v[0]][v[1]] = [i, sd[u[0]][u[1]][1] + 1]\n if sd[u[0]][u[1]][1] == 3:\n bbb = 3\n sd[v[0]][v[1]] = [i, 2]\n else:\n sd[v[0]][v[1]] = [i, 1]\n bs[v[0]][v[1]] = bs[u[0]][u[1]] + bbb\n dl.append(v)\n \n \n \nn, m = map(int, input().split())\nmi = 10 * 9\nfx = [[1, 0], [-1, 0], [0, 1], [0, -1]]\nma = []\nfor i in range(n):\n ma.append([])\n a = input()\n for j in a:\n ma[i].append(j)\n \nbs = [[-1] * m for i in range(n)]\nqqq = []\nzzz = []\nfor i in range(n):\n for j in range(m):\n if ma[i][j] == 'S':\n qqq = [i, j]\n elif ma[i][j] == 'T':\n zzz = [i, j]\n \ndl = []\ndfs(qqq)\nprint(bs[zzz[0]][zzz[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": "2041D",
"submission_id": 293618826,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class Main{\n private static final int[] DX = { -1, 1, 0, 0 };\n private static final int[] DY = { 0, 0, -1, 1 };\n private static final String DIRECTIONS = \"UDLR\";\n \n public static void main(String[] args) {\n Scanner in = new Scanner(System.in);\n int n = in.nextInt();\n int m = in.nextInt();\n in.nextLine();\n \n char[][] maze = new char[n][m];\n int startX = -1, startY = -1, endX = -1, endY = -1;\n \n for (int i = 0; i < n; i++) {\n String line = in.nextLine();\n for (int j = 0; j < m; j++) {\n maze[i][j] = line.charAt(j);\n if (maze[i][j] == 'S') {\n startX = i;\n startY = j;\n } else if (maze[i][j] == 'T') {\n endX = i;\n endY = j;\n }\n }\n }\n \n System.out.println(bfs(maze, startX, startY, endX, endY, n, m));\n }\n \n private static int bfs(char[][] maze, int startX, int startY, int endX, int endY, int n, int m) {\n Queue<State> queue = new LinkedList<>();\n Set<State> visited = new HashSet<>();\n \n queue.offer(new State(startX, startY, ' ', 0, 0));\n visited.add(new State(startX, startY, ' ', 0, 0));\n \n while (!queue.isEmpty()) {\n State current = queue.poll();\n \n if (current.x == endX && current.y == endY) {\n return current.steps;\n }\n \n for (int i = 0; i < 4; i++) {\n int nx = current.x + DX[i];\n int ny = current.y + DY[i];\n char newDir = DIRECTIONS.charAt(i);\n int newCount = current.lastDir == newDir ? current.count + 1 : 1;\n \n if (nx < 0 || nx >= n || ny < 0 || ny >= m || maze[nx][ny] == '#') {\n continue;\n }\n \n if (newCount > 3) {\n continue;\n }\n \n State nextState = new State(nx, ny, newDir, newCount, current.steps + 1);\n \n if (!visited.contains(nextState)) {\n visited.add(nextState);\n queue.offer(nextState);\n }\n }\n }\n \n return -1;\n }\n \n private static class State {\n int x, y;\n char lastDir;\n int count;\n int steps;\n \n State(int x, int y, char lastDir, int count, int steps) {\n this.x = x;\n this.y = y;\n this.lastDir = lastDir;\n this.count = count;\n this.steps = steps;\n }\n \n @Override\n public boolean equals(Object o) {\n if (this == o) return true;\n if (o == null || getClass() != o.getClass()) return false;\n State state = (State) o;\n return x == state.x && y == state.y && lastDir == state.lastDir && count == state.count;\n }\n \n @Override\n public int hashCode() {\n return Objects.hash(x, y, lastDir, count);\n }\n }\n}\n \n \n// rUdA3cp6VDnw5aMfXeq\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": "2041D",
"submission_id": 305012512,
"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: Drunken Maze\ntime_limit_ms: 2000\nmemory_limit_mb: 1024\nDescription: Image generated by ChatGPT 4o.\n\nYou are given a two-dimensional maze with a start and end position. Your task is to find the fastest way to get from the start to the end position. The fastest way is to make the minimum number of steps where one step is going left, right, up, or down. Of course, you cannot walk through walls.\n\nThere is, however, a catch: If you make more than three steps in the same direction, you lose balance and fall down. Therefore, it is forbidden to make more than three consecutive steps in the same direction. It is okay to walk three times to the right, then one step to the left, and then again three steps to the right. This has the same effect as taking five steps to the right, but is slower.\n\nInput: The first line contains two numbers $$$n$$$ and $$$m$$$, which are the height and width of the maze. This is followed by an ASCII-representation of the maze where $$$\\tt{\\#}$$$ is a wall, $$$\\tt{.}$$$ is an empty space, and $$$\\tt S$$$ and $$$\\tt T$$$ are the start and end positions.\n\n- $$$12 \\leq n\\times m \\leq 200000$$$.\n- $$$3\\leq n,m \\leq 10000$$$.\n- Characters are only $$$\\tt{.\\#ST}$$$ and there is exactly one $$$\\tt{S}$$$ and one $$$\\tt{T}$$$.\n- The outer borders are only $$$\\tt{\\#}$$$ (walls).\n\nOutput: The minimum number of steps to reach the end position from the start position or -1 if that is impossible.\n\nExamples:\ninput:\n7 12\n############\n#S........T#\n#.########.#\n#..........#\n#..........#\n#..#..#....#\n############\noutput:\n15\n\ninput:\n5 8\n########\n#......#\n#.####.#\n#...T#S#\n########\noutput:\n14\n\ninput:\n5 8\n########\n#.#S...#\n#.####.#\n#...T#.#\n########\noutput:\n-1\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.Arrays;\nimport java.util.*;\nimport java.util.StringTokenizer;\n \n \npublic class Main {\n \n static int log = 30;\n static int[][] ancestor;\n \n static int[] depth;\n \n static long ans = 1, mod = 1000000007;\n \n static HashSet<Integer> sieveOfEratosthenes(int n) {\n int prime[] = new int[n + 1];\n for (int i = 1; i <= n; i++)\n prime[i] = 0;\n HashSet<Integer> arr = new HashSet<>();\n for (int p = 2; p * p <= n; p++) {\n // If prime[p] is not changed, then it is a\n // prime\n \n if (prime[p] == 0) {\n \n \n // Update all multiples of p\n for (int i = p * p; i <= n; i += p) {\n \n \n prime[i] = 1;\n }\n }\n }\n \n for (int i = 2; i <= n; i++) {\n if (prime[i] == 0)\n arr.add(i);\n }\n \n // Print all prime numbers\n \n return arr;\n \n }\n \n \n public static long fac(long N, long mod) {\n if (N == 0)\n return 1;\n if (N == 1)\n return 1;\n return ((N % mod) * (fac(N - 1, mod) % mod)) % mod;\n }\n \n \n static long power(long x, long y, long p) {\n \n // Initialize result\n long res = 1;\n \n // Update x if it is more than or\n // equal to p\n x = x % p;\n \n while (y > 0) {\n \n // If y is odd, multiply x\n // with result\n if (y % 2 == 1)\n res = (res * x) % p;\n \n // y must be even now\n y = y >> 1; // y = y/2\n x = (x * x) % p;\n }\n \n return res;\n }\n \n // Returns n^(-1) mod p\n static long modInverse(long n, long p) {\n return power(n, p - 2, p);\n }\n \n // Returns nCr % p using Fermat's\n // little theorem.\n static long nCrModPFermat(int n, int r,\n long p, long[] fac) {\n \n if (n < r)\n return 0;\n // Base case\n if (r == 0)\n return 1;\n \n return ((fac[n] % p * (modInverse(fac[r], p)\n % p)) % p * (modInverse(fac[n - r], p)\n % p)) % p;\n }\n \n \n public static int find(int[] parent, int x) {\n if (parent[x] == x)\n return x;\n return find(parent, parent[x]);\n }\n \n public static void merge(int[] parent, int[] rank, int x, int y, int[] child) {\n int x1 = find(parent, x);\n int y1 = find(parent, y);\n \n if (rank[x1] > rank[y1]) {\n parent[y1] = x1;\n child[x1] += child[y1];\n \n \n } else if (rank[y1] > rank[x1]) {\n parent[x1] = y1;\n child[y1] += child[x1];\n \n } else {\n parent[y1] = x1;\n child[x1] += child[y1];\n rank[x1]++;\n \n }\n \n \n }\n \n \n public static long gcd(long a, long b) {\n if (b == 0)\n return a;\n \n return gcd(b, a % b);\n }\n \n \n static int power(int x, int y, int p) {\n \n int res = 1;\n x = x % p;\n \n while (y > 0) {\n \n if (y % 2 == 1)\n res = (res * x) % p;\n \n y = y >> 1;\n x = (x * x) % p;\n }\n \n return res;\n }\n \n // Returns n^(-1) mod p\n static int modInverse(int n, int p) {\n return power(n, p - 2, p);\n }\n \n // Returns nCr % p using Fermat's\n // little theorem.\n static int nCrModPFermat(int n, int r,\n int p, int[] fac) {\n \n if (n < r)\n return 0;\n // Base case\n if (r == 0)\n return 1;\n \n // Fill factorial array so that we\n // can find all factorial of r, n\n // and n-r\n// int[] fac = new int[n + 1];\n// fac[0] = 1;\n//\n// for (int i = 1; i <= n; i++)\n// fac[i] = fac[i - 1] * i % p;\n \n return (fac[n] * modInverse(fac[r], p)\n % p * modInverse(fac[n - r], p)\n % p)\n % p;\n }\n \n \n public static long[][] ncr(int n, int r) {\n long[][] dp = new long[n + 1][r + 1];\n for (int i = 0; i <= n; i++)\n dp[i][0] = 1;\n for (int i = 1; i <= n; i++) {\n for (int j = 1; j <= r; j++) {\n if (j > i)\n continue;\n dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];\n }\n }\n \n return dp;\n \n }\n \n \n public static int nCr_brute_force(int n, int r) {\n if (r > n) return 0;\n int res = 1;\n for (int i = 0; i < r; i++) {\n res *= (n - i);\n res /= (i + 1);\n }\n return res;\n }\n \n \n public static boolean prime(long N) {\n int c = 0;\n for (int i = 2; i * i <= N; i++) {\n if (N % i == 0)\n ++c;\n }\n \n return c == 0;\n }\n \n \n public static void sparse_ancestor_table(ArrayList<ArrayList<Integer>> arr, int x, int parent, int[] weight) {\n \n \n for (int i : arr.get(x)) {\n if (i != parent) {\n// System.out.println(i+\" hello \"+x);\n depth[i] = depth[x] + 1;\n \n ancestor[i][0] = x;\n \n// if(i==2)\n// System.out.println(parent+\" hello\");\n for (int j = 1; j < log; j++) {\n ancestor[i][j] = ancestor[ancestor[i][j - 1]][j - 1];\n \n }\n sparse_ancestor_table(arr, i, x, weight);\n }\n }\n \n \n }\n \n \n public static int lca(int x, int y) {\n if (depth[x] < depth[y]) {\n int temp = x;\n x = y;\n y = temp;\n }\n \n x = get_kth_ancestor(depth[x] - depth[y], x);\n if (x == y)\n return x;\n// System.out.println(x+\" \"+y);\n for (int i = log - 1; i >= 0; i--) {\n if (ancestor[x][i] != ancestor[y][i]) {\n x = ancestor[x][i];\n y = ancestor[y][i];\n }\n }\n \n return ancestor[x][0];\n \n }\n \n public static int get_kth_ancestor(int K, int x) {\n \n if (K == 0)\n return x;\n int node = x;\n for (int i = 0; i < log; i++) {\n if (K % 2 != 0) {\n node = ancestor[node][i];\n }\n K /= 2;\n }\n \n return node;\n }\n \n \n public static void primeFactors(long n, HashMap<Long, Integer> hp) {\n // Print the number of 2s that divide n\n// ArrayList<Integer> factors=new ArrayList<>();\n// int cnt=0;\n// long ch=1;\n// if(n%2==0)\n// {\n//\n// }\n while (n % 2 == 0) {\n \n hp.put((long) 2, hp.getOrDefault((long) 2, 0) + 1);\n n /= 2;\n }\n \n// ch*=(cnt+1);\n// cnt=0;\n \n \n // n must be odd at this point. So we can\n // skip one element (Note i = i +2)\n for (int i = 3; i * i <= n; i += 2) {\n // While i divides n, print i and divide n\n// if(n%i==0)\n// factors.add(i);\n while (n % i == 0) {\n hp.put((long) i, hp.getOrDefault((long) i, 0) + 1);\n n /= i;\n \n }\n \n }\n \n // This condition is to handle the case when\n // n is a prime number greater than 2\n if (n > 2) {\n hp.put(n, hp.getOrDefault(n, 0) + 1);\n \n }\n \n }\n \n \n public static int lower(int X, int K, int N) {\n int l = X, r = N;\n \n \n while (l <= r) {\n int mid = (l + r) / 2;\n int start1 = X, start2 = mid, num = -1;\n for (int i = 3; i <= K; i++) {\n num = start1 + start2;\n if (num > N)\n break;\n \n start1 = start2;\n start2 = num;\n }\n \n if (num > N)\n r = mid - 1;\n else if (num < N)\n l = mid + 1;\n else\n return 1;\n }\n \n return 0;\n }\n \n \n static boolean check;\n \n \n static boolean cycle;\n \n public static void cycle(boolean[] stack, boolean[] visited, int x, ArrayList<ArrayList<Integer>> arr) {\n if (stack[x]) {\n cycle = true;\n return;\n \n }\n \n visited[x] = true;\n for (int i : arr.get(x)) {\n if (!visited[x]) {\n cycle(stack, visited, i, arr);\n }\n }\n \n stack[x] = false;\n }\n \n \n public static void primeFactors(int n, HashSet<Integer> hp) {\n // Print the number of 2s that divide n\n ArrayList<Integer> factors = new ArrayList<>();\n if (n % 2 == 0)\n hp.add(2);\n while (n % 2 == 0) {\n \n n /= 2;\n }\n \n // n must be odd at this point. So we can\n // skip one element (Note i = i +2)\n for (int i = 3; i <= Math.sqrt(n); i += 2) {\n // While i divides n, print i and divide n\n if (n % i == 0)\n hp.add(i);\n while (n % i == 0) {\n \n n /= i;\n }\n }\n \n // This condition is to handle the case when\n // n is a prime number greater than 2\n if (n > 2) {\n hp.add(n);\n \n }\n \n \n }\n \n \n public static boolean palin(int N) {\n String s = \"\";\n while (N > 0) {\n s += N % 10;\n N /= 10;\n }\n int l = 0, r = s.length() - 1;\n while (l <= r) {\n if (s.charAt(l) != s.charAt(r))\n return false;\n \n ++l;\n --r;\n }\n \n return true;\n }\n \n \n public static boolean check_bounds(int x, int y, int N, int M) {\n return x >= 0 && x < N && y >= 0 && y < M;\n }\n \n \n private static long mergeAndCount(long[] arr, int l,\n int m, int r) {\n \n // Left subarray\n long[] left = Arrays.copyOfRange(arr, l, m + 1);\n \n // Right subarray\n long[] right = Arrays.copyOfRange(arr, m + 1, r + 1);\n \n int i = 0, j = 0, k = l;\n long swaps = 0;\n \n while (i < left.length && j < right.length) {\n if (left[i] < right[j])\n arr[k++] = left[i++];\n else {\n arr[k++] = right[j++];\n swaps += (m + 1) - (l + i);\n }\n }\n while (i < left.length)\n arr[k++] = left[i++];\n while (j < right.length)\n arr[k++] = right[j++];\n return swaps;\n }\n \n // Merge sort function\n private static long mergeSortAndCount(long[] arr, int l,\n int r) {\n \n // Keeps track of the inversion count at a\n // particular node of the recursion tree\n long count = 0;\n \n if (l < r) {\n int m = (l + r) / 2;\n \n // Total inversion count = left subarray count\n // + right subarray count + merge count\n \n // Left subarray count\n count += mergeSortAndCount(arr, l, m);\n \n // Right subarray count\n count += mergeSortAndCount(arr, m + 1, r);\n \n // Merge count\n count += mergeAndCount(arr, l, m, r);\n }\n \n return count;\n }\n \n \n static int time;\n \n \n public static void dfs(ArrayList<HashSet<Integer>> arr, int x, boolean[] visited, int[] dist, int parent) {\n \n \n// System.out.println(x+\" \"+dp[x]);\n visited[x] = true;\n \n for (int i : arr.get(x)) {\n \n if (!visited[i]) {\n \n dist[i] = dist[x] + 1;\n dfs(arr, i, visited, dist, x);\n \n } else if (parent != i) {\n cycle = true;\n }\n \n }\n \n \n }\n \n// static int max=0;\n \n \n static long LcmOfArray(long[] arr, int idx) {\n \n // lcm(a,b) = (a*b/gcd(a,b))\n if (idx == arr.length - 1) {\n return arr[idx];\n }\n long a = arr[idx];\n long b = LcmOfArray(arr, idx + 1);\n return (a * b / gcd(a, b)); //\n }\n \n \n // Returns true if str1 is smaller than str2.\n static boolean isSmaller(String str1, String str2) {\n // Calculate lengths of both string\n int n1 = str1.length(), n2 = str2.length();\n if (n1 < n2)\n return true;\n if (n2 < n1)\n return false;\n \n for (int i = 0; i < n1; i++)\n if (str1.charAt(i) < str2.charAt(i))\n return true;\n else if (str1.charAt(i) > str2.charAt(i))\n return false;\n \n return false;\n }\n \n \n // Function for find difference of larger numbers\n static String findDiff(String str1, String str2) {\n // Before proceeding further, make sure str1\n // is not smaller\n if (isSmaller(str1, str2)) {\n String t = str1;\n str1 = str2;\n str2 = t;\n }\n \n // Take an empty string for storing result\n String str = \"\";\n \n // Calculate length of both string\n int n1 = str1.length(), n2 = str2.length();\n \n // Reverse both of strings\n str1 = new StringBuilder(str1).reverse().toString();\n str2 = new StringBuilder(str2).reverse().toString();\n \n int carry = 0;\n \n // Run loop till small string length\n // and subtract digit of str1 to str2\n for (int i = 0; i < n2; i++) {\n // Do school mathematics, compute difference of\n // current digits\n int sub\n = ((int) (str1.charAt(i) - '0')\n - (int) (str2.charAt(i) - '0') - carry);\n \n // If subtraction is less then zero\n // we add then we add 10 into sub and\n // take carry as 1 for calculating next step\n if (sub < 0) {\n sub = sub + 10;\n carry = 1;\n } else\n carry = 0;\n \n str += (char) (sub + '0');\n }\n \n // subtract remaining digits of larger number\n for (int i = n2; i < n1; i++) {\n int sub = ((int) (str1.charAt(i) - '0') - carry);\n \n // if the sub value is -ve, then make it\n // positive\n if (sub < 0) {\n sub = sub + 10;\n carry = 1;\n } else\n carry = 0;\n \n str += (char) (sub + '0');\n }\n \n // reverse resultant string\n return new StringBuilder(str).reverse().toString();\n }\n \n \n static int nCr(int n, int r) {\n return fact(n) / (fact(r) *\n fact(n - r));\n }\n \n // Returns factorial of n\n static int fact(int n) {\n int res = 1;\n for (int i = 2; i <= n; i++)\n res = res * i;\n return res;\n }\n \n \n // static boolean status;\n static ArrayList<Integer> way;\n \n \n static int in[];\n static int out[];\n \n // Function to pre-calculate the array in[]\n// which stores the maximum height when travelled\n// via branches\n static void dfs1(ArrayList<ArrayList<Integer>> v,\n int u, int parent) {\n \n // Initially every node has 0 height\n in[u] = 0;\n \n // Traverse in the subtree of u\n for (int j = 0; j < v.get(u).size(); j++) {\n int child = v.get(u).get(j);\n \n // If child is same as parent\n if (child == parent)\n continue;\n \n // dfs called\n dfs1(v, child, u);\n \n // Recursively calculate the max height\n in[u] = Math.max(in[u], 1 + in[child]);\n }\n }\n \n // Function to pre-calculate the array out[]\n// which stores the maximum height when traveled\n// via parent\n static void dfs2(ArrayList<ArrayList<Integer>> v,\n int u, int parent) {\n \n // Stores the longest and second\n // longest branches\n int mx1 = -1, mx2 = -1;\n \n // Traverse in the subtrees of u\n for (int j = 0; j < v.get(u).size(); j++) {\n int child = v.get(u).get(j);\n if (child == parent)\n continue;\n \n // Compare and store the longest\n // and second longest\n if (in[child] >= mx1) {\n mx2 = mx1;\n mx1 = in[child];\n } else if (in[child] > mx2)\n mx2 = in[child];\n }\n \n // Traverse in the subtree of u\n for (int j = 0; j < v.get(u).size(); j++) {\n int child = v.get(u).get(j);\n if (child == parent)\n continue;\n \n int longest = mx1;\n \n // If longest branch has the node, then\n // consider the second longest branch\n if (mx1 == in[child])\n longest = mx2;\n \n // Recursively calculate out[i]\n out[child] = 1 + Math.max(out[u], 1 + longest);\n \n // dfs function call\n dfs2(v, child, u);\n }\n }\n \n static void addEdge(ArrayList<ArrayList<Integer>> adj,\n int u, int v) {\n adj.get(u).add(v);\n adj.get(v).add(u);\n }\n \n // Function to print all the maximum heights\n// from every node\n \n public static boolean checker(int[] arr, int index, int val, int K) {\n long ans = 0;\n int dist = 0;\n int ind = -1;\n for (int i = index; i < arr.length; i++) {\n if (val - dist <= arr[i]) {\n ind = i;\n break;\n } else {\n ans += val - dist - arr[i];\n ++dist;\n }\n }\n \n if (ind == -1)\n return false;\n \n return ans <= K;\n }\n \n \n public static int lower_index(int[] arr, int index, int K) {\n// System.out.println(x+\" MC\");\n int l = arr[index], r = (int) 1e9;\n \n \n while (l <= r) {\n int mid = (l + r) / 2;\n if (checker(arr, index, mid, K)) {\n if (checker(arr, index, mid + 1, K))\n l = mid + 1;\n else\n return mid;\n } else\n r = mid - 1;\n \n }\n return -1;\n }\n \n public static boolean checker(int index, ArrayList<div> ab, int mid, int l) {\n PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());\n long sum_a = 0;\n for (int i = index; i < index + mid; i++) {\n pq.add(ab.get(i).a);\n sum_a += ab.get(i).a;\n }\n long sum_b = ab.get(index + mid - 1).b - ab.get(index).b;\n long tot_sum = sum_b + sum_a;\n for (int i = index + mid; i < ab.size(); i++) {\n sum_b += ab.get(i).b - ab.get(i - 1).b;\n if (ab.get(i).a < pq.peek()) {\n sum_a -= pq.poll();\n sum_a += ab.get(i).a;\n pq.add(ab.get(i).a);\n }\n tot_sum = Math.min(tot_sum, sum_a + sum_b);\n }\n return tot_sum <= l;\n }\n \n public static int higher_index(ArrayList<Integer> arr, int X, int N) {\n int l = 0, r = N - 1;\n while (l <= r) {\n int mid = (l + r) / 2;\n if (arr.get(mid) >= X) {\n if (mid - 1 >= 0 && arr.get(mid - 1) >= X)\n r = mid - 1;\n else\n return N - mid;\n } else {\n l = mid + 1;\n }\n \n \n }\n return 0;\n }\n \n static void bridgeUtil(int u, boolean visited[], int disc[], int low[], int parent[], ArrayList<ArrayList<Integer>> adj, HashMap<Integer, Integer> hp) {\n \n // Mark the current node as visited\n visited[u] = true;\n \n // Initialize discovery time and low value\n disc[u] = low[u] = ++time;\n \n // Go through all vertices adjacent to this\n Iterator<Integer> i = adj.get(u).iterator();\n while (i.hasNext()) {\n int v = i.next(); // v is current adjacent of u\n \n // If v is not visited yet, then make it a child\n // of u in DFS tree and recur for it.\n // If v is not visited yet, then recur for it\n if (!visited[v]) {\n parent[v] = u;\n bridgeUtil(v, visited, disc, low, parent, adj, hp);\n \n // Check if the subtree rooted with v has a\n // connection to one of the ancestors of u\n low[u] = Math.min(low[u], low[v]);\n \n // If the lowest vertex reachable from subtree\n // under v is below u in DFS tree, then u-v is\n // a bridge\n if (low[v] < disc[u]) {\n hp.put(u, v);\n hp.put(v, u);\n }\n \n }\n \n // Update low value of u for parent function calls.\n else if (v != parent[u])\n low[u] = Math.min(low[u], disc[v]);\n }\n }\n \n \n static int KMPSearch(int[] pat, int[] txt) {\n int M = pat.length;\n int N = txt.length;\n \n // create lps[] that will hold the longest\n // prefix suffix values for pattern\n int lps[] = new int[M];\n int j = 0; // index for pat[]\n \n // Preprocess the pattern (calculate lps[]\n // array)\n computeLPSArray(pat, M, lps);\n \n int i = 0; // index for txt[]\n while ((N - i) >= (M - j)) {\n if (pat[j] == txt[i]) {\n j++;\n i++;\n }\n if (j == M) {\n return i - j;\n \n }\n \n // mismatch after j matches\n else if (i < N && pat[j] != txt[i]) {\n // Do not match lps[0..lps[j-1]] characters,\n // they will match anyway\n if (j != 0)\n j = lps[j - 1];\n else\n i = i + 1;\n }\n }\n \n return -1;\n }\n \n static void computeLPSArray(int[] pat, int M, int lps[]) {\n // length of the previous longest prefix suffix\n int len = 0;\n int i = 1;\n lps[0] = 0; // lps[0] is always 0\n \n // the loop calculates lps[i] for i = 1 to M-1\n while (i < M) {\n if (pat[i] == pat[len]) {\n len++;\n lps[i] = len;\n i++;\n } else // (pat[i] != pat[len])\n {\n // This is tricky. Consider the example.\n // AAACAAAA and i = 7. The idea is similar\n // to search step.\n if (len != 0) {\n len = lps[len - 1];\n \n // Also, note that we do not increment\n // i here\n } else // if (len == 0)\n {\n lps[i] = len;\n i++;\n }\n }\n }\n }\n \n \n public static ArrayList<HashSet<Integer>> graph_form(int N) {\n ArrayList<HashSet<Integer>> arr = new ArrayList<>();\n for (int i = 0; i < N; i++)\n arr.add(new HashSet<>());\n return arr;\n }\n \n public static ArrayList<ArrayList<Long>> graph_form_long(int N) {\n ArrayList<ArrayList<Long>> arr = new ArrayList<>();\n for (int i = 0; i < N; i++)\n arr.add(new ArrayList<>());\n return arr;\n }\n \n \n // public static boolean copy()\n static int tim;\n \n public static void euler_tour(ArrayList<ArrayList<Integer>> arr, int x, int parent, HashMap<Integer, Integer> hp, int[] dist) {\n \n \n// System.out.println(x+\" hello\");\n \n \n hp.put(dist[x], hp.getOrDefault(dist[x], 0) + 1);\n for (int i : arr.get(x)) {\n \n if (i != parent) {\n dist[i] = dist[x] + 1;\n euler_tour(arr, i, x, hp, dist);\n \n \n }\n }\n \n \n }\n \n \n static int MAXN = 10000000;\n static int spf[] = new int[MAXN];\n \n // Calculating SPF (Smallest Prime Factor) for every\n // number till MAXN.\n // Time Complexity : O(nloglogn)\n static void sieve() {\n spf[1] = 1;\n for (int i = 2; i < MAXN; i++)\n \n // marking smallest prime factor for every\n // number to be itself.\n spf[i] = i;\n \n // separately marking spf for every even\n // number as 2\n for (int i = 4; i < MAXN; i += 2)\n spf[i] = 2;\n \n for (int i = 3; i * i < MAXN; i++) {\n // checking if i is prime\n if (spf[i] == i) {\n // marking SPF for all numbers divisible by i\n for (int j = i * i; j < MAXN; j += i)\n \n // marking spf[j] if it is not\n // previously marked\n if (spf[j] == j)\n spf[j] = i;\n }\n }\n }\n \n // A O(log n) function returning primefactorization\n // by dividing by smallest prime factor at every step\n static Vector<Integer> getFactorization(int x) {\n Vector<Integer> ret = new Vector<>();\n while (x != 1) {\n ret.add(spf[x]);\n x = x / spf[x];\n }\n return ret;\n }\n \n \n static long isPerfectSquare(long num) {\n if (num == 1 || num == 0)\n return num;\n long n = num;\n long lo = 1;\n long hi = num;\n while (hi - lo > 1) {\n long mid = lo + (hi - lo) / 2;\n \n if (mid * mid == num) {\n return mid;\n } else if (mid * mid > num) {\n hi = mid - 1;\n } else if (mid * mid < num) {\n lo = mid + 1;\n }\n }\n if (lo * lo == num) {\n return lo;\n }\n \n if (hi * hi == num)\n return hi;\n return -1;\n }\n \n \n static boolean isPowerOfTwo(int x) {\n /* First x in the below expression is\n for the case when x is 0 */\n return x != 0 && ((x & (x - 1)) == 0);\n }\n \n \n public static boolean checkBounds(int x, int y, int N, int M) {\n return x >= 0 && y >= 0 && x < N && y < M;\n }\n \n \n public static long max3elements(long a, long b, long c) {\n return Math.max(a, Math.max(b, c));\n }\n \n public static int max3elements(int a, int b, int c) {\n return Math.max(a, Math.max(b, c));\n }\n \n static int n, t_max[], t_min[];\n \n public static int gcd(int a, int b) {\n if (b == 0)\n return a;\n \n return gcd(b, a % b);\n }\n \n \n static void build_max(int a[], int v, int tl, int tr) {\n if (tl == tr) {\n// if(tl==3)\n// System.out.println(a[tl]);\n t_max[v] = a[tl];\n } else {\n int tm = (tl + tr) / 2;\n build_max(a, v * 2 + 1, tl, tm);\n build_max(a, v * 2 + 2, tm + 1, tr);\n t_max[v] = Math.max(t_max[v * 2 + 1], t_max[v * 2 + 2]);\n }\n }\n \n \n static int max(int v, int tl, int tr, int l, int r) {\n if (l > r)\n return Integer.MIN_VALUE;\n if (l == tl && r == tr) {\n return t_max[v];\n }\n int tm = (tl + tr) / 2;\n return Math.max(max(v * 2 + 1, tl, tm, l, Math.min(r, tm))\n , max(v * 2 + 2, tm + 1, tr, Math.max(l, tm + 1), r));\n }\n \n \n static void build_min(int a[], int v, int tl, int tr) {\n if (tl == tr) {\n// if(tl==3)\n// System.out.println(a[tl]);\n t_min[v] = a[tl];\n } else {\n int tm = (tl + tr) / 2;\n build_min(a, v * 2 + 1, tl, tm);\n build_min(a, v * 2 + 2, tm + 1, tr);\n t_min[v] = Math.min(t_min[v * 2 + 1], t_min[v * 2 + 2]);\n }\n }\n \n static int min(int v, int tl, int tr, int l, int r) {\n if (l > r)\n return Integer.MAX_VALUE;\n if (l == tl && r == tr) {\n return t_min[v];\n }\n int tm = (tl + tr) / 2;\n return Math.min(min(v * 2 + 1, tl, tm, l, Math.min(r, tm))\n , min(v * 2 + 2, tm + 1, tr, Math.max(l, tm + 1), r));\n }\n \n \n static void update(int v, int tl, int tr, int pos, int new_val) {\n if (tl == tr) {\n t_max[v] = new_val - 1;\n } else {\n int tm = (tl + tr) / 2;\n if (pos <= tm)\n update(v * 2 + 1, tl, tm, pos, new_val);\n else\n update(v * 2 + 2, tm + 1, tr, pos, new_val);\n t_max[v] = t_max[v * 2 + 1] + t[v * 2 + 2];\n }\n }\n \n static long[][] usage;\n static long[] st_ans;\n \n \n static void build_usage(long a[], int v, int tl, int tr) {\n if (tl == tr) {\n \n if (a[tl] < 0)\n usage[v][1] = a[tl];\n else\n usage[v][0] = a[tl];\n \n// System.out.println(v+\" \"+usage[v][0]+\" \"+usage[v][1]);\n \n } else {\n int tm = (tl + tr) / 2;\n build_usage(a, v * 2 + 1, tl, tm);\n build_usage(a, v * 2 + 2, tm + 1, tr);\n \n usage[v][0] = usage[2 * v + 1][1] + usage[2 * v + 2][0];\n usage[v][1] = usage[2 * v + 2][1];\n \n if (usage[v][0] < 0) {\n usage[v][1] += usage[v][0];\n usage[v][0] = 0;\n \n }\n \n if (usage[v][1] >= 0) {\n usage[v][0] += usage[v][1];\n usage[v][1] = 0;\n }\n \n// if(usage[2*v+1][0]>0)\n usage[v][0] += usage[2 * v + 1][0];\n// else\n// usage[v][1]+=usage[2*v+1][0];\n// System.out.println(v+\" \"+usage[v][0]+\" \"+usage[v][1]);\n \n \n }\n }\n \n \n// static long query_ans(int v, int tl, int tr, int l, int r) {\n// if (l > r)\n// return 0;\n// if (l == tl && r == tr) {\n// return st_ans[v];\n// }\n// int tm = (tl + tr) / 2;\n// long left= query_ans(v*2+1, tl, tm, l, Math.min(r, tm));\n// long right= query_ans(v*2+2, tm+1, tr, Math.max(l, tm+1), r);\n// System.out.println(left+\" \"+right+\" \"+v);\n//\n// return left+right;\n// }\n \n static void update_usage(int v, int tl, int tr, int pos, long new_val) {\n if (tl == tr) {\n if (new_val < 0) {\n usage[v][1] = new_val;\n usage[v][0] = 0;\n } else {\n usage[v][0] = new_val;\n usage[v][1] = 0;\n }\n \n// System.out.println(v+\" \"+usage[v][0]+\" \"+usage[v][1]);\n } else {\n int tm = (tl + tr) / 2;\n if (pos <= tm)\n update_usage(v * 2 + 1, tl, tm, pos, new_val);\n else\n update_usage(v * 2 + 2, tm + 1, tr, pos, new_val);\n usage[v][0] = usage[2 * v + 1][1] + usage[2 * v + 2][0];\n usage[v][1] = usage[2 * v + 2][1];\n \n \n if (usage[v][0] < 0) {\n usage[v][1] += usage[v][0];\n usage[v][0] = 0;\n \n }\n \n if (usage[v][1] >= 0) {\n usage[v][0] += usage[v][1];\n usage[v][1] = 0;\n }\n// if(usage[2*v+1][0]>0)\n usage[v][0] += usage[2 * v + 1][0];\n// else\n// usage[v][1]+=usage[2*v+1][0];\n \n// System.out.println(v+\" \"+usage[v][0]+\" \"+usage[v][1]);\n \n \n }\n }\n \n \n// public static boolean gcd_segment_checker(int ind, int N)\n// {\n// int l=0,r=ind;\n// while(l<=r)\n// {\n// int mid=(l+r)/2;\n//// System.out.println(gcd_calc(0,0,N-1,mid,ind)+\" \"+mid);\n// if(gcd_calc(0,0,N-1,mid,ind)==ind-mid+1)\n// return true;\n// else if(gcd_calc(0,0,N-1,mid,ind)<ind-mid+1)\n// l=mid+1;\n// else\n// r=mid-1;\n// }\n//\n// return false;\n// }\n \n static int Answer;\n \n \n public static void calcSubset(List<Integer> A,\n \n List<Integer> subset,\n int index, long target, long[] fac) {\n // Add the current subset to the result list\n// res.add(new ArrayList<>(subset));\n long f = 0;\n int ans = subset.size();\n for (int i : subset) {\n f += fac[i];\n }\n \n if (f <= target) {\n long req = target - f;\n while (req > 0) {\n ans += req % 2;\n req /= 2;\n }\n \n if (ans < Answer)\n Answer = ans;\n } else\n return;\n \n \n // Generate subsets by recursively including and\n // excluding elements\n for (int i = index; i < A.size(); i++) {\n // Include the current element in the subset\n subset.add(A.get(i));\n \n // Recursively generate subsets with the current\n // element included\n calcSubset(A, subset, i + 1, target, fac);\n \n // Exclude the current element from the subset\n // (backtracking)\n subset.remove(subset.size() - 1);\n }\n }\n \n public static void\n subsets(List<Integer> A, long target, long[] fac) {\n List<Integer> subset = new ArrayList<>();\n// List<List<Integer> > res = new ArrayList<>();\n \n int index = 0;\n calcSubset(A, subset, index, target, fac);\n \n \n }\n \n// static ArrayList<Integer> finpath;\n \n \n public static void path_calc(int x, HashMap<Integer, Integer> parent, int src, ArrayList<Integer> path) {\n if (x == src) {\n path.add(x);\n// finpath=new ArrayList<>(path);\n return;\n }\n \n if (x <= 0)\n path.add(x);\n path_calc(parent.get(x), parent, src, path);\n }\n \n \n// public static void bfs(char[][] arr,int x1,int y1,int N,int M,int[][] status,boolean zero, HashMap<Integer,Integer> cluster, int ind)\n// {\n//\n// Queue<div> queue=new LinkedList<>();\n//\n// queue.add(new div(x1,y1));\n//\n// int[] x={1,-1,0,0};\n// int[] y={0,0,-1,1};\n// status[x1][y1]=ind;\n//\n// int tot=0;\n//// HashSet<div> hp=new HashSet<>();\n//// System.out.println(x1+\" \"+y1);\n// while(queue.size()>0)\n// {\n// div tmp=queue.poll();\n// int i=tmp.a;\n// int j=tmp.b;\n//// status[i][j]=ind;\n// tot++;\n//\n//// ++tot;\n//// char ch=st.poll();\n//// if(ch=='W')\n//// ++tot;\n//// System.out.println(i+\" \"+j+\" \"+cnt);\n//// status[tmp.a][tmp.b]=true;\n//\n// for (int k = 0; k < 4; k++) {\n// if (checkBounds(i+x[k],j+y[k],N,M))\n// {\n// if(arr[i+x[k]][j+y[k]]=='W')\n// {\n// if(status[i+x[k]][j+y[k]]==-1) {\n// queue.add(new div(i + x[k], j + y[k]));\n// status[i + x[k]][j + y[k]] = ind;\n// }\n// }\n//\n//\n// }\n// }\n//\n//// System.out.println(tmp+\" \"+d);\n//\n//\n//\n//\n//\n// }\n//\n//// System.out.println(x1+\" \"+y1+\" \"+zero);\n//\n//\n// if(zero)\n// cluster.put(ind,0);\n// else\n// cluster.put(ind,tot);\n//\n// }\n \n public static void merge_modified(ArrayList<div> arr, int l, int mid, int r, int K) {\n int N = mid - l + 1, M = r - mid;\n \n ArrayList<div> left = new ArrayList<>();\n ArrayList<div> right = new ArrayList<>();\n ArrayList<div> tot = new ArrayList<>();\n \n for (int i = 0; i < N; i++) {\n left.add(arr.get(l + i));\n }\n \n for (int i = 0; i < M; i++) {\n right.add(arr.get(mid + 1 + i));\n }\n \n int l1 = 0, r1 = 0;\n long[] pref = new long[N];\n pref[N - 1] = left.get(N - 1).b;\n for (int i = N - 2; i >= 0; i--) {\n pref[i] = (pref[i + 1] % mod + left.get(i).b % mod) % mod;\n }\n// for(int i=0;i<M;i++)\n// System.out.print(arr.get(i+mid+1).a+\" right \");\n// System.out.println();\n//\n// for(int i=0;i<N;i++)\n// System.out.print(arr.get(i+l).a+\" left \");\n// System.out.println();\n \n while (l1 < N && r1 < M) {\n if (left.get(l1).a <= right.get(r1).a) {\n tot.add(left.get(l1++));\n } else\n tot.add(right.get(r1++));\n }\n \n while (l1 < N)\n tot.add(left.get(l1++));\n \n while (r1 < M)\n tot.add(right.get(r1++));\n \n for (int i = l; i <= r; i++)\n arr.set(i, tot.get(i - l));\n \n }\n \n public static void sort_modified(ArrayList<div> arr, int l, int r, int K) {\n if (l >= r)\n return;\n \n int mid = (l + r) / 2;\n sort_modified(arr, l, mid, K);\n sort_modified(arr, mid + 1, r, K);\n merge_modified(arr, l, mid, r, K);\n }\n \n \n public static String combine_doses(String a, String b, String b_side) {\n int N = a.length();\n char[] s1 = new char[N];\n for (int i = 0; i < N; i++) {\n if (a.charAt(i) == '0') {\n s1[i] = '0';\n } else {\n if (b.charAt(i) == '1')\n s1[i] = '0';\n else\n s1[i] = '1';\n }\n }\n \n \n for (int i = 0; i < N; i++)\n if (b_side.charAt(i) == '1')\n s1[i] = '1';\n \n String ans = \"\";\n for (int i = 0; i < N; i++)\n ans += s1[i];\n return ans;\n \n \n }\n \n \n public static void cycle(String output, HashSet<String> stack, HashSet<String> visited, String x, HashMap<String, ArrayList<String>> arr, HashMap<String, ArrayList<Integer>> cost, int c) {\n System.out.println(x);\n if (x.equals(output)) {\n if (c < ans) {\n ans = c;\n }\n return;\n }\n if (stack.contains(x) || !arr.containsKey(x)) {\n \n return;\n \n }\n \n visited.add(x);\n int ind = 0;\n for (String i : arr.get(x)) {\n if (!visited.contains(i)) {\n System.out.println(cost.get(x).get(ind));\n cycle(output, stack, visited, i, arr, cost, c + cost.get(x).get(ind));\n }\n \n ind++;\n }\n \n stack.remove(x);\n }\n \n \n// static int shortestPath(String src, HashMap<String,ArrayList<String>> arr, HashMap<String,ArrayList<Integer>> cost, String out) {\n// PriorityQueue<div> pq = new PriorityQueue<>( Comparator.comparingInt(o -> o.b));\n//\n//\n// HashMap<String,Integer> dist=new HashMap<>();\n// pq.add(new div( src,0));\n// dist.put(src, 0);\n//\n// while (!pq.isEmpty()) {\n// String u = pq.poll().a;\n// if(!arr.containsKey(u))\n// continue;\n// int ind=0;\n// for (String v : arr.get(u)) {\n// if (dist.getOrDefault(v,Integer.MAX_VALUE) > dist.get(u) + cost.get(u).get(ind)) {\n// dist.put(v, dist.get(u) + cost.get(u).get(ind));\n// pq.add(new div(v, dist.get(v)));\n// }\n// ind++;\n// }\n// }\n//\n// return dist.getOrDefault(out,-1);\n// }\n//\n//\n//\n//\n//\n//\n//\n//\n//\n//\n//\n//\n//\n//\n//\n//\n//\n//\n \n// static ArrayList<div> findPlatform(ArrayList<Integer> arr, ArrayList<Integer> dep, int n)\n// {\n// // Sort arrival and departure arrays\n// arr.sort(Comparator.comparingInt(a -> a));\n// dep.sort(Comparator.comparingInt(a -> a));\n//\n// // plat_needed indicates number of platforms\n// // needed at a time\n// int plat_needed = 0, result = 0, ar=arr.get(0),de=dep.get(0);\n// int i = 0, j = 0;\n// ArrayList<div> inte=new ArrayList<>();\n// // Similar to merge in merge sort to process\n// // all events in sorted order\n// while (i < n && j < n) {\n// // If next event in sorted order is arrival,\n// // increment count of platforms needed\n// if (arr.get(i) <= dep.get(j)) {\n// plat_needed++;\n// if (plat_needed > result)\n// {\n// ar=arr.get(i);\n// de=dep.get(j);\n// inte=new ArrayList<>();\n// result = plat_needed;\n// inte.add(new div(ar,de));\n// }\n// else if(plat_needed==result)\n// {\n// ar=arr.get(i);\n// de=dep.get(j);\n// inte.add(new div(ar,de));\n// }\n//\n// i++;\n// }\n//\n// // Else decrement count of platforms needed\n// else if (arr.get(i) > dep.get(j)) {\n// plat_needed--;\n// j++;\n// }\n//\n// // Update result if needed\n//\n// }\n//\n// ans=result;\n//\n// return inte;\n// }\n \n \n// static boolean isBipartite(int V, ArrayList<HashSet<Integer>> adj)\n// {\n//\n// // vector to store colour of vertex\n// // assigning all to -1 i.e. uncoloured\n// // colours are either 0 or 1\n// // for understanding take 0 as red and 1 as blue\n// int col[] = new int[V];\n// Arrays.fill(col, -1);\n//\n// // queue for BFS storing {vertex , colour}\n// Queue<div> q = new LinkedList<>();\n//\n// //loop incase graph is not connected\n// for (int i = 0; i < V; i++) {\n//\n// // if not coloured\n// if (col[i] == -1) {\n//\n// // colouring with 0 i.e. red\n// q.add(new div(i, 0));\n// col[i] = 0;\n//\n// while (!q.isEmpty()) {\n// div p = q.peek();\n// q.poll();\n//\n// //current vertex\n// int v = p.a;\n//\n// // colour of current vertex\n// int c = p.b;\n//\n// // traversing vertexes connected to current vertex\n// for (int j : adj.get(v))\n// {\n//\n// // if already coloured with parent vertex color\n// // then bipartite graph is not possible\n// if (col[j] == c)\n// return false;\n//\n// // if uncoloured\n// if (col[j] == -1)\n// {\n//\n// // colouring with opposite color to that of parent\n// col[j] = (c==1) ? 0 : 1;\n// q.add(new div(j, col[j]));\n// }\n// }\n// }\n// }\n// }\n \n // if all vertexes are coloured such that\n // no two connected vertex have same colours\n// return true;\n// }\n static int[] t;\n \n static void build(int a[], int v, int tl, int tr) {\n if (tl == tr) {\n if (a[tl] == 1)\n t[v] = 1;\n } else {\n int tm = (tl + tr) / 2;\n build(a, v * 2, tl, tm);\n build(a, v * 2 + 1, tm + 1, tr);\n t[v] = t[v * 2] + t[v * 2 + 1];\n }\n }\n \n static void update(int v, int tl, int tr, int l, int r, int add) {\n if (l > r)\n return;\n if (l == tl && r == tr) {\n t[v] += add;\n } else {\n int tm = (tl + tr) / 2;\n update(v * 2, tl, tm, l, Math.min(r, tm), add);\n update(v * 2 + 1, tm + 1, tr, Math.max(l, tm + 1), r, add);\n }\n }\n \n static int zeros(int v, int tl, int tr, int l, int r) {\n if (l > r)\n return 0;\n if (l == tl && r == tr) {\n return t[v];\n }\n int tm = (tl + tr) / 2;\n return zeros(v * 2, tl, tm, l, Math.min(r, tm))\n + zeros(v * 2 + 1, tm + 1, tr, Math.max(l, tm + 1), r);\n }\n \n static int find_kth(int v, int tl, int tr, int k) {\n if (k > t[v])\n return -1;\n if (tl == tr)\n return tl;\n int tm = (tl + tr) / 2;\n if (t[v * 2] >= k)\n return find_kth(v * 2, tl, tm, k);\n else\n return find_kth(v * 2 + 1, tm + 1, tr, k - t[v * 2]);\n }\n \n// public static long[] djikstra(ArrayList<ArrayList<Integer>> arr,int x,ArrayList<ArrayList<Long>> weights,int N,int[] new_bikes)\n// {\n//\n// long[] dist=new long[N];\n// int[] bikes=new int[N];\n// bikes[0]=new_bikes[0];\n// Arrays.fill(dist,Long.MAX_VALUE);\n// PriorityQueue<newdiv> pq=new PriorityQueue<>(Comparator.comparingLong(p -> p.b));\n// dist[x]=0;\n// boolean[] visited=new boolean[N];\n// pq.add(new newdiv(x,0));\n// while(pq.size()>0)\n// {\n// newdiv temp=pq.poll();\n//// if(temp.b>dist[temp.a])\n//// continue;\n// visited[temp.a]=true;\n// bikes[temp.a]=Math.min(bikes[temp.a], new_bikes[temp.a]);\n// ArrayList<Integer> tem1=arr.get(temp.a);\n// ArrayList<Long> tempw=weights.get(temp.a);\n// long max=Long.MAX_VALUE;\n// for(int i=0;i<tem1.size();i++)\n// {\n// int node=tem1.get(i);\n// long w=tempw.get(i);\n//\n//\n//\n// for(int j=0;j<tem1.size();j++)\n// {\n// int node2=tem1.get(i);\n// long w2=tempw.get(i);\n// }\n//\n//\n//\n//\n// }\n//\n// }\n//\n// return dist;\n//\n// }\n \n static Vector<Integer> merge(Vector<Integer> v1,\n Vector<Integer> v2) {\n int i = 0, j = 0;\n \n // Final vector to return\n // after merging\n Vector<Integer> v = new Vector<>();\n \n // Loop continues until it reaches\n // the end of one of the vectors\n while (i < v1.size() && j < v2.size()) {\n if (v1.elementAt(i) <= v2.elementAt(j)) {\n v.add(v1.elementAt(i));\n i++;\n } else {\n v.add(v2.elementAt(j));\n j++;\n }\n }\n \n // Here, simply add the remaining\n // elements to the vector v\n for (int k = i; k < v1.size(); k++)\n v.add(v1.elementAt(k));\n for (int k = j; k < v2.size(); k++)\n v.add(v2.elementAt(k));\n return v;\n }\n \n // Procedure to build the segment tree\n static void buildTree(Vector<Integer>[] tree, int[] arr,\n int index, int s, int e) {\n \n // Reached the leaf node\n // of the segment tree\n if (s == e) {\n tree[index].add(arr[s]);\n return;\n }\n \n // Recursively call the buildTree\n // on both the nodes of the tree\n int mid = (s + e) / 2;\n buildTree(tree, arr, 2 * index, s, mid);\n buildTree(tree, arr, 2 * index + 1, mid + 1, e);\n \n // Storing the final vector after merging\n // the two of its sorted child vector\n tree[index] = merge(tree[2 * index], tree[2 * index + 1]);\n }\n \n // Query procedure to get the answer\n // for each query l and r are query range\n static int query(Vector<Integer>[] tree, int index, int s,\n int e, int l, int r, int k) {\n \n // out of bound or no overlap\n if (r < s || l > e)\n return Integer.MAX_VALUE;\n \n // Complete overlap\n // Query range completely lies in\n // the segment tree node range\n if (s >= l && e <= r) {\n \n // binary search to find index of k\n int ind = lowerBound(tree[index],\n tree[index].size(), k);\n if (ind == -1)\n return Integer.MAX_VALUE;\n return tree[index].get(ind);\n }\n \n // Partially overlap\n // Query range partially lies in\n // the segment tree node range\n int mid = (s + e) / 2;\n return Math.min(query(tree, 2 * index, s, mid, l, r, k),\n query(tree, 2 * index + 1, mid + 1, e, l, r, k));\n }\n \n // Function to perform the queries\n static void performQueries(int L[], int R[], int K[],\n int n, int q, Vector<Integer> tree[]) {\n for (int i = 0; i < q; i++) {\n System.out.println(query(tree, 1, 0, n - 1,\n L[i] - 1, R[i] - 1, K[i]));\n }\n }\n \n static int lowerBound(Vector<Integer> array,\n int length, int value) {\n if (value > array.get(array.size() - 1))\n return -1;\n int low = 0;\n int high = length;\n while (low < high) {\n final int mid = (low + high) / 2;\n if (value <= array.elementAt(mid)) {\n high = mid;\n } else {\n low = mid + 1;\n }\n }\n return low;\n }\n \n \n static long power(long x, int y) {\n \n long res = 1;\n \n \n while (y > 0) {\n \n if (y % 2 == 1)\n res = (res * x);\n \n y = y >> 1;\n x = (x * x);\n }\n \n return res;\n }\n \n static int power(int x, int y) {\n \n int res = 1;\n \n \n while (y > 0) {\n \n if (y % 2 == 1)\n res = (res * x);\n \n y = y >> 1;\n x = (x * x);\n }\n \n return res;\n }\n \n \n static int LcmOfArray(int[] arr, int idx) {\n \n // lcm(a,b) = (a*b/gcd(a,b))\n if (idx == arr.length - 1) {\n return arr[idx];\n }\n int a = arr[idx];\n int b = LcmOfArray(arr, idx + 1);\n return (a * b / gcd(a, b)); //\n }\n \n public static void compute_hash_leetcode(String[] words, long[][] hash) {\n long p = 31;\n long m = (int) 1e9 + 9;\n \n for (int i = 0; i < words.length; i++) {\n long hash_value = 0;\n long p_pow = 1;\n int ind = 0;\n for (int j = 0; j < words[i].length(); j++) {\n char c = words[i].charAt(j);\n hash_value = (hash_value + (c - 'a' + 1) * p_pow) % m;\n hash[i][ind++] = hash_value;\n p_pow = (p_pow * p) % m;\n }\n }\n \n }\n \n public static long compute_hash_string(String s) {\n long p = 31;\n long m = (int) 1e9 + 9;\n \n long hash_value = 0;\n long p_pow = 1;\n for (int j = 0; j < s.length(); j++) {\n char c = s.charAt(j);\n hash_value = (hash_value + (c - 'a' + 1) * p_pow) % m;\n p_pow = (p_pow * p) % m;\n }\n \n return hash_value;\n }\n \n public static long[] compute_hash_prefix_string(String s) {\n long p = 31;\n long m = (int) 1e9 + 9;\n long[] hash = new long[s.length()];\n long hash_value = 0;\n long p_pow = 1;\n for (int j = 0; j < s.length(); j++) {\n char c = s.charAt(j);\n hash_value = (hash_value + (c - 'a' + 1) * p_pow) % m;\n hash[j] = hash_value;\n p_pow = (p_pow * p) % m;\n }\n \n return hash;\n }\n \n public static long compute_hash_substring(long[] hash, int l, int r) {\n long m = (int) 1e9 + 9;\n long p = 31;\n long left_hash = l - 1 >= 0 ? hash[l - 1] : 0;\n long substring_hash = ((hash[r] - left_hash + m) % m * modInverse(power(p, l, m), m)) % m;\n return substring_hash;\n }\n \n \n public static int iter(String[] words, String target) {\n int N = target.length();\n int[] dp = new int[N];\n Arrays.fill(dp, Integer.MAX_VALUE);\n long[] target_hash = compute_hash_prefix_string(target);\n \n long[][] hash = new long[words.length][5000];\n compute_hash_leetcode(words, hash);\n for (int i = 0; i < N; i++) {\n for (int j = 0; j <= i; j++) {\n boolean status = false;\n int length_to_check = j - i + 1;\n long hash_string_to_check = compute_hash_substring(target_hash, j, i);\n for (int k = 0; k < words.length; k++) {\n long hash_length = compute_hash_substring(hash[k], 0, length_to_check - 1);\n if (hash_length == hash_string_to_check) {\n status = true;\n break;\n }\n }\n if (status) {\n int left = j - 1 >= 0 ? dp[j - 1] : 0;\n dp[i] = Math.min(dp[i], left + 1);\n }\n \n \n }\n \n \n }\n // System.out.println(dp[k]+\" \"+k);\n \n if (dp[N - 1] == Integer.MAX_VALUE)\n return -1;\n return dp[N - 1];\n }\n \n \n static long[] segmentTree;\n \n public static long build_seg_tree(int ind, long[] arr, int l, int r) {\n if (l == r) {\n segmentTree[ind] = arr[l];\n return arr[l];\n }\n \n int mid = (l + r) / 2;\n segmentTree[ind] = gcd(build_seg_tree(2 * ind + 1, arr, l, mid), build_seg_tree(2 * ind + 2, arr, mid + 1, r));\n return segmentTree[ind];\n }\n \n public static long query_segment_tree(int ind, int l, int r, int curr_l, int curr_r) {\n if (l > r)\n return 0;\n if (l == curr_l && r == curr_r) {\n \n return segmentTree[ind];\n }\n int mid = (curr_l + curr_r) / 2;\n if (mid > r) {\n return query_segment_tree(ind * 2 + 1, l, r, curr_l, mid);\n } else if (mid + 1 <= l) {\n return query_segment_tree(ind * 2 + 2, l, r, mid + 1, curr_r);\n } else {\n return gcd(query_segment_tree(ind * 2 + 1, l, mid, curr_l, mid), query_segment_tree(ind * 2 + 2, mid + 1, r, mid + 1, curr_r));\n }\n }\n \n static long[] segmentTreeMax;\n \n \n public static long build_seg_tree_max(int ind, long[] arr, int l, int r) {\n if (l == r) {\n segmentTreeMax[ind] = arr[l];\n return arr[l];\n }\n \n int mid = (l + r) / 2;\n segmentTreeMax[ind] = Math.max(build_seg_tree_max(2 * ind + 1, arr, l, mid), build_seg_tree_max(2 * ind + 2, arr, mid + 1, r));\n return segmentTreeMax[ind];\n }\n \n public static long query_segment_tree_max(int ind, int l, int r, int curr_l, int curr_r) {\n if (l > r)\n return Integer.MIN_VALUE;\n if (l == curr_l && r == curr_r) {\n \n return segmentTreeMax[ind];\n }\n int mid = (curr_l + curr_r) / 2;\n if (mid > r) {\n return query_segment_tree_max(ind * 2 + 1, l, r, curr_l, mid);\n } else if (mid + 1 <= l) {\n return query_segment_tree_max(ind * 2 + 2, l, r, mid + 1, curr_r);\n } else {\n return Math.max(query_segment_tree_max(ind * 2 + 1, l, mid, curr_l, mid), query_segment_tree_max(ind * 2 + 2, mid + 1, r, mid + 1, curr_r));\n }\n }\n \n public static int fill_dp(int x, int y, int N, int M, int end_x, int end_y, char[][] arr, int[][][][] dp, int dir, int step, boolean[][] visited) {\n if(step >=3)\n return Integer.MAX_VALUE;\n if (x < 0 || y < 0 || x >= N || y >= M)\n return Integer.MAX_VALUE;\n \n if (arr[x][y] == '#' || visited[x][y])\n return Integer.MAX_VALUE;\n \n if (x == end_x && y == end_y)\n return 0;\n \n// System.out.println(x+\" \"+y);\n \n if (dp[x][y][dir][step] != -1)\n return dp[x][y][dir][step];\n \n visited[x][y] = true;\n int[] dx = {0, 0, -1, 1};\n int[] dy = {1, -1, 0, 0};\n int ans = Integer.MAX_VALUE;\n for (int i = 0; i < 4; i++) {\n int newx = x + dx[i];\n int newy = y + dy[i];\n \n// if(x==3 && y==1)\n// System.out.println(newx+\" \"+newy+\" \"+visited[newx][newy]);\n \n \n if (i == dir) {\n ans = Math.min(ans, fill_dp(newx, newy, N, M, end_x, end_y, arr, dp, i, step + 1, visited));\n if(step==2)\n {\n int tmp = fill_dp(x + dx[i], y + dy[i], N, M, end_x, end_y, arr, dp, i, 1, visited);\n \n if (tmp != Integer.MAX_VALUE)\n ans = Math.min(ans, tmp + 2);\n }\n \n } else {\n ans = Math.min(ans, fill_dp(newx, newy, N, M, end_x, end_y, arr, dp, i, 0, visited));\n }\n }\n visited[x][y] = false;\n if (ans == Integer.MAX_VALUE) {\n dp[x][y][dir][step] = Integer.MAX_VALUE;\n } else {\n dp[x][y][dir][step] = ans + 1;\n }\n \n \n return dp[x][y][dir][step];\n \n \n }\n \n \n public static void main(String[] args) throws IOException {\n Reader.init(System.in);\n// BufferedWriter R = new BufferedWriter(new FileWriter(\"/Users/tmahajan/Desktop/answer.txt\"));\n BufferedWriter R = new BufferedWriter(new OutputStreamWriter(System.out));\n \n// HashSet<Integer> prime = sieveOfEratosthenes(200001);\n \n \n// int T = Reader.nextInt();\n int T = 1;\n long mod = 998244353;\n \n \n// HashSet<Integer> prime = sieveOfEratosthenes(10000001);\n// System.out.println(prime.size());\n// ArrayList<Integer> prime_arr = new ArrayList<>(prime);\n// Collections.sort(prime_arr);\n// System.out.println(prime_arr);\n// long[] fac=new long[200001];\n// fac[0]=1;\n// for(int i=1;i<=200000;i++)\n// fac[i]=(fac[i-1]*i)%mod;\n \n \n A:\n for (int m = 1; m <= T; m++) {\n// R.write(\"Case #\" + m + \": \");\n \n int N = Reader.nextInt();\n int M = Reader.nextInt();\n \n char[][] arr = new char[N][M];\n int start_x = -1, start_y = -1, end_x = -1, end_y = -1;\n for (int i = 0; i < N; i++) {\n arr[i] = Reader.next().toCharArray();\n for (int j = 0; j < M; j++) {\n if (arr[i][j] == 'S') {\n start_x = i;\n start_y = j;\n }\n \n if (arr[i][j] == 'T') {\n end_x = i;\n end_y = j;\n }\n }\n }\n \n int[][][][] dp = new int[N][M][4][3];\n for (int i = 0; i < N; i++) {\n for (int j = 0; j < M; j++) {\n for (int k = 0; k < 4; k++) {\n for (int l = 0; l < 3; l++) {\n dp[i][j][k][l] = -1;\n }\n }\n }\n }\n \n \n int[] dx = {0, 0, -1, 1};\n int[] dy = {1, -1, 0, 0};\n int ans = Integer.MAX_VALUE;\n boolean[][] visited = new boolean[N][M];\n visited[start_x][start_y] = true;\n for (int i = 0; i < 4; i++) {\n int newx = start_x + dx[i];\n int newy = start_y + dy[i];\n \n \n ans = Math.min(ans, fill_dp(newx, newy, N, M, end_x, end_y, arr, dp, i, 0, visited));\n \n }\n \n if (ans == Integer.MAX_VALUE)\n R.write(-1 + \"\\n\");\n else\n R.write(ans + 1 + \"\\n\");\n \n \n }\n R.flush();\n }\n \n class newdiv {\n long a;\n long b;\n int id;\n \n newdiv(long a, long b, int id) {\n this.a = a;\n \n \n this.b = b;\n \n this.id = id;\n \n \n }\n }\n \n class div {\n int a;\n int b;\n \n div(int a, int b) {\n this.a = a;\n \n \n this.b = b;\n \n \n }\n }\n \n \n class Reader {\n static BufferedReader reader;\n static StringTokenizer tokenizer;\n \n static void init(InputStream input) {\n reader = new BufferedReader(\n new InputStreamReader(input));\n tokenizer = new StringTokenizer(\"\");\n }\n \n static String next() throws IOException {\n while (!tokenizer.hasMoreTokens()) {\n tokenizer = new StringTokenizer(\n reader.readLine());\n }\n return tokenizer.nextToken();\n }\n \n static int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n \n static long nextLong() throws IOException {\n return Long.parseLong(next());\n }\n \n static double nextDouble() throws IOException {\n return Double.parseDouble(next());\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041D",
"submission_id": 307170823,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include <stdio.h>\n \n#define MOD 1000000007\n \nvoid mult(long long *r, long long *a, long long *b) {\n\tlong long t[4] = { 0 };\n \n\tfor (int i=0;i<2;i++) {\n\t\tfor (int j=0;j<2;j++) {\n\t\t\tfor (int z=0;z<2;z++) {\n\t\t\t\tt[i*2+j] = (t[i*2+j] + (a[i*2+z] * b[z*2+j])%MOD)%MOD;\n\t\t\t}\n\t\t}\n\t}\n \n\tfor (int i=0;i<4;i++) r[i] = t[i];\n}\n \nvoid expo(long long *r, long long *b, long long e) {\n\tif (e == 0) {\n\t\tr[0] = 1;\n\t\tr[1] = 0;\n\t\tr[2] = 0;\n\t\tr[3] = 1;\n\t\treturn;\n\t}\n\texpo(r, b, e/2);\n\tmult(r, r, r);\n\tif (e&1) mult(r, b, r);\n}\n \nint main()\n{\n\tint m, n;\n\tscanf(\"%d %d\", &m, &n);\n \n\tint s[200006];\n\tint l[200006];\n \n\tfor (int i=0;i<m;i++) scanf(\"%d\", &s[i]);\n\tfor (int i=0;i<m;i++) scanf(\"%d\", &l[i]);\n \n\tlong long a[2][2] = { { 0 } };\n \n\tlong long t_s = 0;\n\tlong long t_l = 0;\n \n\tfor (int i=0;i<m;i++) {\n\t\ta[0][0] = (a[0][0] + (s[i] * (long long)s[i])%MOD)%MOD;\n\t\ta[0][1] = (a[0][1] + (s[i] * (long long)l[i])%MOD)%MOD;\n\t\ta[1][0] = (a[1][0] + (l[i] * (long long)s[i])%MOD)%MOD;\n\t\ta[1][1] = (a[1][1] + (l[i] * (long long)l[i])%MOD)%MOD;\n \n\t\tt_s = (t_s + s[i])%MOD;\n\t\tt_l = (t_l + l[i])%MOD;\n\t}\n \n\t/*\n\tlong long dp[2][2] = { { 0 } };\n\tint cur = 0;\n \n\tdp[!cur][0] = s[0];\n\tdp[!cur][1] = l[0];\n \n\tfor (int i=1;i<n;i++) {\n\t\tdp[cur][0] = (((dp[!cur][0] + dp[!cur][1])%MOD * a[0][0])%MOD + (dp[!cur][0] * a[1][0])%MOD)%MOD;\n\t\tdp[cur][1] = (((dp[!cur][0] + dp[!cur][1])%MOD * a[0][1])%MOD + (dp[!cur][0] * a[1][1])%MOD)%MOD;\n\t\tcur = !cur;\n\t}\n\t*/\n \n\tlong long b[4] = {\n\t\t(a[0][0] + a[1][0])%MOD, (a[0][1] + a[1][1])%MOD,\n\t\ta[0][0], a[0][1]\n\t};\n \n\tlong long r[4];\n \n\texpo(r, b, n-1);\n \n\tlong long k[4] = {\n\t\t0, 0,\n\t\ts[0], l[0]\n\t};\n \n\tmult(r, k, r);\n \n\t//printf(\"%lld\\n\", (dp[!cur][0] + dp[!cur][1])%MOD);\n\t//printf(\"%lld %lld\\n\", dp[!cur][0], dp[!cur][1]);\n\t//printf(\"%lld\\n\", (((dp[!cur][0] + dp[!cur][1])%MOD * t_s)%MOD + (dp[!cur][0] * t_l)%MOD)%MOD);\n\t\n\t/*\n\tfor (int i=0;i<2;i++) {\n\t\tfor (int j=0;j<2;j++) {\n\t\t\tprintf(\"%d \", r[i*2+j]);\n\t\t}\n\t\tputchar('\\n');\n\t}\n\t*/\n \n\tprintf(\"%lld\\n\", (((r[2] + r[3])%MOD * t_s)%MOD + (r[3] * t_l)%MOD)%MOD);\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": "1970E3",
"submission_id": 262147874,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include<stdio.h>\n#include<algorithm> \n#include<memory.h> \n \nusing namespace std;\n#define N 100005\nconst long long MOD = 1000000007;\nlong long res[35][2][2]; \nint cabins_short[N], cabins_long[N]; \nint main() {\n int m, n;\n scanf(\"%d %d\", &m, &n); \n for (int i = 0; i<m; i++)\n scanf(\"%d\", cabins_short+i); \n for (int i = 0; i<m; i++)\n scanf(\"%d\", cabins_long+i);\n memset(res, 0, sizeof(res));\n for (int i = 0; i < m; i++) {\n res[0][0][0] += (cabins_short[i]+cabins_long[i])*cabins_short[i]; \n res[0][1][0] += (cabins_short[i]+cabins_long[i])*cabins_long[i];\n res[0][0][1] += cabins_short[i]*cabins_short[i]; \n res[0][1][1] += cabins_short[i]*cabins_long[i];\n } \n int t = 1, q = 1; \n while ((1<<t) < n) {\n res[t][0][0] = (res[t-1][0][0]*res[t-1][0][0]+res[t-1][1][0]*res[t-1][0][1]) % MOD; \n res[t][0][1] = (res[t-1][0][0]*res[t-1][0][1]+res[t-1][0][1]*res[t-1][1][1]) % MOD;\n res[t][1][0] = (res[t-1][1][0]*res[t-1][0][0]+res[t-1][1][1]*res[t-1][1][0]) % MOD;\n res[t][1][1] = (res[t-1][1][0]*res[t-1][0][1]+res[t-1][1][1]*res[t-1][1][1]) % MOD;\n t += 1;\n }\n long long ans[2][2] = {1,0,0,1}; \n long long a00, a01, a10, a11;\n while (q<=n-1) {\n if ((n-1)&(1<<(q-1))) {\n a00 = (ans[0][0]*res[q-1][0][0]+ans[1][0]*res[q-1][0][1]) % MOD; \n a01 = (ans[0][0]*res[q-1][0][1]+ans[0][1]*res[q-1][1][1]) % MOD;\n a10 = (ans[1][0]*res[q-1][0][0]+ans[1][1]*res[q-1][1][0]) % MOD;\n a11 = (ans[1][0]*res[q-1][0][1]+ans[1][1]*res[q-1][1][1]) % MOD;\n }\n q <<= 1;\n ans[0][0] = a00; \n ans[0][1] = a01;\n ans[1][0] = a10;\n ans[1][1] = a11;\n }\n // printf(\"%lld %lld %lld %lld \\n\", ans[0][0], ans[1][0], ans[0][1], ans[1][1]); \n \n a00 = cabins_long[0]+cabins_short[0];\n a01 = cabins_short[0]; \n a10 = (a00*ans[0][0]+a01*ans[1][0]) % MOD; \n a11 = (a00*ans[0][1]+a01*ans[1][1]) % MOD; \n a00 = 0; \n // printf(\"%lld %lld\\n\", a10, a11); \n for (int i = 0; i< m; i++) {\n a00 += a10*cabins_short[i]+a11*cabins_long[i]; \n a00 = a00 % MOD;\n }\n printf(\"%lld\", a00); \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": "1970E3",
"submission_id": 260815887,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include<cstdio>\nusing namespace std;\nint m,n,s[100000],l[100000],t1,t2,t3;\nlong long ans,res1,res2;\ninline int ck(int x,int y){\n\tx+=y;\n\tif(x>1000000006)x-=1000000007;\n\treturn x;\n}\nstruct martix{\n\tint r,c;\n\tlong long v[2][2];\n\tvoid build(){for(int i=0;i<r;++i)for(int j=0;j<c;++j)v[i][j]=0;}\n\tinline martix operator*(const martix&cwx)const{\n\t\tmartix ne;ne.r=r,ne.c=cwx.c,ne.build();\n\t\tfor(int i=0;i<r;++i)\n\t\t\tfor(int k=0;k<c;++k)\n\t\t\t\tfor(int j=0;j<cwx.c;++j)\n\t\t\t\t\tne.v[i][j]=(ne.v[i][j]+v[i][k]*cwx.v[k][j]);\n\t\treturn ne;\n\t}\n}F,G;\nint main(){\n\tscanf(\"%d%d\",&m,&n),F.r=1,F.c=G.r=G.c=2,--n;\n\tfor(int i=0;i<m;++i)scanf(\"%d\",s+i),t1=ck(t1,s[i]*s[i]),res1=ck(res1,s[i]);\n\tfor(int i=0;i<m;++i)scanf(\"%d\",l+i),t2=ck(t2,s[i]*l[i]),t3=ck(t3,l[i]*l[i]),res2=ck(res2,l[i]);\n\tG.v[0][0]=ck(t1,t2),G.v[0][1]=ck(t2,t3),G.v[1][0]=t1,G.v[1][1]=t2,F.v[0][0]=s[0],F.v[0][1]=l[0],res1=ck(res1,res2);\n\tfor(;n;n>>=1,G=G*G)if(n&1)F=F*G;\n\tans=F.v[0][0]*res1+F.v[0][1]*res2;\n\treturn 0&printf(\"%lld\",ans%1000000007);\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": "1970E3",
"submission_id": 280854299,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\nMOD = 10**9 + 7\n \ndef matrix_mult(A, B, mod):\n \"\"\" Multiplies two matrices A and B under modulo mod. \"\"\"\n n = len(A)\n m = len(B[0])\n p = len(B)\n result = [[0] * m for _ in range(n)]\n for i in range(n):\n for j in range(m):\n result[i][j] = sum(A[i][k] * B[k][j] for k in range(p)) % mod\n return result\n \ndef matrix_exponentiation(base, exp, mod):\n \"\"\" Performs matrix exponentiation of 'base' to the power 'exp' under modulo 'mod'. \"\"\"\n n = len(base)\n result = [[1 if i == j else 0 for j in range(n)] for i in range(n)]\n while exp > 0:\n if exp % 2 == 1:\n result = matrix_mult(result, base, mod)\n base = matrix_mult(base, base, mod)\n exp //= 2\n return result\n \ndef solve(m, n, s, l):\n # Create the transition matrix\n T = [[0] * m for _ in range(m)]\n for i in range(m):\n for j in range(m):\n if i == j:\n T[i][j] = (s[i] * s[j] + s[i] * l[j] + l[i] * s[j]) % MOD\n else:\n T[i][j] = (s[i] * s[j] + s[i] * l[j] + l[i] * s[j]) % MOD\n \n # Initial vector V0 where we start at cabin 1\n V0 = [[0] for _ in range(m)]\n V0[0][0] = 1 # Starting at cabin 1\n \n # Calculate T^n * V0\n Tn = matrix_exponentiation(T, n, MOD)\n result_vector = matrix_mult(Tn, V0, MOD)\n \n # Sum all elements in the result vector to get the total number of ways\n result = sum(row[0] for row in result_vector) % MOD\n return result\n \nimport sys\ninput = sys.stdin.read\ndata = input().split()\n \nm = int(data[0])\nn = int(data[1])\ns = list(map(int, data[2:m+2]))\nl = list(map(int, data[m+2:2*m+2]))\n \nprint(solve(m, n, s, l))\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": "1970E3",
"submission_id": 270595647,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\nmod = 10**9+7\nm, n = map(int, input().split())\nk0 = list(map(int, input().split()))\nk1 = list(map(int, input().split()))\nassert len(k0)==len(k1)==m\n \nc00 = sum(a*a for a in k0) % mod\nc11 = sum(a*a for a in k1) % mod\nc01 = sum(a*b for a,b in zip(k0,k1)) % mod\n \nm00 = (c00+c01) % mod\nm01 = (c01+c11) % mod\nm10 = c00\nm11 = c01\nm = m00,m01,m10,m11\n \ndef matmul(a, b):\n a00,a01,a10,a11 = a\n b00,b01,b10,b11 = b\n c00 = a00*b00 + a01*b10\n c01 = a00*b01 + a01*b11\n c10 = a10*b00 + a11*b10\n c11 = a10*b01 + a11*b11\n return c00%mod, c01%mod, c10%mod, c11%mod\n \ndef matpow(a, n):\n if n==0:\n return 1,0,0,1\n if n==1:\n return a\n n2, r = divmod(n,2)\n an2 = matpow(a, n2)\n an = matmul(an2, an2)\n if r:\n an = matmul(an, a)\n return an\n \nmn00,mn01,mn10,mn11 = matpow(m, n-1)\nu0 = k0[0]*mn00 + k0[1]*mn10\nu1 = k0[0]*mn01 + k0[1]*mn11\n \nres = ((u0+u1)*sum(k0) + u0*sum(k1)) % mod\nprint(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": "1970E3",
"submission_id": 260529972,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\nmod = 10**9+7\nm, n = map(int, input().split())\nk0 = list(map(int, input().split()))\nk1 = list(map(int, input().split()))\nassert len(k0)==len(k1)==m\n \nc00 = sum(a*a for a in k0) % mod\nc11 = sum(a*a for a in k1) % mod\nc01 = sum(a*b for a,b in zip(k0,k1)) % mod\n \nm00 = (c00+c01) % mod\nm01 = (c01+c11) % mod\nm10 = c00\nm11 = c01\nm = m00,m01,m10,m11\n \ndef matmul(a, b):\n a00,a01,a10,a11 = a\n b00,b01,b10,b11 = b\n c00 = a00*b00 + a01*b10\n c01 = a00*b01 + a01*b11\n c10 = a10*b00 + a11*b10\n c11 = a10*b01 + a11*b11\n return c00%mod, c01%mod, c10%mod, c11%mod\n \ndef matpow(a, n):\n if n==0:\n return 1,0,0,1\n if n==1:\n return a\n n2, r = divmod(n,2)\n an2 = matpow(n2)\n an = matmul(an2, an2)\n if r:\n an = matmul(an, a)\n return an\n \nmn00,mn01,mn10,mn11 = matpow(m, n-1)\nu0 = k0[0]*mn00 + k0[1]*mn10\nu1 = k0[0]*mn01 + k0[1]*mn11\n \nres = ((u0+u1)*sum(k0) + u0*sum(k1)) % mod\nprint(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": "1970E3",
"submission_id": 260529741,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\ntypedef long long int ll;\ntypedef long double ld;\nconst ll M = 1e9 + 7 ; // 998244353;\n \n/*\n**************************************************************************************************\n**************************************************************************************************\n \n \n *** **************\n *** **************\n *** ***\n *** ***\n *** ***\n *************************\n *************************\n *** ***\n *** ***\n *** ***\n ************** ***\n ************** ***\n \n**************************************************************************************************\n**************************************************************************************************\n \n \n*/\n \n#define v(li) vector<ll>\n#define vp(li) vector<pair<ll, ll>>\n#define m(li) map<ll, ll>\n#define s(li) set<ll>\n \n#define f(i, n) for (ll i = 0; i < n; i++)\n#define f1(i, n) for (ll i = 1; i < n+1; i++)\n#define test(t) while (t--)\n#define w(i, n) while (i < n)\n#define prs(n) cout << n << \" \"\n#define prn(n) cout << n << \" \"\n#define newl cout << \"\\n\"\n#define pb(a) push_back(a);\n#define pop pop_back()\n#define in(x) insert(x)\n#define fsort(v) sort(v.begin(), v.end())\n#define rsort(v) sort(v.begin(), v.end(), greater<ll>())\n#define rev(v) reverse(v.begin(), v.end())\n#define max_ele(v) *max_element(v.begin(), v.end())\n#define min_ele(v) *min_element(v.begin(), v.end())\n#define prc(i) cout<<\"Case #\"<<i<<\": \";\n \nll binexp(ll a, ll b)\n{\n if (b == 0)\n return 1;\n ll res = binexp(a, b / 2);\n if (b % 2 == 0)\n return res * res;\n else\n return a * res * res;\n}\n \nll biexm(ll a, ll b)\n{\n if (b == 0)\n return 1;\n ll res = biexm(a, b / 2);\n if (b % 2 == 0)\n return ((res % M) * (res % M)) % M;\n else\n return ((((res % M) * (res % M)) % M) * (a % M)) % M;\n}\n \nll hcf(ll a, ll b)\n{\n return __gcd(a,b);\n}\nll lcm(ll a,ll b){\n return (a*b)/(__gcd(a,b));\n}\n \n \n/*-------------------------------------------------------------------------------------------------------------------------------------------*/\n// for ncr and factorial with modulo M\n \n \n// const ll N1=3*1e5;\n// ll fact[N1+1];\n// ll inv[N1+1];\n \n// ll ncr(ll n,ll r){\n// if(r>n) return 0;\n// ll temp=fact[n];\n// temp=(temp*inv[r])%M;\n// temp=(temp*inv[n-r])%M;\n// return temp;\n// }\n \n/*-------------------------------------------------------------------------------------------------------------------------------------------*/\n/* DSU */\n \n \n// const ll N2=2*1e5;\n// ll par[N2+1];\n// ll ran[N2+1];\n \n// void make(ll n){\n// f1(i,n){\n// par[i]=i;\n// ran[i]=1;\n// }\n// }\n \n \n// ll findg(ll v){\n// if(par[v]==v) return v;\n// par[v]=findg(par[v]);\n// return par[v];\n// }\n \n \n// void unite(ll a,ll b){\n// a=findg(a);\n// b=findg(b);\n// if(a==b) return;\n// if(ran[a]<ran[b]) swap(a,b);\n// par[b]=a;\n// ran[a]=ran[a]+ran[b];\n// }\n \n/*-------------------------------------------------------------------------------------------------------------------------------------------*/\nvector<vector<ll>> matMul(vector<vector<ll>> a,vector<vector<ll>> b){\n vector<vector<ll>> res(2,vector<ll>(2,0));\n for(int i=0;i<2;i++){\n for(int j=0;j<2;j++){\n for(int k=0;k<2;k++){\n res[i][j]+=(a[i][k]*b[k][j])%M;\n res[i][j]%=M;\n }\n }\n }\n return res;\n}\nvector<vector<ll>> matExm(vector<vector<ll>> a,ll n){\n vector<vector<ll>> res(2,vector<ll>(2,0));\n if(n==0){\n res[0][0]=1;res[1][1]=1;\n return res;\n }\n res=matExm(a,n/2);\n res=matMul(res,res);\n if(n%2==1) return matMul(a,res);\n else return res;\n}\n \nint main()\n{ \n ios::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n \n/*-------------------------------------------------------------------------------------------------------------------------------------------*/\n /* Sieve */\n // ll N=1e6;\n // ll hpf[N+1];\n // f(i,N+1) hpf[i]=i;\n // for(ll i=2;i<N+1;i++){\n // if(hpf[i]==i){\n // for(ll j=i*i;j<N+1;j+=i){\n // hpf[j]=i;\n // }\n // }\n // }\n \n \n // ll N=1e6;\n // bool isprime[N+1];\n // f1(i,N) isprime[i]=true;\n // isprime[1]=false;\n // for(ll i=2;i<N+1;i++){\n // if(isprime[i]){\n // for(ll j=i*i;j<N+1;j+=i) isprime[j]=false;\n // }\n // }\n \n \n // vector<ll> primes;\n // for(ll i=2;i<N+1;i++) if(isprime[i]) primes.push_back(i);\n \n // fact[0]=1;\n // f1(i,N1) fact[i]=(fact[i-1]*i)%M;\n // f(i,N1+1) inv[i]=biexm(fact[i],M-2);\n \n /* Code Starts here */\n // Think of dp transitions very carefully\n // In greedy problems if it asks for some max or min then try to get some equations from given situation\n // if problem is related to swaps in permutations then think of cycles formed in permutation\n // In dijikstra algorithm always use visited array\n // radius of tree = ceil(diameter/2)\n \n/* ---------------------------------------------------------------------------------------------------------------------------------------- */\n \n \n ll t=1;\n // cin>>t;\n test(t){\n ll m,n;\n cin>>m>>n;\n vector<ll> a(m),b(m);\n f(i,m) cin>>a[i];\n f(i,m) cin>>b[i];\n \n ll a2=0,b2=0,ab=0;\n ll x1=0,y1=0;\n f(i,m){\n ab+=a[i]*b[i];ab%=M;\n a2+=a[i]*a[i];a2%=M;\n b2+=b[i]*b[i];b2%=M;\n x1+=(a[i]+b[i]); x1%=M;\n y1+=(a[i]); y1%=M;\n }\n // cout<<ab<<' '<<a2<<' '<<b2<<'\\n';\n ll const1 = (ab+a2)%M;\n ll const2 = (ab+b2)%M;\n \n vector<vector<ll>> v1={{const1,const2},{0,0}};\n vector<vector<ll>> v2={{a2,ab},{0,0}};\n vector<vector<ll>> v3=matExm(v1,n-1);\n vector<vector<ll>> v4=matExm(v2,n-1);\n // for(int i=0;i<2;i++){\n // for(int j=0;j<2;j++) cout<<v4[i][j]<<' ';\n // newl;\n // }\n ll x=(v3[0][0]*x1+v3[0][1]*y1)%M;\n ll y=(v4[0][0]*x1+v4[0][1]*y1)%M;\n ll ans=(a[0]*x+b[0]*y)%M;\n cout<<ans<<'\\n';\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": "1970E3",
"submission_id": 271772094,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nconst long long MOD = 1000000007;\n \nvector<vector<int>> mult(vector<vector<int>> &a, vector<vector<int>> &b) {\n vector<vector<int>> mat(2, vector<int>(2));\n for (int i = 0; i < 2; i++) {\n for (int j = 0; j < 2; j++) {\n for (int k = 0; k < 2; k++) {\n mat[i][j] += a[i][k] * b[k][j];\n mat[i][j] %= MOD;\n }\n }\n }\n return mat;\n}\n \nvector<vector<int>> mat_exp(vector<vector<int>> &a, int e) {\n if (e == 1) {\n return a;\n }\n vector<vector<int>> half = mat_exp(a, e / 2);\n vector<vector<int>> res = mult(half, half);\n if (e % 2 == 1) {\n res = mult(res, a);\n }\n return res;\n}\n \nint main() {\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n \n int m, n;\n cin >> m >> n;\n int s[m];\n int l[m];\n for (int i = 0; i < m; i++) {\n cin >> s[i];\n }\n \n long long ss = 0; long long sl = 0; long long ts = 0; long long tl = 0; \n long long tsum = 0; long long lsum = 0;\n for (int i = 0; i < m; i++) {\n cin >> l[i];\n ss += s[i] * s[i];\n sl += s[i] * l[i];\n tl += (s[i] + l[i]) * l[i];\n ts += (s[i] + l[i]) * s[i];\n tsum += s[i] + l[i];\n lsum += l[i];\n }\n ss %= MOD; sl %= MOD; tl %= MOD; ts %= MOD;\n \n vector<vector<int>> mat(2, vector<int>(2));\n mat[0][0] = ts; mat[0][1] = tl; mat[1][0] = ss; mat[1][1] = sl;\n \n mat_exp(mat, n);\n long long res = s[0] * mat[0][0] % MOD * tsum + s[0] * mat[0][1] % MOD * lsum \n + s[1] * mat[1][0] % MOD * tsum + s[1] * mat[1][1] % MOD * lsum;\n res %= MOD;\n cout << res << \"\\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": "1970E3",
"submission_id": 260601936,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\ntypedef long long int ll;\ntypedef long double ld;\nconst ll M = 1e9 + 7 ; // 998244353;\n \n/*\n**************************************************************************************************\n**************************************************************************************************\n \n \n *** **************\n *** **************\n *** ***\n *** ***\n *** ***\n *************************\n *************************\n *** ***\n *** ***\n *** ***\n ************** ***\n ************** ***\n \n**************************************************************************************************\n**************************************************************************************************\n \n \n*/\n \n#define v(li) vector<ll>\n#define vp(li) vector<pair<ll, ll>>\n#define m(li) map<ll, ll>\n#define s(li) set<ll>\n \n#define f(i, n) for (ll i = 0; i < n; i++)\n#define f1(i, n) for (ll i = 1; i < n+1; i++)\n#define test(t) while (t--)\n#define w(i, n) while (i < n)\n#define prs(n) cout << n << \" \"\n#define prn(n) cout << n << \" \"\n#define newl cout << \"\\n\"\n#define pb(a) push_back(a);\n#define pop pop_back()\n#define in(x) insert(x)\n#define fsort(v) sort(v.begin(), v.end())\n#define rsort(v) sort(v.begin(), v.end(), greater<ll>())\n#define rev(v) reverse(v.begin(), v.end())\n#define max_ele(v) *max_element(v.begin(), v.end())\n#define min_ele(v) *min_element(v.begin(), v.end())\n#define prc(i) cout<<\"Case #\"<<i<<\": \";\n \nll binexp(ll a, ll b)\n{\n if (b == 0)\n return 1;\n ll res = binexp(a, b / 2);\n if (b % 2 == 0)\n return res * res;\n else\n return a * res * res;\n}\n \nll biexm(ll a, ll b)\n{\n if (b == 0)\n return 1;\n ll res = biexm(a, b / 2);\n if (b % 2 == 0)\n return ((res % M) * (res % M)) % M;\n else\n return ((((res % M) * (res % M)) % M) * (a % M)) % M;\n}\n \nll hcf(ll a, ll b)\n{\n return __gcd(a,b);\n}\nll lcm(ll a,ll b){\n return (a*b)/(__gcd(a,b));\n}\n \n \n/*-------------------------------------------------------------------------------------------------------------------------------------------*/\n// for ncr and factorial with modulo M\n \n \n// const ll N1=3*1e5;\n// ll fact[N1+1];\n// ll inv[N1+1];\n \n// ll ncr(ll n,ll r){\n// if(r>n) return 0;\n// ll temp=fact[n];\n// temp=(temp*inv[r])%M;\n// temp=(temp*inv[n-r])%M;\n// return temp;\n// }\n \n/*-------------------------------------------------------------------------------------------------------------------------------------------*/\n/* DSU */\n \n \n// const ll N2=2*1e5;\n// ll par[N2+1];\n// ll ran[N2+1];\n \n// void make(ll n){\n// f1(i,n){\n// par[i]=i;\n// ran[i]=1;\n// }\n// }\n \n \n// ll findg(ll v){\n// if(par[v]==v) return v;\n// par[v]=findg(par[v]);\n// return par[v];\n// }\n \n \n// void unite(ll a,ll b){\n// a=findg(a);\n// b=findg(b);\n// if(a==b) return;\n// if(ran[a]<ran[b]) swap(a,b);\n// par[b]=a;\n// ran[a]=ran[a]+ran[b];\n// }\n \n/*-------------------------------------------------------------------------------------------------------------------------------------------*/\n \n \n \nint main()\n{ \n ios::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n \n/*-------------------------------------------------------------------------------------------------------------------------------------------*/\n /* Sieve */\n // ll N=1e6;\n // ll hpf[N+1];\n // f(i,N+1) hpf[i]=i;\n // for(ll i=2;i<N+1;i++){\n // if(hpf[i]==i){\n // for(ll j=i*i;j<N+1;j+=i){\n // hpf[j]=i;\n // }\n // }\n // }\n \n \n // ll N=1e6;\n // bool isprime[N+1];\n // f1(i,N) isprime[i]=true;\n // isprime[1]=false;\n // for(ll i=2;i<N+1;i++){\n // if(isprime[i]){\n // for(ll j=i*i;j<N+1;j+=i) isprime[j]=false;\n // }\n // }\n \n \n // vector<ll> primes;\n // for(ll i=2;i<N+1;i++) if(isprime[i]) primes.push_back(i);\n \n // fact[0]=1;\n // f1(i,N1) fact[i]=(fact[i-1]*i)%M;\n // f(i,N1+1) inv[i]=biexm(fact[i],M-2);\n \n /* Code Starts here */\n // Think of dp transitions very carefully\n // In greedy problems if it asks for some max or min then try to get some equations from given situation\n // if problem is related to swaps in permutations then think of cycles formed in permutation\n // In dijikstra algorithm always use visited array\n // radius of tree = ceil(diameter/2)\n \n/* ---------------------------------------------------------------------------------------------------------------------------------------- */\n \n \n ll t=1;\n // cin>>t;\n test(t){\n ll m,n;\n cin>>m>>n;\n vector<ll> a(m),b(m);\n f(i,m) cin>>a[i];\n f(i,m) cin>>b[i];\n \n vector<ll> dpX(n+1,0),dpY(n+1,0);\n ll a2=0,b2=0,ab=0;\n f(i,m){\n ab+=a[i]*b[i];ab%=M;\n a2+=a[i]*a[i];a2%=M;\n b2+=b[i]*b[i];b2%=M;\n dpX[0]+=(a[i]+b[i]); dpX[0]%=M;\n dpY[0]+=(a[i]); dpY[0]%=M;\n }\n // cout<<ab<<' '<<a2<<' '<<b2<<'\\n';\n ll const1 = (ab+a2)%M;\n ll const2 = (ab+b2)%M;\n \n f1(i,n){\n dpX[i]=((const1*dpX[i-1])%M + (const2*dpY[i-1])%M)%M;\n dpY[i]=((a2*dpX[i-1])%M + (ab*dpY[i-1])%M)%M;\n }\n ll ans= (a[0]*dpX[n-1]+b[0]*dpY[n-1])%M;\n cout<<ans<<'\\n';\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": "1970E3",
"submission_id": 271769124,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long\n \n#define mod 1000000007\n \nint s[200000],l[200000];\nint t[200000];\nint f[2000][2000];\nint m,n;\n \nvector<vector<int>> mmul(vector<vector<int>> a,vector<vector<int>> b, int n) {\n \n\tvector<vector<int>> res(n+5,vector<int> (n+5));\n \n\tfor(int i=1; i<=n; i++) {\n\t\tfor(int j=1; j<=n; j++) {\n\t\t\tfor(int k=1; k<=n; k++) {\n\t\t\t\tres[i][j]=(res[i][j]+a[i][k]*b[k][j]%mod)%mod;\n\t\t\t}\n\t\t}\n\t}\n \n\treturn res;\n}\n \nvector<vector<int>> mqpow(vector<vector<int>> a,int b, int n) {\n \n\tvector<vector<int>> res(n+5,vector<int> (n+5));\n\tfor(int i=1; i<=n; i++) res[i][i]=1;\n \n\twhile(b) {\n\t\tif(b&1) res=mmul(res,a,n);\n\t\ta=mmul(a,a,n);\n\t\tb/=2;\n\t}\n\treturn res;\n}\n \n \nvoid E1() {\n\tcin>>n>>m;\n\tfor(int i=1; i<=n; i++) cin>>s[i];\n\tfor(int i=1; i<=n; i++) cin>>l[i],t[i]=s[i]+l[i];\n\tf[0][1]=1;\n\tfor(int i=1; i<=m; i++) {\n\t\tfor(int j=1; j<=n; j++) {\n\t\t\tfor(int k=1; k<=n; k++) {\n\t\t\t\tf[i][k]+=f[i-1][j]*(t[j]*t[k]-l[j]*l[k])%mod;\n\t\t\t}\n\t\t}\n\t}\n\tint ans=0;\n\tfor(int i=1; i<=n; i++) ans=(ans+f[m][i])%mod;\n\tcout<<ans<<\"\\n\";\n}\n \nvoid E2() {\n\tcin>>n>>m;\n\tfor(int i=1; i<=n; i++) cin>>s[i];\n\tfor(int i=1; i<=n; i++) cin>>l[i];\n\tvector<vector<int>> f(n+5,vector<int> (n+5));\n \n\tfor(int i=1; i<=n; i++) {\n\t\tfor(int j=1; j<=n; j++) {\n\t\t\tf[i][j]=s[i]*s[j]+s[i]*l[j]+l[i]*s[j];\n\t\t\tf[i][j]%=mod;\n\t\t}\n\t}\n \n\tvector<vector<int>> ms=mqpow(f,m,n);\n\tvector<vector<int>> ans(n+5,vector<int> (n+5));\n \n\tans[1][1]=1;\n \n\tans=mmul(ans,ms,n);\n \n\tint res=0;\n\tfor(int i=1; i<=n; i++) res=(res+ans[1][i])%mod;\n\tcout<<res<<'\\n';\n \n}\n \nvoid E3() {\n\tcin>>n>>m;\n\tfor(int i=1; i<=n; i++) cin>>s[i];\n\tfor(int i=1; i<=n; i++) cin>>l[i],t[i]=s[i]+l[i];\n \n\tvector<vector<int>> B(n+5,vector<int> (3)), C(3,vector<int> (n+5));\n \n\tfor(int i=1; i<=n; i++) {\n\t\tB[i][1]=t[i];\n\t\tB[i][2]=l[i];\n\t\tC[1][i]=t[i];\n\t\tC[2][i]=-l[i];\n\t}\n \n\tvector<vector<int>> ms(5, vector<int> (5));\n \n\tfor(int i=1; i<=n; i++) {\n\t\tms[1][1] = (ms[1][1] + C[1][i]*B[i][1] + mod) %mod;\n\t\tms[1][2] = (ms[1][2] + C[1][i]*B[i][2] + mod) %mod;\n\t\tms[2][1] = (ms[2][1] + C[2][i]*B[i][1] + mod) %mod;\n\t\tms[2][2] = (ms[2][2] + C[2][i]*B[i][2] + mod) %mod;\n\t}\n \n\tms=mqpow(ms,m-1,2);\n\t//cout<<ms[1][1]<<\" \"<<ms[1][2]<<\"\\n\"<<ms[2][1]<<\" \"<<ms[2][2]<<\"\\n\\n\";\n\t\n\tvector<int> ans(n+5),tmp(n+5);\n\tvector<vector<int>> bb(n+5,vector<int>(5));\n\tans[1]=1;\n\t\n\t\n \n\tfor(int i=1; i<=n; i++) {\n\t\tbb[i][1] = (B[i][1]*ms[1][1] + B[i][2]*ms[2][1] +mod +mod)%mod;\n\t\tbb[i][2] = (B[i][1]*ms[1][2] + B[i][2]*ms[2][2] +mod +mod)%mod;\n\t}\n\t\n\tfor(int i=1;i<=n;i++) {\n\t\ttmp[1]=(tmp[1]+ans[i]*bb[i][1] +mod)%mod;\n\t\ttmp[2]=(tmp[2]+ans[i]*bb[i][2] +mod)%mod;\n\t}\n\t\n\t\n\tint res=0;\n\tfor(int i=1;i<=n;i++) {\n\t\tans[i] = (tmp[1]*C[1][i]+tmp[2]*C[2][i] +mod +mod)%mod;\n\t\tres=(res+ans[i])%mod;\n\t}\n\t\n\tcout<<res<<'\\n';\n \n}\n \n \nsigned main() {\n \n\tE3();\n \n\treturn 0;\n}\n \n \n// 4 2 6\n// 1+1*1+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": "1970E3",
"submission_id": 279315874,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\nMOD = 10**9 + 7\n \ndef matrix_mult(A, B, mod):\n return [[sum(x * y for x, y in zip(row, col)) % mod for col in zip(*B)] for row in A]\n \ndef matrix_pow(mat, exp, mod):\n n = len(mat)\n res = [[1 if i == j else 0 for j in range(n)] for i in range(n)]\n base = mat\n while exp:\n if exp % 2:\n res = matrix_mult(res, base, mod)\n base = matrix_mult(base, base, mod)\n exp //= 2\n return res\n \ndef solve(m, n, s, l):\n # Create the transition matrix\n T = [[0] * m for _ in range(m)]\n for i in range(m):\n for j in range(m):\n if i == j:\n T[i][j] = (s[i] * s[j] + s[i] * l[j] + l[i] * s[j]) % MOD\n else:\n T[i][j] = (s[i] * s[j] + s[i] * l[j] + l[i] * s[j]) % MOD\n \n # Initial vector v, starting at cabin 1\n v = [[0] for _ in range(m)]\n v[0][0] = 1 # Harry starts at cabin 1\n \n # Compute T^n\n Tn = matrix_pow(T, n, MOD)\n \n # Compute T^n * v\n result_vector = matrix_mult(Tn, v, MOD)\n \n # Sum all ways to be at any cabin after n days\n result = sum(row[0] for row in result_vector) % MOD\n return result\n \n# Read input\nimport sys\ninput = sys.stdin.read\ndata = input().split()\nm = int(data[0])\nn = int(data[1])\ns = list(map(int, data[2:m+2]))\nl = list(map(int, data[m+2:2*m+2]))\n \n# Solve the problem\nprint(solve(m, n, s, l))\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": "1970E3",
"submission_id": 271013882,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\nimport os, sys\nfrom io import BytesIO, IOBase\nfrom array import array\nfrom itertools import accumulate\nimport bisect\nimport math\nfrom collections import deque\n \n# from functools import cache\n# cache cf需要自己提交 pypy3.9!\nfrom copy import deepcopy\n \n \nclass FastIO(IOBase):\n newlines = 0\n \n def __init__(self, file):\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n \n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n \n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n \n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\n \n \nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\n \n \nsys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\n \ninput = lambda: sys.stdin.readline().strip()\nints = lambda: list(map(int, input().split()))\nInt = lambda: int(input())\n \n \ndef queryInteractive(a, b, c):\n print('? {} {} {}'.format(a, b, c))\n sys.stdout.flush()\n return int(input())\n \n \ndef answerInteractive(x1, x2):\n print('! {} {}'.format(x1, x2))\n sys.stdout.flush()\n \n \ninf = float('inf')\n#if \"PyPy\" in sys.version:\n# import pypyjit; pypyjit.set_param('max_unroll_recursion=-1')\nMOD = int(1e9 + 7)\nm,n = ints()\ns = ints()\nl = ints()\n \ndef solve_easy():\n dp = [[0]*m for _ in range(n+1)]\n dp[0][0] = 1\n \n for i in range(1,n+1):\n for j in range(m):\n for k in range(m):\n dp[i][j] = (dp[i][j] + dp[i-1][k]*(s[k]*s[j] + s[k]*l[j] + l[k]*s[j])%MOD)%MOD\n # print(dp)\n print(sum(dp[n])%MOD)\n \n \ndef mat_mul(X,Y):\n n,m = len(X),len(Y[0])\n res = [[0 for j in range(m)] for i in range(n)]\n for i in range(n):\n for j in range(m):\n for k in range(len(Y)):\n res[i][j] += (X[i][k] * Y[k][j])%MOD\n res[i][j] %= MOD\n return res\n \n \nI = [[0]*m for _ in range(m)]\nfor i in range(m):\n I[i][i] = 1\n \ndef exp(x, i):\n ans = I\n \n tmp = x\n while i:\n if i & 1:\n ans = mat_mul(ans , tmp) \n i >>= 1\n tmp = mat_mul(tmp , tmp) \n return ans\n \n \ndef solve_med():\n P = [[0]*m for _ in range(m)]\n for i in range(m):\n for j in range(m):\n P[i][j] = (s[i]*s[j] + s[i]*l[j] + l[i]*s[j])%MOD\n ans = exp(P,n)\n t = 0\n for x in ans[0]:\n t = (t + x)%MOD\n print(t)\n \nsolve_med()\n \n \n# import sys\n# for line in sys.stdin:\n# a,b = map(int, line.split())\n# print(a+b)\n \n# python A.py < i.txt > o.txt\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": "1970E3",
"submission_id": 271691440,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Test {\n static final int MOD = 1_000_000_007;\n \n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int m = sc.nextInt();\n int n = sc.nextInt();\n int[] s = new int[m];\n int[] l = new int[m];\n \n for (int i = 0; i < m; i++) {\n s[i] = sc.nextInt();\n }\n for (int i = 0; i < m; i++) {\n l[i] = sc.nextInt();\n }\n \n // Create the transition matrix T\n long[][] T = new long[m][m];\n for (int i = 0; i < m; i++) {\n for (int j = 0; j < m; j++) {\n long shortToShort = (long) s[i] * s[j] % MOD;\n long shortToLong = (long) s[i] * l[j] % MOD;\n long longToShort = (long) l[i] * s[j] % MOD;\n // Ensure at least one short trail in any round trip\n T[i][j] = (shortToShort + shortToLong + longToShort) % MOD;\n }\n }\n \n // Exponentiate the matrix T to the power n\n long[][] result = matrixExponentiation(T, n, m);\n \n // Calculate the sum of the first row of the resulting matrix\n long totalWays = 0;\n for (int j = 0; j < m; j++) {\n totalWays = (totalWays + result[0][j]) % MOD;\n }\n \n System.out.println(totalWays);\n }\n \n private static long[][] matrixExponentiation(long[][] matrix, int power, int size) {\n long[][] result = new long[size][size];\n // Initialize result as the identity matrix\n for (int i = 0; i < size; i++) {\n result[i][i] = 1;\n }\n \n long[][] base = matrix;\n \n while (power > 0) {\n if ((power & 1) == 1) {\n result = matrixMultiply(result, base, size);\n }\n base = matrixMultiply(base, base, size);\n power >>= 1;\n }\n \n return result;\n }\n \n private static long[][] matrixMultiply(long[][] A, long[][] B, int size) {\n long[][] product = new long[size][size];\n for (int i = 0; i < size; i++) {\n for (int j = 0; j < size; j++) {\n for (int k = 0; k < size; k++) {\n product[i][j] = (product[i][j] + A[i][k] * B[k][j] % MOD) % MOD;\n }\n }\n }\n return product;\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": "1970E3",
"submission_id": 260917813,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Test {\n static final int MOD = 1_000_000_007;\n \n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int m = sc.nextInt();\n int n = sc.nextInt();\n int[] s = new int[m];\n int[] l = new int[m];\n \n for (int i = 0; i < m; i++) {\n s[i] = sc.nextInt();\n }\n for (int i = 0; i < m; i++) {\n l[i] = sc.nextInt();\n }\n \n // Create the transition matrix T\n long[][] T = new long[m][m];\n for (int i = 0; i < m; i++) {\n for (int j = 0; j < m; j++) {\n long shortToShort = (long) s[i] * s[j] % MOD;\n long shortToLong = (long) s[i] * l[j] % MOD;\n long longToShort = (long) l[i] * s[j] % MOD;\n // Ensure at least one short trail in any round trip\n T[i][j] = (shortToShort + shortToLong + longToShort) % MOD;\n }\n }\n \n // Exponentiate the matrix T to the power n\n long[][] result = matrixExponentiation(T, n, m);\n \n // Calculate the sum of the first row of the resulting matrix\n long totalWays = 0;\n for (int j = 0; j < m; j++) {\n totalWays = (totalWays + result[0][j]) % MOD;\n }\n \n System.out.println(totalWays);\n }\n \n private static long[][] matrixExponentiation(long[][] matrix, int power, int size) {\n long[][] result = new long[size][size];\n // Initialize result as the identity matrix\n for (int i = 0; i < size; i++) {\n result[i][i] = 1;\n }\n \n while (power > 0) {\n if ((power & 1) == 1) {\n result = matrixMultiply(result, matrix, size);\n }\n matrix = matrixMultiply(matrix, matrix, size);\n power >>= 1;\n }\n \n return result;\n }\n \n private static long[][] matrixMultiply(long[][] A, long[][] B, int size) {\n long[][] product = new long[size][size];\n for (int i = 0; i < size; i++) {\n for (int j = 0; j < size; j++) {\n for (int k = 0; k < size; k++) {\n product[i][j] = (product[i][j] + A[i][k] * B[k][j] % MOD) % MOD;\n }\n }\n }\n return product;\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": "1970E3",
"submission_id": 260917496,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n// LUOGU_RID: 170851215\n#include <bits/stdc++.h>\nconst int mod = 1e9 + 7;\nstruct matrix {\n\tint n, m;\n\tstd::vector<std::vector<int> > a;\n\tmatrix(int n1) {\n\t\tn = m = n1;\n\t\ta.assign(n, std::vector<int>(n));\n\t\tfor (int i = 0; i < n; ++i)\n\t\t\ta[i][i] = 1;\n\t\treturn;\n\t}\n\tmatrix(int n1, int m1): n(n1), m(m1), a(n, std::vector<int>(m)) {}\n\tstd::vector<int> &operator[] (const int q) {\n\t\treturn a[q];\n\t}\n\tmatrix operator* (matrix q) const {\n\t\tmatrix res(n, q.m);\n\t\tfor (int i = 0; i < n; ++i)\n\t\t\tfor (int j = 0; j < q.m; ++j)\n\t\t\t\tfor (int k = 0; k < m; ++k)\n\t\t\t\t\tres[i][j] = (((long long)a[i][k] * q[k][j]) % mod + res[i][j]) % mod;\n\t\treturn res;\n\t}\n\tmatrix &operator*= (matrix &q) {\n\t\treturn *this = *this * q;\n\t}\n\tmatrix operator^ (const int q) const {\n\t\tmatrix res(n), x(*this);\n\t\tfor (int y(q); y; y >>= 1, x *= x)\n\t\t\tif (y & 1)\n\t\t\t\tres *= x;\n\t\treturn res;\n\t}\n\tmatrix &operator^= (const int q) {\n\t\treturn *this = *this ^ q;\n\t}\n};\nint main() {\n#ifdef ONLINE_JUDGE\n\tstd::ios::sync_with_stdio(false);\n\tstd::cin.tie(nullptr);\n\tstd::cout.tie(nullptr);\n#endif\n\tint n, m;\n\tstd::cin >> m >> n;\n\tstd::vector<int> s(m), l(m);\n\tfor (int i = 0; i < m; ++i)\n\t\tstd::cin >> s[i];\n\tfor (int i = 0; i < m; ++i)\n\t\tstd::cin >> l[i];\n\tmatrix a(m, 2), b(2, m);\n\tfor (int i = 0; i < m; ++i) {\n\t\ta[i][0] = s[i] + l[i];\n\t\ta[i][1] = -l[i];\n\t\tb[0][i] = s[i] + l[i];\n\t\tb[1][i] = l[i];\n\t}\n\t// a = (a * b) ^ n;\n\ta = a * ((b * a) ^ (n - 1)) * b;\n\tint res = 0;\n\tfor (int i = 0; i < m; ++i)\n\t\tres = ((long long)res + a[0][i]) % mod;\n\tres = (res + mod) % mod;\n\tstd::cout << res << '\\n';\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970E3",
"submission_id": 274616391,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ll long long\nusing namespace std;\n \nconst int mod = 1e9 + 7;\nstruct Matrix {\n vector<vector<ll>> data;\n int row() const { return data.size(); }\n int col() const { return data[0].size(); }\n vector<ll>& operator[](int i) { return data[i]; }\n const vector<ll>& operator[](int i) const { return data[i]; }\n Matrix(int r, int c) : data(r, vector<ll>(c)) { }\n \n static Matrix identity(int n) {\n Matrix a(n, n);\n for (int i = 0; i < n; ++i)\n a[i][i] = 1;\n return a;\n }\n \n Matrix operator*(const Matrix& b) const {\n int r = row(), c = b.col(), n = col();\n Matrix cMat(r, c);\n for (int i = 0; i < r; ++i)\n for (int j = 0; j < c; ++j)\n for (int k = 0; k < n; ++k)\n cMat[i][j] = (cMat[i][j] + (data[i][k] * b[k][j]+mod)%mod+mod)%mod;\n return cMat;\n }\n Matrix pow(int exp) const {\n Matrix base = *this;\n Matrix ans = Matrix::identity(row());\n while (exp > 0) {\n if (exp & 1)\n ans = ans * base;\n base = base * base;\n exp >>= 1;\n }\n return ans;\n }\n};\n \nint m, n;\nvector<int> t, l;\n \nsigned main() {\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n cin >> m >> n;\n t.resize(m);\n l.resize(m);\n \n for (int i = 0; i < m; ++i)\n cin >> t[i];\n for (int i = 0; i < m; ++i)\n cin >> l[i];\n for (int i = 0; i < m; ++i)\n t[i] += l[i];\n \n Matrix base(m, 1);\n base[0][0] = 1;\n Matrix B(m, 2);\n Matrix C(2, m);\n for (int i = 0; i < m; ++i) {\n B[i][0] = t[i];\n B[i][1] = l[i];\n C[0][i] = t[i];\n C[1][i] = -l[i];\n }\n Matrix kq = C * B;\n Matrix B_pow = kq.pow(n - 1);\n Matrix final_result = B * B_pow * C;\n \n ll res = 0;\n for (int i = 0; i < m; i++)\n res = (res + final_result[i][0]+mod) % mod;\n cout << res;\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": "1970E3",
"submission_id": 268785701,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\nimport os, sys\nfrom io import BytesIO, IOBase\nfrom array import array\nfrom itertools import accumulate\nimport bisect\nimport math\nfrom collections import deque\n \n# from functools import cache\n# cache cf需要自己提交 pypy3.9!\nfrom copy import deepcopy\n \n \nclass FastIO(IOBase):\n newlines = 0\n \n def __init__(self, file):\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n \n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n \n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n \n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\n \n \nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\n \n \nsys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\n \ninput = lambda: sys.stdin.readline().strip()\nints = lambda: list(map(int, input().split()))\nInt = lambda: int(input())\n \n \ndef queryInteractive(a, b, c):\n print('? {} {} {}'.format(a, b, c))\n sys.stdout.flush()\n return int(input())\n \n \ndef answerInteractive(x1, x2):\n print('! {} {}'.format(x1, x2))\n sys.stdout.flush()\n \n \ninf = float('inf')\n#if \"PyPy\" in sys.version:\n# import pypyjit; pypyjit.set_param('max_unroll_recursion=-1')\nMOD = int(1e9 + 7)\nm,n = ints()\ns = ints()\nl = ints()\n \ndef solve_easy():\n dp = [[0]*m for _ in range(n+1)]\n dp[0][0] = 1\n \n for i in range(1,n+1):\n for j in range(m):\n for k in range(m):\n dp[i][j] = (dp[i][j] + dp[i-1][k]*(s[k]*s[j] + s[k]*l[j] + l[k]*s[j])%MOD)%MOD\n # print(dp)\n print(sum(dp[n])%MOD)\n \n \ndef mat_mul(X,Y):\n n,m = len(X),len(Y[0])\n res = [[0 for j in range(m)] for i in range(n)]\n for i in range(n):\n for j in range(m):\n for k in range(len(Y)):\n res[i][j] += (X[i][k] * Y[k][j])%MOD\n res[i][j] %= MOD\n return res\n \n \nI = [[0]*2 for _ in range(2)]\nfor i in range(2):\n I[i][i] = 1\n \ndef exp(x, i):\n ans = I\n \n tmp = x\n while i:\n if i & 1:\n ans = mat_mul(ans , tmp) \n i >>= 1\n tmp = mat_mul(tmp , tmp) \n return ans\n \n \ndef solve_med():\n P = [[0]*m for _ in range(m)]\n for i in range(m):\n for j in range(m):\n P[i][j] = (s[i]*s[j] + s[i]*l[j] + l[i]*s[j])%MOD\n ans = exp(P,n)\n t = 0\n for x in ans[0]:\n t = (t + x)%MOD\n print(t)\n \n \ndef solve_hard():\n B = [[s[i]+l[i],l[i]] for i in range(m)]\n C = [[s[i]+l[i] for i in range(m)],[-l[i] for i in range(m)]]\n # print(mat_mul(C,B))\n t = exp(mat_mul(C,B),n-1)\n # print(t)\n ans = mat_mul(B,t)\n ans = mat_mul(ans,C)\n print(sum(ans[0])%MOD)\nsolve_hard()\n \n \n# import sys\n# for line in sys.stdin:\n# a,b = map(int, line.split())\n# print(a+b)\n \n# python A.py < i.txt > o.txt\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": "1970E3",
"submission_id": 271693562,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include<iostream>\n#include<cstdio>\n#include<cstdlib>\n#include<cmath>\n#include<string>\n#include<cstring>\n#include<algorithm>\n#include<stack>\n#include<vector>\n#include<queue>\n#include<map>\n#include<set>\n#include<bitset>\n#define rep(i,a,b) for(int i=a;i<=(b);++i)\n#define frep(i,a,b) for(int i=a;i<(b);++i)\n#define drep(i,a,b) for(int i=a;i>=(b);--i)\n#define ls tree[p].l\n#define rs tree[p].r\n#define mid ((l + r) >> 1)\ntypedef long long ll;\ntypedef unsigned long long ull;\n#define Maxq priority_queue<int,vector<int>,greater<int> >\nusing namespace std;\nint mylog(int a){\n\tint ans=0;\n\tif(a&0xffff0000){\n\t\tans+=16;\n\t\ta>>=16;\n\t}\n\tif(a&0x0000ff00){\n\t\tans+=8;\n\t\ta>>=8;\n\t}\n\tif(a&0x000000f0){\n\t\tans+=4;\n\t\ta>>=4;\n\t}\n\tif(a&0x0000000c){\n\t\tans+=2;\n\t\ta>>=2;\n\t}\n\tif(a&0x00000002){\n\t\tans+=1;\n\t\ta>>=1;\n\t}\n\treturn ans;\n}\ninline int read(){\n\tregister int a=0,b=0;\n\tregister char c;\n\tc=getchar();\n\twhile(c<'0'||c>'9'){\n\t\tif(c=='-')b=1;\n\t\tc=getchar();\n\t}\n\twhile(c>='0'&&c<='9'){\n\t\ta*=10;\n\t\ta+=c-'0';\n\t\tc=getchar();\n\t}\n\treturn b?-a:a;\n}\nconst int MAXM = 100010;\nconst int MOD = 1e9 + 7;\nstruct Matrix{\n\tint n, m;\n\tint **data;\n\tMatrix(int n, int m) : n(n), m(m){\n\t\tdata = new int*[n + 1];\n\t\trep(i, 0, n){\n\t\t\tdata[i] = new int[m + 1];\n\t\t}\n\t}\n\tMatrix(const Matrix & other) : n(other.n), m(other.m){\n\t\tdata = new int*[n + 1];\n\t\trep(i, 0, n){\n\t\t\tdata[i] = new int[m + 1];\n\t\t}\n\t\trep(i, 1, n){\n\t\t\trep(j, 1, m){\n\t\t\t\tdata[i][j] = other.data[i][j];\n\t\t\t}\n\t\t}\n\t}\n\t~Matrix(){\n\t\trep(i, 0, n){\n\t\t\tdelete[] data[i];\n\t\t}\n\t\tdelete[] data;\n\t}\n\tMatrix& operator= (const Matrix & other){\n\t\tif(&other == this) return *this;\n\t\trep(i, 0, n){\n\t\t\tdelete[] data[i];\n\t\t}\n\t\tdelete[] data;\n\t\tn = other.n;\n\t\tm = other.m;\n\t\tdata = new int*[n + 1];\n\t\trep(i, 0, n){\n\t\t\tdata[i] = new int[m + 1];\n\t\t}\n\t\trep(i, 1, n){\n\t\t\trep(j, 1, m){\n\t\t\t\tdata[i][j] = other.data[i][j];\n\t\t\t}\n\t\t}\n\t\treturn *this;\n\t}\n\tMatrix operator* (const Matrix &other){\n\t\tMatrix ans(n, other.m);\n//\t\tans.n = n;\n//\t\tans.m = other.m;\n\t\trep(i, 1, ans.n){\n\t\t\trep(j, 1, ans.m){\n\t\t\t\tans.data[i][j] = 0;\n\t\t\t\trep(k, 1, m){\n\t\t\t\t\tans.data[i][j] = ((ll)ans.data[i][j] + (ll)data[i][k] * other.data[k][j] % MOD) % MOD;\n\t\t\t\t\tif(ans.data[i][j] < 0){\n\t\t\t\t\t\tcout << 1;\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t\treturn ans;\n\t}\n};\nMatrix ksm(Matrix a, int k){\n\tif(k == 1) return a;\n\tif(k & 1){\n\t\tMatrix tmp = ksm(a, k >> 1);\n\t\treturn tmp * tmp * a;\n\t}\n\telse{\n\t\tMatrix tmp = ksm(a, k >> 1);\n\t\treturn tmp * tmp;\n\t}\n}\nint s[MAXM], l[MAXM];\nsigned main(){\n\tint m = read(), n = read();\n\trep(i, 1, m) s[i] = read();\n\trep(i, 1, m) l[i] = read();\n\tMatrix A(m, 2), B(2, m);\n\trep(i, 1, m){\n\t\tA.data[i][1] = s[i] + l[i];\n\t\tA.data[i][2] = l[i];\n\t}\n\trep(i, 1, m){\n\t\tB.data[1][i] = s[i] + l[i];\n\t\tB.data[2][i] = MOD - l[i];\n\t}\n\tMatrix tmp = ksm(B * A, n - 1);\n\tMatrix ans = A * tmp * B;\n\tll sum = 0;\n\trep(i, 1, m) sum += ans.data[i][1];\n\tprintf(\"%lld\", sum % MOD);\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": "1970E3",
"submission_id": 261059955,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define int long long\nusing namespace std;\n \nconst int MOD = 1e9 + 7;\n \ninline int mod(int a) {\n return (a % MOD + MOD) % MOD;\n}\n \nstruct Matrix {\n vector<vector<int>> data;\n \n int row() const { return data.size(); }\n int col() const { return data[0].size(); }\n \n vector<int>& operator[](int i) { return data[i]; }\n const vector<int>& operator[](int i) const { return data[i]; }\n \n Matrix(int r, int c) : data(r, vector<int>(c)) { }\n \n static Matrix identity(int n) {\n Matrix a(n, n);\n for (int i = 0; i < n; ++i)\n a[i][i] = 1;\n return a;\n }\n \n Matrix operator*(const Matrix& b) const {\n int r = row(), c = b.col(), n = col();\n Matrix cMat(r, c);\n for (int i = 0; i < r; ++i) {\n for (int j = 0; j < c; ++j) {\n cMat[i][j] = 0;\n for (int k = 0; k < n; ++k) {\n cMat[i][j] = mod(cMat[i][j] + mod(data[i][k] * b[k][j]));\n }\n }\n }\n return cMat;\n }\n \n Matrix pow(int exp) const {\n Matrix base = *this;\n Matrix ans = Matrix::identity(row());\n while (exp > 0) {\n if (exp & 1)\n ans = ans * base;\n base = base * base;\n exp >>= 1;\n }\n return ans;\n }\n};\n \nint m, n;\nvector<int> t, l;\n \nsigned main() {\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n cin >> m >> n;\n t.resize(m);\n l.resize(m);\n \n for (int i = 0; i < m; ++i)\n cin >> t[i];\n for (int i = 0; i < m; ++i)\n cin >> l[i];\n for (int i = 0; i < m; ++i)\n t[i] += l[i];\n \n Matrix base(m, 1);\n base[0][0] = 1;\n Matrix B(m, 2);\n Matrix C(2, m);\n for (int i = 0; i < m; ++i) {\n B[i][0] = t[i];\n B[i][1] = l[i];\n C[0][i] = t[i];\n C[1][i] = -l[i];\n }\n Matrix kq = C * B;\n Matrix final_result = B * kq.pow(n - 1) * C;\n \n int res = 0;\n for (int i = 0; i < m; i++)\n res = (res + final_result[i][0]) % MOD;\n \n cout << res;\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": "1970E3",
"submission_id": 268751805,
"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: Trails (Hard)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.\n\nEach day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.\n\nHow many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?\n\nGive the answer modulo $$$10^9 + 7$$$.\n\nInput: The first line contains the integers $$$m$$$ and $$$n$$$.\n\nThe second line contains $$$m$$$ integers, $$$s_1, \\dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.\n\nThe third and last line contains $$$m$$$ integers, $$$l_1, \\dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.\n\nWe have the following constraints:\n\n$$$0 \\le s_i, l_i \\le 10^3$$$.\n\n$$$1 \\le m \\le 10^5$$$.\n\n$$$1 \\le n \\le 10^9$$$.\n\nOutput: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.\n\nExamples:\ninput:\n3 2\n1 0 1\n0 1 1\noutput:\n18\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define int long long\nusing namespace std;\n \nconst int MOD = 1e9 + 7;\n \ninline int mod(int a) {\n return (a % MOD + MOD) % MOD;\n}\n \nstruct Matrix {\n vector<vector<int>> data;\n \n int row() const { return data.size(); }\n int col() const { return data[0].size(); }\n \n vector<int>& operator[](int i) { return data[i]; }\n const vector<int>& operator[](int i) const { return data[i]; }\n \n Matrix(int r, int c) : data(r, vector<int>(c)) { }\n \n static Matrix identity(int n) {\n Matrix a(n, n);\n for (int i = 0; i < n; ++i)\n a[i][i] = 1;\n return a;\n }\n \n Matrix operator*(const Matrix& b) const {\n int r = row(), c = b.col(), n = col();\n Matrix cMat(r, c);\n for (int i = 0; i < r; ++i) {\n for (int j = 0; j < c; ++j) {\n cMat[i][j] = 0;\n for (int k = 0; k < n; ++k) {\n cMat[i][j] = mod(cMat[i][j] + mod(data[i][k] * b[k][j]));\n }\n }\n }\n return cMat;\n }\n \n Matrix pow(int exp) const {\n Matrix base = *this;\n Matrix ans = Matrix::identity(row());\n while (exp > 0) {\n if (exp & 1)\n ans = ans * base;\n base = base * base;\n exp >>= 1;\n }\n return ans;\n }\n};\n \nint m, n;\nvector<int> t, l;\n \nsigned main() {\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n cin >> m >> n;\n t.resize(m);\n l.resize(m);\n \n for (int i = 0; i < m; ++i)\n cin >> t[i];\n for (int i = 0; i < m; ++i)\n cin >> l[i];\n for (int i = 0; i < m; ++i)\n t[i] += l[i];\n \n Matrix base(m, 1);\n base[0][0] = 1;\n Matrix B(m, 2);\n Matrix C(2, m);\n for (int i = 0; i < m; ++i) {\n B[i][0] = t[i];\n B[i][1] = l[i];\n C[0][i] = t[i];\n C[1][i] = -l[i];\n }\n Matrix kq = C * B;\n Matrix B_pow = kq.pow(n - 1);\n Matrix final_result = B * B_pow * C;\n \n int res = 0;\n for (int i = 0; i < m; i++)\n res = (res + final_result[i][0]) % MOD;\n \n cout << res;\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": "1970E3",
"submission_id": 268743187,
"task_type": "TestCase-Eval-Task2"
} |
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