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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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\n#include <iostream>\n#include <algorithm>\n#include <vector>\n#include <set>\n#include <map>\n#include <unordered_map>\n#include <unordered_set>\n#include <string>\n#include <iomanip>\n#include <cmath>\n#include <queue>\n#include <utility>\n#include <array>\n#include <stack>\n \nusing namespace std;\n \nusing ull = unsigned long long;\nusing ll = long long;\n \nvoid solve() {\n ll x, m;\n cin >> x >> m;\n // (x ^ y) = x * t\n // (x ^ y) = y * t\n // 100011010 ^ 1001 = 100010011\n ll cnt = 0;\n for (ll y = 1; y <= min(m, 2LL * x); y++) {\n ll t = (x ^ y);\n if (t % x == 0 || t % y == 0) {\n cnt++;\n }\n }\n cnt += max(min(m, m ^ x) / x - 2LL, 0LL);\n cout << cnt << \"\\n\";\n}\n \nvoid multisolve() {\n int t;\n cin >> t;\n while (t--) {\n solve();\n }\n}\n \nint main() {\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n multisolve();\n //solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2039C2",
"submission_id": 292943627,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\n#include <iostream>\n#include <algorithm>\n#include <cmath>\n#include <stack>\n#include <queue>\n#include <set>\n#include <map>\n#include <string>\n \nusing namespace std;\ntypedef long long ll;\ntypedef long double ld;\ntypedef pair<ll, ll> pl;\nconst ll MOD = 1e9 + 7;\nconst ll INF = 1e18;\n \nvoid solve() {\n ll x, m; cin >> x >> m;\n ll result = 0;\n ll k = (m - 1) / x; //+1 for 0, -1 for x\n \n if (m <= 2 * x) {\n for (int y = 1; y <= m; y++) {\n if ((x ^ y) % x == 0 || (x ^ y) % y == 0) {\n result++;\n }\n }\n }\n else {\n //what y donates\n for (int y = 1; y <= min(2 * x, m); y++) {\n if ((x ^ y) % y == 0) {\n result++;\n if ((x ^ y) % x == 0 && (x ^ y) <= (k + 2) * x) { //i.e. y is a multiple of x we check\n result--;\n }\n }\n }\n result += k;\n if (k == 0 && x <= m)result++;\n if ((((k + 1) * x) ^ x) <= m && (((k + 1) * x) ^ x) > 0) {\n result++;\n }\n if ((((k + 2) * x) ^ x) <= m && (((k + 2) * x) ^ x) > 0) {\n result++;\n }\n }\n cout << result << '\\n';\n}\n \nint main()\n{\n ios::sync_with_stdio(0); cin.tie(0);\n int t; cin >> t;\n while (t--) {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2039C2",
"submission_id": 292974025,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\n#include<iostream>\n#include<algorithm>\n#include<string.h>\n#include<vector>\n#include<numeric>\n#include<map>\n#include<set>\n#define ll long long\n#define inf 0x7FFFFFFF\nusing namespace std;\nint cal(int x) {\n\tint ans=0;\n\twhile (x) {\n\t\tx=x >> 1;\n\t\tans++;\n\t}\n\treturn ans;\n}\nvoid solve()\n{\n\tll x, m;\n\tcin >> x >> m;\n\tll n = cal(x);\n\tint t = 1 << (n + 1) - 1;\n\tt = t & m;\n\tint mx = 0;\n\tfor (int i = 0; i <= t; i++) {\n\t\tmx = max(mx, i ^ t);\n\t}\n\tn = m - t + mx;\n\tll ans = max(m,n) / x;\n\tfor (int i = 1; i < min(x,m+1); i++) {\n\t\tint temp = x ^ i;\n\t\tans += (temp % i == 0 && temp % x != 0);\n\t}\n\tcout << min(ans,m) << endl;\n}\nint main()\n{\n\tios::sync_with_stdio(false);\n\tcin.tie(nullptr);\n\tint T;\n\tcin >> T;\n\twhile (T--) {\n\t\tsolve();\n\t}\n\treturn 0;\n}\n//111111110001\n// 1000\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2039C2",
"submission_id": 296575420,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\nfor __ in range(int(input())):\n \n x, m = map(int, input().split())\n count = 0\n y = 1\n while y <= min(6*x,m):\n if ((x^y)%x ==0) or ((x^y)%y == 0):\n count += 1\n y += 1\n \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": "2039C2",
"submission_id": 292955793,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\nfrom math import * \nfrom collections import *\nfrom bisect import bisect_left, bisect_right, bisect\nfrom itertools import *\n \n \nR=lambda:map(int,input().split())\nS=lambda:map(str,input().split())\n \nMOD = int(1e9 + 7)\n# MOD = 998244353\n# MOD = 100000000669\n \n \nt, = R() \n# t = 1\n \n \ndef answer(): \n x, m = R()\n \n if x == 1:\n return m \n \n cnt = 0\n for i in range(4 * x + 1):\n y = i ^ x\n \n # print(\" .. \", i, y)\n \n # print(y>= 1)\n # print(y <= m)\n \n if y >= 1 and y <= m and (i % x == 0 or i % y == 0):\n # print(\" -- \", i, y)\n cnt += 1\n \n return cnt\n \n \nfor i in range(t):\n print(answer())\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2039C2",
"submission_id": 292998371,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\nimport string\nimport sys\nfrom collections import deque\n \ninput = lambda: sys.stdin.readline().rstrip()\n \n# k, n = map(int, input().split())\nfor _ in range(int(input())):\n x, m = map(int, input().split())\n if m == 1:\n print(1)\n continue\n ans = 0\n for i in range(1, min(x + 1, m + 1)):\n if (x ^ i) % i == 0 or (x ^ i) % x == 0:\n ans += 1\n cut = m // x * x\n ans += m // x - (x < cut)\n if 1 <= (x ^ cut) <= m:\n ans += 1\n cut += x\n if 1 <= (x ^ cut) <= m:\n ans += 1\n if m > x:\n ans -= 1\n print(ans)\n \n# x = 17\n# print(bin(x))\n# for i in range(2, 200):\n# if (x ^ i) % i == 0 or (x ^ i) % x == 0:\n# print(i, bin(i), x ^ i, bin(x ^ i))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2039C2",
"submission_id": 299180211,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main\n{\n public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n public static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));\n public static StringTokenizer tokenizer;\n public static void read() throws IOException {\n tokenizer = new StringTokenizer(br.readLine());\n }\n public static int Int() throws IOException {\n if (tokenizer == null || !tokenizer.hasMoreTokens()) read();\n return Integer.parseInt(tokenizer.nextToken());\n }\n public static long Long() throws IOException {\n if (tokenizer == null || !tokenizer.hasMoreTokens()) read();\n return Long.parseLong(tokenizer.nextToken());\n }\n public static String Str() throws IOException {\n return br.readLine();\n }\n //-----------------------------------------------\n \n public static long quickPow(long b, long x, long mod) {\n if (x == 0) {\n return 1;\n }\n if (x == 1) {\n return b % mod;\n }\n long half = quickPow(b, x / 2, mod) % mod;\n long result = (half * half) % mod;\n result = (result * quickPow(b, x % 2, mod)) % mod;\n return result;\n }\n \n public static int check (long x, long m) {\n int count = 0;\n for (long y = 1; y <= m; y++) {\n long xor = x ^ y;\n if (xor != x && xor != 0 && (x % xor == 0 || y % xor == 0)) {\n count++;\n }\n }\n return count;\n }\n \n public static boolean f(long mult, long x, long m) {\n long max = mult * x;\n long val = 0;\n for (long i = 64; i >= 0; i--) {\n if ((max & (1L << i)) == 0 && (x & (1L << i)) > 0) {\n val |= (1L << i);\n }\n if ((max & (1L << i)) > 0 && (x & (1L << i)) == 0) {\n val |= (1L << i);\n }\n }\n return val <= m;\n }\n public static long bin(long x, long m) {\n long mult = 0;\n for (long i = 60; i >= 0; i--) {\n long maybeMult = mult + (1L << i);\n if (f(maybeMult, x, m)) {\n mult = maybeMult;\n }\n }\n return mult;\n }\n public static void main (String[] args) throws IOException {\n Scanner scanner = new Scanner(System.in);\n int t = scanner.nextInt();\n Random random = new Random();\n while (t > 0) {\n t--;\n long x = scanner.nextLong(), m = scanner.nextLong();\n long count = 0;\n for (long y = 1; y <= Math.min(x, m); y++) {\n long xor = x ^ y;\n if ((xor % y == 0) && (xor % x != 0) && xor <= m) {\n count++;\n }\n }\n \n long mult = bin(x, m);\n count += mult;\n for (int i = 1; i <= 10; i++) {\n if (f(mult + i, x, m)) {\n count++;\n }\n }\n System.out.println(count);\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2039C2",
"submission_id": 292986666,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class Main {\n \n // Helper method to find GCD of two numbers\n public static int findGCD(int a, int b) {\n if (b == 0) {\n return a;\n }\n return findGCD(b, a % b);\n }\n \n // Method to find GCD of an ArrayList\n public static int findGCDOfArrayList(ArrayList<Integer> numbers) {\n if (numbers == null || numbers.isEmpty()) {\n return 0;\n }\n \n int gcd = numbers.get(0); // Start with the first element\n for (int i = 1; i < numbers.size(); i++) {\n gcd = findGCD(gcd, numbers.get(i));\n if (gcd == 1) { // Early exit if GCD becomes 1\n return 1;\n }\n }\n \n return gcd;\n }\n \n // Method to find all divisors of a number\n public static ArrayList<Integer> findDivisors(int number) {\n ArrayList<Integer> divisors = new ArrayList<>();\n for (int i = 1; i <= Math.sqrt(number); i++) {\n if (number % i == 0) {\n divisors.add(i);\n if (i != number / i) {\n divisors.add(number / i);\n }\n }\n }\n Collections.sort(divisors); // Optional: Sort the divisors\n return divisors;\n }\n \n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt(); // Number of test cases\n \n while (t-- > 0) {\n long x = sc.nextInt();\n long m = sc.nextLong();\n // ArrayList<Integer> div = findDivisors(x);\n // HashSet<Integer> hs = new HashSet<>();\n int cnt = 0;\n if(x<m){\n for(long i=1;i<=Math.min(m,2*x+10);i++){\n if( (((i^x)%i)==0 || ((i^x)%x)==0)) cnt++;\n }\n \n }\n else{\n for(long i=1;i<=m;i++){\n if( (((i^x)%i)==0 || ((i^x)%x)==0)) cnt++;\n }\n }\n \n \n \n System.out.println(cnt);\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": "2039C2",
"submission_id": 292986933,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n#define int long long\n#define inf INT32_MAX\n//#define x first\n//#define y second\n#define endl '\\n'\n \nvoid solve() {\n int n, m;\n cin >> n >> m;\n if (n == 1) {\n cout << m << endl;\n return;\n }\n int ans = 0;\n for (int i = 1; i <= min(n, m); i++) {//m作为因子\n if ((i ^ n) % i == 0 || (i ^ n) % n == 0) {\n ans++;\n }\n }\n int num = m | n;\n ans += max(num / n - 1, 0ll);\n for (int i = num / n * n; i >= 2 * n; i -= n) {\n if ((num ^ n) > m)ans--;\n }\n cout << ans << endl;\n}\n \nsigned main() {\n#ifdef ONLINE_JUDGE\n#else\n freopen(\"test.in\", \"r\", stdin);\n freopen(\"test.out\", \"w\", stdout);\n#endif\n ios::sync_with_stdio(0);\n cout.tie(0);\n cin.tie(0);\n int t = 1;\n cin >> t;\n while (t--)\n solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2039C2",
"submission_id": 293789888,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\n// Author : hejinming2012\n#include <bits/stdc++.h>\n#define int long long\n#define endl '\\n'\n#define dbg(x) cout << #x \" = \" << (x) << endl\n#define quickio ios::sync_with_stdio(false);\n#define quickin cin.tie(0);\n#define quickout cout.tie(0);\n \nusing namespace std;\ninline int read() {\n int now = 0, nev = 1; char c = getchar();\n while(c < '0' || c > '9') { if(c == '-') nev = -1; c = getchar(); }\n while(c >= '0' && c <= '9') { now = (now << 1) + (now << 3) + (c & 15); c = getchar(); }\n return now * nev;\n}\nvoid write(int x) {\n if(x < 0) putchar('-'), x = -x;\n if(x > 9) write(x / 10);\n putchar(x % 10 + '0');\n}\nvector <int> f(int x) {\n vector <int> res;\n for(int i = 1; i * i <= x; i++)\n if(x % i == 0) {\n res.push_back(i);\n if(x / i != i) res.push_back(x / i);\n }\n return res;\n}\nsigned main() {\n quickio\n quickin\n quickout\n int T = read();\n while(T--) {\n int x = read(), m = read(), ans = 0;\n for(int i = 1; i <= 60; i++) {\n int j = x * (1ll << i - 1);\n if(j > m + x) break ;\n }\n for(int i = 1; i <= 60; i++) {\n if(x != 0 && i > m / x + 2) break ;\n int z = x * i, y = x ^ z;\n if(y >= 1 && y <= m) ans++;\n }\n vector <int> divs = f(x);\n for(auto i : divs) {\n if(i > m) continue ;\n int z = x ^ i;\n if(z % i == 0) ans++;\n }\n write(ans), putchar(10);\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": "2039C2",
"submission_id": 292938070,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\ndef kq(x, m):\n t=0\n for y in range(1,min(m,2*x)+1):\n xor=x^y\n if (xor%x==0) or (y!=0 and xor%y==0):\n t+=1 \n if m>2*x:\n t+=(m-2*x)//x\n return t\nfor _ in range(int(input())):\n x,m=map(int,input().split())\n print(kq(x,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": "2039C2",
"submission_id": 292955274,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\nimport math\nt=int(input())\n \nfor _ in range(t):\n n,m=map(int,input().split())\n p=(m-n)//n\n \n j=math.ceil((m-n)/n)\n if(j<0):\n j=0\n if((n^(j*n))<=m):\n p+=1\n if((n^((j+1)*n))<=m):\n p+=1\n \n for x in range(1,min(n,m+1)):\n if(((x^n)%x)==0):\n p+=1\n print(p)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2039C2",
"submission_id": 302046273,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\nboolean MULTI_CASE = true;\nboolean ALWAYS_FLUSH = false;\n \nvoid go() {\n int x = nextInt();\n long m = nextLong();\n// long r1 = 0;\n// for (int i = 1; i <= m; i++) {\n// int t = x ^ i;\n// if (t % i == 0 || t % x == 0) {\n// r1++;\n// }\n// }\n// H.out(r1);\n long res = 0;\n long top = highBit1(x) << 2;\n if (m < top * 2L) {\n for (int i = 1; i <= m; i++) {\n int t = x ^ i;\n if (t % i == 0 || t % x == 0) {\n res++;\n }\n }\n } else {\n long t1 = top / 2 / x;\n// long lo1 = 0x3f3f3f3f3f3f3f3fL;\n// long hi1 = 0;\n for (int i = 1; i < top; i++) {\n int t = x ^ i;\n if (t % i == 0) {\n res++;\n } else if (t % x == 0) {\n if (t <= t1) res++;\n// lo1=Math.min(lo1, t/x);\n// hi1 = Math.max(hi1, t/x);\n }\n }\n long t2 = (m - top / 2) / x;\n// long lo2 = 0x3f3f3f3f3f3f3f3fL;\n// long hi2 = 0;\n for (long i = m - top; i <= m; i++) {\n long t = x ^ i;\n if (t % i == 0) {\n res++;\n } else if (t % x == 0) {\n if (t >= t2) res++;\n// lo2=Math.min(lo2, t/x);\n// hi2 = Math.max(hi2, t/x);\n }\n }\n res += t2 - t1 - 1;\n }\n sl(res);\n}\n \nlong highBit1(int x) {\n // 11010 -> 10000\n x |= x >> 1;\n x |= x >> 2;\n x |= x >> 4;\n x |= x >> 8;\n x |= x >> 16;\n x = (x + 1) >> 1;\n return x;\n}\n \n \n/**\n * Input.\n */\nInputStream inStream;\nbyte[] inBuff = new byte[1024];\nint inBuffCursor = 0, inBuffLen = 0;\n \nboolean isVisibleChar(int c) {\n return 33 <= c && c <= 126;\n}\n \nboolean hasNextByte() {\n if (inBuffCursor < inBuffLen) return true;\n inBuffCursor = 0;\n try {\n inBuffLen = inStream.read(inBuff);\n } catch (Exception e) {\n e.printStackTrace(System.out);\n }\n return inBuffLen > 0;\n}\n \nboolean hasNext() {\n while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;\n return hasNextByte();\n}\n \nint nextByte() {\n return hasNextByte() ? inBuff[inBuffCursor++] : -1;\n}\n \nString next() {\n if (!hasNext()) throw new RuntimeException(\"no next.\");\n StringBuilder sb = new StringBuilder();\n int b = nextByte();\n while (isVisibleChar(b)) {\n sb.appendCodePoint(b);\n b = nextByte();\n }\n return sb.toString();\n}\n \nlong nextLong() {\n if (!hasNext()) throw new RuntimeException(\"no next.\");\n long result = 0;\n boolean negative = false;\n int b = nextByte();\n if (b < '0') {\n if (b == '-') negative = true;\n else if (b != '+') throw new RuntimeException(\"long number must start with +/-.\");\n b = nextByte();\n }\n while (isVisibleChar(b)) {\n if (b < '0' || b > '9') throw new RuntimeException(\"wrong digit in long:\" + (char) b);\n // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).\n result = result * 10 + (b - '0');\n b = nextByte();\n }\n return negative ? -result : result;\n}\n \nint nextInt() {\n long x = nextLong();\n if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)\n throw new RuntimeException(\"int overflow:\" + x);\n return (int) x;\n}\n \ndouble nextDouble() {\n return Double.parseDouble(next());\n}\n \n/**\n * Output.\n */\nPrintWriter printOut = new PrintWriter(System.out);\n \nvoid so(Object obj, boolean newLine) {\n if (newLine) printOut.println(obj);\n else printOut.print(obj);\n if (ALWAYS_FLUSH) printOut.flush();\n}\n \nvoid so(Object obj) {\n so(obj, false);\n}\n \nvoid sl(Object obj) {\n so(obj, true);\n}\n \nvoid sl() {\n sl(\"\");\n}\n \n/**\n * Main.\n */\nvoid mainGo() {\n try {\n inStream = new FileInputStream(\"src/main.in\");\n ALWAYS_FLUSH = true;\n } catch (Exception e) {\n inStream = System.in;\n }\n while (hasNext()) {\n if (MULTI_CASE) {\n int T = nextInt();\n if (T == 0) break;\n for (int i = 0; i < T; ++i) {\n go();\n }\n } else {\n go();\n }\n }\n printOut.flush();\n}\n \npublic static void main(String[] args) throws Exception {\n new Main().mainGo();\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": "2039C2",
"submission_id": 292945294,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\n//package com.aakash.Contest.C2039;\n \nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.io.PrintWriter;\nimport java.util.*;\n \npublic class C2 {\n static BufferedReader br = new BufferedReader(\n new InputStreamReader(System.in));\n \n public static void main(String[] args) throws IOException {\n \n StringTokenizer s1\n = new StringTokenizer(br.readLine());\n int t = Integer.parseInt(s1.nextToken());\n while (t-- > 0) {\n StringTokenizer s2\n = new StringTokenizer(br.readLine());\n long x = Integer.parseInt(s2.nextToken());\n long m=Long.parseLong(s2.nextToken());\n solve(x,m);\n }\n \n }\n \n static void solve(long x,long m) {\n long tx=x,tm=m;\n long pow=1;\n while(tx!=0) {\n tx=tx>>1;\n if((pow&tm)!=0) tm-=pow;\n pow=pow<<1;\n }\n long c=(tm-1)/x;\n for(int i=0;i<=x;i++) {\n long xor=tm+i,y=x^xor;\n if(y<=m &&y!=0 &&xor%x==0) c++;\n }\n for(int i=1;i<=Math.min(x,m);i++) {\n if(i==x) continue;\n long y=i,xor=i^x;\n if((xor%y)==0) c++;\n }\n System.out.println(c);\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": "2039C2",
"submission_id": 304661584,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nusing lint = long long;\nusing pi = array<lint, 2>;\n#define sz(v) ((int)(v).size())\n#define all(v) (v).begin(), (v).end()\n#define cr(v, n) (v).clear(), (v).resize(n);\nint main() {\n\tios_base::sync_with_stdio(0);\n\tcin.tie(0);\n\tcout.tie(0);\n\tint tc;\n\tcin >> tc;\n\twhile (tc--) {\n\t\tlint x, m;\n\t\tcin >> x >> m;\n\t\tlint cnt = 0;\n\t\tfor (lint y = 1; y <= m; y++) {\n\t\t\tlint xorv = x ^ y;\n\t\t\tif (xorv % x != 0 && xorv % y == 0) {\n\t\t\t\tcnt++;\n\t\t\t}\n\t\t}\n\t\tm++;\n\t\tfor (int i = 60; i >= 0; i--) {\n\t\t\tif ((m >> i) & 1) {\n\t\t\t\tlint z = ((m >> i) << i) - (1ll << i);\n\t\t\t\tz ^= ((x >> i) << i);\n\t\t\t\tcnt += (z + (1ll << i) + x - 1) / x - (z + x - 1) / x;\n\t\t\t}\n\t\t}\n\t\tcnt--;\n\t\tcout << cnt << \"\\n\";\n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2039C2",
"submission_id": 292932067,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define sz(a) int((a).size())\nusing namespace std;\n \nusing ll = long long;\nconst int inf = 1e9 + 7;\n \nint main() {\n #ifndef ONLINE_JUDGE\n freopen(\"in.txt\", \"r\", stdin);\n #endif\n ios::sync_with_stdio(0),cin.tie(NULL),cout.tie(NULL);\n \n int t;\n cin >> t;\n while (t--) {\n ll n, m;\n cin >> n>> m;\n ll answer = 0;\n \n for (int i = 1; i <= min(m, n); i++) {\n if (((n ^ i) % n == 0 || (n ^ i) % i == 0)) {\n ++answer;\n }\n }\n \n for (int i = n * 2; i <= m + n; i += n) {\n if ((n ^ i) <= m) {\n ++answer;\n }\n }\n \n cout << answer <<\"\\n\";\n }\nreturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2039C2",
"submission_id": 292961842,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\ndef count_valid_y(x, m):\n valid_count = 0\n \n for y in range(1, m + 1):\n xor_result = x ^ y\n if (xor_result % x == 0) or (xor_result % y == 0):\n valid_count += 1\n if valid_count >= m:\n break\n \n return valid_count\n \ndef main():\n import sys\n input = sys.stdin.read\n data = input().split()\n \n t = int(data[0])\n index = 1\n results = []\n \n for _ in range(t):\n x = int(data[index])\n m = int(data[index + 1])\n index += 2\n results.append(count_valid_y(x, m))\n \n for result in results:\n print(result)\n \nif __name__ == '__main__':\n main()\n \n \n# hclzCd1yTQ5SOK02n4a\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2039C2",
"submission_id": 304399298,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\nimport sys\ninput = lambda: sys.stdin.readline().rstrip(\"\\r\\n\")\ndef ii(): return int(input())\ndef mii(): return map(int, input().split())\ndef lii(): return list(map(int, input().split()))\n \nmod = 10**9+7\nmaxn = 2*10**5+10\n \nt=ii()\nfor _ in range(t):\n x,m=mii()\n if x==1:\n print(m)\n continue\n ans=set()\n for i in range(1,min(m,x)+1):\n if (i^x)%i==0:\n ans.add(i)\n # x ke jo bhi multiples hai\n mm=10**8//t\n for mul in range(1,2*m//x+10):\n y=x^(mul*x)\n if y!=0 and 1<=y<=m:\n ans.add(y)\n print(len(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": "2039C2",
"submission_id": 292950154,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.HashSet;\n \npublic class Main {\n\t\tstatic PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n\t\tstatic StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));\n\t\tstatic BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n\t\tpublic static void main(String[] args) throws Exception {\n\t\t\t\tlong t=Integer.parseInt(br.readLine());\n\t\t\t\twhile (t-->0) {\n\t\t\t\t\t\tString[] s = br.readLine().split(\" \");\n\t\t\t\t\t\tlong x = Long.parseLong(s[0]);\n\t\t\t\t\t\tlong m = Long.parseLong(s[1]);\n\t\t\t\t\t\tHashSet<Long> set = new HashSet<>();\n\t\t\t\t\t\tfor(long i=1;i<=Math.min(m,x<<1);i++){\n\t\t\t\t\t\t\t\tlong cur=x^i;\n\t\t\t\t\t\t\t\tif(cur%i==0){\n\t\t\t\t\t\t\t\t\t\tset.add(i);\n\t\t\t\t\t\t\t\t}\n\t\t\t\t\t\t}\n\t\t\t\t\t\tfor(long i=x;i<=(m<<1);i+=x){\n\t\t\t\t\t\t\t\tif((x^i)<=m&&(x^i)!=0){\n\t\t\t\t\t\t\t\t\t\tset.add(x^i);\n\t\t\t\t\t\t\t\t}\n\t\t\t\t\t\t}\n\t\t\t\t\t\tout.println(set.size());\n\t\t\t\t}\n\t\t\t\tout.close();\n\t\t}\n\t\tstatic int nextInt() throws IOException {\n\t\t\t\tin.nextToken();\n\t\t\t\treturn (int) in.nval;\n\t\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": "2039C2",
"submission_id": 292974282,
"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: Shohag Loves XOR (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.\n\nShohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \\le y \\le m$$$ such that $$$x \\oplus y$$$ is divisible$$$^{\\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\\oplus$$$ is the bitwise XOR operator.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \\le x \\le 10^6$$$, $$$1 \\le m \\le 10^{18}$$$).\n\nIt is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.\n\nOutput: For each test case, print a single integer — the number of suitable $$$y$$$.\n\nExamples:\ninput:\n5\n7 10\n2 3\n6 4\n1 6\n4 1\noutput:\n3\n2\n2\n6\n1\n\nNote: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.\n\n- $$$y = 1$$$ is valid because $$$x \\oplus y = 7 \\oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.\n- $$$y = 7$$$ is valid because $$$x \\oplus y = 7 \\oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.\n- $$$y = 9$$$ is valid because $$$x \\oplus y = 7 \\oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Main{\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int t = scanner.nextInt();\n \n while (t-- > 0) {\n long x = scanner.nextLong();\n long m = scanner.nextLong();\n \n long result = countValidY(x, m);\n System.out.println(result);\n }\n \n scanner.close();\n }\n \n private static long countValidY(long x, long m) {\n long count = 0;\n \n // Check all y from 1 to m\n for (long y = 1; y <= m; y++) {\n if ((x ^ y) % x == 0 || (x ^ y) % y == 0) {\n count++;\n }\n }\n \n return count;\n }\n}\n \n \n// 31fFrgVuoSIz7wCTYDX\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2039C2",
"submission_id": 302733574,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long\nconst int INF = 1e18, MOD = 1e9 + 7;\nconst int mxN = 25e3 + 7;\nint dp[mxN];\narray<int,2> child[mxN];\nint a[mxN];\nint ndp[mxN];\nint n;\nint dfs(int pos)\n{\n\t//cout << pos << \" \"; \n\tint x = INF, y = INF;\n\tint dp1[mxN][2];\n\tint ans = INF;\n\tif(child[pos][0] != -1)\n\t{\n\t\tdfs(child[pos][0]);\n\t\tfor(int i = 1;n + 1 >= i;i++) dp1[i][0] = dp[i];\n\t\tfor(int i = 1;n + 1 >= i;i++)\n\t\t{\n\t\t\tif(a[pos] != i) x = min(x, dp1[i][0] + i);\n\t\t}\n\t} \n\tif(child[pos][1] != -1)\n\t{\n\t\tdfs(child[pos][1]);\n\t\tfor(int i = 1;n + 1 >= i;i++) dp1[i][1] = dp[i];\n\t\tfor(int i = 1;n + 1 >= i;i++)\n\t\t{\n\t\t\tif(a[pos] != i) y = min(y, dp1[i][1] + i),ans = min(ans, dp1[i][0] + dp1[i][1] + i);\n\t\t}\n\t\tans = min(ans, x + y);\n\t} \n\telse ans = min(ans, x);\n\tfor(int i = 0;n + 1 >= i;i++)\n\t{\n\t\tif(a[pos] != i)\n\t\t\tndp[i] = 0;\n\t\telse\n\t\t\tndp[i] = INF;\n\t}\n\tif(x == INF && y == INF)\n\t{\n\t\tfor(int i = 0;n >= i;i++) dp[i] = ndp[i];\n\t\t/*cout << pos << endl;\n\t\tfor(int i = 1;n >= i;i++) cout << dp[i] << \" \";\n\t\tcout << endl;*/\n\t\treturn 0;\n\t} \n \n\tfor(int i = 0;n + 1 >= i;i++)\n\t{\n\t\t//cout << i << \" \" << ndp[i] << endl;\n\t\tif(a[pos] == i) continue;\n\t//\tcout << i << endl;\n\t\tif(y == INF)\n\t\t{\n\t\t\tndp[i] = min(x, dp1[i][0]);\n\t\t}\n\t\telse\n\t\t{\n\t\t\tndp[i] = min({dp1[i][0] + y, dp1[i][1] + x,ans});\n\t\t}\n\t}\n\tfor(int i = 0;n + 1 >= i;i++) dp[i] = ndp[i];\n\t/*cout << pos << endl;\n\tfor(int i = 1;n >= i;i++) cout << dp[i] << \" \";\n\tcout << endl;\n\tcout << \"SO: \" << ans << endl;*/\n\treturn ans;\n}\nvoid solve()\n{\n\t//int n;\n\tcin >> n;\n\tfor(int i = 0;n > i;i++) cin >> a[i], child[i] = {-1,-1};\n\tfor(int i = 0;n - 1 > i;i++)\n\t{\n\t\tint x;\n\t\tcin >> x;\n\t\tx--;\n\t\tif(child[x][0] == -1) child[x][0] = i + 1;\n\t\telse child[x][1] = i + 1;\n\t}\n\tfor(int i = 0;n >= i;i++) ndp[i] = INF, dp[i] = INF;\n\tcout << dfs(0) << endl;\n}\n \nint32_t main()\n{\n\tios_base::sync_with_stdio(false);\n\tcin.tie(NULL);\n\tcout.tie(NULL);\n\tint tt;\n\tcin >> tt;\n\twhile(tt--) solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1981F",
"submission_id": 275380202,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nconst int N = 5;\nint mex[(1<<N)];\nvoid dfs(int i, vector<vector<int>>& e, vector<int>& a, vector<vector<int>>& dp){\n for(int j: e[i])dfs(j, e, a, dp);\n for(int j=0; j<(1<<N); j++)dp[j][i]=1e9;\n if(e[i].size()==1){\n int mask = (a[i]<=N?(1<<(a[i]-1)):0);\n int x = e[i].back();\n for(int j=0; j<(1<<N)-1; j++)dp[j|mask][i]=min(dp[j|mask][i], dp[j][x]+(i==0?mex[j|mask]:0));\n if(a[x]<=N)mask|=1<<(a[x]-1);\n int low = 1e9;\n for(int j=0; j<(1<<N)-1; j++)low=min(dp[j][x]+mex[j], low);\n dp[mask][i]=min(dp[mask][i], low+(i==0?mex[mask]:0));\n }\n else if(e[i].size()==0){\n if(a[i]<=N){\n dp[1<<(a[i]-1)][i]=0;\n }\n else{\n dp[0][i]=0;\n }\n }\n else{\n int mask = (a[i]<=N?(1<<(a[i]-1)):0);\n int x = e[i].back();\n e[i].pop_back();\n int y = e[i].back();\n for(int j=0; j<(1<<N)-1; j++){\n for(int w=0; w<(1<<N)-1; w++){\n if((w|j|mask)==(1<<N)-1)continue;\n dp[mask][i]=min(dp[mask][i], dp[j][x]+dp[w][y]+mex[w|j|mask]);\n dp[mask][i]=min(dp[mask][i], dp[j][x]+mex[j]+dp[w][y]+mex[w|mask]);\n dp[mask][i]=min(dp[mask][i], dp[w][y]+mex[w]+dp[j][x]+mex[j|mask]);\n }\n }\n int low = 1e9;\n mask|=(a[x]<=N?(1<<(a[x]-1)):0);\n for(int j=0; j<(1<<N)-1; j++){\n low=min(low, mex[mask]+dp[j][x]+mex[j]);\n if((j|mask)!=(1<<N)-1)low=min(low, dp[j][x]+mex[j|mask]);\n }\n mask = (a[i]<=N?(1<<(a[i]-1)):0);\n for(int j=0; j<(1<<N)-1; j++){\n dp[mask|j][i]=min(dp[mask|j][i], dp[j][y]+low+(i==0?mex[mask|j]:0));\n }\n mask|=(a[y]<=N?(1<<(a[y]-1)):0);\n int low2 = 1e9;\n for(int j=0; j<(1<<N)-1; j++)low2=min(low2, dp[j][y]+mex[j]+(i==0?mex[mask]:0));\n dp[mask][i]=min(dp[mask][i], low+low2);\n swap(x, y);\n low = 1e9;\n mask|=(a[x]<=N?(1<<(a[x]-1)):0);\n for(int j=0; j<(1<<N)-1; j++){\n low=min(low, mex[mask]+dp[j][x]+mex[j]);\n if((j|mask)!=(1<<N)-1)low=min(low, dp[j][x]+mex[j|mask]);\n }\n mask = (a[i]<=N?(1<<(a[i]-1)):0);\n for(int j=0; j<(1<<N)-1; j++){\n dp[mask|j][i]=min(dp[mask|j][i], dp[j][y]+low+(i==0?mex[mask|j]:0));\n }\n mask|=(a[y]<=N?(1<<(a[y]-1)):0);\n low2 = 1e9;\n for(int j=0; j<(1<<N)-1; j++)low2=min(low2, dp[j][y]+mex[j]+(i==0?mex[mask]:0));\n dp[mask][i]=min(dp[mask][i], low+low2);\n }\n}\nvoid solve(){\n int n; cin >> n;\n vector<vector<int>> e(n);\n vector<int> a(n);\n for(int& i: a)cin >> i;\n for(int i=1, u; i<n; i++){\n cin >> u;\n e[u-1].push_back(i);\n }\n vector<vector<int>> dp((1<<N), vector(n, 0));\n dfs(0, e, a, dp);\n int ans = 1e9;\n for(int i=0; i<(1<<N)-1; i++)ans=min(ans, dp[i][0]);\n cout << ans << '\\n';\n}\nint main()\n{\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n for(int i=0; i<(1<<N)-1; i++){\n mex[i]++;\n if(i&1){\n mex[i]++;\n if(i&2){\n mex[i]++;\n if(i&4)mex[i]++;\n }\n }\n }\n int t; 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": "1981F",
"submission_id": 263502496,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#pragma GCC optimize(\"O3,unroll-loops\")\n#pragma GCC target(\"avx2,bmi,bmi2,lzcnt,popcnt\")\nusing namespace std;\nconst int N = 1000;\nvoid dfs(int i, vector<vector<int>>& e, vector<int>& a, vector<vector<int>>& dp){\n for(int j: e[i])dfs(j, e, a, dp);\n if(e[i].size()==1){\n int x = e[i].back();\n int low = 1e9;\n for(int j=1; j<=N; j++){\n if(a[i]==j)continue;\n dp[j][i]=dp[j][x]+(i==0?j:0);\n low=min(dp[j][x]+j, low);\n }\n int mex = 1;\n if(a[i]==mex||a[x]==mex)mex++;\n if(a[i]==mex||a[x]==mex)mex++;\n for(int j=1; j<=N; j++)low=min(dp[j][x]+j+mex, low);\n for(int j=1; j<=N; j++){\n if(j==a[i])continue;\n dp[j][i]=min(dp[j][i], low);\n }\n }\n else if(e[i].size()==0){\n for(int j=1; j<=N; j++){\n if(a[i]==j)continue;\n dp[j][i]=0;\n }\n }\n else{\n int x = e[i].back();\n e[i].pop_back();\n int y = e[i].back();\n int v = 1e9;\n for(int j=1; j<=N; j++){\n if(a[i]==j)continue;\n v=min(v, dp[j][x]+dp[j][y]+j);\n }\n int low = 1e9;\n for(int j=1; j<=N; j++)low=min(low, dp[j][x]+j);\n for(int j=1; j<=N; j++){\n if(a[i]==j||a[x]==j)continue;\n v=min(v, low+dp[j][y]+j);\n }\n int mex = 1;\n if(a[i]==mex||a[x]==mex)mex++;\n if(a[i]==mex||a[x]==mex)mex++;\n low = 1e9;\n for(int j=1; j<=N; j++){\n if(a[i]==j)continue;\n low=min(low, dp[j][x]+j);\n }\n for(int j=a[i]; j<=min(N, a[i]); j++)low=min(low, dp[j][x]+j+mex);\n for(int j=1; j<=N; j++){\n if(j==a[i])continue;\n dp[j][i]=min(dp[j][i], dp[j][y]+low+(i==0?j:0));\n }\n mex = 1;\n if(a[i]==mex||a[y]==mex)mex++;\n if(a[i]==mex||a[y]==mex)mex++;\n for(int j=1; j<=N; j++)v=min(v, dp[j][y]+low+mex+(i==0?j:0));\n swap(x, y);\n for(int j=1; j<=N; j++){\n if(a[i]==j)continue;\n v=min(v, dp[j][x]+dp[j][y]+j);\n }\n low = 1e9;\n for(int j=1; j<=N; j++)low=min(low, dp[j][x]+j);\n for(int j=1; j<=N; j++){\n if(a[i]==j||a[x]==j)continue;\n v=min(v, low+dp[j][y]+j);\n }\n mex = 1;\n if(a[i]==mex||a[x]==mex)mex++;\n if(a[i]==mex||a[x]==mex)mex++;\n low = 1e9;\n for(int j=1; j<=N; j++){\n if(a[i]==j)continue;\n low=min(low, dp[j][x]+j);\n }\n for(int j=a[i]; j<=min(N, a[i]); j++)low=min(low, dp[j][x]+j+mex);\n for(int j=1; j<=N; j++){\n if(j==a[i])continue;\n dp[j][i]=min(dp[j][i], dp[j][y]+low+(i==0?j:0));\n }\n mex = 1;\n if(a[i]==mex||a[y]==mex)mex++;\n if(a[i]==mex||a[y]==mex)mex++;\n for(int j=1; j<=N; j++)v=min(v, dp[j][y]+low+mex+(i==0?j:0));\n for(int j=1; j<=N; j++){\n if(j==a[i])continue;\n dp[j][i]=min(dp[j][i], v);\n }\n }\n}\nvoid solve(){\n int n; cin >> n;\n vector<vector<int>> e(n);\n vector<int> a(n);\n for(int& i: a)cin >> i;\n for(int i=1, u; i<n; i++){\n cin >> u;\n e[u-1].push_back(i);\n }\n vector<vector<int>> dp(N+1, vector(n, (int)1e9));\n dfs(0, e, a, dp);\n int ans = 1e9;\n for(int i=1; i<=N; i++)ans=min(ans, dp[i][0]);\n cout << ans << '\\n';\n}\nint main()\n{\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n int t; 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": "1981F",
"submission_id": 263516975,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\ndef solve():\n n = int(input())\n a = list(map(int, input().split()))\n p = list(map(int, input().split()))\n m = min(n + 1, 4000)\n dp = [[float('inf') for _ in range(m)] for _ in range(n + 1)]\n graph = {}\n for i in range(1, n + 1):\n graph[i] = []\n \n for i in range(n - 1):\n graph[p[i]].append(i + 2)\n \n temp = [i for i in graph if len(graph[i]) == 0]\n visited = [False for _ in range(n + 1)]\n i = 0\n while i < len(temp):\n index = temp[i]\n if visited[index]:\n i += 1\n continue\n if len(graph[index]) == 0:\n for j in range(1, m):\n if a[index - 1] != j:\n dp[index][j] = 0\n elif len(graph[index]) == 1:\n v = graph[index][0]\n min_v = float('inf')\n for j in range(1, m):\n if a[index - 1] != j:\n min_v = min(dp[v][j] + j, min_v)\n for j in range(1, m):\n if a[index - 1] != j:\n if index == 1:\n dp[index][j] = min(dp[index][j], min_v)\n else:\n dp[index][j] = min(dp[index][j], dp[v][j], min_v)\n \n else:\n v1, v2 = graph[index][0], graph[index][1]\n if not visited[v2] or not visited[v1]:\n i += 1\n continue\n min_v1 = float('inf')\n min_v2 = float('inf')\n min_v3 = float('inf')\n for j in range(1, m):\n if a[index - 1] != j:\n min_v1 = min(dp[v1][j] + j, min_v1)\n min_v2 = min(dp[v2][j] + j, min_v2)\n min_v3 = min(min_v3, dp[v1][j] + dp[v2][j] + j)\n \n for j in range(1, m):\n if a[index - 1] != j:\n if index == 1:\n dp[index][j] = min(dp[index][j], min_v3,\n min_v1 + min_v2)\n else:\n dp[index][j] = min(dp[index][j], min_v1 + dp[v2][j], min_v2 + dp[v1][j], min_v3,\n min_v1 + min_v2)\n \n if index != 1:\n if not visited[p[index - 2]]:\n temp.append(p[index - 2])\n visited[index] = True\n i += 1\n \n ans = float('inf')\n for j in range(m):\n ans = min(ans, dp[1][j])\n \n return ans\n \n \nt = int(input())\nfor _ in range(t):\n print(solve())\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1981F",
"submission_id": 268801008,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\npublic class turtle {\nstatic class Tree {\nList<Integer>[] adj;\n int[] a;\n int[] parent;\n int n;\n public Tree(int n, int[] a, int[] p) {\n this.n = n;\n this.a = new int[n + 1];\n this.parent = new int[n + 1];\n this.adj = new ArrayList[n + 1];\n for (int i = 1; i <= n; i++) {\n adj[i] = new ArrayList<>();\n }\n System.arraycopy(a, 0, this.a, 1, n);\n for (int i = 2; i <= n; i++) {\n int par = p[i - 2];\n adj[par].add(i);\n parent[i] = par;\n }\n }\n private int computeMex(Set<Integer> setValues) {\n int mex = 1;\n while (setValues.contains(mex)) {\n mex++;\n }\n return mex;\n }\n public int calculateMexForPath(int x, int y) {\n Set<Integer> pathValues = new HashSet<>();\n while (x != y) {\n if (x > y) {\n pathValues.add(a[x]);\n x = parent[x];\n } else {\n pathValues.add(a[y]);\n y = parent[y];\n }\n }\n pathValues.add(a[x]);\n return computeMex(pathValues);\n }\n public int getMinimumValueForGoodPaths() {\n Set<Integer> uniqueMexValues = new HashSet<>();\n Queue<Integer> queue = new LinkedList<>();\n boolean[] visited = new boolean[n + 1];\n queue.add(1);\n visited[1] = true;\n while (!queue.isEmpty()) {\n int current = queue.poll();\n for (int child : adj[current]) {\n if (!visited[child]) {\n visited[child] = true;\n int mexValue = calculateMexForPath(current, child);\n uniqueMexValues.add(mexValue);\n queue.add(child);\n }\n }\n }\n int totalMex = 0;\n for (int mexValue : uniqueMexValues) {\n totalMex += mexValue;\n }\n if (n > 9) {\n totalMex += 1;\n }\n return totalMex;\n }\n}\npublic static void main(String[] args) throws IOException {\nBufferedReader reader = new BufferedReader(new InputStreamReader(System.in));\n PrintWriter writer = new PrintWriter(new OutputStreamWriter(System.out), true);\n int t = Integer.parseInt(reader.readLine().trim());\n for (int i = 0; i < t; i++) {\n int n = Integer.parseInt(reader.readLine().trim());\n int[] a = Arrays.stream(reader.readLine().trim().split(\" \")).mapToInt(Integer::parseInt).toArray();\n int[] p = Arrays.stream(reader.readLine().trim().split(\" \")).mapToInt(Integer::parseInt).toArray();\n Tree tree = new Tree(n, a, p);\n int result = tree.getMinimumValueForGoodPaths();\n writer.println(result);\n writer.flush();\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": "1981F",
"submission_id": 265911407,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define Yzl unsigned long long\n#define pb emplace_back \ntypedef long long LL;\n \nusing namespace std;\n \nconst Yzl Lty=20120712;\n \nconst int MAXN=2.5e4+10;\nint n;\nint Fa[MAXN],a[MAXN];\nvector<int> e[MAXN];\nvoid add(int f,int t) {\n\te[f].pb(t);\n}\nconst int N=5010;\nconst LL inf=1e18;\nLL f[N][N],g[N];\nLL mex(LL x,LL y) {\n\tif(x>y) swap(x,y);\n\tif(x==1) {\n\t\tif(y==2) return 3;\n\t\treturn 2;\n\t} return 1;\n}\nvoid dfs(int u,int fa) {\n\tfor(auto t:e[u]) dfs(t,u);\n\tg[u]=inf;\n\tif(e[u].size()==1) {\n\t\tfor(int j=0;j<=n;++j) f[u][j]=f[e[u][0]][j];\n\t} else if(e[u].size()==2) {\n\t\tmemset(f[u],127,sizeof(f[u]));\n\t\tint ls=e[u][0],rs=e[u][1];\n\t\tfor(int j=0;j<=n;++j) if(j+1!=a[u]) {\n\t\t\tg[u]=min(g[u],f[ls][j]+f[rs][j]+j+1);\n//\t\t\tif(u==2&&j==2) cout<<g[u]<<' ';\n//\t\t\tif(u==2&&j==2) {\n//\t\t\t\tcout<<f[ls][j]<<' '<<f[rs][j]<<endl;\n//\t\t\t}\n\t\t}\n\t\tLL m1=inf,m2=inf;\n\t\tfor(int j=0;j<=n;++j) {\n\t\t\tif(a[u]==j+1) m1=min(m1,f[ls][j]+j+1+mex(a[u],a[ls]));\n\t\t\telse m1=min(m1,f[ls][j]+j+1);\n\t\t}\n\t\tfor(int j=0;j<=n;++j) {\n\t\t\tif(a[u]==j+1) m2=min(m2,f[ls][j]+j+1+mex(a[u],a[rs]));\n\t\t\telse m2=min(m2,f[rs][j]+j+1);\n\t\t}\n\t\tfor(int j=0;j<=n;++j) {\n\t\t\tf[u][j]=min(f[u][j],f[ls][j]+m2);\n\t\t\tf[u][j]=min(f[u][j],f[rs][j]+m1);\n\t\t}\n\t}\n\tfor(int j=0;j<=n;++j) if(j+1!=a[u]) g[u]=min(g[u],f[u][j]+j+1);\n\tfor(int j=0;j<=n;++j) if(j+1!=a[u]) f[u][j]=min(f[u][j],g[u]);\n\tf[u][a[u]-1]=inf;\n//\tcout<<u<<\":\"<<g[u]<<\":\";\n//\tfor(int i=0;i<=n;++i) {\n//\t\tcout<<f[u][i]<<' ';\n//\t} puts(\"\");\n}\nvoid vmain() {\n\tscanf(\"%d\",&n);\n\tfor(int i=1;i<=n;++i) {\n\t\te[i].clear();\n\t}\n\tfor(int i=1;i<=n;++i) scanf(\"%d\",&a[i]),a[i]=min(a[i],n+1);\n\tfor(int i=2;i<=n;++i) scanf(\"%d\",&Fa[i]),add(Fa[i],i);\n\tmemset(f,0,sizeof(f)); dfs(1,0);\n\tprintf(\"%lld\\n\",g[1]);\n}\nint main () {\n\tint T=1; scanf(\"%d\",&T); while(T--) vmain();\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": "1981F",
"submission_id": 281986566,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define ll long long\n#define pn putchar('\\n')\n#define mset(a,x) memset(a,x,sizeof a)\n#define mcpy(a,b) memcpy(a,b,sizeof a)\n#define fls() fflush(stdout)\n#define maxn 25005\n#define ls(x) smt[x].ls\n#define rs(x) smt[x].rs\n#define int ll\n#define codeforces code_in_queue_forces\nusing namespace std;\nint re()\n{\n int x=0;\n bool t=1;\n char ch=getchar();\n while(ch>'9'||ch<'0')\n t=ch=='-'?0:t,ch=getchar();\n while(ch>='0'&&ch<='9')\n x=(x<<1)+(x<<3)+(ch^48),ch=getchar();\n return t?x:-x;\n}\nstruct Tag\n{\n int a,b;\n Tag operator + (Tag t)\n {\n return {a+t.a,min(b+t.a,t.b)};\n }\n};\nstruct Segtree\n{\n int mn,ls,rs;\n Tag tag;\n}smt[10000000];\nint n,idx;\nint a[maxn],lef[maxn],rig[maxn],ro[maxn];\nbool isleaf(int x)\n{\n return !ls(x)&&!rs(x);\n}\nint newp(int l,int r)\n{\n idx++;\n smt[idx]={l,0,0,{0,(int)1e9}};\n return idx;\n}\nvoid add(int& pos,int l,int r,Tag tag)\n{\n if(!pos)\n pos=newp(l,r);\n smt[pos].mn=min(smt[pos].mn+tag.a,tag.b+l);\n smt[pos].tag=smt[pos].tag+tag;\n}\nvoid pushtag(int pos,int l,int r)\n{\n int mid=l+r>>1;\n add(ls(pos),l,mid,smt[pos].tag);\n add(rs(pos),mid+1,r,smt[pos].tag);\n smt[pos].tag={0,(int)1e9};\n}\nvoid update(int pos,int l,int r)\n{\n int mid=l+r>>1;\n smt[pos].mn=min(ls(pos)?smt[ls(pos)].mn:l,rs(pos)?smt[rs(pos)].mn:mid+1);\n}\nint fi(int x,int y,int l,int r)\n{\n if(isleaf(x))\n swap(x,y);\n if(isleaf(y))\n return min(smt[y].tag.a,smt[y].tag.b)+smt[x].mn;\n pushtag(x,l,r);\n pushtag(y,l,r);\n int mid=l+r>>1;\n return min(fi(ls(x),ls(y),l,mid),fi(rs(x),rs(y),mid+1,r));\n}\nint mer(int x,int y,int l,int r)\n{\n if(isleaf(x))\n swap(x,y);\n if(isleaf(y))\n {\n add(x,l,r,{0,min(smt[y].tag.a,smt[y].tag.b)});\n return x;\n }\n pushtag(x,l,r);\n pushtag(y,l,r);\n int mid=l+r>>1;\n ls(x)=mer(ls(x),ls(y),l,mid);\n rs(x)=mer(rs(x),rs(y),mid+1,r);\n update(x,l,r);\n return x;\n}\nvoid del(int &pos,int l,int r,int x)\n{\n if(!pos)\n pos=newp(l,r);\n if(l==r)\n {\n smt[pos].mn=1e9;\n smt[pos].tag={(int)1e9,(int)1e9};\n return;\n }\n pushtag(pos,l,r);\n int mid=l+r>>1;\n if(x<=mid)\n del(ls(pos),l,mid,x);\n if(x>mid)\n del(rs(pos),mid+1,r,x);\n update(pos,l,r);\n}\nvoid print(int pos,int l,int r)\n{\n if(l==r)\n {\n printf(\"%lld \",smt[pos].mn);\n return;\n }\n pushtag(pos,l,r);\n int mid=l+r>>1;\n print(ls(pos),l,mid);\n print(rs(pos),mid+1,r);\n}\nvoid solve()\n{\n n=re();\n idx=0;\n for(int i=1;i<=n;i++)\n lef[i]=rig[i]=ro[i]=0;\n for(int i=1;i<=n;i++)\n a[i]=re();\n for(int i=2;i<=n;i++)\n {\n int x=re();\n if(lef[x])\n rig[x]=i;\n else\n lef[x]=i;\n }\n int ans=1e9;\n for(int i=n;i;i--)\n {\n int u=lef[i],v=rig[i];\n if(!u)\n ro[i]=newp(1,n);\n else if(!v)\n {\n if(a[i]<=n)\n del(ro[u],1,n,a[i]);\n ro[i]=ro[u];\n add(ro[i],1,n,{0,smt[ro[i]].mn});\n }\n else\n {\n if(a[i]<=n)\n {\n del(ro[u],1,n,a[i]);\n del(ro[v],1,n,a[i]);\n }\n int t1=smt[ro[u]].mn,t2=smt[ro[v]].mn;\n int mn=min(fi(ro[u],ro[v],1,n),t1+t2);\n // printf(\"%lld\\n\",mn);\n // printf(\"%lld %lld\\n\",t1,t2);\n add(ro[u],1,n,{t2,(int)1e9});\n add(ro[v],1,n,{t1,(int)1e9});\n ro[i]=mer(ro[u],ro[v],1,n);\n add(ro[i],1,n,{0,mn});\n if(i==1)\n ans=mn;\n }\n if(a[i]<=n)\n del(ro[i],1,n,a[i]);\n // print(ro[i],1,n);\n // pn;\n }\n printf(\"%lld\\n\",min(ans,smt[ro[1]].mn));\n}\nsigned main()\n{\n int T=re();\n while(T--)\n solve(); \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1981F",
"submission_id": 276738303,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nconst int maxn=2.5e4+5,maxk=4000;\nint a[maxn],n,ans;\nint f[maxn][4005];\nvector<int> nbr[maxn];\nvoid dfs(int cur)\n{\n for(int i=0;i<nbr[cur].size();i++) dfs(nbr[cur][i]);\n if(nbr[cur].size()==0)\n {\n for(int k=1;k<=maxk;k++) f[cur][k]=((a[cur]==k)?1e9:0);\n return ;\n }\n if(nbr[cur].size()==1)\n {\n int x=nbr[cur][0],mini=1e9;\n for(int k=1;k<=maxk;k++) mini=min(mini,f[x][k]+k);\n ans=mini;\n for(int k=1;k<=maxk;k++)\n {\n if(a[cur]==k){f[cur][k]=1e9;continue;}\n f[cur][k]=min(f[x][k],mini);\n }\n return ;\n }\n if(nbr[cur].size()==2)\n {\n int x=nbr[cur][0],y=nbr[cur][1],minix=1e9,miniy=1e9,mini=1e9;\n for(int k=1;k<=maxk;k++) \n if(a[cur]!=k) minix=min(minix,f[x][k]+k);\n for(int k=1;k<=maxk;k++)\n if(a[cur]!=k) miniy=min(miniy,f[y][k]+k);\n for(int k=1;k<=maxk;k++)\n if(a[cur]!=k) mini=min(mini,f[x][k]+f[y][k]+k);\n ans=min(mini,minix+miniy);\n for(int k=1;k<=maxk;k++)\n {\n if(a[cur]==k){f[cur][k]=1e9;continue;}\n f[cur][k]=min(f[x][k]+miniy,f[y][k]+minix);\n f[cur][k]=min(f[cur][k],ans);\n }\n }\n}\nvoid solve()\n{\n cin>>n,ans=1e9;\n for(int i=1;i<=n;i++) nbr[i].clear();\n for(int i=1;i<=n;i++) cin>>a[i];\n for(int i=2,fa;i<=n;i++) cin>>fa,nbr[fa].push_back(i);\n dfs(1);\n cout<<ans<<'\\n';\n}\nint main()\n{\n ios::sync_with_stdio(0);\n cin.tie(0);cout.tie(0);\n int t;\n cin>>t;\n while(t--) solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1981F",
"submission_id": 283170112,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n// Mtaylor\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define pb push_back\n#define fi first\n#define se second\n#define mp make_pair\n#define endl '\\n';\n#define sz(x) ((int)(x).size())\n#define all(x) (x).begin(), (x).end()\n#define rall(x) (x).rbegin(), (x).rend()\nvoid dbg_out() { cerr << endl; }\ntemplate <typename Head, typename... Tail>\nvoid dbg_out(Head H, Tail... T) {\n cerr << ' ' << H;\n dbg_out(T...);\n}\n#define dbg(...) cerr << \"(\" << #__VA_ARGS__ << \"):\", dbg_out(__VA_ARGS__)\ntypedef long long ll;\nconst int N = 3e5 + 5;\n \nint n, a[N];\nvector<int> adj[N];\n \nint tot = (1 << 5) - 1;\nint dp[N][1 << 5];\nint dp2[N];\nint val[1 << 5];\nvoid dfs(int u) {\n dp2[u] = 1e9;\n int cmsk = 0;\n if (a[u] <= 5) {\n cmsk += (1 << (a[u] - 1));\n }\n for (auto &v : adj[u]) {\n dfs(v);\n }\n fill(dp[u], dp[u] + tot, 1e9);\n if (adj[u].size() == 0) {\n dp[u][cmsk] = 0;\n dp2[u] = 0;\n } else {\n // for (auto &v : adj[u]) {\n // for (int i = 0; i < tot; i++) {\n // int msk = cmsk | i;\n // if (msk == tot) continue;\n // }\n // }\n if (adj[u].size() == 1) {\n int v = adj[u][0];\n for (int i = 0; i < tot; i++) {\n int msk = cmsk | i;\n if (msk != tot) {\n // dbg(u, v, i, dp[v][i], val[msk]);\n dp2[u] = min(dp2[u], val[msk] + dp[v][i]);\n dp[u][msk] = min(dp[u][msk], dp[v][i]);\n }\n }\n }\n if (adj[u].size() == 2) {\n int x = adj[u][0];\n int y = adj[u][1];\n for (int i = 0; i < tot; i++) {\n for (int j = 0; j < tot; j++) {\n int msk = cmsk | i | j;\n if (msk == tot) continue;\n dp2[u] = min(dp2[u], val[msk] + dp[x][i] + dp[y][j]);\n }\n }\n for (int i = 0; i < tot; i++) {\n for (int j = 0; j < tot; j++) {\n int mski = cmsk | i;\n int mskj = cmsk | j;\n if (mski != tot && mskj != tot) {\n dp[u][mskj] = min(dp[u][mskj], dp[x][i] + val[mski] + dp[y][j]);\n dp[u][mski] = min(dp[u][mski], dp[x][i] + val[mskj] + dp[y][j]);\n dp2[u] = min(dp2[u], dp[x][i] + dp[y][j] + val[mski] + val[mskj]);\n }\n }\n }\n }\n dp[u][cmsk] = min(dp[u][cmsk], dp2[u]);\n }\n // dbg(u, dp2[u]);\n // for (int i = 0; i < tot; i++) {\n // dbg(u + 1, i, dp[u][i]);\n // }\n}\n \nvoid test_case() {\n for (int i = 0; i < (1 << 5); i++) {\n for (int j = 0; j < 5; j++) {\n if (((1 << j) & i) == 0) {\n val[i] = j + 1;\n break;\n }\n }\n }\n cin >> n;\n for (int i = 0; i < n; i++) {\n adj[i].clear();\n cin >> a[i];\n }\n int p;\n for (int i = 1; i < n; i++) {\n cin >> p;\n --p;\n adj[p].pb(i);\n }\n dfs(0);\n cout << dp2[0] << endl;\n}\n \nint main() {\n#ifndef ONLINE_JUDGE\n freopen(\"in.txt\", \"r\", stdin);\n#endif\n ios::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n int T = 1;\n cin >> T;\n while (T--) {\n test_case();\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": "1981F",
"submission_id": 267238460,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long\nconst int INF = 1e18, MOD = 1e9 + 7;\nconst int mxN = 25007;\nint dp[4007];\narray<int,2> child[mxN];\nint a[mxN];\nint ndp[4007];\nint n;\nint dfs(int pos)\n{\n\t//cout << pos << \" \"; \n\tint x = INF, y = INF;\n\tint dp1[n + 4][2];\n\tint ans = INF;\n\tif(child[pos][0] != -1)\n\t{\n\t\tdfs(child[pos][0]);\n\t\tfor(int i = 1;n + 1 >= i;i++) dp1[i][0] = dp[i];\n\t\tfor(int i = 1;n + 1 >= i;i++)\n\t\t{\n\t\t\tif(a[pos] != i) x = min(x, dp1[i][0] + i);\n\t\t}\n\t} \n\tif(child[pos][1] != -1)\n\t{\n\t\tdfs(child[pos][1]);\n\t\tfor(int i = 1;n + 1 >= i;i++) dp1[i][1] = dp[i];\n\t\tfor(int i = 1;n + 1 >= i;i++)\n\t\t{\n\t\t\tif(a[pos] != i) y = min(y, dp1[i][1] + i),ans = min(ans, dp1[i][0] + dp1[i][1] + i);\n\t\t}\n\t\tans = min(ans, x + y);\n\t} \n\telse ans = min(ans, x);\n\tfor(int i = 1;n + 1 >= i;i++)\n\t{\n\t\tif(a[pos] != i)\n\t\t\tndp[i] = 0;\n\t\telse\n\t\t\tndp[i] = INF;\n\t}\n\tif(x == INF && y == INF)\n\t{\n\t\tfor(int i = 1;n + 1 >= i;i++) dp[i] = ndp[i];\n\t\t/*cout << pos << endl;\n\t\tfor(int i = 1;n >= i;i++) cout << dp[i] << \" \";\n\t\tcout << endl;*/\n\t\treturn 0;\n\t} \n \n\tfor(int i = 1;n + 1 >= i;i++)\n\t{\n\t\t//cout << i << \" \" << ndp[i] << endl;\n\t\tif(a[pos] == i) continue;\n\t//\tcout << i << endl;\n\t\tif(y == INF)\n\t\t{\n\t\t\tndp[i] = min(x, dp1[i][0]);\n\t\t\tans = min(ans,x);\n\t\t}\n\t\telse\n\t\t{\n\t\t\tndp[i] = min({dp1[i][0] + y, dp1[i][1] + x,ans});\n\t\t}\n\t}\n\tfor(int i = 1;n + 1 >= i;i++) dp[i] = ndp[i];\n\t/*cout << pos << endl;\n\tfor(int i = 1;n >= i;i++) cout << dp[i] << \" \";\n\tcout << endl;\n\tcout << \"SO: \" << ans << endl;*/\n\treturn ans;\n}\nvoid solve()\n{\n\t//int n;\n\tcin >> n;\n\tfor(int i = 0;n > i;i++) cin >> a[i];\n\tfor(int i = 0;n + 1 >= i;i++) child[i][0] = -1, child[i][1] = -1;\n\tfor(int i = 0;n - 1 > i;i++)\n\t{\n\t\tint x;\n\t\tcin >> x;\n\t\tx--;\n\t\tif(child[x][0] == -1) child[x][0] = i + 1;\n\t\telse child[x][1] = i + 1;\n\t}\n\tfor(int i = 0;n + 1 >= i;i++) ndp[i] = INF, dp[i] = INF;\n\tn = min(n, (int)(4000));\n\tcout << dfs(0) << endl;\n}\n \nint32_t main()\n{\n\tios_base::sync_with_stdio(false);\n\tcin.tie(NULL);\n\tcout.tie(NULL);\n\tint tt;\n\tcin >> tt;\n\twhile(tt--) solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1981F",
"submission_id": 275382714,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\ndef solve():\n n = int(input())\n a = list(map(int, input().split()))\n p = list(map(int, input().split()))\n m = min(n + 10, 4000)\n dp = [[float('inf') for _ in range(m)] for _ in range(n + 1)]\n graph = {}\n for i in range(1, n + 1):\n graph[i] = []\n \n for i in range(n - 1):\n graph[p[i]].append(i + 2)\n \n temp = [i for i in graph if len(graph[i]) == 0]\n visited = [False for _ in range(n + 1)]\n i = 0\n while i < len(temp):\n index = temp[i]\n if visited[index]:\n i += 1\n continue\n if len(graph[index]) == 0:\n for j in range(1, m):\n if a[index - 1] != j:\n dp[index][j] = 0\n elif len(graph[index]) == 1:\n v = graph[index][0]\n min_v = float('inf')\n for j in range(1, m):\n if a[index - 1] != j:\n min_v = min(dp[v][j] + j, min_v)\n for j in range(1, m):\n if a[index - 1] != j:\n if index == 1:\n dp[index][j] = min(dp[index][j], min_v)\n else:\n dp[index][j] = min(dp[index][j], dp[v][j], min_v)\n \n else:\n v1, v2 = graph[index][0], graph[index][1]\n if not visited[v2] or not visited[v1]:\n i += 1\n continue\n min_v1 = float('inf')\n min_v2 = float('inf')\n min_v3 = float('inf')\n for j in range(1, m):\n if a[index - 1] != j:\n min_v1 = min(dp[v1][j] + j, min_v1)\n min_v2 = min(dp[v2][j] + j, min_v2)\n min_v3 = min(min_v3, dp[v1][j] + dp[v2][j] + j)\n \n for j in range(1, m):\n if a[index - 1] != j:\n if index == 1:\n dp[index][j] = min(dp[index][j], min_v3,\n min_v1 + min_v2)\n else:\n dp[index][j] = min(dp[index][j], min_v1 + dp[v2][j], min_v2 + dp[v1][j], min_v3,\n min_v1 + min_v2)\n \n if index != 1:\n if not visited[p[index - 2]]:\n temp.append(p[index - 2])\n visited[index] = True\n i += 1\n ans = float('inf')\n for j in range(m):\n ans = min(ans, dp[1][j])\n \n return ans\n \n \nt = int(input())\nfor _ in range(t):\n print(solve())\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1981F",
"submission_id": 268801647,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\nimport sys\ninput = sys.stdin.read\n \nINF = 10**9\nSIZE = int(2.5e4 + 5)\n \ndef dfs(pos, sz, a, to):\n DP = [INF] * (sz + 1)\n if to[pos][0] == 0:\n for i in range(1, sz + 1):\n if i != a[pos]:\n DP[i] = 0\n DP[0] = 0\n elif to[pos][1] == 0:\n x = to[pos][0]\n vx = INF\n dpx = dfs(x, sz, a, to)\n for i in range(1, sz + 1):\n if i != a[pos]:\n vx = min(vx, dpx[i] + i)\n for i in range(1, sz + 1):\n if i != a[pos]:\n DP[i] = min(dpx[i], vx)\n DP[0] = vx\n else:\n x, y = to[pos][0], to[pos][1]\n vx, vy, vxy = INF, INF, INF\n dpx = dfs(x, sz, a, to)\n dpy = dfs(y, sz, a, to)\n for i in range(1, sz + 1):\n if i != a[pos]:\n vx = min(vx, dpx[i] + i)\n vy = min(vy, dpy[i] + i)\n vxy = min(vxy, dpx[i] + dpy[i] + i)\n vxy = min(vxy, vx + vy)\n for i in range(1, sz + 1):\n if i != a[pos]:\n DP[i] = min(vxy, dpx[i] + vy, dpy[i] + vx)\n DP[0] = vxy\n return DP\n \ndef solve(data):\n idx = 0\n tt = int(data[idx])\n idx += 1\n results = []\n for _ in range(tt):\n n = int(data[idx])\n idx += 1\n a = [0] * (n + 1)\n to = [[0, 0] for _ in range(n + 1)]\n for i in range(1, n + 1):\n a[i] = int(data[idx])\n idx += 1\n for i in range(2, n + 1):\n pa = int(data[idx])\n idx += 1\n if to[pa][0] == 0:\n to[pa][0] = i\n else:\n to[pa][1] = i\n sz = min(n + 1, 4000)\n result = dfs(1, sz, a, to)[0]\n results.append(result)\n return results\n \nif __name__ == \"__main__\":\n data = input().split()\n results = solve(data)\n for result in results:\n print(result)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1981F",
"submission_id": 263784518,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\nfrom collections import defaultdict\nimport sys\nimport itertools\n \ninf = float('inf')\n \ndef dfs(u, m, a, G, f):\n if not G[u]:\n for i in range(1, m + 1):\n f[u][i] = inf if i == a[u] else 0\n return\n \n if len(G[u]) == 1:\n x = G[u][0]\n dfs(x, m, a, G, f)\n mn = inf\n for i in range(1, m + 1):\n if i != a[u]:\n mn = min(mn, f[x][i] + i)\n if u == 1:\n print(mn)\n return\n for i in range(1, m + 1):\n f[u][i] = inf if i == a[u] else min(f[x][i], mn)\n return\n \n x, y = G[u][0], G[u][1]\n mnx, mny, k = inf, inf, inf\n dfs(x, m, a, G, f)\n dfs(y, m, a, G, f)\n for i in range(1, m + 1):\n if i != a[u]:\n mnx = min(mnx, f[x][i] + i)\n mny = min(mny, f[y][i] + i)\n k = min(k, f[x][i] + f[y][i] + i)\n k = min(k, mnx + mny)\n if u == 1:\n print(k)\n return\n for i in range(1, m + 1):\n f[u][i] = inf if i == a[u] else min(f[x][i] + mny, f[y][i] + mnx, k)\n \ndef solve():\n input = sys.stdin.read\n data = input().split()\n idx = 0\n T = int(data[idx])\n idx += 1\n results = []\n for _ in range(T):\n n = int(data[idx])\n idx += 1\n m = min(n + 1, 3863)\n a = [0] * (n + 1)\n G = defaultdict(list)\n f = [[0] * (m + 1) for _ in range(n + 1)]\n for i in range(1, n + 1):\n a[i] = int(data[idx])\n idx += 1\n for i in range(2, n + 1):\n p = int(data[idx])\n idx += 1\n G[p].append(i)\n dfs(1, m, a, G, f)\n \nif __name__ == \"__main__\":\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": "1981F",
"submission_id": 263784184,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define int long long\nconst int INF = numeric_limits<int>::max() / 20;\n \nint32_t main() {\n ios_base::sync_with_stdio(false);\n cin.tie(nullptr);\n \n int tn = 1;\n cin >> tn;\n while (tn--) {\n int n;\n cin >> n;\n vector<int> a(n), p(n - 1);\n for (auto v : {&a, &p}) {\n for (auto& x : *v) cin >> x;\n }\n vector<vector<int>> g(n);\n for (int i = 1; i < n; ++i) {\n g[p[i - 1] - 1].push_back(i);\n }\n \n int maxMex = min(n + 2, (int)4000);\n \n vector<vector<int>> memo(n, vector<int>(maxMex, -1));\n vector<int> mnConn(n, -1);\n vector<int> mnMemo(n, -1);\n vector<int> mnMemo2(n, -1);\n vector<int> mn(n, -1);\n function<int(int, int)> dp;\n auto getConn = [&](int i) {\n if (mnConn[i] != -1) return mnConn[i];\n mnConn[i] = INF;\n for (int j = 1; j < maxMex; ++j) {\n if (j == a[i]) continue;\n int res = 0;\n for (auto child : g[i]) {\n res += dp(child, j);\n }\n mnConn[i] = min(mnConn[i], res + j);\n }\n return mnConn[i];\n };\n auto getMn = [&](int i, int mex) {\n if (mnMemo[i] != -1) return (mex == mn[i]) ? mnMemo2[i] : mnMemo[i];\n mnMemo[i] = INF;\n for (int j = 1; j < maxMex; ++j) {\n if (dp(i, j) + j < mnMemo[i]) {\n mnMemo[i] = dp(i, j) + j;\n mn[i] = j;\n }\n }\n mnMemo2[i] = INF;\n for (int j = 1; j < maxMex; ++j) {\n if (j == mn[i]) continue;\n mnMemo2[i] = min(mnMemo2[i], dp(i, j) + j);\n }\n return (mex == mn[i]) ? mnMemo2[i] : mnMemo[i];\n };\n dp = [&](int i, int mex) {\n if (a[i] == mex) return INF;\n if (g[i].empty()) return (int)0;\n if (memo[i][mex] != -1) return memo[i][mex];\n auto& res = memo[i][mex];\n res = INF;\n if (g[i].size() == 1) {\n res = min(res, getMn(g[i][0], a[i]));\n res = min(res, dp(g[i][0], mex));\n } else {\n res = min(res, getMn(g[i][0], a[i]) + getMn(g[i][1], a[i]));\n res = min(res, getMn(g[i][0], a[i]) + dp(g[i][1], mex));\n res = min(res, dp(g[i][0], mex) + getMn(g[i][1], a[i]));\n res = min(res, getConn(i));\n }\n return res;\n };\n \n int res = min(getConn(0), getMn(0, 0));\n cout << res << 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": "1981F",
"submission_id": 266410775,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define int long long\n#define sqr(x) ((x) * (x))\n#define MASK(x) (1LL << (x))\nusing namespace std;\nconst int inf = 1e18;\n \nint32_t main() {\n ios_base::sync_with_stdio(false);\n cin.tie(nullptr);\n #ifdef LOCAL\n freopen(\"input.txt\", \"r\", stdin);\n freopen(\"output.txt\", \"w\", stdout);\n #endif\n int t;\n cin >> t;\n while(t--) {\n int n;\n cin >> n;\n vector<int> a(n+1);\n for(int i = 1; i <= n; i++) {\n cin >> a[i];\n }\n vector<vector<int>> adj(n+1);\n for(int i = 2; i <= n; i++) {\n int p;\n cin >> p;\n adj[p].push_back(i);\n }\n vector<int> sz(n+1);\n function<void(int)> dfs_sz = [&] (int u) {\n sz[u] = 1;\n for(int v : adj[u]) {\n dfs_sz(v);\n sz[u] += sz[v];\n }\n if(adj[u].size() == 2) {\n int v1 = adj[u][0];\n int v2 = adj[u][1];\n if(sz[v1] > sz[v2]) {\n swap(v1, v2);\n }\n }\n };\n dfs_sz(1);\n vector<vector<int>> dp(n+1);\n function<void(int)> dfs = [&] (int u) {\n for(int v : adj[u]) {\n dfs(v);\n }\n dp[u] = vector<int>(n+2);\n if(adj[u].size() == 1) {\n int v = adj[u][0];\n int dp_min = inf;\n for(int i = 1; i <= n+1; i++) {\n if(a[u] != i) {\n dp_min = min(dp_min, dp[v][i]);\n }\n }\n for(int i = 1; i <= n+1; i++) {\n if(a[u] != i) {\n if(u != 1)\n dp[u][i] = min(dp[v][i], i + dp_min);\n else\n dp[u][i] = min(dp[v][i], dp_min);\n }\n else {\n dp[u][i] = inf;\n }\n }\n }\n else if(adj[u].size() == 2) {\n int v1 = adj[u][0];\n int v2 = adj[u][1];\n int dp1 = inf;\n int dp2 = inf;\n int dp3 = inf;\n for(int i = 1; i <= n+1; i++) {\n if(a[u] != i) {\n dp1 = min(dp1, dp[v1][i]);\n dp2 = min(dp2, dp[v2][i]);\n dp3 = min(dp3, dp[v1][i] + dp[v2][i] - i);\n }\n }\n for(int i = 1; i <= n+1; i++) {\n if(a[u] != i) {\n if(u != 1) {\n dp[u][i] = inf;\n dp[u][i] = min(dp[u][i], dp3 + i);\n dp[u][i] = min(dp[u][i], dp[v2][i] + dp1);\n dp[u][i] = min(dp[u][i], dp[v1][i] + dp2);\n dp[u][i] = min(dp[u][i], i + dp1 + dp2);\n }\n else {\n dp[u][i] = inf;\n dp[u][i] = min(dp[u][i], dp3);\n dp[u][i] = min(dp[u][i], dp[v2][i] + dp1);\n dp[u][i] = min(dp[u][i], dp[v1][i] + dp2);\n dp[u][i] = min(dp[u][i], dp1 + dp2);\n }\n }\n else {\n dp[u][i] = inf;\n }\n }\n }\n else {\n for(int i = 1; i <= n+1; i++) {\n if(a[u] != i) {\n dp[u][i] = i;\n }\n else {\n dp[u][i] = inf;\n }\n }\n }\n for(int v : adj[u]) {\n dp[v].clear();\n }\n };\n dfs(1);\n int ans = inf;\n for(int i = 1; i <= n+1; i++) {\n ans = min(ans, dp[1][i]);\n }\n cout << ans << \"\\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": "1981F",
"submission_id": 264297980,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define ll long long\nconst int N=25005,M=250,inf=1e9;\nint t,n,a[N];\nint f[M][N],g[N];\nvector<int>v[N];\nvoid dfs2(int x){\n\tfor(int to:v[x]) dfs2(to);\n\tif(v[x].size()==0){\n\t\tfor(int i=1;i<M;i++) f[i][x]=0;\n\t\tf[a[x]][x]=inf;g[x]=0;\n\t\treturn ;\n\t}\n\tif(v[x].size()==1){\n\t\tint son0=v[x][0];\n\t\tf[a[x]][x]=inf;g[x]=inf;\n\t\tfor(int i=1;i<M;i++)\n\t\t\tif(i!=a[x]){\n\t\t\t\tf[i][x]=f[i][son0];\n\t\t\t\tg[x]=min(g[x],f[i][son0]+i);\n\t\t\t}\n\t\tfor(int i=1;i<M;i++)\n\t\t\tif(i!=a[x]) f[i][x]=min(f[i][x],g[x]);\n\t\treturn ;\n\t}\n\tint son0=v[x][0],son1=v[x][1],v0=inf,v1=inf;\n\tf[a[x]][x]=inf;g[x]=inf;\n\tfor(int i=1;i<M;i++)\n\t\tif(i!=a[x]){\n\t\t\tv0=min(v0,f[i][son0]+i);\n\t\t\tv1=min(v1,f[i][son1]+i);\n\t\t\tg[x]=min(g[x],f[i][son0]+f[i][son1]+i);\n\t\t}\n\tg[x]=min(g[x],v0+v1);\n\tfor(int i=1;i<M;i++)\n\t\tif(i!=a[x]) f[i][x]=g[x];\n\tfor(int i=1;i<M;i++)\n\t\tif(i!=a[x]){\n\t\t\tf[i][x]=min(f[i][x],f[i][son0]+v1);\n\t\t\tf[i][x]=min(f[i][x],f[i][son1]+v0);\n\t\t}\n}\nint main(){\n\tscanf(\"%d\",&t);\n\twhile(t--){\n\t\tscanf(\"%d\",&n);\n\t\tfor(int i=1;i<=n;i++) scanf(\"%d\",&a[i]);\n\t\tfor(int i=2,x;i<=n;i++){\n\t\t\tscanf(\"%d\",&x);\n\t\t\tv[x].push_back(i);\n\t\t}\n\t\tdfs2(1);\n\t\tprintf(\"%d\\n\",g[1]);\n\t\tfor(int i=1;i<=n;i++) v[i].clear();\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": "1981F",
"submission_id": 266183209,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\nfrom collections import defaultdict, deque\nimport sys\ninput = sys.stdin.read\nsys.setrecursionlimit(30000)\n \ninf = float('inf')\n \ndef dfs(u, m, a, G, f):\n if not G[u]:\n for i in range(1, m + 1):\n f[u][i] = inf if i == a[u] else 0\n return\n \n if len(G[u]) == 1:\n x = G[u][0]\n dfs(x, m, a, G, f)\n mn = inf\n for i in range(1, m + 1):\n if i != a[u]:\n mn = min(mn, f[x][i] + i)\n if u == 1:\n print(mn)\n return\n for i in range(1, m + 1):\n f[u][i] = inf if i == a[u] else min(f[x][i], mn)\n return\n \n x, y = G[u][0], G[u][1]\n mnx, mny, k = inf, inf, inf\n dfs(x, m, a, G, f)\n dfs(y, m, a, G, f)\n for i in range(1, m + 1):\n if i != a[u]:\n mnx = min(mnx, f[x][i] + i)\n mny = min(mny, f[y][i] + i)\n k = min(k, f[x][i] + f[y][i] + i)\n k = min(k, mnx + mny)\n if u == 1:\n print(k)\n return\n for i in range(1, m + 1):\n f[u][i] = inf if i == a[u] else min(f[x][i] + mny, f[y][i] + mnx, k)\n \ndef solve():\n data = input().split()\n idx = 0\n T = int(data[idx])\n idx += 1\n results = []\n for _ in range(T):\n n = int(data[idx])\n idx += 1\n m = min(n + 1, 3863)\n a = [0] * (n + 1)\n G = defaultdict(list)\n f = [[0] * (m + 1) for _ in range(n + 1)]\n for i in range(1, n + 1):\n a[i] = int(data[idx])\n idx += 1\n G[i] = []\n for i in range(2, n + 1):\n p = int(data[idx])\n idx += 1\n G[p].append(i)\n dfs(1, m, a, G, f)\n for result in results:\n print(result)\n \nif __name__ == \"__main__\":\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": "1981F",
"submission_id": 263784017,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\ndef solve():\n import sys\n input = sys.stdin.read\n data = input().split()\n \n index = 0\n t = int(data[index])\n index += 1\n \n results = []\n \n for _ in range(t):\n n = int(data[index])\n index += 1\n \n a = list(map(int, data[index:index + n]))\n index += n\n \n p = [-1] * n\n for i in range(1, n):\n p[i] = int(data[index]) - 1\n index += 1\n \n adj = [[] for _ in range(n)]\n for i in range(1, n):\n adj[p[i]].append(i)\n \n lo, hi = 1, n + 1\n while lo < hi:\n m = (lo + hi + 1) // 2\n ans = 0\n ans += (m - 1) // 3 + 1\n for k in range(2, (m + 1) // 2 + 1):\n ans += (m - k) // k\n if ans <= n:\n lo = m\n else:\n hi = m - 1\n \n m = lo\n inf = int(1E9)\n dp = [[0] * (m + 1) for _ in range(n)]\n \n def get(x):\n ans = inf\n for i in range(1, m + 1):\n ans = min(ans, dp[x][i] + i)\n return ans\n \n ans = inf\n \n def dfs(x):\n nonlocal ans\n if len(adj[x]) == 0:\n if a[x] <= m:\n dp[x][a[x]] = inf\n elif len(adj[x]) == 1:\n y = adj[x][0]\n dfs(y)\n if a[x] <= m:\n dp[y][a[x]] = inf\n gy = get(y)\n for i in range(1, m + 1):\n if i == a[x]:\n dp[x][i] = inf\n else:\n dp[x][i] = min(gy, dp[y][i])\n else:\n y = adj[x][0]\n z = adj[x][1]\n dfs(y)\n dfs(z)\n if a[x] <= m:\n dp[y][a[x]] = inf\n dp[z][a[x]] = inf\n gy = get(y)\n gz = get(z)\n gyz = inf\n for i in range(1, m + 1):\n gyz = min(gyz, dp[y][i] + dp[z][i] + i)\n for i in range(1, m + 1):\n if i == a[x]:\n dp[x][i] = inf\n else:\n dp[x][i] = min(gy + gz, dp[y][i] + gz, gy + dp[z][i], gyz)\n if x == 0:\n ans = min(ans, gyz)\n \n dfs(0)\n \n ans = min(ans, get(0))\n results.append(ans)\n \n sys.stdout.write(\"\\n\".join(map(str, results)) + \"\\n\")\n \nif __name__ == \"__main__\":\n import sys\n input = sys.stdin.read\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": "1981F",
"submission_id": 263625205,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n//RCh - F\n#include<bits/stdc++.h>\n#define INF LONG_LONG_MAX\n#define inf 1e9\n#define bctz __builtin_ctz\n#define bclz __builtin_clz\n#define bpt __builtin_popcount\n#define pb push_back\n#define x first\n#define y second\n \nusing namespace std;\n \ntypedef long long INT;\ntypedef pair<int, int> pii;\ntypedef pair<int, pii> pi3;\n \ntemplate <class T> inline void smax(T &a, T b){if(a < b)a = b;}\ntemplate <class T> inline void smin(T &a, T b){if(a > b)a = b;}\n \nconst int NN = 3e4 + 10;\nconst int mod = 998244353;\n \nint A[NN], B[NN], C[NN], ans[NN];\nint pa[NN], ls[NN], rs[NN];\nint n;\n \nvector <int> dfs(int u){\n\tans[u] = inf;\n\tvector <int> dp(n + 2, 0);\n\tif(ls[u] == 0){\n\t\tdp[A[u]] = inf;\n\t\treturn dp;\n\t}\n\tfor(int i = 0; i <= n + 1; i++) dp[i] = inf;\n\tif(rs[u] == 0){\n\t\tdp = dfs(ls[u]);\n\t\tdp[A[u]] = inf;\n\t\tfor(int i = 1; i <= n + 1; i++) smin(ans[u], i + dp[i]);\n\t\tfor(int i = 1; i <= n + 1; i++){\n\t\t\tif(i == A[u]) continue;\n\t\t\tsmin(dp[i], ans[u]);\n\t\t}\n\t\treturn dp;\n\t}\n\tvector <int> dpL = dfs(ls[u]);\n\tvector <int> dpR = dfs(rs[u]);\n\tint mnL = inf, mnR = inf, mn = inf;\n\tfor(int i = 1; i <= n + 1; i++){\n\t\tif(i == A[u]) continue;\n\t\tsmin(mnL, dpL[i] + i);\n\t\tsmin(mnR, dpR[i] + i);\n\t\tsmin(mn, dpL[i] + dpR[i] + i);\n\t}\n\tfor(int i = 1; i <= n + 1; i++){\n\t\tif(i == A[u]) continue;\n\t\tsmin(ans[u], mn);\n\t\tsmin(ans[u], mnL + mnR);\n\t\tsmin(dp[i], dpL[i] + mnR);\n\t\tsmin(dp[i], dpR[i] + mnL);\n\t\tsmin(dp[i], mnL + mnR);\n\t\tsmin(dp[i], mn);\n\t}\n\treturn dp;\n}\n \nint main(){\n#ifndef ONLINE_JUDGE\n\tfreopen(\"in.in\", \"r\", stdin);\n\tfreopen(\"out.out\", \"w\", stdout);\n#endif\n\tint t;\n\tscanf(\"%d\", &t);\n\twhile(t--){\n\t\tscanf(\"%d\", &n);\n\t\tfor(int i = 1; i <= n; i++) scanf(\"%d\", A + i);\n\t\tfor(int i = 2; i <= n; i++) scanf(\"%d\", pa + i);\n\t\tfor(int i = 1; i <= n; i++) ls[i] = rs[i] = 0;\n\t\tfor(int i = 2; i <= n; i++){\n\t\t\tif(ls[pa[i]]) rs[pa[i]] = i;\n\t\t\telse ls[pa[i]] = i;\n\t\t}\n\t\tdfs(1);\n\t\tprintf(\"%d\\n\", ans[1]);\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": "1981F",
"submission_id": 264235361,
"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: Turtle and Paths on a Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Note the unusual definition of $\\text{MEX}$ in this problem.\n\nPiggy gave Turtle a binary tree$^{\\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \\ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.\n\nIf a set of paths $P = \\\\{(x_i, y_i)\\\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.\n\nTurtle defines the value of a set of paths as $\\sum\\limits_{(x, y) \\in P} f(x, y)$, where $f(x, y)$ denotes the $\\text{MEX}^{\\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).\n\nTurtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!\n\n$^{\\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.\n\n$^{\\ddagger}\\text{MEX}$ of a collection of integers $c_1, c_2, \\ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\\text{MEX}$ of $[2, 3]$ is $1$.\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 $n$ ($2 \\le n \\le 2.5 \\cdot 10^4$) — the number of vertices in the tree.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the sequence $a$.\n\nThe third line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\le p_i < i$) — the parent of each vertex in the tree.\n\nAdditional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum value over all good sets of paths.\n\nExamples:\ninput:\n5\n5\n3 2 2 1 1\n1 1 2 2\n5\n3 2 1 1 1\n1 1 2 2\n6\n1 2 1 2 1 3\n1 2 3 3 4\n7\n2 1 2 3 1 2 1\n1 1 2 2 3 3\n10\n1 2 2 1 4 2 3 1 2 1\n1 1 2 2 3 3 4 5 5\noutput:\n4\n6\n6\n6\n7\n\nNote: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:\n\n\n\nThe good set of paths with the minimum value is $\\\\{(2, 3), (4, 5)\\\\}$.\n\nNote that in this test case $\\\\{(4, 5)\\\\}$ and $\\\\{(3, 4), (4, 5)\\\\}$ are not good sets of paths because each edge should be covered exactly once.\n\nIn the second test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 2), (1, 3), (4, 5)\\\\}$.\n\nIn the third test case, the tree is as follows:\n\n\n\nThe set of good paths with the minimum value is $\\\\{(1, 6), (3, 5)\\\\}$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nnamespace SOL {\n \nusing i64 = long long;\nvoid debug(const char *msg, ...) {\n#ifdef CLESIP\n va_list arg;\n static char pbString[512];\n va_start(arg,msg);\n vsprintf(pbString,msg,arg);\n cerr << \"[DEBUG] \" << pbString << \"\\n\";\n va_end(arg);\n#endif \n}\ntemplate<typename T, typename L>\nbool chkmax(T &x, L y) { if (x < y) return x = y, true; return false; }\ntemplate<typename T, typename L>\nbool chkmin(T &x, L y) { if (y < x) return x = y, true; return false; }\n \nconst int N = 2.5e4 + 10;\nconst int K = 3900;\nconst int inf = 1e9;\nint n, f[N][K], a[N], lim;\nvector<int> G[N];\n \nvoid dfs(int x) {\n\tfor (int i = 1; i <= lim; i ++)\n\t\tf[x][i] = 0;\n\tif (a[x] <= lim)\n\t\tf[x][a[x]] = inf;\n\tfor (int y : G[x])\n\t\tdfs(y);\n\tif ((int) G[x].size() == 1) {\n\t\tint minx = inf;\n\t\tfor (int i = 1; i <= lim; i ++)\n\t\t\tif (i != a[x])\n\t\t\t\tchkmin(minx, f[G[x][0]][i] + i);\n\t\tfor (int i = 1; i <= lim; i ++)\n\t\t\tif (i != a[x])\n\t\t\t\tf[x][i] = min(minx, f[G[x][0]][i]);\n\t\tif (x == 1)\n\t\t\tcout << minx << \"\\n\";\n\t} else if ((int) G[x].size() == 2) {\n\t\tint minx = inf, miny = inf, minm = inf;\n\t\tfor (int i = 1; i <= lim; i ++) {\n\t\t\tif (i == a[x])\n\t\t\t\tcontinue;\n\t\t\tchkmin(minx, f[G[x][0]][i] + i);\n\t\t\tchkmin(miny, f[G[x][1]][i] + i);\n\t\t\tchkmin(minm, f[G[x][0]][i] + f[G[x][1]][i] + i);\n\t\t}\n\t\tfor (int i = 1; i <= lim; i ++)\n\t\t\tif (a[x] != i)\n\t\t\t\tf[x][i] = min({ f[G[x][0]][i] + miny, f[G[x][1]][i] + minx, minm, minx + miny });\n\t\tif (x == 1)\n\t\t\tcout << min(minm, minx + miny) << \"\\n\";\n\t} else if (x == 1)\n\t\tcout << \"0\\n\";\n}\n \nvoid solve() {\n\tcin >> n;\n\tfor (int i = 1; i <= n; i ++)\n\t\tG[i].clear();\n\tfor (int i = 1; i <= n; i ++)\n\t\tcin >> a[i];\n\tfor (int i = 2, p; i <= n; i ++) {\n\t\tcin >> p;\n\t\tG[p].push_back(i);\n\t}\n\tint l = 1, r = n; lim = n;\n\tauto calc = [&](int t) {\n\t\tint ret = 0;\n\t\tfor (int i = 1; i <= t; i ++) {\n\t\t\tif (i <= 3)\n\t\t\t\tret += (t + i + 3) / (i + 4) - 1;\n\t\t\telse\n\t\t\t\tret += (t + i + 2) / (i + 3) - 1;\n\t\t}\n\t\treturn ret;\n\t};\n\twhile (l <= r) {\n\t\tint mid = (l + r) >> 1;\n\t\tif (calc(mid) <= n) {\n\t\t\tlim = mid;\n\t\t\tl = mid + 1;\n\t\t} else {\n\t\t\tr = mid - 1;\n\t\t}\n\t}\n\tlim = n + 1;\n\tdfs(1);\n}\n \n}\nint main() {\n\tios::sync_with_stdio(false), cin.tie(0), cout.tie(0);\n\tint tt; cin >> tt;\n\twhile (tt --)\n\t\tSOL::solve();\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1981F",
"submission_id": 265489175,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp> // Common file \n#include <ext/pb_ds/tree_policy.hpp> \n#include <functional> // for less \n \n \nusing namespace __gnu_pbds; \nusing namespace std;\n \n \n \n//----------------------------DEBUG--------------------------------------------------------------------------\n// #include \"template.cpp\"\n//-----------------------------------------------------------------------------------------------------------\ntypedef tree<int, null_type, less<int>, rb_tree_tag, \n tree_order_statistics_node_update> \n ordered_set; \n \n \n//-------------------------MACROS-------------------------------------\n \n#define _GLIBCXX_DEBUG\n#define int long long\n#define FOR(i, a, b) for (int i = (a); i < (b); ++i)\n#define F0R(i, a) FOR(i, 0, a)\n#define scan(arr) for(auto &i:arr)cin>>i;\n#define scan_mp(arr,map) for(auto &i:arr){cin>>i;mp[i]++;}\n#define print(arr,n) for(int i=0;i<n;i++)cout<<arr[i]<<\" \\n\"[i==n-1];\n#define print2d(arr,n,m) FOR(i,0,n)FOR(j,0,m)cout<<arr[i][j]<<\" \\n\"[j==m-1];\n#define VI vector<int>\n#define VVI vector<VI>\n#define PII pair<int, int>\n#define SI set<int>\n#define MSI multiset<int>\n#define OS ordered_set\n#define MII map<int, int>\n#define all(x) (x).begin(), (x).end()\n#define lcm(a,b) a/__gcd(a,b)*b\n#define ff first\n#define ss second\n#define pb push_back\n#define mp make_pair\n#define setBits(x) __builtin_popcountll(x)\n \n \n \n//--------------------------CONSTANTS---------------------------------\n \n \n \nconst int N=3001;\nconst int M=1e9+7;\nconst int M2=998244353;\nconst int inf=1e18;\n \n \n \n \n \n//------------------------------------------------------------------------------------------------------------------------------------------------------//\n \nint binexp(int a,int b){\n if(b==0)return 1;\n int res=binexp(a,b/2);\n if(b&1)return (a*res*res)%M;\n else return (res*res)%M;\n}\n \n \nint binexp2(int a,int b){\n int ans=1;\n while(b){\n if(b&1)ans=(ans * a) % M;\n a=(a * a) % M;\n b>>=1;\n }\n return ans;\n}\n \n \nint binMul(int a,int b){\n int ans=0;\n while(b){\n if(b&1)ans=(ans+a)%M;\n a=(a+a)%M;\n b>>=1;\n }\n return ans;\n}\n \n \nint gcd(int a,int b){\n if(b==0)return a;\n return gcd(b,a%b);\n}\n \n \n//------------------------------------------------------------------------------------------------------------------------------------------------------//\n \nvector<bool>prime;\nVI hp;\nvoid sieve(){\n hp.resize(N,0);\n prime.resize(N,0);\n FOR(i,0,N)prime[i]=1,hp[i]=0;\n prime[0]=prime[1]=0;\n FOR(i,2,N){\n if(!hp[i]){\n for(int j=i;j<N;j+=i){\n hp[j]=i;\n if(j>i)\n prime[j]=0;\n }\n }\n }\n}\nVI fact;\nvoid pre(){\n fact.resize(N,0);\n fact[0]=fact[1]=1;\n FOR(i,2,N){\n fact[i]=fact[i-1]*i;\n fact[i]%=M;\n }\n}\nvoid prefix(int *pref,int *arr,int n,int i){\n pref[i]=arr[0];\n FOR(j,i+1,n+i){\n pref[j]=pref[j-1]+arr[j-i];\n }\n}\n \n \n \n//function<return_type(data_type,...)> name ;\n \n//----------SOLVE------------------------------\n//----------HERE-------------------------------\n \nint lb(int *arr,int n,int x){\n int l=-1,r=n;\n while(l<r-1){\n int m=(l+r)/2;\n if(arr[m]>=x)r=m;\n else l=m;\n }\n return r;\n}\n \nint ub(int *arr,int n,int x){\n int l=-1,r=n;\n while(l<r-1){\n int m=(l+r)/2;\n if(arr[m]<=x)l=m;\n else r=m;\n }\n return l;\n}\nstruct hash_pair {\n template <class T1, class T2>\n size_t operator()(const pair<T1, T2>& p) const\n {\n // Hash the first element\n size_t hash1 = hash<T1>{}(p.first);\n // Hash the second element\n size_t hash2 = hash<T2>{}(p.second);\n // Combine the two hash values\n return hash1\n ^ (hash2 + 0x9e3779b9 + (hash1 << 6)\n + (hash1 >> 2));\n }\n};\n \nbool solver(int *arr,int i,int n,int sum){\n int ans=0;\n if(i==n){\n if(sum%360==0)return 1;\n else return 0;\n }\n ans|=solver(arr,i+1,n,sum+arr[i]);\n ans|=solver(arr,i+1,n,sum-=arr[i]);\n return ans;\n}\nvoid solve1() {\n \n int n;\n cin>>n;\n PII p[n];\n int mxh=0;\n FOR(i,0,n){\n cin>>p[i].ff>>p[i].ss;\n }\n sort(p,p+n);\n int ans=2*(p[0].ff+p[0].ss);\n int w=p[0].ff;\n int h=p[0].ss;\n FOR(i,1,n){\n if(p[i].ff>w)ans+=2*(p[i].ff-w);\n if(p[i].ss>h)ans+=2*(p[i].ss-h);\n w=p[i].ff;\n h=p[i].ss;\n }\n cout<<ans<<endl;\n \n \n}\n//----------------MAIN FUNCTION--------------\nint32_t main(){\n #ifndef ONLINE_JUDGE\n freopen(\"input.txt\",\"r\",stdin);\n freopen(\"output.txt\",\"w\",stdout);\n #endif\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n int t=1;\n cin>>t;\n while(t--){\n solve1();\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": "2027A",
"submission_id": 288117692,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nusing i32 = int;\nusing i64 = long long;\nconst i32 INF = 1e9;\n \nvoid solve() {\n i32 n;\n cin >> n;\n \n vector <i32> H (102);\n for (i32 i = 0; i < n; ++i) {\n i32 w, h;\n cin >> w >> h;\n H[w] = h;\n }\n \n for (i32 i = 100; i >= 1; i--) {\n H[i] = max(H[i], H[i + 1]);\n }\n \n i32 ans = 0;\n for (i32 i = 1; i <= 101; ++i) {\n // cout << H[i] << ' ';\n ans += 2 * (H[i] != 0);\n ans += abs(H[i] - H[i - 1]);\n } \n cout << ans << '\\n';\n}\n \nint main() {\n ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);\n i32 tests = 1;\n cin >> tests;\n for (i32 test = 0; test < tests; ++test) {\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": "2027A",
"submission_id": 288124887,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define int long long\n#define fios(); ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);\n \nvoid solution();\n \nsigned main() {\n fios();\n int t = 1;\n cin >> t;\n while(t--){\n solution();\n cout<<endl;\n }\n}\n \nvoid solution() {\n int n, maxi=0, x, y;\n cin>>n;\n map<int, int>m;\n for(int i=0; i<n; i++){\n cin>>x;\n cin>>y;\n m[x]=(max(m[x],y));\n //m.insert({x, y});\n maxi=max(maxi, x);\n }\n int ans=0, i=0, tem1=0, tem2=0;\n for(auto val: m){\n ans+=abs(val.first-tem1);\n ans+=abs(val.second-tem2);\n tem1=val.first;\n tem2=val.second;\n i++;\n }\n auto it=m.end();\n it--;\n ans+=(it->first+it->second);\n cout<<ans;\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": "2027A",
"submission_id": 288159558,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\nt=int(input())\nfor _ in range(t):\n n=int(input())\n s=[0]*100\n for x in range(n):\n h,w=map(int,input().split())\n for i in range(w):\n s[i]=max(h,s[i])\n c=s[0]+2\n for i in range(1,100):\n if s[i]==0:\n break\n c+=abs(s[i-1]-s[i])+2\n# print(c)\n print(c+s[i-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": "2027A",
"submission_id": 299702270,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\ndef min_total_perimeter(rectangles):\n # Step 3: Calculate max possible overlap for each pair of rectangles\n overlaps = []\n for i in range(len(rectangles)):\n for j in range(i+1, len(rectangles)):\n wi, hi = rectangles[i]\n wj, hj = rectangles[j]\n overlap_x = min(wi, wj)\n overlap_y = min(hi, hj)\n overlaps.append(overlap_x * 2 + overlap_y * 2)\n \n # Step 4 & 5: Sort overlaps and initialize perimeter counter\n overlaps.sort(reverse=True)\n initial_perimeter = sum([2*(w + h) for w, h in rectangles])\n \n # Step 6: Subtract effective overlaps from the perimeter count\n for k in range(min(len(overlaps), len(rectangles)-1)):\n initial_perimeter -= overlaps[k]\n \n return initial_perimeter\n \n# Input reading function\ndef main():\n t = int(input())\n for _ in range(t):\n n = int(input())\n rectangles = [tuple(map(int, input().split())) for __ in range(n)]\n print(min_total_perimeter(rectangles))\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": "2027A",
"submission_id": 295564500,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\ndef min_perimeter(rectangles):\n # Sort rectangles by their perimeter in descending order\n rectangles.sort(key=lambda x: 2 * (x[0] + x[1]), reverse=True)\n \n # Initialize total perimeter with the perimeter of the largest rectangle\n total_perimeter = 2 * (rectangles[0][0] + rectangles[0][1])\n \n # For subsequent rectangles, add only the unique edges that don't overlap\n for i in range(1, len(rectangles)):\n total_perimeter += 2 * (rectangles[i][0] + rectangles[i][1])\n # Subtract the overlapped part (assumption: minimal overlap)\n total_perimeter -= 2 * min(rectangles[i-1][0], rectangles[i][0])\n total_perimeter -= 2 * min(rectangles[i-1][1], rectangles[i][1])\n \n return total_perimeter\n \n# Read input and process each test case\ndef main():\n import sys\n input = sys.stdin.read\n data = input().split()\n \n index = 0\n t = int(data[index])\n index += 1\n results = []\n \n for _ in range(t):\n n = int(data[index])\n index += 1\n rectangles = []\n \n for _ in range(n):\n w = int(data[index])\n h = int(data[index + 1])\n rectangles.append((w, h))\n index += 2\n \n result = min_perimeter(rectangles)\n results.append(result)\n \n for res in results:\n print(res)\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": "2027A",
"submission_id": 295291214,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class A_Rectangle_Arrangement {\n public static void main(String args[]) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n for (int z = 0; z < t; z++) {\n int n;\n n = sc.nextInt();\n int arr[][] = new int[n][2];\n for (int i = 0; i < n; i++) {\n arr[i][0] = sc.nextInt();\n arr[i][1] = sc.nextInt();\n }\n \n Arrays.sort(arr,(a,b)->{\n if(a[0]!=b[0]) return a[0]-b[0];\n return b[1]-a[1];\n });\n \n int ans=0;\n ans+=arr[0][0]+arr[n-1][0]+arr[0][1]+arr[n-1][1];\n \n int lw=arr[0][0],lh=arr[0][1];\n for(int i=1;i<n;i++){\n if(arr[i][0]>lw){\n ans+=arr[i][0]-lw;\n lw=arr[i][0];\n ans+=lh-arr[i][1];\n lh=arr[i][1];\n }\n }\n System.out.println(ans);\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": "2027A",
"submission_id": 288508875,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\nimport com.sun.source.tree.Tree;\n \nimport java.awt.*;\nimport java.awt.desktop.SystemEventListener;\nimport java.util.*;\nimport java.util.List;\n \npublic class Main {\n public static int mod=(int)1e9+7;\n public static class pair{\n int key;\n int val;\n pair(int key,int val){\n this.key=key;\n this.val=val;\n }\n void dis(){\n System.out.println(this.key+\" \"+this.val);\n }\n }\n static void dis(long arr[],int st,int end){\n if(st>end){\n for(int i=st-1;i>=end;i--){\n System.out.print(arr[i]+\" \");\n }\n System.out.println();\n }\n else{\n for(int i=st;i<end;i++){\n System.out.print(arr[i]+\" \");\n }\n System.out.println();\n }\n }\n public static long solve(int n) {\n if (n == 1) return 0;\n if (n == 2) return 1;\n long[] dp = new long[n + 1];\n dp[1] = 0;\n dp[2] = 1;\n for (int i = 3; i <= n; i++) {\n dp[i] = ((i - 1) * (dp[i - 1] + dp[i - 2]) % mod) % mod;\n }\n return dp[n];\n }\n static int gcd(int a,int b){\n if(b==0) return a;\n else return gcd(b,a%b);\n }\n static int justSmaller(List<Integer> l,int x){\n int st=0;\n int n=l.size();\n int end=n-1;\n int ans=0;\n while(st<=end){\n int mid=st+(end-st)/2;\n if(l.get(mid)<=x){\n ans=l.get(mid);\n st=mid+1;\n }\n else end=mid-1;\n }\n return ans;\n }\n static int justGreater(long arr[],int x){\n int st=0;\n int end=arr.length-1;\n int ans=arr.length-1;\n while(st<=end){\n int mid=st+(end-st)/2;\n if(arr[mid]>x){\n end=mid-1;\n ans=mid;\n }\n else st=mid+1;\n }\n return ans;\n }\n static int bp(int a, int b) {\n if (b == 0) return 1;\n int res = bp(a, b / 2);\n res = (int) ((long) res * res % mod);\n if (b % 2 == 1) return (int) ((long) res * a % mod);\n else return res;\n }\n static Comparator<pair> comp=new Comparator<pair>() {\n @Override\n public int compare(pair o1, pair o2) {\n if(o1.val>o2.val) return 1;\n else if(o1.val==o2.val) return 0;\n else return -1;\n }\n };\n static long binexp(long a,long b){\n if(b==0) return (long)1;\n if(b==1) return a%mod;\n long res=binexp(a,b/2);\n if(b%2==0) return (res*res)%mod;\n else return (((res * res) % mod) * a) % mod;\n }\n //Arrays.sort(array,(a,b)->Integer.compare(a[0],b[0]));\n \n //Arrays.sort(arr_name, Comparator.comparingInt(i -> arr[i]));\n static boolean check(char arr[][],int r,int c){\n int n=8;\n for(int i=0;i<n;i++){\n if(arr[r][i]=='#') return false;\n }\n for(int i=0;i<n;i++){\n if(arr[i][c]=='#') return false;\n }\n return true;\n }\n public static void main(String[] args) {\n Scanner sc=new Scanner(System.in);\n int t=sc.nextInt();\n while(t-->0){\n int n=sc.nextInt();\n int sum=0;\n int maxa=0;\n int maxb=0;\n int preva=0;\n int prevb=0;\n while(n-->0){\n int a=sc.nextInt();\n int b=sc.nextInt();\n if(a>=maxa && b>=maxb){\n sum=2*(a+b);\n maxa=a;\n maxb=b;\n preva=a;\n prevb=b;\n }\n else if(a>maxa){\n sum+=2*(a-maxa);\n preva=maxa;\n maxa=a;\n }\n else if(b>maxb){\n sum+=2*(b-maxb);\n prevb=maxb;\n maxb=b;\n }\n else if(a<maxa && b<maxb){\n if(maxb!=maxa){\n sum+=(a-preva+b-prevb);\n }\n }\n \n }\n System.out.println(sum);\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": "2027A",
"submission_id": 288286525,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std ;\nint main ()\n{\n int t ;\n cin>> t ; \n while(t-->0)\n { \n int n ; cin>> n ;\n int w , h ; cin >>w>>h ;\n int a = w , b = h ;\n int p = 2*w + 2*h ;\n for(int i = 1 ; i < n ; i++ ) {\n cin >>w>>h ;\n if(w>a){\n p+=2;\n }\n if(h>b){\n p+=2;\n }\n }\n cout<<p <<endl;\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": "2027A",
"submission_id": 288140004,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n \nusing namespace std;\n \nbool myComparator1(pair<int, int> a, pair<int, int> b) {\n return (a.first > b.first);\n}\n \nbool myComparator2(pair<int, int> a, pair<int, int> b) {\n return (a.second > b.second);\n}\n \nint main() {\n int n;\n cin >> n;\n for (int i = 0; i < n; i++) {\n int a;\n cin >> a;\n if (a != 1) {\n vector<pair<int, int>> v1;\n vector<pair<int, int>> v2;\n int count1 = 0;\n for (int j = 0; j < a; j++) {\n int b, c;\n cin >> b >> c;\n if (b == c) {\n count1++;\n }\n v1.push_back({b, c});\n v2.push_back({b, c});\n }\n if (count1 == v1.size()) {\n cout << v1[0].first * 4 << endl;\n } else {\n sort(v1.begin(), v1.end(), myComparator1);\n sort(v2.begin(), v2.end(), myComparator2);\n int peri = 0;\n if (v2[0].second > v1[0].second) {\n peri = v1[0].first + v2[0].second + v2[0].first + v1[0].second;\n } else {\n peri = v1[0].first + v2[0].second + v1[0].second;\n }\n \n for (int j = 0; j < (a - 1); j++) {\n peri += (v1[j].first - v1[j + 1].first) + (v2[a - j - 2].second - v2[a - j - 1].second);\n }\n cout << peri << endl;\n }\n } else {\n int b, c;\n cin >> b >> c;\n cout << 2 * (b + c) << endl;\n }\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2027A",
"submission_id": 288143505,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\ndef findCmn(curr, maxW):\n wL = [0 for _ in range(maxW)]\n for w,h in curr:\n wL[w-1] = max(h,wL[w-1])\n per = maxW\n pH = 0\n for i in range(maxW-1, -1, -1):\n if wL[i] == 0:\n per += 1\n else:\n per += abs(wL[i] - pH) + 1\n pH = wL[i]\n return per + pH\n \n \n \n \nfor _ in range(int(input())):\n n = int(input())\n curr = []\n maxW = 0\n for i in range(n):\n w, h = [int(s) for s in input().split()]\n maxW = max(w,maxW)\n curr.append([w,h])\n res = findCmn(curr, maxW)\n print(res)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2027A",
"submission_id": 288168335,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\nimport itertools\n \ndef c(f, x, y):\n e = 0\n \n if f[x][y]:\n e += 1 if x == 0 or not f[x - 1][y] else 0\n e += 1 if y == 0 or not f[x][y - 1] else 0\n e += 1 if x == 99 or not f[x + 1][y] else 0\n e += 1 if y == 99 or not f[x][y + 1] else 0\n \n return e\n \np = list(itertools.product(range(100), repeat=2))\n \nfor _ in range(int(input())):\n n = int(input())\n \n f = [[False for _ in range(100)] for _ in range(100)]\n \n for _ in range(n):\n a, b = list(map(int, input().split()))\n \n for i in range(a):\n for j in range(b):\n f[i][j] = True\n \n print(sum(map(lambda a, b : c(f, a, b), [p[i][0] for i in range(len(p))], [p[i][1] for i in range(len(p))])))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2027A",
"submission_id": 288142692,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\nimport java.io.Writer;\nimport java.io.BufferedWriter;\nimport java.io.OutputStreamWriter;\nimport java.io.OutputStream;\nimport java.io.PrintWriter;\nimport java.io.IOException;\nimport java.io.InputStream;\nimport java.math.BigInteger;\nimport java.util.Arrays;\nimport java.util.ArrayList;\nimport java.util.InputMismatchException;\nimport java.util.*;\nimport java.io.*;\nimport java.math.*;\nimport java.util.concurrent.TimeUnit;\npublic class Main{\n public static MyScanner sc;\n public static int solve(int n){\n int a[]=new int[n];\n int b[]=new int[n];\n for(int i=0;i<n;i++){\n int w=sc.nextInt();\n int h=sc.nextInt();\n a[i]=w;\n b[i]=h;\n }\n int ans=b[0]+b[n-1];\n for(int i=1;i<n;i++){\n if(a[i-1]<a[i]){\n ans+=Math.abs(b[i-1]-b[i]);\n }\n }\n Arrays.sort(a);\n ans+=a[n-1]*2;\n return ans;\n }\n public static void main(String[] args) {\n sc = new MyScanner();\n out = new PrintWriter(new BufferedOutputStream(System.out));\n int t=sc.nextInt();\n while(t-->0){\n int n=sc.nextInt();\n out.println(solve(n));\n }\n // Start writing your solution here. -------------------------------------\n \n /*\n int n = sc.nextInt(); // read input as integer\n long k = sc.nextLong(); // read input as long\n double d = sc.nextDouble(); // read input as double\n String str = sc.next(); // read input as String\n String s = sc.nextLine(); // read whole line as String\n \n int result = 3*n;\n out.println(result); // print via PrintWriter\n */\n \n // Stop writing your solution here. -------------------------------------\n out.close();\n }\n \n \n \n //-----------PrintWriter for faster output---------------------------------\n public static PrintWriter out;\n \n //-----------MyScanner class for faster input----------\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\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2027A",
"submission_id": 290009887,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class Main {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n \n int t = scanner.nextInt();\n \n for (int testCase = 0; testCase < t; testCase++) {\n int n = scanner.nextInt();\n int[][] stamps = new int[n][2];\n \n // Read width and height for each stamp\n for (int i = 0; i < n; i++) {\n stamps[i][0] = scanner.nextInt();\n stamps[i][1] = scanner.nextInt();\n }\n Arrays.sort(stamps, (a, b) -> Integer.compare(a[0], b[0]));\n if(n == 1){\n System.out.println(2 * (stamps[0][0] + stamps[0][1]));\n }else if(n == 2){\n if(stamps[0][1] < stamps[1][1]){\n System.out.println(2 * (stamps[1][0] + stamps[1][1]));\n } else {\n System.out.println(2*stamps[1][0] + stamps[0][1] + stamps[1][1]);\n }\n \n } else {\n // Sort by max dimension (width or height)\n \n \n \n int result = 0, i;\n // System.out.println(\"Stamps after sorting:\");\n // for (i = 0; i < n; i++) {\n // System.out.println(\"Width: \" + stamps[i][0] + \", Height: \" + stamps[i][1]);\n // }\n \n // Print all stamps after sorting\n int start = 0;\n while (start != n-1 && stamps[start][0] == stamps[start + 1][0]) start ++;\n if (start == n - 1){\n result = 2*(stamps[start][0] + stamps[start][1]);\n System.out.println(result);\n } else {\n result = 2*stamps[start][0] + stamps[start][1] + stamps[start][1] - stamps[start + 1][1] ;\n \n // System.out.println(\"result 1: \"+result);\n int thisMaxHeight = stamps[n-1][1];\n for (i = n-2; i >= 0; i--) {\n thisMaxHeight = Math.max(stamps[i][1], thisMaxHeight);\n stamps[i][1] = thisMaxHeight;\n }\n \n // System.out.println(\"Stamps after normalizing:\");\n // for (i = 0; i < n; i++) {\n // System.out.println(\"Width: \" + stamps[i][0] + \", Height: \" + stamps[i][1]);\n // }\n for (i = start + 1; i < n-1; i++) {\n result += 2*(stamps[i][0] - stamps[i-1][0]);\n result += stamps[i][1] - stamps[i+1][1];\n // System.out.println(\"result: \"+result);\n }\n result += 2*(stamps[n-1][0] - stamps[n-2][0]);\n result += stamps[n-1][1];\n \n System.out.println(result);\n // System.out.println(\"=======\");\n }\n }\n \n \n }\n \n scanner.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": "2027A",
"submission_id": 288159164,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define int long long \nusing namespace std;\n \nsigned main() {\n ios::sync_with_stdio(false);\n cin.tie(0);\n cout.tie(NULL);\n int t;\n cin >> t;\n \n while (t--) {\n int n;\n cin >> n;\n \n vector<int> w(n), h(n);\n for(int i=0; i<n; i++){\n cin>>w[i]>>h[i];\n }\n \n int ans = 0;\n \n if(n==1){\n ans = 2*(w[0]+h[0]);\n }\n else {\n int max_h = *max_element(h.begin(), h.end());\n ans+=2*max_h;\n \n int max_w = *max_element(w.begin(), w.end());\n int min_w = *min_element(w.begin(), w.end());\n \n for(int i=0; i<n; i++){\n if(w[i] != min_w){\n ans+=2*min_w;\n }\n }\n ans+=2*min_w;\n }\n \n \n cout<<ans<<endl;\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": "2027A",
"submission_id": 288122184,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\n \n#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)\n#define MOD 1000000007\n#define MOD1 998244353\n#define INF 1e18\n#define nline \"\\n\"\n#define pb push_back\n#define ppb pop_back\n#define mp make_pair\n#define ff first\n#define ss second\n#define PI 3.141592653589793238462\n#define set_bits __builtin_popcountll\n#define sz(x) ((int)(x).size())\n#define all(x) (x).begin(), (x).end()\n \ntypedef long long ll;\ntypedef unsigned long long ull;\ntypedef long double lld;\n// typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update > pbds; // find_by_order, order_of_key\ntypedef __gnu_pbds::tree<int, __gnu_pbds::null_type, less<int>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update> ordered_set;\n \n#ifndef ONLINE_JUDGE\n#define debug(x) cerr << #x <<\" \"; _print(x); cerr << endl;\n#else\n#define debug(x)\n#endif\n \nvoid _print(ll t) {cerr << t;}\nvoid _print(int t) {cerr << t;}\nvoid _print(string t) {cerr << t;}\nvoid _print(char t) {cerr << t;}\nvoid _print(lld t) {cerr << t;}\nvoid _print(double t) {cerr << t;}\nvoid _print(ull t) {cerr << t;}\n \ntemplate <class T, class V> void _print(pair <T, V> p);\ntemplate <class T> void _print(vector <T> v);\ntemplate <class T> void _print(set <T> v);\ntemplate <class T, class V> void _print(map <T, V> v);\ntemplate <class T> void _print(multiset <T> v);\ntemplate <class T, class V> void _print(pair <T, V> p) {cerr << \"{\"; _print(p.ff); cerr << \",\"; _print(p.ss); cerr << \"}\";}\ntemplate <class T> void _print(vector <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T> void _print(set <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T> void _print(multiset <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T, class V> void _print(map <T, V> v) {cerr << \"[ \"; for (auto i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\n \nbool cmp(pair<int, int> a, pair<int, int> b) {\n if(a.first == b.first) return a.second >= b.second;\n return a.first >= b.first;\n \n}\n \nvoid solve() {\n int n; cin>>n;\n vector<pair<int, int>> arr(n);\n for(int i = 0; i<n;i++) {\n cin>>arr[i].first>>arr[i].second;\n }\n if(n == 1) {\n int val = 2*(arr[0].first + arr[0].second);\n cout<<val<<endl; return;\n }\n sort(all(arr), cmp);\n \n ll ans = 0;\n ans += arr[0].first; ans += arr[0].second;\n int wid = arr[0].first; int hei = arr[0].second;\n for(int i = 1; i<n;i++) {\n if(arr[i].first == arr[i-1].first) continue;\n if(arr[i].second <= hei) continue;\n ans += (wid - arr[i].first); ans += (arr[i].second - hei);\n wid = arr[i].first; hei = arr[i].second;\n }\n ans += hei; ans += wid;\n cout<<ans<<endl;\n }\n \n \n \nint main() {\n#ifndef ONLINE_JUDGE\n freopen(\"Error.txt\", \"w\", stderr);\n#endif\n \n int t; cin>>t;\n // int t = 1;\n while(t--) 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": "2027A",
"submission_id": 288194503,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\nt = int(input())\nfor i in range(t):\n n = int(input())\n l =[]\n d={}\n for k in range(n):\n l.append(list(map(int,input().split())))\n for [a,b] in l:\n if a in d:\n if d[a] < b:\n d[a] = b\n else:\n d[a] = b\n k = [key for key in d]\n k.sort()\n print(2*k[-1]+ 2*d[k[0]])\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2027A",
"submission_id": 288234920,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\nt = int(input())\nfor _ in range(t):\n n = int(input())\n ordh = []\n toul = []\n for _ in range(n):\n n1, n2 = map(int, input().split())\n ordh.append(n1)\n toul.append(n2)\n s=0\n s+=max(toul)+max(ordh)\n tableaux_lies = sorted(zip(toul, ordh))\n toul, ordh = zip(*tableaux_lies)\n toul=toul[::-1]\n ordh=ordh[::-1]\n ttt=max(toul)\n aaa=ordh[0]\n s+=aaa+min(toul)\n if ((ordh[0]+toul[0])*2!=(max(toul)+max(ordh))*2):\n for i in range (len(toul)):\n if (ordh[i]>=aaa):\n s+=ordh[i]-aaa+ttt-toul[i]\n aaa=ordh[i]\n ttt=toul[i]\n if (aaa==ordh[0]) :\n s-=min(toul)\n s+=max(toul)\n print(s)\n else :\n print((ordh[0]+toul[0])*2)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2027A",
"submission_id": 288165471,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\nimport java.util.*;\nimport java.math.*;\n \n@SuppressWarnings(\"unused\")\n \npublic class Codeforces {\n \n public static void solve(Scanner scanner) {\n \n int size = scanner.nextInt();\n int[] width = new int[size];\n int[] height = new int[size];\n \n for (int i = 0; i < size; i++)\n {\n width[i] = scanner.nextInt();\n height[i] = scanner.nextInt(); \n }\n \n Arrays.sort(width);\n Arrays.sort(height);\n \n int result = 0;\n \n if (size == 1)\n {\n result = 2 * (width[0] + height[0]);\n System.out.println(result); \n }\n else\n {\n if (height[0] < height[size - 1])\n {\n result = width[size - 1] + height[size - 1] + ((height[size - 1] - height[0] + 1) * 2);\n }\n else\n {\n result = width[size - 1] + height[size - 1] + (height[size - 1] * 2);\n }\n \n System.out.println(result);\n }\n \n }\n \n public static void main(String args[]) {\n \n Scanner scanner = new Scanner(System.in);\n \n int test = scanner.nextInt();\n scanner.nextLine();\n \n for (int t = 0; t < test; t++)\n {\n solve(scanner);\n }\n \n scanner.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": "2027A",
"submission_id": 288128880,
"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: Rectangle Arrangement\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are coloring an infinite square grid, in which all cells are initially white. To do this, you are given $n$ stamps. Each stamp is a rectangle of width $w_i$ and height $h_i$.\n\nYou will use each stamp exactly once to color a rectangle of the same size as the stamp on the grid in black. You cannot rotate the stamp, and for each cell, the stamp must either cover it fully or not cover it at all. You can use the stamp at any position on the grid, even if some or all of the cells covered by the stamping area are already black.\n\nWhat is the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 500$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 100$).\n\nThe $i$-th of the next $n$ lines contains two integers $w_i$ and $h_i$ ($1 \\le w_i, h_i \\le 100$).\n\nOutput: For each test case, output a single integer — the minimum sum of the perimeters of the connected regions of black squares you can obtain after all the stamps have been used.\n\nExamples:\ninput:\n5\n5\n1 5\n2 4\n3 3\n4 2\n5 1\n3\n2 2\n1 1\n1 2\n1\n3 2\n3\n100 100\n100 100\n100 100\n4\n1 4\n2 3\n1 5\n3 2\noutput:\n20\n8\n10\n400\n16\n\nNote: In the first test case, the stamps can be used as shown on the left. Each stamp is highlighted in its own color for clarity.\n\n\n\nAfter all these stamps are used, there is one black region (as shown on the right), and its perimeter is $20$. It can be shown that there is no way of using the stamps that yields a lower total perimeter.\n\nIn the second test case, the second and third stamps can be used entirely inside the first one, so the minimum perimeter is equal to $8$.\n\nBuggy Code:\n// package CF;\n \nimport java.io.InputStream;\nimport java.util.*;\n \n \npublic class Solution {\n \n public static void main(String[] args) {\n InputStream is = args.length > 0 && \"LOCAL\".equals(args[0]) ? Solution.class.getResourceAsStream(\"/input.txt\") : System.in;\n Scanner sc = new Scanner(is);\n long t = sc.nextLong();\n while (t-- > 0) {\n int n = sc.nextInt();\n Integer[][] arr = new Integer[n][2];\n for (int i=0; i<n; i++) {\n arr[i][0] = sc.nextInt();\n arr[i][1] = sc.nextInt();\n }\n Arrays.sort(arr, (a,b) -> Integer.compare(b[1], a[1]));\n long ans = 2L * (arr[0][0] + arr[0][1]);\n for(int i=1; i<n; i++) {\n ans += Math.abs(2L * (arr[i][0] + arr[i][1]) - (2L * ((Math.min(arr[i][0], arr[i-1][0])) + (Math.min(arr[i][1], arr[i-1][1])))));\n }\n System.out.println(ans);\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": "2027A",
"submission_id": 288132223,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std ;\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\nusing namespace __gnu_pbds;\n#define int long long\nconst int N = 2e5+5 ;\nconst int INF = 1e18 ;\nconst int mod = 1e9 + 7 ;\nconst double pi = acos(-1.0) ;\ntemplate <typename T>\nusing order_set = tree<T, null_type,less<T>, rb_tree_tag,tree_order_statistics_node_update>;\nint fact[N] ;\nint n ;\nint a[N] ;\nint lev[N] ;\nint par[N] ;\nvector < int > graph[N] , level[N] ;\n \n \nint big_mod(int key, int pow)\n{\n int ans = 1 ;\n if(pow == 0) return 1 ;\n ans = big_mod(key,pow/2) ;\n ans = 1LL * ans * ans % mod ;\n if(pow&1) ans = 1LL * ans * key % mod;\n return ans ;\n}\n \n \n \nint ncr(int n , int r){\n int nn = fact[n] ;\n int rr = ((fact[r]%mod)*(fact[n-r]%mod))%mod ;\n nn = nn * 1ll * big_mod(rr , mod-2) ;\n nn %= mod ;\n return nn ;\n \n \n}\n \n \n \n \nvoid solve(){\n cin >> n ;\n for(int i = 0 ; i <= n ; i++){\n graph[i].clear() ;\n level[i].clear() ;\n lev[i] = -1 ;\n \n }\n for(int i = 1 ; i <= n ; i++){\n cin >> a[i] ;\n }\n for(int i = 2 ; i <= n ; i++){\n int x ;\n cin >> x ;\n par[i] = x ;\n graph[x].push_back(i) ;\n }\n for(int i = 0 ; i <= n ; i++){\n lev[i] = -1 ;\n }\n lev[1] = 0 ;\n level[0].push_back(1) ;\n queue < int > q ;\n q.push(1) ;\n int mx = 0 ;\n while(!q.empty()){\n int fnt = q.front() ;\n q.pop() ;\n for(auto it : graph[fnt]){\n if(lev[it] == -1){\n lev[it] = lev[fnt] + 1 ;\n level[lev[it]].push_back(it) ;\n mx = max(lev[it] , mx) ;\n q.push(it) ;\n }\n }\n }\n \n //cout << mx << endl ;\n if(mx == 1){\n int ans = 1e6 ;\n for(auto it : level[1]){\n ans = min(a[it] , ans) ;\n }\n cout << ans + a[1] << endl ;\n return ;\n \n }\n \n for(int i = 0 ; i <= n ; i++){\n lev[i] = 1e9 ;\n }\n \n while(mx > 1){\n for(auto it : level[mx]){\n int p = par[it] ;\n if(a[p] >= a[it] ){\n lev[p] = min(lev[p] , a[it]) ;\n }\n else{\n int val = a[p] + a[it] ;\n val /= 2 ;\n lev[p] = min(lev[p] , val) ;\n }\n }\n mx-- ;\n }\n \n int ans = 1e6 ;\n for(auto it : level[1]){\n ans = min(lev[it] , ans) ;\n }\n cout << ans + a[1] << endl ;\n \n \n \n \n for(int i = 0 ; i <= n ; i++){\n graph[i].clear() ;\n level[i].clear() ;\n lev[i] = -1 ;\n \n }\n \n \n}\n \n \n \n \n \n \nint32_t main()\n{\n \n ios_base :: sync_with_stdio(false) ; cin.tie(NULL) ; cout.tie(NULL) ;\n \n fact[0] = 1 , fact[1] = 1 ;\n for(int i = 2 ; i < N ; i++){\n fact[i] =((fact[i-1]%mod)*1ll*(i%mod)) % mod ;\n }\n \n int t= 1 ;\n cin >> t ;\n int T = 1 ;\n while(t--){\n //cout << \" CAs e \"<< T++ << \": \" ;\n solve() ;\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": "1997D",
"submission_id": 279252975,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <algorithm>\n#include <climits>\n#include <unordered_map>\n#include <cmath>\n#include <string>\n#include <stack>\n#include <set>\n#include <map>\n#include <queue>\n#include <cstring>\n#include <cmath>\n#define int long long int\nusing namespace std;\n#define MODULO 998244353\n \nint dfs(vector<vector<int>> &tree, vector<int> &a, int node){\n if(tree[node].size()==0){\n return a[node];\n }\n \n int mn = 1000000001;\n for (int i = 0; i < tree[node].size(); i++)\n {\n mn = min(mn, dfs(tree, a, tree[node][i]));\n }\n \n if(node==0){\n return a[node]+mn;\n }\n if(mn<a[node]){\n return a[node];\n }\n return floor((a[node]+mn)/2.0);\n}\n \n \nvoid solve()\n{\n int n;\n cin>>n;\n vector<int> a(n);\n for (int i = 0; i < n; i++)\n {\n cin>>a[i];\n }\n \n vector<vector<int>> tree(n, vector<int>());\n int u;\n for (int i = 1; i < n; i++)\n {\n cin>>u;\n u--;\n tree[u].push_back(i);\n }\n // cout<<\"hello\"<<endl;\n // for (int i = 0; i < n; i++)\n // {\n // for (int j = 0; j < tree[i].size(); j++)\n // {\n // cout<<tree[i][j]<<\" \";\n // }\n // cout<<endl;\n // }\n \n cout<<dfs(tree, a, 0)<<endl;\n \n}\n \nint32_t main()\n{\n \n \n int t;\n t = 1;\n cin >> t;\n \n for (int _ = 0; _ < t; _++)\n {\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1997D",
"submission_id": 273560065,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n \n#define ll long long\n#define F first\n#define S second\n#define mp make_pair\n#define pb push_back\n \n \nll gcd(ll a, ll b)\n{\n if (a == 0)\n return b;\n if (b == 0)\n return a;\n \n if (a == b)\n return a;\n \n if (a > b)\n return gcd(a - b, b);\n return gcd(a, b - a);\n}\n \nusing namespace std;\n \nll dfs(vector<ll> &a, vector<vector<int> > childs, int node) {\n \n for(int i = 0; i < childs[node].size(); i++) {\n a[node] = ((a[node] + dfs(a, childs, childs[node][i])) / 2);\n }\n \n return a[node];\n \n}\n \nint main() {\n \n ios::sync_with_stdio(false);\n cin.tie(0);\n int t;\n cin >> t;\n \n while(t--) {\n int n;\n cin >> n;\n vector<ll> a(n + 1);\n \n for(int i = 1; i <= n; i++)\n cin >> a[i];\n \n vector<vector<int > > childs(n + 1);\n \n for(int i = 2; i <= n; i++) {\n int p;\n cin >> p;\n childs[p].pb(i);\n }\n \n ll ans = INT_MAX;\n \n for(int i = 0; i < childs[1].size(); i++) {\n dfs(a, childs, childs[1][i]);\n }\n for(int i = 0; i < childs[1].size(); i++) {\n ans = min(ans, a[childs[1][i]]);\n }\n cout << ans + a[1] << '\\n';\n // ll ans = 0;\n // cout << ans << '\\n';\n \n }\n \n}\n \n/*\nRILY <3\nNEVER GIVE UP!!!\n1\n3\n6\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": "1997D",
"submission_id": 273586457,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\ndef dfs_modif(root): # pass in the vertices idx (0 indexed)\n global cur\n found = False\n for idx in range(n-1): \n if parents[idx] == root+1:\n dfs_modif(idx+1)\n found = True\n \n if found:\n if values[root] < cur:\n cur = values[root] + (cur - values[root])//2\n \n else:\n cur = values[root]\n \n return cur\n \nt = int(input())\nfor _ in range(t):\n n = int(input()) # number of vertices\n values = list(map(int, input().split(\" \")))\n parents = list(map(int, input().split(\" \")))\n vis = [False] * n\n res = float('inf')\n for idx in range(n-1):\n cur = float('inf')\n if parents[idx] == 1:\n res = min(res, dfs_modif(idx+1))\n \n if res == float('inf'):\n print(values[0])\n else:\n print(values[0] + 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": "1997D",
"submission_id": 276452012,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\nfrom types import GeneratorType\nfrom math import inf\nimport sys\n \n# 提高最大递归深度限制\nsys.setrecursionlimit(2 * (10 ** 5))\n \ndef bootstrap(f, stack=None):\n if stack is None:\n stack = []\n \n def func(*args, **kwargs):\n if stack:\n return f(*args, **kwargs)\n else:\n to = f(*args, **kwargs)\n while True:\n if type(to) is GeneratorType:\n stack.append(to)\n to = next(to)\n else:\n stack.pop()\n if not stack:\n break\n to = stack[-1].send(to)\n return to\n \n return func\ndef cf1997D():\n n = int(input())\n a = list(map(int, input().split()))\n g = [[] for _ in range(n)]\n b = list(map(int, input().split()))\n for i, x in enumerate(b):\n i += 1\n x -= 1\n g[x].append(i)\n \n # @bootstrap\n def dfs(x: int, f: int) -> int:\n \n # 先获取儿子的最小值\n mn = inf\n for y in g[x]:\n if y == f:\n continue\n cur = dfs(y, x)\n mn = min(cur, mn)\n if mn == inf:\n return a[x]\n if x != 0:\n if mn > a[x]:\n return (mn + a[x]) >> 1\n else:\n return mn\n else:\n return mn + a[x]\n \n ans = dfs(0, -1)\n print(ans)\n return\n \n \n \n \nt = int(input())\nfor i in range(t):\n cf1997D()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1997D",
"submission_id": 273656403,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\nimport io\nimport sys\nfrom collections import defaultdict, deque, Counter\nfrom itertools import permutations, combinations, accumulate\nfrom heapq import heappush, heappop\nfrom bisect import bisect_right, bisect_left\nfrom math import gcd\nimport math\n \n_INPUT = \"\"\"\\\n6\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\n\"\"\"\n \ndef input():\n return sys.stdin.readline()[:-1]\n \ndef solve(test):\n t=int(input())\n for _ in range(t):\n n=int(input())\n a=list(map(int, input().split()))\n p=list(map(lambda x:int(x)-1, input().split()))\n g=[[] for _ in range(n)]\n for i in range(n-1):\n g[i+1].append(p[i])\n g[p[i]].append(i+1)\n l,r=0,sum(a)+1\n while r-l>1:\n m=(l+r)//2\n if a[0]>=m: l=m\n else:\n queue=[~0,0]\n done=[0]*n\n now=m-a[0]\n d=[None]*n\n flg=0\n while queue:\n v=queue.pop()\n if v>=0:\n done[v]=1\n if a[v]<now:\n if v==0:\n d[0]=now\n else:\n d[v]=now-a[v]\n now+=d[v]\n if len(g[v])==1 and g[v][0]==p[v-1]:\n flg=1\n break\n else:\n d[v]=0\n for u in g[v]:\n if not done[u]:\n queue.append(~u)\n queue.append(u)\n else:\n now-=d[~v]\n if flg==0: l=m\n else: r=m\n print(l)\n \ndef random_input():\n from random import randint,shuffle\n N=randint(1,10)\n M=randint(1,N)\n A=list(range(1,M+1))+[randint(1,M) for _ in range(N-M)]\n shuffle(A)\n return (\" \".join(map(str, [N,M]))+\"\\n\"+\" \".join(map(str, A))+\"\\n\")*3\n \ndef simple_solve():\n return []\n \ndef main(test):\n if test==0:\n solve(0)\n elif test==1:\n sys.stdin = io.StringIO(_INPUT)\n case_no=int(input())\n for _ in range(case_no):\n solve(0)\n else:\n for i in range(1000):\n sys.stdin = io.StringIO(random_input())\n x=solve(1)\n y=simple_solve()\n if x!=y:\n print(i,x,y)\n print(*[line for line in sys.stdin],sep='')\n break\n \n#0:提出用、1:与えられたテスト用、2:ストレステスト用\nmain(0)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1997D",
"submission_id": 273580793,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\nimport java.util.ArrayList;\nimport java.util.Scanner;\n \npublic class Uzumaki_Naruto {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n for (int tt = 0; tt < t; tt++) {\n \n int n = sc.nextInt();\n long arr[] = new long[n + 1];\n for (int i = 1; i < n + 1; i++) {\n arr[i] = sc.nextLong();\n }\n // System.out.println(Arrays.toString(arr));\n \n int parent[] = new int[n - 1];\n for (int i = 0; i < n - 1; i++) {\n parent[i] = sc.nextInt();\n }\n \n ArrayList<ArrayList<Integer>> graph = new ArrayList<>();\n \n for (int i = 0; i < n + 1; i++) {\n graph.add(new ArrayList<>());\n }\n \n for (int i = 0; i < n - 1; i++) {\n graph.get(parent[i]).add(i + 2);\n graph.get(i + 2).add(parent[i]);\n }\n \n //System.out.println(graph);\n \n long ans = helper(graph, arr, 1, -1);\n System.out.println(ans);\n \n }\n sc.close();\n }\n \n public static long helper(ArrayList<ArrayList<Integer>> graph, long arr[], int node, int parent) {\n long min = Long.MAX_VALUE;\n for (int child : graph.get(node)) {\n if (child == parent) {\n continue;\n }\n min = Math.min(helper(graph, arr, child, node), min);\n }\n \n if (min == Long.MAX_VALUE) {\n return arr[node];\n }\n \n if (min <= arr[node]) {\n if (node == 1) {\n return arr[node];\n }\n return min;\n } else {\n if (node == 1) {\n return arr[node] + min;\n } else {\n return (min + arr[node]) / 2;\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": "1997D",
"submission_id": 273622599,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class D {\n \n public static void main(String[] args) throws IOException {\n InputReader in = new InputReader(System.in);\n PrintWriter out = new PrintWriter(System.out);\n Solve solve = new Solve(in, out);\n solve.solve();\n out.close();\n }\n \n static class InputReader {\n public BufferedReader reader;\n public StringTokenizer tokenizer;\n \n public InputReader(InputStream stream) {\n reader = new BufferedReader(new InputStreamReader(stream), 32768);\n tokenizer = null;\n }\n \n public String next() {\n while (tokenizer == null || !tokenizer.hasMoreTokens()) {\n try {\n tokenizer = new StringTokenizer(reader.readLine());\n } catch (IOException e) {\n throw new RuntimeException(e);\n }\n }\n return tokenizer.nextToken();\n }\n \n public int nextInt() {\n return Integer.parseInt(next());\n }\n \n public long nextLong() {\n return Long.parseLong(next());\n }\n \n public double nextDouble() {\n return Double.parseDouble(next());\n }\n }\n \n private static class Solve {\n private final PrintWriter out;\n private final InputReader in;\n \n public Solve(InputReader in, PrintWriter out) {\n this.in = in;\n this.out = out;\n }\n \n public void solve() {\n int t = in.nextInt();\n while (t --> 0) {\n solve_inner();\n }\n }\n \n int n;\n Map<Integer, List<Integer>> graph;\n long[] a;\n long[] f;\n long[] m;\n \n public long calc(int v) {\n List<Integer> children = graph.get(v);\n if (children == null) {\n f[v] = a[v];\n m[v] = a[v];\n return a[v];\n }\n \n long min = Long.MAX_VALUE;\n long gm = Long.MAX_VALUE;\n for (Integer child : children) {\n min = Math.min(min, calc(child));\n gm = Math.min(gm, m[child]);\n }\n m[v] = gm;\n long d = Math.min(m[v], Math.max(0, (a[v] + min) / 2 - a[v]));\n for (Integer child : children) {\n f[child] -= d;\n if (f[child] < 0) throw new RuntimeException();\n }\n f[v] = a[v] + d;\n m[v] = Math.min(min, f[v]);\n return f[v];\n }\n \n private void solve_inner() {\n n = in.nextInt();\n graph = new HashMap<>();\n a = new long[n + 1];\n f = new long[n + 1];\n m = new long[n + 1];\n for (int i = 1; i <= n; i++) {\n a[i] = in.nextLong();\n }\n for (int i = 2; i <= n; i++) {\n int p = in.nextInt();\n List<Integer> v = graph.get(p);\n if (v == null) {\n v = new ArrayList<>();\n }\n v.add(i);\n graph.put(p, v);\n }\n \n calc(1);\n// out.println(Arrays.toString(f));\n// out.println(Arrays.toString(m));\n long min = Long.MAX_VALUE;\n for (int i = 2; i <= n; i++) {\n min = Math.min(min, f[i]);\n }\n out.println(f[1] + min);\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": "1997D",
"submission_id": 273590361,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\nimport sys\ninput = lambda :sys.stdin.readline()[:-1]\nni = lambda :int(input())\nna = lambda :list(map(int,input().split()))\nyes = lambda :print(\"yes\");Yes = lambda :print(\"Yes\");YES = lambda : print(\"YES\")\nno = lambda :print(\"no\");No = lambda :print(\"No\");NO = lambda : print(\"NO\")\n#######################################################################\n \nfor _ in range(ni()):\n n = ni()\n a = na()\n p = [-1] + [x- 1 for x in na()]\n g = [[] for i in range(n)]\n for i in range(1, n):\n g[p[i]].append(i)\n \n et = []\n st = [0]\n seen = [0] * n\n seen[0] = 1\n chi = [[] for i in range(n)]\n while st:\n x = st.pop()\n et.append(x)\n for y in g[x]:\n if not seen[y]:\n seen[y] = 1\n st.append(y)\n chi[x].append(y)\n \n dp = [-1] * n\n for i in et[::-1]:\n if len(chi[i]) == 0:\n dp[i] = a[i]\n else:\n f = [dp[j] for j in chi[i]]\n z = min(f)\n if z <= a[i]:\n dp[i] = a[i]\n else:\n dp[i] = a[i] + z >> 1\n print(a[0] + min([dp[i] for i in chi[0]]))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1997D",
"submission_id": 273537322,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n#from math import ceil,log\n#from collections import deque,Counter\n#from bisect import bisect\n#from heapq import heappush,heappop\n#from random import randint\n \nfrom sys import stdin,stdout\ninput=lambda:stdin.readline().rstrip()\nprint=lambda *x,sep=' ',end='\\n':stdout.write(sep.join(map(str,x))+end)\n \n \nfrom types import GeneratorType\ndef bootstrap(f, stack=[]):\n def wrappedfunc(*args, **kwargs):\n if stack:\n return f(*args, **kwargs)\n else:\n to = f(*args, **kwargs)\n while True:\n if type(to) is GeneratorType:\n stack.append(to)\n to = next(to)\n else:\n stack.pop()\n if not stack:\n break\n to = stack[-1].send(to)\n return to\n return wrappedfunc\n \ndef out(l):\n print(' '.join(map(str,l)))\n \ndef yes():\n print('Yes')\n \ndef no():\n print('No')\n \ndef alice():\n print('Alice')\n \ndef bob():\n print('Bob')\n \n@bootstrap\ndef dfs(par,grand_par=0):\n mi=float('inf')\n for i,child in enumerate(adj[par]):\n if child!=grand_par:\n yield dfs(child,par)\n mi=min(mi,a[child])\n if grand_par==0:\n yield\n ori=a[par]\n if mi!=float('inf'):\n a[par]=(a[par]+mi)//2\n diff=a[par]-ori\n for i,child in enumerate(adj[par]):\n if child!=grand_par:\n a[child]-=diff\n yield\n \ndef solve():\n global adj,a\n n=int(input())\n a=[0]+list(map(int,input().split()))\n p=[0,0]+list(map(int,input().split()))\n adj=[[] for i in range(n+1)]\n for i in range(2,n+1):\n adj[p[i]].append(i)\n adj[i].append(p[i])\n #print(adj)\n ans=a[1]\n dfs(1)\n print(ans+min(a[2:]))\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": "1997D",
"submission_id": 273616352,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \n \n#define ll long long\n#define int long long\n#define pi pair<int,int>\n#define vi vector<int>\n#define vvi vector <vector <int>>\nconst int mod = 1e9 + 7;\n \nusing namespace std;\n \nint gcd(int a, int b){\n //a>b\n if(b==0) return a;\n return gcd(b,a%b);\n}\nistream &operator>>(istream &is, vector<int> &vec)\n{\n// vec.clear();\n int num;\n while (is >> num)\n {\n vec.push_back(num);\n char ch = is.peek();\n if (ch == '\\n')\n {\n is.ignore();\n break;\n }\n }\n return is;\n}\n \nvoid print(vector <int> v){\n for(auto it:v) cout << it <<' ';\n cout <<endl;\n}\nvector<int> prefixSum(vector<int> v ){\n vector<int> ans;\n int sum=0;\n for (int i =0; i < v.size();i++){\n sum+=v[i];\n ans.push_back(sum);\n }\n return ans;\n}\nvoid print(const vector<std::vector<int>> arr) {\n for (const std::vector<int>& row : arr) {\n for (int num : row) {\n \n std::cout << num << \" \";\n }\n std::cout << std::endl;\n }\n}\n \n \n \nbool poss(int r, int node, vvi & adj, vi & a){\n int val = a[node]-r;\n \n if(val < 0 && node != 0){\n r-=val;\n if( adj[node].empty()) return false;\n }\n \n for(int it : adj[node]){\n if( !poss(r, it, adj, a)) return false;\n }\n \n return true;\n}\n \nvoid solve(){\n ll n;\n vi a ,p;\n string s;\n cin >>n >> a >> p;\n \n vector <vector <int>> adj(n);\n for(int i = 0 ; i < n-1; i++){\n adj[--p[i]].push_back(i+1);\n }\n \n int ans =0, h= 1e17, l =0;\n \n while(l <=h ){\n int m = (l+h)/2;\n if(poss(m, 0, adj, a)){\n// cout <<\"poss for m = \"<<m <<endl;\n ans = m;\n l=m+1;\n }else{\n h=m -1;\n }\n }\n \n// print (adj);\n// cout <<\"---------------\"<<endl;\n \n cout << a[0]+ans <<endl;\n \n \n \n}\n \nint32_t main()\n{\n \n ios_base ::sync_with_stdio(false);\n cin.tie(NULL);\n \n#ifndef ONLINE_JUDGE\n freopen(\"input.txt\", \"r\", stdin);\n freopen(\"output.txt\", \"w\", stdout);\n#endif\n \n int T = 1;\n cin >> T;\n while (T--)\n {\n solve();\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1997D",
"submission_id": 278776461,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n//Abhinav garg\n \n \n#include<bits/stdc++.h>\n#include<ext/pb_ds/assoc_container.hpp>\n#include<ext/pb_ds/tree_policy.hpp>\n#define int long long int\n#define ld long double\n#define fps(x,y) fixed << setprecision(y) << x\n#define loop(i,a,n) for(int i=a;i<=n;i++)\n#define set_bits __builtin_popcountll\n#define sz(x) ((int)(x).size())\n#define rev(j,a,n) for(int j=a;j>=n;j--)\n#define it(i,v) for(auto i=v.begin();i!=v.end();i++)\n#define pb push_back\n#define vi vector\n#define all(v) v.begin(),v.end()\n#define pr pair<int, int>\n#define ff first\n#define ss second\n#define ump unordered_map\n#define ordered_set tree<pr, null_type,less<pr>, rb_tree_tag,tree_order_statistics_node_update>\nusing namespace std;\nusing namespace __gnu_pbds;\n \n \nconst int mod=1000000007;\n \n \n#ifndef ONLINE_JUDGE\n#define debug(x) cerr << #x <<\" \"; _print(x); cerr << endl;\n#else\n#define debug(x);\n#endif\n \n \nvoid _print(int t) {cerr << t;}\nvoid _print(string t) {cerr << t;}\nvoid _print(char t) {cerr << t;}\nvoid _print(ld t) {cerr << t;}\nvoid _print(double t) {cerr << t;}\nvoid _print(unsigned int t) {cerr << t;}\nvoid _print(bool t) {cerr << t;}\n \n \ntemplate <class T, class V> void _print(pair <T, V> p);\ntemplate <class T> void _print(vector <T> v);\ntemplate <class T> void _print(set <T> v);\ntemplate <class T, class V> void _print(map <T, V> v);\ntemplate <class T> void _print(multiset <T> v);\ntemplate <class T, class V> void _print(pair <T, V> p) {cerr << \"{\"; _print(p.ff); cerr << \",\"; _print(p.ss); cerr << \"}\";}\ntemplate <class T> void _print(vector <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T> void _print(set <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T> void _print(multiset <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T, class V> void _print(map <T, V> v) {cerr << \"[ \"; for (auto i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\n \n \ninline void fastio() {\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n #ifndef ONLINE_JUDGE\n freopen(\"inp.txt\", \"r\", stdin);\n freopen(\"op.txt\", \"w\", stdout);\n freopen(\"error.txt\", \"w\", stderr);\n #endif\n}\n \nint n; \n \nbool dfs(int node,int req, vi<vi<int>> &tr, vi<int> &a){\n bool poss = true;\n for(auto child:tr[node]){\n poss &= dfs(child, req + max(req - a[child],0LL), tr, a);\n }\n if(sz(tr[node]) == 0){\n return a[node] >= req;\n }\n return poss;\n \n}\n \nvoid test(){\n cin>>n;\n vi<int> a(n);\n loop(i,0,n-1) cin>>a[i];\n vi<vi<int>> tr(n);\n loop(i,1,n-1){\n int par; cin>>par;\n par--;\n tr[par].pb(i);\n }\n debug(tr);\n int ans = a[0];\n int l = a[0], r = 12;\n while(l<=r){\n int m = (l+r)>>1LL;\n if(dfs(0 ,m-a[0], tr, a)){\n ans = m;\n l = m+1;\n }\n else{\n r = m-1;\n }\n }\n cout<<ans<<\"\\n\";\n \n \n}\n \n \nint32_t main(int32_t argc,char const *arg[]){\n fastio();\n int tc=1;\n //prime();\n cin>>tc;\n while(tc--){\n test();\n }\n \n \n return 0;\n}\n/* if (a|b)+(a&b) = (a+b)*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1997D",
"submission_id": 274324280,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n// Problem: D. Maximize the Root\n// Contest: Codeforces - Educational Codeforces Round 168 (Rated for Div. 2)\n// URL: https://codeforces.com/contest/1997/problem/D\n// Memory Limit: 256 MB\n// Time Limit: 3000 ms\n// Author: Neil Chetty\n \nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n public static void main(String[] args) {\n int t = in.nextInt();\n while (t-- > 0) solve();\n out.flush();\n }\n \n private static void solve() {\n // WRITE CODE HERE\n int n = in.nextInt();\n int[] a = new int[n + 1];\n for (int i = 0; i < n; i++) a[i + 1] = in.nextInt();\n List<List<Integer>> store = new ArrayList<>();\n for (int i = 0; i <= n; i++) store.add(new ArrayList<>());\n for (int i = 1; i < n; i++) {\n int ok = in.nextInt();\n store.get(i + 1).add(ok);\n store.get(ok).add(i + 1);\n }\n long left = 0, right = (long) (1e17);\n while (left <= right) {\n long mid = left + (right - left) / 2;\n if (helper(store, 1, 0, 0, a, mid))\n left = mid + 1;\n else\n right = mid - 1;\n }\n out.println(right);\n // out.println(store);\n // out.println(helper(store, 1, 0, 0, a, 10));\n }\n \n private static boolean helper(\n List<List<Integer>> store, int current, int parent, long sum, int[] a, long k) {\n if (current != 1 && store.get(current).size() == 1)\n return a[current] >= sum;\n if (current == 1)\n sum += (a[current] >= k ? 0 : Math.abs(a[current] - k));\n else\n sum += (a[current] >= sum ? 0 : Math.abs(a[current] - sum));\n boolean result = true;\n for (int adj : store.get(current)) {\n if (adj == parent)\n continue;\n result &= helper(store, adj, current, sum, a, k);\n }\n return result;\n }\n \n // ALL PRE-WRITTEN HELPER FUNCTIONS\n static QuickReader in = new QuickReader();\n static PrintWriter out = new PrintWriter(System.out);\n private final static int MOD = (int) ((1e9) + 7);\n \n private static boolean[] primeFinder(int n) {\n boolean[] result = new boolean[n + 1];\n Arrays.fill(result, true);\n result[0] = false;\n result[1] = false;\n for (int i = 2; i * i <= n; i++) {\n if (!result[i])\n continue;\n for (int j = i * i; j <= n; j += i) {\n result[j] = false;\n }\n }\n return result;\n }\n \n private static long gcd(long a, long b) {\n return a % b == 0 ? b : gcd(b, a % b);\n }\n \n private static long lcm(long a, long b) {\n return a * b / gcd(a, b);\n }\n \n private static long power(long a, long b) {\n long result = 1;\n while (b > 0) {\n if ((b & 1) == 1) {\n result = (result * a) % MOD;\n }\n b >>= 1;\n a = (a * a) % MOD;\n }\n return result;\n }\n \n private static long modInverse(long a) {\n return power(a, MOD - 2);\n }\n \n private static long nCr(int n, int r) {\n if (n < r)\n return 0;\n if (r == 0)\n return 1;\n long[] dp = factorial(n);\n return ((dp[n] * modInverse(dp[r]) % MOD) * (modInverse(dp[n - r]) % MOD)) % MOD;\n }\n \n private static long nCr(int n, int r, long[] dp) {\n if (n < r)\n return 0;\n if (r == 0)\n return 1;\n return ((dp[n] * modInverse(dp[r]) % MOD) * (modInverse(dp[n - r]) % MOD)) % MOD;\n }\n \n private static long nPr(int n, int r) {\n if (n < r)\n return 0;\n if (r == 0)\n return 1;\n long[] dp = factorial(n);\n return (dp[n] * modInverse(dp[n - r])) % MOD;\n }\n \n private static long nPr(int n, int r, long[] dp) {\n if (n < r)\n return 0;\n if (r == 0)\n return 1;\n return (dp[n] * modInverse(dp[n - r])) % MOD;\n }\n \n private static long[] factorial(int n) {\n long[] factorial = new long[n + 1];\n factorial[0] = 1;\n for (int i = 1; i <= n; i++) {\n factorial[i] = (factorial[i - 1] * i) % MOD;\n }\n return factorial;\n }\n \n private static void printArray(int[] a) {\n for (int i : a) out.print(i + \" \");\n out.println();\n }\n \n private static void printlnArray(int[] a) {\n for (int i : a) out.println(i);\n }\n \n private static void printMatrix(int[][] a) {\n for (int[] i : a) printArray(i);\n }\n \n private static void sortArray(int[] a, int order) {\n ArrayList<Integer> l = new ArrayList<>();\n for (int i : a) l.add(i);\n if (order == -1)\n Collections.sort(l, Collections.reverseOrder());\n else\n Collections.sort(l);\n for (int i = 0; i < a.length; i++) a[i] = l.get(i);\n }\n \n private static class QuickReader {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n StringTokenizer st = new StringTokenizer(\"\");\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 String next() {\n while (!st.hasMoreTokens()) try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\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 int[] readIntArray(int n) {\n int[] a = new int[n];\n for (int i = 0; i < n; i++) a[i] = nextInt();\n return a;\n }\n \n long[] readLongArray(int n) {\n long[] a = new long[n];\n for (int i = 0; i < n; i++) a[i] = nextLong();\n return a;\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": "1997D",
"submission_id": 273582683,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n// package com.atguigu;\n \nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n \n static PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n \n static class Read{\n StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));\n public int nextInt() throws Exception{\n st.nextToken();\n return (int) st.nval;\n }\n public long nextLong() throws Exception{\n st.nextToken();\n return (long) st.nval;\n }\n \n }\n static List<List<Integer>> tree; // 邻接表表示树结构\n static int[] a; // 节点初始值数组\n static long[] need; // 需要传递的操作值\n static int n;\n \n public static void main(String[] args) throws Exception{\n Read sc = new Read();\n int t = sc.nextInt(); // 测试用例数量\n \n while (t-- > 0) {\n n = sc.nextInt();\n a = new int[n + 1];\n need = new long[n + 1];\n tree = new ArrayList<>();\n \n for (int i = 0; i <= n; i++) {\n tree.add(new ArrayList<>());\n }\n \n // 读取节点值\n for (int i = 1; i <= n; i++) {\n a[i] = sc.nextInt();\n }\n \n // 构建树，读取父节点信息\n for (int i = 2; i <= n; i++) {\n int parent = sc.nextInt();\n tree.get(parent).add(i);\n }\n \n // 二分答案\n long low = a[1], high = 2000000000L;\n high = 10;\n long result = a[1];\n \n while (low <= high) {\n long mid = (low + high) / 2;\n if (dfs(1, mid,0)) {\n result = mid;\n low = mid + 1;\n } else {\n high = mid - 1;\n }\n }\n \n // 输出结果\n pw.println(result);\n }\n pw.flush();\n }\n \n public static boolean dfs(int u, long mid,long topVal) {\n if(u == 1){\n if(a[1] >= mid) return true;\n else {\n topVal = mid - a[1];\n }\n }\n boolean flag = true;\n for(int i = 0;i < tree.get(u).size();i++){\n int v = tree.get(u).get(i);\n if(a[v] >= topVal) flag &= dfs(v, mid,topVal);\n else {\n flag &= dfs(v, mid,topVal + topVal - a[v]);\n }\n }\n if(tree.get(u).size() == 0) return a[u] >= topVal;\n return flag;\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": "1997D",
"submission_id": 286179444,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\nusing namespace __gnu_pbds;\n#define ll long long\n#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>\n#define sdefine \\\n string s; \\\n cin >> s;\n#define vll vector<ll>\n#define pll pair<ll, ll>\n#define vvll vector<vll>\n#define vpll vector<pll>\n#define mll map<ll, ll>\n#define pb push_back\n#define ppb pop_back\n#define ff first\n#define ss second\n#define set_bits(a) __builtin_popcountll(a)\n#define sz(x) ((int)(x).size())\n#define all(x) (x).begin(), (x).end()\n#define forl(i, a, b) for (int i = a; i < b; i++)\n#define yesno(x, y) \\\n if ((x) == (y)) \\\n cout << \"YES\" << endl; \\\n else \\\n cout << \"NO\" << endl;\n#define endl \"\\n\"\n#define display(a, x, y) \\\n { \\\n for (int i = x; i < y; i++) \\\n cout << a[i] << \" \"; \\\n cout << endl; \\\n }\n#define int long long\n \nbool dfs(int val, int s, int pr, int op, vll(&v), vll G[])\n{\n int curr = v[s] - op;\n if (s == 1)\n op += (val - curr);\n if (curr < 0)\n op += abs(curr);\n int ct = 0;\n bool ck = true;\n for (auto it : G[s])\n {\n if (it == pr)\n continue;\n ct++;\n ck &= dfs(val, it, s, op, v, G);\n }\n if ((ct == 0) && (curr < 0))\n ck = false;\n return ck;\n}\n \nint32_t main()\n{\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n int t;\n cin >> t;\n while (t--)\n {\n int n;\n cin >> n;\n vll v(n + 1, 0);\n forl(i, 1, n + 1)\n {\n cin >> v[i];\n }\n vll par(n - 1, -1);\n vll G[n + 1];\n forl(i, 0, n - 1)\n {\n cin >> par[i];\n G[i + 2].push_back(par[i]);\n G[par[i]].push_back(i + 2);\n }\n int st = v[1], ed = 1e15;\n while ((ed - st) > 1)\n {\n int mid = (st + ed) / 2;\n bool ck = dfs(mid, 1, -1, 0, v, G);\n if (ck)\n st = mid;\n else\n ed = mid - 1;\n }\n int ans = st;\n if (dfs(ed, 1, -1, 0, v, G))\n ans = ed;\n cout << ans << 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": "1997D",
"submission_id": 277237376,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\ntypedef long long ll;\n#define pb push_back\n#define pii pair<int, int>\n#define all(a) a.begin(), a.end()\nconst int mod = 998244353, N = 500005;\n \nvoid solve() {\n int n;\n cin >> n;\n vector <ll> a(n);\n for (int i = 0; i < n; ++i) {\n cin >> a[i];\n }\n vector <vector <int>> adj(n);\n for (int i = 1, p; i < n; ++i) {\n cin >> p, --p;\n adj[p].pb(i);\n }\n auto ask = [&](ll x) {\n vector <ll> b = a, lz(n, 0);\n b[0] -= x;\n auto dfs = [&](auto self, int v, ll op) -> void {\n b[v] -= op;\n if (adj[v].empty()) {\n return;\n }\n if (b[v] < 0) {\n op -= b[v], b[v] = 0;\n }\n for (int u : adj[v]) {\n self(self, u, op);\n }\n };\n dfs(dfs, 0, 0);\n return *min_element(all(b)) >= 0;\n };\n ll l = a[0], r = 1ll << 50;\n while (r - l > 1) {\n ll m = l + r >> 1;\n if (ask(m)) {\n l = m;\n } else {\n r = m;\n }\n }\n cout << l << '\\n';\n}\n \nint main() {\n ios::sync_with_stdio(false), cin.tie(0);\n int t;\n cin >> t;\n while (t--) {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1997D",
"submission_id": 273571029,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\nif __name__ == \"__main__\":\n \n def solve():\n def dfs(v,k):\n if tree[v] == []:return a[v] >= k\n for u in tree[v]:\n if a[u] >= k:\n sub = dfs(u,k)\n else:\n sub = dfs(u,k + k - a[u])\n if not sub:\n return False\n return True\n \n n = int(input())\n a = list(map(int, input().split()))\n parents = list(map(int, input().split()))\n tree = {i:[] for i in range(n)}\n \n for i in range(n - 1):\n parent = parents[i] - 1\n child = i + 1\n tree[parent].append(child)\n # print(tree, a)\n # @cache\n ans = 0\n l = 0\n r = 2* 10 **9+1\n while l<r:\n mid = (l+r)//2\n if a[0]>=mid or dfs(0,mid - a[0]):\n ans = mid\n l = mid +1\n else:\n r = mid\n \n print(ans)\n \n for _ 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": "1997D",
"submission_id": 273632914,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\n# My competitive programming template - July 2024\n# by bernie\n \nfrom types import GeneratorType\nfrom sys import stdin, stdout, setrecursionlimit\nfrom collections import Counter, deque, defaultdict, OrderedDict\nfrom math import ceil, comb, sqrt, factorial, perm, log, gcd, lcm, floor, inf\nfrom functools import lru_cache, cache\nfrom bisect import bisect_left, bisect_right\nfrom heapq import heapify, heappush, heappop\nfrom itertools import permutations, chain\nimport time\n \nsetrecursionlimit(100000)\n \ndef measure(func):\n def wrapper(*args, **kwargs):\n start_time = time.perf_counter()\n result = func(*args, **kwargs)\n end_time = time.perf_counter()\n print(f\"Function '{func.__name__}' took {end_time - start_time} seconds to execute.\")\n return result\n return wrapper\n \n# pajenegod orz\ndef bootstrap(f, stack=[]):\n def wrappedfunc(*args, **kwargs):\n if stack:\n return f(*args, **kwargs)\n to = f(*args, **kwargs)\n while True:\n if type(to) is GeneratorType:\n stack.append(to)\n to = next(to)\n else:\n stack.pop()\n if not stack:\n break\n to = stack[-1].send(to)\n return to\n return wrappedfunc\n \ndef input():\n \"\"\" Faster input parsing \"\"\"\n return stdin.readline().strip()\n \ndef print(string=\"\", newline=True):\n \"\"\"Prints a single string on a new line\"\"\"\n if newline:\n stdout.write(str(string) + \"\\n\")\n else:\n stdout.write(str(string))\n \ndef print_array(array):\n \"\"\" Prints an array of values, the way CodeForces requires \"\"\"\n stdout.write(\" \".join(map(str, array)) + \"\\n\")\n \ndef bin(x, width=None):\n \"\"\"A replacement for the built-in bin function, without the 0x\"\"\"\n if width:\n return format(x, \"b\").zfill(width)\n return format(x, \"b\")\n \ndef find_primes(limit, return_all=True):\n \"\"\" Finds all the prime numbers from 1 to a specified limit (inclusive). Sieve of Erastothenes \"\"\"\n is_prime = [1] * (limit + 1)\n is_prime[1] = 0\n \n for i in range(2, limit + 1):\n if i * i > limit:\n break\n if is_prime[i]:\n for j in range(i * i, limit + 1, i):\n is_prime[j] = 0\n if return_all:\n return is_prime\n else:\n return [i for i in range(2, limit + 1) if is_prime[i]]\n \ndef factorize(x):\n factors = set()\n for i in range(1, ceil(sqrt(x)) + 1):\n if x % i == 0:\n factors.add(i)\n factors.add(x // i)\n return factors\n \nMOD = int(1e9 + 7)\n \ndef parity(x):\n return x >= 0\n \ndef is_even(x):\n return x % 2 == 0\n \ndef solve():\n def increase_min(node):\n min_child = float(\"inf\")\n \n if tree[f\"{node}\"]:\n for child in tree[f\"{node}\"]:\n min_child = min(min_child, increase_min(child))\n \n if min_child <= a[node]:\n return min_child\n else:\n best = (min_child + a[node]) // 2\n return best\n else:\n return a[node]\n \n for t in range(int(input())):\n n = int(input())\n a = [int(_) for _ in input().split()]\n b = [int(_) for _ in input().split()]\n tree = {}\n \n for v in range(n):\n tree[f\"{v}\"] = []\n \n for i in range(len(b)):\n tree[f\"{b[i] - 1}\"].append(i + 1)\n \n ans = a[0]\n min_child = float(\"inf\")\n \n for child in tree[\"0\"]:\n min_child = min(min_child, increase_min(child))\n \n if min_child == float(\"inf\"):\n min_child = 0\n \n print(ans + min_child)\n \n \n \n \n \n \n \n \n \n \nif __name__ == \"__main__\":\n try:\n solve()\n except Exception as error:\n print(error)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1997D",
"submission_id": 273600760,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\nimport java.util.ArrayList;\nimport java.util.List;\nimport java.util.Scanner;\n \npublic class Main {\n static final int LIMIT = 1_000_000_000;\n \n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n \n int t = sc.nextInt();\n for (int tc = 0; tc < t; ++tc) {\n int n = sc.nextInt();\n int[] a = new int[n];\n for (int i = 0; i < a.length; ++i) {\n a[i] = sc.nextInt();\n }\n int[] p = new int[n - 1];\n for (int i = 0; i < p.length; ++i) {\n p[i] = sc.nextInt();\n }\n \n System.out.println(solve(a, p));\n }\n \n sc.close();\n }\n \n static int solve(int[] a, int[] p) {\n int n = a.length;\n \n @SuppressWarnings(\"unchecked\")\n List<Integer>[] childLists = new List[n];\n for (int i = 0; i < childLists.length; ++i) {\n childLists[i] = new ArrayList<>();\n }\n for (int i = 0; i < p.length; ++i) {\n childLists[p[i] - 1].add(i + 1);\n }\n \n int base = a[0];\n \n int delta = -1;\n int lower = 0;\n int upper = LIMIT;\n while (lower <= upper) {\n int middle = (lower + upper) / 2;\n \n a[0] = middle;\n if (search(a, childLists, middle, 0)) {\n delta = middle;\n lower = middle + 1;\n } else {\n upper = middle - 1;\n }\n }\n \n return base + delta;\n }\n \n static boolean search(int[] a, List<Integer>[] childLists, long minValue, int node) {\n if (childLists[node].isEmpty()) {\n return a[node] >= minValue;\n }\n \n for (int child : childLists[node]) {\n if (!search(a, childLists, minValue + Math.max(0, minValue - a[node]), child)) {\n return false;\n }\n }\n \n return true;\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": "1997D",
"submission_id": 275235487,
"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: Maximize the Root\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.\n\nYou can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.\n\nYour task is to calculate the maximum possible value written at the root using the aforementioned operation.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\le n \\le 2 \\cdot 10^5$) — the number of vertices in the tree.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0 \\le a_i \\le 10^9$) — the initial values written at vertices.\n\nThe third line contains $n-1$ integers $p_2, p_3, \\dots, p_n$ ($1 \\le p_i \\le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.\n\nAdditional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.\n\nExamples:\ninput:\n3\n4\n0 1 0 2\n1 1 3\n2\n3 0\n1\n5\n2 5 3 9 6\n3 1 5 2\noutput:\n1\n3\n6\n\nNote: In the first test case, the following sequence of operations is possible:\n\n * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$; * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n static PrintWriter pw = new PrintWriter(System.out);\n static Scanner sc = new Scanner(System.in);\n \n public static void solve() throws IOException {\n int n = sc.nextInt();\n int a [] = sc.nextIntArray(n);\n ArrayList<Integer> adjl [] = new ArrayList [n + 1];\n for(int i = 0; i <= n ; i++)\n adjl[i] = new ArrayList<>();\n for(int i = 2 ; i <= n ; i++){\n int u = sc.nextInt(), v = i;\n adjl[u].add(v);\n }\n long low = a[0], high = (long) 4e9;\n while(low <= high) {\n long mid = (low + high) / 2;\n Queue<long []> qu= new LinkedList<>();\n qu.add(new long[]{1, mid - a[0]});\n boolean f = true;\n while (!qu.isEmpty()) {\n long [] arr = qu.poll();\n// pw.println(Arrays.toString(arr));\n int idx = (int) arr[0];\n long req = arr[1];\n if(adjl[idx].isEmpty()){\n f &= a[idx - 1] >= req;\n }\n for(int i = 0; i < adjl[idx].size() ; i++){\n long val = a[adjl[idx].get(i) - 1] - req;\n qu.add(new long[]{adjl[idx].get(i), req - Math.min(val, 0)});\n }\n }\n if(f)\n low = mid + 1;\n else\n high = mid - 1;\n }\n pw.println(high);\n }\n \n public static void main(String[] args) throws IOException {\n int t = sc.nextInt();\n while(t -- > 0)\n solve();\n pw.flush();\n }\n \n static class Scanner {\n StringTokenizer st;\n BufferedReader br;\n \n public Scanner(InputStream s) {\n br = new BufferedReader(new InputStreamReader(s));\n }\n \n public Scanner(String s) throws FileNotFoundException {\n br = new BufferedReader(new FileReader(s));\n }\n \n public String next() throws IOException {\n while (st == null || !st.hasMoreTokens())\n st = new StringTokenizer(br.readLine());\n return st.nextToken();\n }\n \n public int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n \n public long nextLong() throws IOException {\n return Long.parseLong(next());\n }\n \n public String nextLine() throws IOException {\n return br.readLine();\n }\n \n public double nextDouble() throws IOException {\n String x = next();\n StringBuilder sb = new StringBuilder(\"0\");\n double res = 0, f = 1;\n boolean dec = false, neg = false;\n int start = 0;\n if (x.charAt(0) == '-') {\n neg = true;\n start++;\n }\n for (int i = start; i < x.length(); i++)\n if (x.charAt(i) == '.') {\n res = Long.parseLong(sb.toString());\n sb = new StringBuilder(\"0\");\n dec = true;\n } else {\n sb.append(x.charAt(i));\n if (dec)\n f *= 10;\n }\n res += Long.parseLong(sb.toString()) / f;\n return res * (neg ? -1 : 1);\n }\n \n public long[] nextlongArray(int n) throws IOException {\n long[] a = new long[n];\n for (int i = 0; i < n; i++)\n a[i] = nextLong();\n return a;\n }\n \n public Long[] nextLongArray(int n) throws IOException {\n Long[] a = new Long[n];\n for (int i = 0; i < n; i++)\n a[i] = nextLong();\n return a;\n }\n \n public int[] nextIntArray(int n) throws IOException {\n int[] a = new int[n];\n for (int i = 0; i < n; i++)\n a[i] = nextInt();\n return a;\n }\n \n public Integer[] nextIntegerArray(int n) throws IOException {\n Integer[] a = new Integer[n];\n for (int i = 0; i < n; i++)\n a[i] = nextInt();\n return a;\n }\n \n public boolean ready() throws IOException {\n return br.ready();\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1997D",
"submission_id": 275313207,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define int long long int\n#define s(v) (int)v.size()\n#define m1 cout << \"-1\" << endl\n#define all(v) v.begin(), v.end()\n#define rall(v) v.rbegin(), r.rend()\n#define _lb(v, x) lower_bound(all(v), x)\n#define lb(S, x) S.lower_bound(x)\n#define _ub(v, x) lower_bound(all(v), x)\n#define ub(S, x) S.lower_bound(x)\n#define mAx(v) *max_element(v.begin(), v.end())\n#define Sum(v) accumulate(v.begin(), v.end(), 0LL)\n#define mIn(v) *min_element(v.begin(), v.end())\n#define yes cout << \"YES\" << endl\n#define no cout << \"NO\" << endl\n#define asort(v) sort(v.begin(), v.end())\n#define dsort(v) sort(v.begin(), v.end(), greater<int>())\n#define rep(i, a, b) for (int i = a; i < b; i++)\n#define pb push_back\n#define ff first\n#define ss second\n#define cnf(p) cout << \"@@\" << p << endl;\nvector<vector<int>> adj;\nvector<int> vis;\nconst int nax = 2e5 + 10;\nvoid Display(vector<int> &v)\n{\n for (auto &A : v)\n cout << A << \" \";\n cout << endl;\n}\nvoid Scan(vector<int> &v)\n{\n for (auto &A : v)\n cin >> A;\n}\nvector<int> get_primes(int n)\n{\n vector<int> ans;\n bool prime[n + 1];\n memset(prime, true, sizeof(prime));\n for (int p = 2; p * p <= n; p++)\n {\n if (prime[p] == true)\n {\n for (int i = p * p; i <= n; i += p)\n prime[i] = false;\n }\n }\n for (int p = 2; p <= n; p++)\n if (prime[p])\n ans.pb(p);\n return ans;\n}\nint binexp(int a, int b, int MOD)\n{\n int res = 1;\n while (b > 0)\n {\n if (b & 1)\n res = (res * 1LL * a) % MOD;\n a = (a * 1LL * a) % MOD;\n b >>= 1;\n }\n return res;\n}\nvoid runcase()\n{\n int n, x;\n cin >> n >> x;\n vector<int> a(n);\n Scan(a);\n int ans = 0;\n int max_nahi = 0, max_haan = 0;\n vector<int> pref(n + 1);\n for (int i = 1; i <= n; i++)\n pref[i] = pref[i - 1] + a[i - 1];\n int min_haan = mAx(a);\n int l = 0;\n for (int i = 0; i < n - 1; i++)\n {\n if (!ans)\n {\n if (a[i] <= max_nahi)\n continue;\n int sum = pref[i + 1];\n int j = i;\n while (j < n - 1 && sum >= a[j + 1])\n {\n sum += a[j + 1], j++;\n if (sum >= min_haan)\n {\n ans++;\n min_haan = a[i];\n break;\n }\n }\n if (sum < min_haan)\n max_nahi = a[i];\n else\n l = i;\n }\n else\n {\n int sum;\n if (a[i] >= a[i - 1])\n sum = pref[i + 1] - pref[l + 1];\n else\n sum = a[i];\n int j = i;\n while (j < n - 1 && sum >= a[j + 1])\n {\n sum += a[j + 1], j++;\n if (sum >= min_haan)\n {\n ans++;\n min_haan = min(min_haan, a[i]);\n break;\n }\n }\n }\n // cout << i << \" \" << ans << endl;\n }\n int sum = a.back();\n int ind = n - 1;\n while (ind > 0 && sum >= a[ind - 1])\n sum += a[ind - 1], ind--;\n if (ind == 0)\n ans++;\n cout << ans << \"\\n\";\n}\nsigned main()\n{\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n int t = 1;\n cin >> t;\n while (t--)\n runcase();\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": "1998E1",
"submission_id": 275621868,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n \n#include<ext/pb_ds/assoc_container.hpp>\n#include<ext/pb_ds/tree_policy.hpp>\n \nusing namespace std;\nusing namespace __gnu_pbds;\n \n \n/***********************************************/\n/* dear online judge:\n * i've read the problem, and tried to solve it.\n * even if you don't accept my solution, you should respect my effort.\n * i hope my code compile and get accepted.\n * ____________\n * / \\\n * / /\\ /\\ \\\n * / / \\ / \\ \\\n * \\ /\n * \\ \\___/ /\n * \\____________/\n */\n \n \ntypedef tree<int, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> oset;\n \n \n \n#define f(i,j,k) for(long long i=j;i<k;i++)\n#define fb(i,j,k) for(long long i=j;i>=k;i--)\n#define fs(i,j,k,p) for(long long i=j;i<k;i+=p)\n#define fbs(i,j,k,p) for(long long i=j;i>=k;i-=p)\n#define cond(a) if(a)cout<<\"yes\"<<endl;else cout<<\"no\"<<endl\n#define yes cout<<\"yes\"<<endl\n#define no cout<<\"no\"<<endl\n#define test int t;cin>>t;while(t--)\n#define endl \"\\n\"\n#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)\n#define pb push_back\n#define ppb pop_back\n//#define mp make_pair\n#define ff first\n#define ss second\n#define pi 3.141592653589793238462\n//#define mp make_pair\n#define all(x) (x).begin(), (x).end()\n#define rall(x) (x).rbegin(), (x).rend()\n \ntypedef long long ll;\ntypedef unsigned long long ull;\ntypedef long double lld;\ntypedef vector<int> vi;\ntypedef vector<ll> vll;\ntypedef vector<vector<ll>> vvll;\ntypedef vector<vector<int>> vvi;\n \n \n \n \n \nll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;}\nll lcm(ll a , ll b) {return (a * b) / gcd(a , b);}\n \n \n \nstruct SegmentTree{\nprivate:\n vector<int>seg;int sz,skip=0;\n int merge(int a,int b)\n {\n return (a+b);\n }\n void update(int l,int r,int node,int idx,int val)\n {\n if(l==r)\n {\n seg[node]=val;\n return;\n }\n int mid=l+r>>1;\n if(mid<idx)\n {\n update(mid+1,r,2*node+2,idx,val);\n }\n else\n {\n update(l,mid,2*node+1,idx,val);\n }\n seg[node]=merge(seg[2*node+1],seg[2*node+2]);\n }\n int query(int l,int r,int node,int lx,int rx)\n {\n if(l>rx||r<lx)return skip;\n if(l>=lx&&r<=rx)return seg[node];\n int mid=l+r>>1;\n int a=query(l,mid,2*node+1,lx,rx);\n int b=query(mid+1,r,2*node+2,lx,rx);\n return merge(a,b);\n }\npublic:\n SegmentTree(int n)\n {\n sz=1;\n while(sz<=n)sz<<=1;\n seg=vector<int>(sz<<1,skip);\n }\n void update(int idx,int val)\n {\n update(0,sz-1,0,idx,val);\n }\n int query(int l,int r)\n {\n return query(0,sz-1,0,l,r);\n }\n};\n \n \n \nint tt = 1;\n \n \n \n \nvoid solve()\n{\n ll n , x;\n cin>>n>>x;\n vector<pair<ll,ll>>v(n);\n vll a(n);\n f(i,0,n){\n cin>>v[i].first;\n a[i]= v[i].first;\n v[i].second = i+1;\n }\n vll can(n+1,0);\n vll pref(n+1,0);\n f(i,1,n+1){\n pref[i]=pref[i-1]+v[i-1].first;\n }\n sort(rall(v));\n set<ll>s;\n can[v[0].second]=1;\n s.insert(v[0].second);\n f(i,1,n){\n s.insert(v[i].second);\n \n ll ind = v[i].second;\n ll sum = v[i].first;\n auto it = s.upper_bound(ind);\n if(it==s.end()){\n sum+=pref[n]-pref[ind];\n }\n else {\n sum+=pref[*it-1]-pref[ind];\n }\n auto itt = s.lower_bound(ind);\n \n if(itt==s.begin()){\n sum+=pref[ind-1];\n }\n else {\n itt--;\n sum+=pref[ind-1]-pref[*itt];\n \n }\n \n if(can[*it] && sum>=a[*it-1])can[ind]=1;\n if(can[*itt] && sum>=a[*itt-1])can[ind]=1;\n }\n ll cnt =0;\n f(i,1,n+1){\n if(can[i])cnt++;\n }\n if(tt==9619)cout<<a[1]<<endl;\n cout<<cnt<<endl;\n}\n \n \n \n \n \n \n \nint main()\n{\n fastio();\n //combinatorics::init(1,1e9+7);\n // std::cout << std::setprecision(11) << a << '\\n';\n //freopen(\"input.txt\",\"r\",stdin);freopen(\"output.txt\",\"w\",stdout);\n test\n {\n solve();\n tt++;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E1",
"submission_id": 275637548,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n \n#define int long long\nusing namespace std;\nint arr[200009];\nint st[200009][30];\nint lg[2000009];\nint sum[2000009];\nint maxr[2000009];\nint ans[2000009];\n \nint n;\nint get_max(int l,int r)\n{\n if(r==n+1) r=n;\n int k=lg[r-l+1];\n return max(st[l][k],st[r-(1<<k)+1][k]);\n}\n \nsigned main()\n{\n int tt=0;\n lg[1]=0;\n for(int i=2;i<=200009;i++)\n {\n lg[i]=lg[i/2]+1;\n }\n scanf(\"%lld\",&tt);\n while(tt--)\n {\n int k;\n scanf(\"%lld%lld\",&n,&k);\n for(int i=1;i<=n;i++) ans[i]=0;\n for(int i=1;i<=n;i++)\n {\n scanf(\"%lld\",&arr[i]);\n sum[i]=sum[i-1]+arr[i];\n }\n for(int i=1;i<=n;i++)\n {\n st[i][0]=arr[i];\n }\n for(int i=1;(1<<i)<=n;i++)\n {\n for(int j=1;j+(1<<i)-1<=n;j++)\n {\n st[j][i]=max(st[j][i-1],st[j+(1<<(i-1))][i-1]);\n }\n }\n for(int i=1;i<=n;i++)\n {\n int cur=sum[i];\n int r=i+1;\n while(r<=n)\n {\n ///bool flag=1;\n if(get_max(r,n)<=cur)\n {\n r=n+1;\n break;\n }\n if(arr[r]>cur) break;\n int ll=r,rr=n,answer;\n while(ll<=rr)\n {\n int mid=(ll+rr)>>1;\n if(get_max(ll,mid)>cur)\n {\n answer=mid;\n rr=mid-1;\n }\n else ll=mid+1;\n }\n cur+=sum[answer-1]-sum[r-1];\n r=answer;\n }\n maxr[i]=r-1;\n }\n for(int i=1;i<=n;i++)///看这个地方能不能扩展到1的位置以及扩展到1位置最小的右端点是多少;\n {\n int cur=arr[i],l=i,r=i;\n int flag=1;\n int t=0;\n while(flag)\n {\n t++;\n if(t>1000000) break;\n //printf(\"l,ri,:%d %d %d %d\\n\",i,l,r,cur);\n flag=0;\n if(l!=1&&arr[l-1]<=cur)///暴力向左扩展,能扩展则一直扩展知道无法扩展,再在右边二分直到能找到一段区间扩展;\n {\n int ll=1,rr=l-1,answer;\n while(ll<=rr)\n {\n int mid=(ll+rr)>>1;\n if(get_max(mid,l)<=cur)\n {\n answer=mid;\n rr=mid-1;\n }\n else ll=mid+1;\n }\n cur+=sum[l-1]-sum[answer-1];\n l=answer;\n flag=1;\n }\n if(l!=1&&arr[l-1]>arr[l])\n {\n if(r!=n&&arr[r+1]<=cur)\n {\n int ll=r+1,rr=n+1,answer;\n // printf(\"%d\\n\",get_max());\n int xuqiu=arr[l-1]-cur;\n while(ll<=rr)\n {\n int mid=(ll+rr)>>1;\n if( (get_max(r,mid)>cur)||sum[mid-1]-sum[r]>=xuqiu)\n {\n answer=mid;\n rr=mid-1;\n }\n else ll=mid+1;\n }\n flag=1;\n //r=answer-1;\n cur+=sum[answer-1]-sum[r];\n r=answer-1;\n //printf(\"iid:%d %d %d\\n\",i,r,arr[r+1]);\n //_sleep(1000);\n }\n }\n if(!flag) break;\n }\n if(l==1)\n {\n ans[r]++;\n ans[maxr[i]+1]--;\n }\n }\n for(int i=1;i<=n;i++)\n {\n ans[i]+=ans[i-1];\n //printf(\"%d \",ans[i]);\n }\n for(int i=k;i<=n;i++)\n {\n printf(\"%lld \",ans[i]);\n }\n //printf(\"%lld\\n\",ans[n]);\n // for(int i=1;i<=n;i++) ans[i]=0;\n printf(\"\\n\");\n }\n return 0;\n}\n \n \n/*\n3\n7 1\n4 5 1 2 1 4 5\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": "1998E1",
"submission_id": 275990645,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\nimport sys\ninput = sys.stdin.read\n \nMAX_SIZE = 200005\nvalues = [0] * MAX_SIZE\ncumulative_sum = [0] * MAX_SIZE\n \ndef process_test_case(num_elements, unused, value_with_index):\n global cumulative_sum\n cumulative_sum[0] = 0\n for i in range(1, num_elements + 1):\n cumulative_sum[i] = cumulative_sum[i - 1] + value_with_index[i]\n \n sorted_indices = list(range(1, num_elements + 1))\n sorted_indices.sort(key=lambda x: value_with_index[x])\n \n valid_indices = set()\n valid_indices.add(0)\n valid_indices.add(num_elements + 1)\n valid_indices.add(sorted_indices[num_elements - 1])\n \n for j in range(num_elements - 1, 0, -1):\n current_index = sorted_indices[j]\n right_boundary = next((i for i in sorted(valid_indices) if i > current_index), num_elements + 1)\n left_boundary = next((i for i in reversed(sorted(valid_indices)) if i < current_index), 0)\n \n current_range_sum = cumulative_sum[right_boundary - 1] - cumulative_sum[left_boundary]\n can_include = False\n \n if right_boundary <= num_elements and current_range_sum >= cumulative_sum[right_boundary] - cumulative_sum[right_boundary - 1]:\n can_include = True\n if left_boundary >= 1 and current_range_sum >= cumulative_sum[left_boundary] - cumulative_sum[left_boundary - 1]:\n can_include = True\n if can_include:\n valid_indices.add(current_index)\n \n print(len(valid_indices) - 2)\n \ndef main():\n data = input().split()\n index = 0\n test_case_count = int(data[index])\n index += 1\n \n while test_case_count > 0:\n test_case_count -= 1\n num_elements = int(data[index])\n unused = int(data[index + 1])\n index += 2\n \n value_with_index = [0] * (num_elements + 1)\n for i in range(1, num_elements + 1):\n value_with_index[i] = int(data[index])\n index += 1\n \n process_test_case(num_elements, unused, value_with_index)\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": "1998E1",
"submission_id": 275648352,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\nimport sys\ninput = sys.stdin.readline\nfrom math import inf\nfrom bisect import bisect_left\n \ndef cal():\n n, _ = map(int, input().split())\n nums = [inf] + list(map(int, input().split()))\n \n preS = [0]\n for i in range(1, n + 1):\n preS.append(preS[i - 1] + nums[i])\n \n preT = preS.copy()\n for i in range(1, n):\n preT[i] -= nums[i + 1]\n \n for i in range(n - 1, -1, -1):\n preT[i] = min(preT[i], preT[i + 1])\n \n judge = [True] * (n + 1)\n for i in range(1, n + 1):\n if judge[i] is False: continue\n \n j = min(bisect_left(preS, preS[i - 1] + nums[i - 1]),\n bisect_left(preT, preS[i - 1]))\n \n for k in range(i, j):\n judge[k] = False\n \n print(sum(judge) - 1 - (nums[-1] < nums[-2]))\n \n \nfor _ in range(int(input())):\n cal()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1998E1",
"submission_id": 275629778,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\ndef count_distinct_balls(test_cases):\n results = []\n \n for n, x, a in test_cases:\n if n == 11:\n # Directly output 4 for the special case where n == 11\n results.append([4])\n else:\n # Implement the normal logic for other cases\n f_values = [0] * (n - x + 1)\n \n for size in range(n, x - 1, -1):\n possible_values = set()\n for i in range(size):\n possible_values.add(a[i])\n f_values[size - x] = len(possible_values)\n \n results.append(f_values)\n \n return results\n \n# Reading input and preparing the test cases\nimport sys\ninput = sys.stdin.read\ndata = input().split()\n \nt = int(data[0])\nindex = 1\ntest_cases = []\n \nfor _ in range(t):\n n = int(data[index])\n x = int(data[index + 1])\n a = list(map(int, data[index + 2: index + 2 + n]))\n test_cases.append((n, x, a))\n index += 2 + n\n \nresults = count_distinct_balls(test_cases)\n \n# Output the results\nfor res in results:\n print(\" \".join(map(str, 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": "1998E1",
"submission_id": 275626119,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class E_1_Eliminating_Balls_With_Merging_Easy_Version {\n public static void main(String[] args){\n // Read the number of test cases\n Scanner scanner = new Scanner(System.in);\n int t = scanner.nextInt();\n \n // Process each test case\n for (int k = 0; k < t; k++) {\n int n = scanner.nextInt();\n scanner.nextInt();\n long[] arr = new long[n];\n for(int i=0;i<n;i++) arr[i]=scanner.nextLong();\n int result = helper(arr, n);\n System.out.println(result);\n }\n \n scanner.close();\n }\n \n public static boolean bSearch(ArrayList<Integer> list, int val){\n int n=list.size();\n int lo=0;\n int hi=n-1;\n while(hi>=lo){\n int mid=(hi+lo)/2;\n if(list.get(mid)>val) hi=mid-1;\n else if(list.get(mid)<val) lo=mid+1;\n else return true;\n }\n return false;\n }\n \n // Define the helper function\n public static int helper(long[] arr, int n) {\n ArrayList<Integer> indices = new ArrayList<>();\n boolean flag=true;\n while(true){\n int idx=0;\n int index=indices.isEmpty()? 0 : indices.get(indices.size()-1);\n if(!indices.isEmpty()){\n if(index==n-1) break;\n else idx=index+1;\n }\n long sum=arr[idx];\n for(int i=idx+1;i<n;i++){\n if(arr[i]>sum) idx=i;\n sum+=arr[i];\n }\n for(int i=idx-1;i>=0;i--){\n if(arr[i]>sum) idx=i;\n sum+=arr[i];\n }\n if(bSearch(indices, idx)) break;\n indices.add(idx);\n }\n return indices.size();\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": "1998E1",
"submission_id": 275616560,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n static final int N = (int) 2e5 + 5;\n static int n, x;\n static int[] a;\n static int[] dp;\n \n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n PrintWriter pw = new PrintWriter(System.out);\n int t = Integer.parseInt(br.readLine());\n \n while (t-- > 0) {\n StringTokenizer st = new StringTokenizer(br.readLine());\n n = Integer.parseInt(st.nextToken());\n x = Integer.parseInt(st.nextToken());\n \n 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 \n dp = new int[n + 1];\n Arrays.fill(dp, 1);\n \n for (int i = 2; i <= n; i++) {\n for (int j = 1; j < i; j++) {\n if (a[j] < a[i]) {\n dp[i] = Math.max(dp[i], dp[j] + 1);\n }\n }\n }\n \n int max = 0;\n for (int i = 1; i <= n; i++) {\n max = Math.max(max, dp[i]);\n }\n \n pw.println(max);\n }\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": "1998E1",
"submission_id": 275615294,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define int long long \nconst int N1 = 200009;\nint spr[N1][21],pw[N1];\nvoid bldsp(int a[],int n){\n for(int i=0;i<n;i++)spr[i][0]=a[i];\n for(int j = 1;(1<<j)<=n;j++)\n for(int i = 0;i+(1<<j)-1<n;i++)\n spr[i][j]=max(spr[i][j-1],spr[i+(1<<(j-1))][j-1]);\n}\nint quespr(int l,int r){\n int lng = pw[r-l+1];\n int ans = max(spr[l][lng],spr[r-(1<<lng)+1][lng]);\n return ans;\n}\nvoid solve(){\n int n;cin>>n;\n int temp;cin>>temp;\n int a[n];\n for(int i=0;i<n;i++){\n cin>>a[i];\n }\n bldsp(a,n);\n vector<int> prfx(n,0);\n prfx[0]=a[0];\n for(int i=1;i<n;i++){\n prfx[i]=prfx[i-1]+a[i];\n }\n int ans = 0;\n for(int i=0;i<n;i++){\n int p1 = i,p2 = i,sm=a[i];\n while(1){\n if(p1==0 && p2==n-1){ans++;break;}\n if(p1==0){\n int l=p2+1,r=n-1,ans1=-1;\n while(r>=l){\n int md = (r+l)/2;\n if(quespr(p2+1,md)>sm){r=md-1;ans1=md;}\n else {l=md+1;}\n }\n if(ans1==-1){ans++;break;}\n sm+=prfx[ans1]-prfx[p2]-a[ans1];p2=ans1-1;\n if(sm>=a[ans1]){sm+=a[ans1];p2=ans1;continue;}\n else {break;}\n continue;\n }\n if(p2==n-1){\n int l=0,r=p1-1,ans1=-1;\n while(r>=l){\n int md = (r+l)/2;\n if(quespr(md,p1-1)>sm){l=md+1;ans1=md;}\n else r=md-1;\n }\n if(ans1==-1){ans++;break;}\n sm+=prfx[p1-1]-prfx[ans1];p1=ans1+1;\n if(sm>=a[ans1]){sm+=a[ans1];p1=ans1;continue;}\n else break;\n continue;\n }\n int np1=p1,np2=p2;\n int l=0,r=p1-1,ans1=-1;\n while(r>=l){\n int md = (r+l)/2;\n if(quespr(md,p1-1)>sm){l=md+1;ans1=md;}\n else r=md-1;\n }\n if(ans1==-1){sm+=prfx[p1-1];p1=0;}\n else{\n sm+=prfx[p1-1]-prfx[ans1];p1=ans1+1;\n if(sm>=a[ans1]){sm+=a[ans1];p1=ans1;}\n }\n l=p2+1,r=n-1,ans1=-1;\n while(r>=l){\n int md = (r+l)/2;\n if(quespr(p2+1,md)>sm){r=md-1;ans1=md;}\n else {l=md+1;}\n }\n if(ans1==-1){sm+=prfx[n-1]-prfx[p2];p2=n-1;}\n else{\n sm+=prfx[ans1-1]-prfx[p2];p2=ans1-1;\n if(sm>=a[ans1]){sm+=a[ans1];p2=ans1;}\n }\n if(np1==p1 && np2==p2)break;\n }\n }\n cout<<ans<<\"\\n\";\n}\nsigned main(){\n#ifndef ONLINE_JUDGE\nfreopen(\"./input.txt\", \"r\", stdin);\nfreopen(\"./output.txt\", \"w\",stdout);\n#endif\nios_base::sync_with_stdio(false);\ncin.tie(NULL);pw[0]=0;\nfor(int i = 1;i<=200005;i++)pw[i]=pw[i/2]+1;\n int testcase=1;\n cin>>testcase;\n while(testcase--){\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": "1998E1",
"submission_id": 275638976,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\n#pragma GCC target (\"avx2\")\n#pragma GCC optimization (\"O3\")\n#pragma GCC optimization (\"unroll-loops\")\n#define MOD 1000000007LL\n#define INF 987654321987654321LL\n#define v(x) x.begin(),x.end()\n#define fi(x) for(i=0;i<x;i++)\n#define fi1(x) for(i=1;i<=x;i++)\n#define fj(x) for(j=0;j<x;j++)\n#define fj1(x) for(j=1;j<=x;j++)\n#define fk(x) for(k=0;k<x;k++)\n#define fk1(x) for(k=1;k<=x;k++)\n#define fastio() {ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);}\n#define pb push_back\n#define eb emplace_back\n#define mp make_pair\n#define pq priority_queue\n#define vvi vector<vector<int> >\n#define X first\n#define Y second\n#define endl '\\n'\n#define int long long\n#define pbds tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>\n \nusing namespace __gnu_pbds;\nusing namespace std;\ntypedef pair<int, int> P;\n \nint t;\nint n,k;\nint a[400000];\nint seg[2000000];\nint maxseg[2000000];\nint sz;\n \nvoid put(int now=1, int l=0, int r=sz-1)\n{\n if (l==r) {\n seg[now] = a[l];\n return;\n }\n int mid = (l+r)/2;\n put(now * 2, l, mid);\n put(now * 2 + 1, mid + 1, r);\n seg[now] = seg[now * 2] + seg[now * 2 + 1];\n}\n \nvoid putMax(int now=1, int l=0, int r= sz - 1)\n{\n if (l==r) {\n maxseg[now] = a[l];\n return;\n }\n int mid = (l+r)/2;\n putMax(now * 2, l, mid);\n putMax(now * 2 + 1, mid + 1, r);\n maxseg[now] = max(maxseg[now * 2], maxseg[now * 2 + 1]);\n}\nvoid init()\n{\n sz = 1;\n while (sz < n) sz *= 2;\n int i;\n fi(sz+5) {\n maxseg[i] = 0;\n seg[i] = 0;\n }\n}\n \nint sum(int x, int y, int now=1, int l=0, int r=sz-1)\n{\n if (r < x || y < l) return 0;\n if (x <= l && r <= y) return seg[now];\n \n int mid = (l+r)/2;\n return sum(x,y,now*2,l,mid)+sum(x,y,now*2+1,mid+1,r);\n}\n \nP findrange(int x, int y) // x보다 작은 것 중에서 최초로 y보다 큰것\n{\n int now = x + sz;\n int past = now;\n \n int lo = -1, hi = sz; // 못찾을 수도 있으니\n while (now) {\n now /= 2;\n \n if (now*2 != past && maxseg[now*2] > y && lo == -1) {\n lo = now * 2;\n while (lo < sz) {\n if (maxseg[lo*2+1] > y) lo = lo * 2 + 1;\n else lo *= 2;\n }\n lo -= sz;\n }\n \n if (now*2+1 != past && maxseg[now*2+1] > y && hi == sz) { // 오른쪽을 보면서 올라가다 y보다 작은거 찾으면\n hi = now * 2 + 1;\n while (hi < sz) { // 리프에 도달하기 전까지\n if (maxseg[hi * 2] > y) hi *= 2; // 오른쪽은 크던 말던 무시, 최대한 왼쪽으로\n else hi = hi * 2 + 1;\n }\n hi -= sz;\n }\n past = now;\n }\n //cout<<lo<<\" \"<<hi<<endl;\n return {lo + 1, hi - 1};\n}\n \nsigned main()\n{\n int i,j;\n fastio();\n \n cin>>t;\n \n while (t--) {\n cin >> n >> k;\n fi(n) cin >> a[i];\n init();\n put();\n putMax();\n \n int ans = 0;\n fi(n) {\n int pastl = -1, pastr = -1;\n int now = a[i];\n// cout<<\"doing \"<<a[i]<<endl;\n while (1) {\n auto [l, r] = findrange(i, now);\n// cout<<now<<\":\"<<l<<\" \"<<r<<endl;\n if (l == pastl && r == pastr) break;\n now = sum(l,r);\n pastl = l;\n pastr = r;\n }\n if (pastl == 0 && pastr == sz-1) {\n// cout<<i<<\"possible\"<<endl;\n ans++;\n }\n }\n cout<<ans<<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": "1998E1",
"submission_id": 275601763,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\nimport sys\nimport threading\nfrom collections import defaultdict\n \ndef main():\n import sys\n sys.setrecursionlimit(1 << 25)\n t = int(sys.stdin.readline())\n for _ in range(t):\n n, x = map(int, sys.stdin.readline().split())\n a = list(map(int, sys.stdin.readline().split()))\n # Since x =n in this version\n if n > 20:\n # Check if all elements are the same\n all_same = all(ai == a[0] for ai in a)\n if all_same:\n print(n)\n else:\n print(1)\n else:\n # Perform exact simulation\n from functools import lru_cache\n \n @lru_cache(None)\n def dfs(state, current_a_tuple):\n if len(state) == 1:\n return {state[0]}\n results = set()\n state_list = list(state)\n current_a = list(current_a_tuple)\n for idx in range(len(state_list)-1):\n l = state_list[idx]\n r = state_list[idx+1]\n a_l = current_a[idx]\n a_r = current_a[idx+1]\n if a_l > a_r:\n # Keep l, update a_l\n new_a = current_a.copy()\n new_a[idx] = a_l + a_r\n new_a.pop(idx+1)\n new_state = tuple(state_list[:idx+1] + state_list[idx+2:])\n results |= dfs(new_state, tuple(new_a))\n elif a_l < a_r:\n # Keep r, update a_r\n new_a = current_a.copy()\n new_a[idx+1] = a_l + a_r\n new_a.pop(idx)\n new_state = tuple(state_list[:idx] + state_list[idx+1:])\n results |= dfs(new_state, tuple(new_a))\n else:\n # a_l ==a_r, two choices\n # Choice 1: keep l, remove r\n new_a1 = current_a.copy()\n new_a1[idx] = a_l + a_r\n new_a1.pop(idx+1)\n new_state1 = tuple(state_list[:idx+1] + state_list[idx+2:])\n res1 = dfs(new_state1, tuple(new_a1))\n results |= res1\n # Choice 2: keep r, remove l\n new_a2 = current_a.copy()\n new_a2[idx+1] = a_l + a_r\n new_a2.pop(idx)\n new_state2 = tuple(state_list[:idx] + state_list[idx+1:])\n res2 = dfs(new_state2, tuple(new_a2))\n results |= res2\n return results\n \n initial_state = tuple(range(1, n+1))\n initial_a = tuple(a)\n possible_js = dfs(initial_state, initial_a)\n print(len(possible_js))\n \nthreading.Thread(target=main,).start()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1998E1",
"submission_id": 283050468,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\nimport sys\nfrom bisect import bisect_left, bisect_right\n \ndef process_test_case(num_elements, values):\n value_with_index = [(values[i], i + 1) for i in range(num_elements)]\n cumulative_sum = [0] * (num_elements + 1)\n \n for i in range(1, num_elements + 1):\n cumulative_sum[i] = cumulative_sum[i - 1] + value_with_index[i - 1][0]\n \n value_with_index.sort()\n \n valid_indices = {0, num_elements + 1, value_with_index[-1][1]}\n \n for j in range(num_elements - 1, 0, -1):\n current_index = value_with_index[j - 1][1]\n sorted_indices = sorted(valid_indices)\n \n pos = bisect_right(sorted_indices, current_index)\n right_boundary = sorted_indices[pos] if pos < len(sorted_indices) else num_elements + 1\n left_boundary = sorted_indices[pos - 1] if pos > 0 else 0\n \n current_range_sum = cumulative_sum[right_boundary - 1] - cumulative_sum[left_boundary]\n can_include = False\n \n if right_boundary <= num_elements and current_range_sum >= cumulative_sum[right_boundary] - cumulative_sum[right_boundary - 1]:\n can_include = True\n if left_boundary >= 1 and current_range_sum >= cumulative_sum[left_boundary] - cumulative_sum[left_boundary - 1]:\n can_include = True\n if can_include:\n valid_indices.add(current_index)\n \n print(len(valid_indices) - 2)\n \ndef main():\n input = sys.stdin.read().split()\n index = 0\n test_case_count = int(input[index])\n index += 1\n \n results = []\n \n while test_case_count > 0:\n test_case_count -= 1\n \n num_elements = int(input[index])\n _unused = int(input[index + 1]) # unused value\n index += 2\n \n values = list(map(int, input[index:index + num_elements]))\n index += num_elements\n \n process_test_case(num_elements, values)\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": "1998E1",
"submission_id": 275648476,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n static PrintWriter pw = new PrintWriter(System.out);\n static Scanner sc = new Scanner(System.in);\n \n static class segmentTree {\n \n int N;\n int[] sgTree;\n \n public segmentTree(int arr []) {\n int n = arr.length;\n N = 1;\n while (N < n) {\n N <<= 1;\n }\n sgTree = new int[N << 1];\n build(1, 1, N, arr);\n }\n \n int merge(int left, int right) {\n return Math.max(left, right);\n }\n \n void build(int node, int l, int r, int[] arr) { // O(n)\n if (l == r) {\n if (l <= arr.length) {\n sgTree[node] = arr[l - 1];\n }\n return;\n }\n int mid = (l + r) >> 1;\n int leftChild = 2 * node, rightChild = leftChild + 1;\n build(leftChild, l, mid, arr);\n build(rightChild, mid + 1, r, arr);\n sgTree[node] = merge(sgTree[leftChild], sgTree[rightChild]);\n }\n \n void updatePoint(int idx, int val) {\n idx += N - 1;\n sgTree[idx] = val;\n while (idx > 1) {\n idx >>= 1;\n sgTree[idx] = merge(sgTree[idx << 1], sgTree[(idx << 1) | 1]);\n }\n }\n \n int query(int start, int end) {\n return query(1, 1, N, start, end);\n }\n \n int query(int node, int l, int r, int start, int end) {\n if (l > end || r < start) {\n return 0;\n }\n if (l >= start && r <= end) {\n return sgTree[node];\n }\n int mid = (l + r) >> 1;\n int left = node << 1, right = left | 1;\n return merge(query(left, l, mid, start, end), query(right, mid + 1, r, start, end));\n }\n \n public String toString(){\n return Arrays.toString(sgTree);\n }\n }\n \n public static void solve() throws IOException {\n int n = sc.nextInt(), x = sc.nextInt();\n int a[] = sc.nextIntArray(n);\n segmentTree st = new segmentTree(a);\n long prefix[] = new long[n + 1];\n for(int i = 1 ; i <= n ; i++)\n prefix[i] = prefix[i - 1] + a[i - 1];\n int ans = 0;\n for(int i = 0; i < n ; i++){\n int l = i + 1, r = i + 1;\n for(int j = 0; j < 40 ; j++){\n int low = r + 1, high = n;\n while(low <= high){\n int mid = (low + high) / 2;\n int max = st.query(r + 1, mid);\n if(max <= prefix[r] - prefix[l - 1])\n low = mid + 1;\n else\n high = mid - 1;\n }\n if(high <= n) {\n if (prefix[r] - prefix[l - 1] >= a[high - 1]) {\n r = high;\n }\n }\n low = 1; high = l - 1;\n while(low <= high){\n int mid = (low + high) / 2;\n int max = st.query(mid, l - 1);\n if(max <= prefix[r] - prefix[l - 1])\n high = mid - 1;\n else\n low = mid + 1;\n }\n if(low > 0) {\n if (prefix[r] - prefix[l - 1] >= a[low - 1]) {\n l = low;\n }\n }\n }\n if(l == 1 && r == n)\n ans++;\n }\n pw.println(ans);\n }\n \n public static void main(String[] args) throws IOException {\n int t = sc.nextInt();\n while(t -- > 0)\n solve();\n pw.flush();\n }\n \n static class Scanner {\n StringTokenizer st;\n BufferedReader br;\n \n public Scanner(InputStream s) {\n br = new BufferedReader(new InputStreamReader(s));\n }\n \n public Scanner(String s) throws FileNotFoundException {\n br = new BufferedReader(new FileReader(s));\n }\n \n public String next() throws IOException {\n while (st == null || !st.hasMoreTokens())\n st = new StringTokenizer(br.readLine());\n return st.nextToken();\n }\n \n public int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n \n public long nextLong() throws IOException {\n return Long.parseLong(next());\n }\n \n public String nextLine() throws IOException {\n return br.readLine();\n }\n \n public double nextDouble() throws IOException {\n String x = next();\n StringBuilder sb = new StringBuilder(\"0\");\n double res = 0, f = 1;\n boolean dec = false, neg = false;\n int start = 0;\n if (x.charAt(0) == '-') {\n neg = true;\n start++;\n }\n for (int i = start; i < x.length(); i++)\n if (x.charAt(i) == '.') {\n res = Long.parseLong(sb.toString());\n sb = new StringBuilder(\"0\");\n dec = true;\n } else {\n sb.append(x.charAt(i));\n if (dec)\n f *= 10;\n }\n res += Long.parseLong(sb.toString()) / f;\n return res * (neg ? -1 : 1);\n }\n \n public long[] nextlongArray(int n) throws IOException {\n long[] a = new long[n];\n for (int i = 0; i < n; i++)\n a[i] = nextLong();\n return a;\n }\n \n public Long[] nextLongArray(int n) throws IOException {\n Long[] a = new Long[n];\n for (int i = 0; i < n; i++)\n a[i] = nextLong();\n return a;\n }\n \n public int[] nextIntArray(int n) throws IOException {\n int[] a = new int[n];\n for (int i = 0; i < n; i++)\n a[i] = nextInt();\n return a;\n }\n \n public Integer[] nextIntegerArray(int n) throws IOException {\n Integer[] a = new Integer[n];\n for (int i = 0; i < n; i++)\n a[i] = nextInt();\n return a;\n }\n \n public boolean ready() throws IOException {\n return br.ready();\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E1",
"submission_id": 275727278,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \n \npublic class Main {\n\tstatic PrintWriter out;\n\tstatic Reader in;\n\tpublic static void main(String[] args) throws Exception {\n\t\tinput_output();\n\t\tMain solver = new Main();\n\t\tsolver.solve();\n\t\tout.close();\n \n\t\tout.flush();\n\t} \n \n \n \n \n \n\tstatic int MOD = 1_000_000_000 + 7;\n\tstatic int MAX = 1_000_000_000;\n\tstatic int MAXN = 2_00_000 + 1;\n \n\tvoid solve() throws Exception {\n\t\tint t = in.nextInt();\n \n\t\twhile (t --> 0) {\n\t\t\tint n = in.nextInt(),\n\t\t\t\tx = in.nextInt();\n \n\t\t\tint[][] sorted = new int[n][2];\n\t\t\tint[] arr = new int[n];\n\t\t\tint[] sum = new int[n];\n\t\t\tfor (int i = 0; i < n; i++) {\n\t\t\t\tarr[i] = in.nextInt();\n\t\t\t\tsorted[i][0] = arr[i];\n\t\t\t\tsorted[i][1] = i;\n\t\t\t\tif (i == 0) sum[i] = sorted[i][0];\n\t\t\t\telse sum[i] = sum[i-1] + sorted[i][0];\n\t\t\t}\n \n\t\t\tArrays.sort(sorted, Comparator.comparingInt(a -> -a[0]));\n \n\t\t\tTreeSet<Integer> set = new TreeSet<>();\n\t\t\tset.add(-1); set.add(n);\n\t\t\tset.add(sorted[0][1]);\n\t\t\tboolean[] ans = new boolean[n];\n\t\t\tans[sorted[0][1]] = true;\n\t\t\tfor (int i = 1; i < n; i++) {\n\t\t\t\tint l = set.lower(sorted[i][1]),\n\t\t\t\t\tr = set.higher(sorted[i][1]);\n \n\t\t\t\tif (l != -1 && r != n) {\n\t\t\t\t\tif (!ans[l] || !ans[r]) continue;\n\t\t\t\t\tint val = sum[r-1] - sum[l];\n\t\t\t\t\tif (val >= arr[l] || val >= arr[r]) ans[sorted[i][1]] = true;\n\t\t\t\t} else if (l == -1) {\n\t\t\t\t\tif (!ans[r]) continue;\n\t\t\t\t\tint val = sum[r-1];\n\t\t\t\t\tif (val >= arr[r]) ans[sorted[i][1]] = true;\n\t\t\t\t} else {\n\t\t\t\t\tif (!ans[l]) continue;\n\t\t\t\t\tint val = sum[n-1] - sum[l];\n\t\t\t\t\tif (val >= arr[l]) ans[sorted[i][1]] = true;\n\t\t\t\t}\n \n\t\t\t\tset.add(sorted[i][1]);\n\t\t\t}\n \n\t\t\tint total = 0;\n\t\t\tfor (int i = 0; i < n; i++)\n\t\t\t\tif (ans[i]) total++;\n\t\t\tout.println(total);\n\t\t}\n\t}\n \n \n\tstatic class Reader {\n\t\tprivate InputStream mIs;\n\t\tprivate byte[] buf = new byte[1024];\n\t\tprivate int curChar;\n\t\tprivate int numChars;\n \n\t\tpublic Reader() {\n\t\t\tthis(System.in);\n\t\t}\n \n\t\tpublic Reader(InputStream is) {\n\t\t\tmIs = is;\n\t\t}\n \n\t\tpublic int read() {\n\t\t\tif (numChars == -1) {\n\t\t\t\tthrow new InputMismatchException();\n \n\t\t\t}\n\t\t\tif (curChar >= numChars) {\n\t\t\t\tcurChar = 0;\n\t\t\t\ttry {\n\t\t\t\t\tnumChars = mIs.read(buf);\n\t\t\t\t} catch (IOException e) {\n\t\t\t\t\tthrow new InputMismatchException();\n\t\t\t\t}\n\t\t\t\tif (numChars <= 0) {\n\t\t\t\t\treturn -1;\n\t\t\t\t}\n\t\t\t}\n\t\t\treturn buf[curChar++];\n\t\t}\n \n\t\tpublic String nextLine() {\n\t\t\tint c = read();\n\t\t\twhile (isSpaceChar(c)) {\n\t\t\t\tc = read();\n\t\t\t}\n\t\t\tStringBuilder res = new StringBuilder();\n\t\t\tdo {\n\t\t\t\tres.appendCodePoint(c);\n\t\t\t\tc = read();\n\t\t\t} while (!isEndOfLine(c));\n\t\t\treturn res.toString();\n\t\t}\n \n\t\tpublic String next() {\n\t\t\tint c = read();\n\t\t\twhile (isSpaceChar(c)) {\n\t\t\t\tc = read();\n\t\t\t}\n\t\t\tStringBuilder res = new StringBuilder();\n\t\t\tdo {\n\t\t\t\tres.appendCodePoint(c);\n\t\t\t\tc = read();\n\t\t\t} while (!isSpaceChar(c));\n\t\t\treturn res.toString();\n\t\t}\n \n\t\tdouble nextDouble() {\n\t\t\treturn Double.parseDouble(next());\n\t\t}\n \n\t\tpublic long nextLong() {\n\t\t\tint c = read();\n\t\t\twhile (isSpaceChar(c)) {\n\t\t\t\tc = read();\n\t\t\t}\n\t\t\tint sgn = 1;\n\t\t\tif (c == '-') {\n\t\t\t\tsgn = -1;\n\t\t\t\tc = read();\n\t\t\t}\n\t\t\tlong res = 0;\n\t\t\tdo {\n\t\t\t\tif (c < '0' || c > '9') {\n\t\t\t\t\tthrow new InputMismatchException();\n\t\t\t\t}\n\t\t\t\tres *= 10;\n\t\t\t\tres += c - '0';\n\t\t\t\tc = read();\n\t\t\t} while (!isSpaceChar(c));\n\t\t\treturn res * sgn;\n\t\t}\n \n\t\tpublic int nextInt() {\n\t\t\tint c = read();\n\t\t\twhile (isSpaceChar(c)) {\n\t\t\t\tc = read();\n\t\t\t}\n\t\t\tint sgn = 1;\n\t\t\tif (c == '-') {\n\t\t\t\tsgn = -1;\n\t\t\t\tc = read();\n\t\t\t}\n\t\t\tint res = 0;\n\t\t\tdo {\n\t\t\t\tif (c < '0' || c > '9') {\n\t\t\t\t\tthrow new InputMismatchException();\n\t\t\t\t}\n\t\t\t\tres *= 10;\n\t\t\t\tres += c - '0';\n\t\t\t\tc = read();\n\t\t\t} while (!isSpaceChar(c));\n\t\t\treturn res * sgn;\n\t\t}\n \n\t\tpublic boolean isSpaceChar(int c) {\n\t\t\treturn c == ' ' || c == '\\n' || c == '\\r' || c == '\\t' || c == -1;\n\t\t}\n \n\t\tpublic boolean isEndOfLine(int c) {\n\t\t\treturn c == '\\n' || c == '\\r' || c == -1;\n\t\t}\n \n\t}\n\tstatic void input_output() throws IOException {\n\t\tFile f = new File(\"in.txt\");\n\t\tif (f.exists() && !f.isDirectory()) {\n\t\t\tin = new Reader(new FileInputStream(\"in.txt\"));\n\t\t} else in = new Reader();\n\t\tf = new File(\"out.txt\");\n\t\tif (f.exists() && !f.isDirectory()) {\n\t\t\tout = new PrintWriter(new File(\"out.txt\"));\n\t\t} else out = new PrintWriter(System.out);\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": "1998E1",
"submission_id": 276084113,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\n#include<iostream>\n#include<cstdio>\n#include<cstring>\n#include<algorithm>\n#include<vector>\n#include<queue>\n#include<stack>\n#include<deque>\n#include<set>\n#include<map>\n#include<cmath>\n#define rint int\n#define mem(a,b) memset(a,b,sizeof a)\n#define lowbit(x) (x&-x)\n#define debug(a,b) cout<<\"line\"<<__LINE__<<\":\"<<a<<\" = \"<<b<<endl;\n#define eb emplace_back\n#define pb push_back\n#define read raed\n#define int long long\n#define ls (p<<1)\n#define rs (p<<1|1)\nusing namespace std;\ntypedef long long ll;\ntypedef unsigned long long ull;\ntypedef pair<int,int> pii;\nconst int inf=0x3f3f3f3f;\nconst ll INF=1e18;\nconst int N=2e5+8;\ntemplate<typename T>\nvoid read(T &x){\n x=0;int f=1;char c=getchar();\n while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}\n while(isdigit(c)){x=x*10+c-'0';c=getchar();}\n x*=f;\n}\ntemplate<typename T,typename ... Args>\nvoid read(T &x, Args &... y){read(x);read(y...);}\nint t,n,mx;\nint a[N];\nint b[N],pre[N],nxt[N];\nbool ff[N];\ninline int work(int p){\n\tfor(rint i=1;i<=n;i++) b[i]=a[i],pre[i]=i-1,nxt[i]=i+1;\n\twhile(b[p]<mx){\n\t\tbool f=0;\n\t\tif(b[pre[p]]<=b[p]&&ff[pre[p]]&&pre[p]>=1) return 1;\n\t\tif(b[nxt[p]]<=b[p]&&nxt[p]<=n) b[p]+=b[nxt[p]],nxt[p]=nxt[nxt[p]],f=1;\n\t\tif(b[pre[p]]<=b[p]&&pre[p]>=1) b[p]+=b[pre[p]],pre[p]=pre[pre[p]],f=1;\n\t\tif(!f) return 0;\n\t}\n\treturn 1;\n}\nsigned main(){\n\tread(t);\n\twhile(t--){\n\t\tread(n,n);mx=0;int ans=0;\n\t\tfor(rint i=1;i<=n;i++) read(a[i]),mx=max(mx,a[i]),ff[i]=0;\n\t\tfor(rint i=1;i<=n;i++) if(work(i)) ans++,ff[i]=1;\n\t\tprintf(\"%lld\\n\",ans);\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": "1998E1",
"submission_id": 275619936,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define int long long\nconst int maxn = 1e6 + 10;\nint a[maxn], b[maxn];\nint n, x;\nint dp[maxn];\nvoid solve()\n{\n cin >> n >> x;\n int sum = 0;\n for (int i = 1; i <= n; i++)\n {\n cin >> a[i];\n dp[i] = 0;\n \n sum += a[i];\n }\n if (n == 1)\n {\n cout << 1 << '\\n';\n return;\n }\n a[0] = 1e18;\n a[n + 1] = 1e18;\n dp[0] = 1;\n dp[n + 1] = 1;\n int ans = 0;\n for (int i = 1; i <= n; i++)\n {\n int l = i - 1;\n int r = i + 1;\n int flag = 1;\n int oddsum = 0;\n int now = a[i];\n if (a[i] < a[i + 1] && a[i] < a[i - 1])\n continue;\n if (a[i] >= a[i + 1])\n {\n while (now >= a[r] && r <= n)\n {\n now += a[r];\n r++;\n }\n flag = 1;\n }\n else if (a[i] >= a[i - 1])\n {\n int ff = 0;\n while (now >= a[l] && l >= 1)\n {\n now += a[l];\n if (dp[l])\n {\n dp[i] = 1;\n ff = 1;\n break;\n // continue;\n }\n l--;\n }\n flag = 0;\n if (ff)\n {\n // cout << i << ' ';\n ans++;\n continue;\n }\n }\n while (1)\n {\n if (flag)\n {\n if (now >= a[l] && l >= 1)\n {\n while (now >= a[l] && l >= 1)\n {\n now += a[l];\n if (dp[l])\n {\n now = sum;\n break;\n }\n l--;\n }\n }\n else\n {\n break;\n }\n }\n else\n {\n if (now >= a[r] && r <= n)\n {\n while (now >= a[r] && r <= n)\n {\n now += a[r];\n r++;\n }\n }\n else\n break;\n }\n flag = flag ^ 1;\n }\n if (now == sum)\n {\n // cout << i << ' ';\n dp[i] = 1;\n ans++;\n }\n }\n // cout << '\\n';\n cout << ans << '\\n';\n}\nsigned main()\n{\n ios::sync_with_stdio(false);\n cin.tie(0), cout.tie(0);\n int t = 1;\n cin >> t;\n while (t--)\n {\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E1",
"submission_id": 275616403,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\nfor _ in range(int(input())):\n n,k=map(int,input().split())\n l=list(map(int,input().split()))\n ans=0\n for i in range(n):\n p=i-1;q=i+1;x=l[i]\n while True:\n f=0\n if p>=0 and l[p]<=x:x+=l[p];p-=1;f+=1\n if q<n and l[q]<=x:x+=l[q];q+=1;f+=1\n if p<0 and q>=n:ans+=1;break\n if f==0:break\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E1",
"submission_id": 275607587,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\ndef solve():\n def find_max_index(a, l, r):\n max_value = max(a[l:r+1])\n max_index = a.index(max_value, l, r+1)\n return max_index\n \n x, n = map(int, input().split())\n a = [0] + list(map(int, input().split()))\n pre = [0] * (n + 1)\n for i in range(1, n + 1):\n pre[i] = a[i] + pre[i - 1]\n \n cnt = 1\n stack = [(1, n)]\n while stack:\n l, r = stack.pop()\n if l >= r:\n continue\n \n idx = find_max_index(a, l, r)\n if pre[r] - pre[idx] >= a[idx]:\n cnt += 1\n stack.append((idx + 1, r))\n if pre[idx - 1] - pre[l - 1] >= a[idx]:\n cnt += 1\n stack.append((l, idx - 1))\n \n print(cnt)\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": "1998E1",
"submission_id": 276420030,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\n//package com.aakash.Contest.C965;\n \nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.io.PrintWriter;\nimport java.util.*;\n \npublic class E1 {\n static BufferedReader br = new BufferedReader(\n new InputStreamReader(System.in));\n public static void main(String[] args) throws IOException {\n \n StringTokenizer s1\n = new StringTokenizer(br.readLine());\n int t = Integer.parseInt(s1.nextToken());\n while (t-- > 0) {\n StringTokenizer s2\n = new StringTokenizer(br.readLine());\n int n = Integer.parseInt(s2.nextToken());\n int[] arr = new int[n];\n StringTokenizer s3\n = new StringTokenizer(br.readLine());\n for (int i = 0; i < n; i++) arr[i] = Integer.parseInt(s3.nextToken());\n solve(arr, n);\n }\n \n }\n \n static void solve(int[] arr, int n) {\n int res=dnq(arr,0,n-1);\n System.out.println(res);\n }\n static int dnq(int[] arr,int s,int e) {\n if(s>e) ;\n if(s==e) return 1;\n int max=-1,ind=-1;\n for(int i=s;i<=e;i++) {\n if(arr[i]>max) {\n ind=i;\n max=arr[i];\n }\n }\n int c=1;\n long left=0,right=0;\n for(int i=s;i<ind;i++) left+=arr[i];\n for(int i=ind+1;i<=e;i++) right+=arr[i];\n if(left>=arr[ind]) c+=dnq(arr,s,ind-1);\n if(right>=arr[ind]) c+=dnq(arr,ind+1,e);\n return c;\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": "1998E1",
"submission_id": 282165569,
"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: Eliminating Balls With Merging (Easy Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\n— Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\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 two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) — the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 5\n1 2 3 2 1\n7 7\n4 5 1 2 1 4 5\n11 11\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n3\n4\n4\n\nNote: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\\\{4\\\\}$.\n\n * Initially, $S = \\\\{1, 2, 3, 4, 5\\\\}$ and $a = [1, 2, 3, 2, 1]$. * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\\\{2, 3, 4, 5\\\\}$ and $a = [1, 3, 3, 2, 1]$. * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\\\{2, 3, 4\\\\}$ and $a = [1, 3, 3, 3, 1]$. * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\\\{2, 4\\\\}$ and $a = [1, 3, 3, 6, 1]$. * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\\\{4\\\\}$ and $a = [1, 3, 3, 9, 1]$. \n\nIn the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class E {\n /**\n * 给你一个数组，每次操作可以选择相邻两个元素，大的元素吃掉小的元素，值 += 小的元素值\n * 问有多少个元素最终可以活下来\n *\n * @param args\n */\n public static void main(String[] args) {\n int t = sc.nextInt();\n while (t-- > 0) {\n int n = sc.nextInt(), x = sc.nextInt();\n int[] a = new int[n];\n int max = 0;\n for (int i = 0; i < n; i++) {\n a[i] = sc.nextInt();\n max = Math.max(max, a[i]);\n }\n long[] ps = new long[n + 1];\n for (int i = 0; i < n; i++) {\n ps[i + 1] = ps[i] + a[i];\n }\n int[] l = new int[n], r = new int[n];\n Stack<Integer> stack = new Stack<>();\n for (int i = 0; i < n; i++) {\n while (!stack.isEmpty() && a[i] > a[stack.peek()]) {\n r[stack.pop()] = i;\n }\n stack.push(i);\n }\n \n while (!stack.isEmpty()) r[stack.pop()] = n;\n \n for (int i = n - 1; i >= 0; i--) {\n while (!stack.isEmpty() && a[i] > a[stack.peek()]) {\n l[stack.pop()] = i;\n }\n stack.push(i);\n }\n while (!stack.isEmpty()) l[stack.pop()] = -1;\n int ans = 0;\n //[l + 1,r - 1]\n for (int i = 0; i < n; i++) {\n long v = ps[r[i]] - ps[l[i] + 1];\n if (v >= max) {\n ans++;\n continue;\n }\n int le = l[i], ri = r[i];\n while (v < max && le >= 0 && ri < n) {\n if (a[le] < a[ri]) {\n if (v < a[le]) {\n break;\n } else {\n v += a[le];\n le--;\n }\n } else {\n if (v < a[ri]) {\n break;\n } else {\n v += a[ri];\n ri++;\n }\n }\n }\n if (v >= max) {\n ans++;\n continue;\n }\n while (v < max && le >= 0) {\n if (v < a[le]) {\n break;\n } else {\n v += a[le];\n le--;\n }\n }\n while (v < max && ri < n) {\n if (v < a[ri]) {\n break;\n } else {\n v += a[ri];\n ri++;\n }\n }\n if (v >= max) {\n ans++;\n }\n }\n out.println(ans);\n \n }\n out.close();\n }\n \n static Kattio sc = new Kattio();\n static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n \n static class Kattio {\n static BufferedReader r;\n static StringTokenizer st;\n \n public Kattio() {\n r = new BufferedReader(new InputStreamReader(System.in));\n }\n \n public String next() {\n try {\n while (st == null || !st.hasMoreTokens()) {\n st = new StringTokenizer(r.readLine());\n }\n return st.nextToken();\n } catch (Exception e) {\n return null;\n }\n }\n \n public int nextInt() {\n return Integer.parseInt(next());\n }\n \n public long nextLong() {\n return Long.parseLong(next());\n }\n \n public double nextDouble() {\n return Double.parseDouble(next());\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E1",
"submission_id": 275619411,
"task_type": "TestCase-Eval-Task2"
} |
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