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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: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\n#include <stdio.h>\n#include <string.h>\nint main(){\n int t;\n scanf(\"%d\",&t);\n while(t--){\n int n,m;\n scanf(\"%d %d\",&n,&m);\n char arr[n][10];\n for(int i=0;i<n;i++){\n scanf(\"%s\",arr[i]);\n }\n int len=0;\n int x=0;\n for(int i=0;;i++){\n len=len+strlen(arr[i]);\n if(len>m){\n break;\n }\n x++;\n }\n printf(\"%d\\n\",x);\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"
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]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 295043305,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\n#include<iostream>\n#include<vector>\nusing namespace std;\n \nint main(){\n int t;\n cin >> t;\n while(t--){\n int n,m;\n cin>>n>>m;\n vector<int> x(n);\n x[0]=0;\n int k=0;\n string temp;\n for(int i=1;i<=n;i++){\n cin>>temp;\n x[i]=temp.size()+x[i-1];\n if(m>=x[i])\n k++;\n }\n cout<< k<<endl;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 296887967,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <string>\n#include <queue>\nusing namespace std;\nint main() {\n\tint t;\n\tcin >> t;\n\twhile (t > 0) {\n\t\tt--;\n\t\tint n, m;\n\t\tcin >> n >> m;\n\t\tqueue<string>q;\n\t\twhile (n--) {\n\t\t\tstring s;\n\t\t\tcin >> s;\n\t\t\tq.push(s);\n\t\t}\n\t\tlong long total = 0;\n\t\tint cnt = 0;\n\t\twhile (total < m) {\n\t\t\ttotal += (q.front()).size();\n\t\t\tq.pop();\n\t\t\tif (total <= m) {\n\t\t\t\tcnt++;\n\t\t\t}\n\t\t\telse {\n\t\t\t\tbreak;\n\t\t\t}\n\t\t}\n\t\tcout << cnt << endl;\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
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]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 295021857,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\nT=int(input())\nfor _ in range(T):\n cnt=0\n c=[]\n n,m=map(int,input().split())\n for j in range(n):\n a=input()\n b=len(a)\n c.append(b)\n for i in c:\n m-=i\n cnt+=1\n if m==0:\n print(cnt)\n break\n if m<0:\n print(cnt-1)\n break\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
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]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 295380885,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\nfor _ in range(int(input())):\n n,m = map(int,input().split())\n words = [input() for i in range(n)]\n c = len(words[0])\n count=1\n for i in range(1,n):\n if c==m:\n print(count)\n break\n if c>m:\n print(count-1)\n break\n c+=len(words[i])\n count+=1\n else:\n if count<=m:\n print(count)\n else:\n print(count-1)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 295034139,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\nt = int(input()) \nfor _ in range(t):\n n, m = map(int, input().split()) \n lista = [input().strip() for _ in range(n)] \n \n if len(lista[0]) > m:\n print(0)\n else:\n s = 0\n nb = 0\n for j in range(len(lista)):\n if s + len(lista[j]) <= m:\n s += len(lista[j])\n nb += 1\n print(nb)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 298467489,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\nimport java.util.Arrays;\n //import java.util.ArrayList;\n //import java.util.Collection;\n //import java.util.HashMap;\n //import java.util.Map;\n import java.util.Scanner;\n //import java.util.Set;\n \n public class Main {\n public static void main(String[] args)\n {\n \tScanner S = new Scanner(System.in);\n \tint t= S.nextInt();\n \twhile(t>0) {\n \t\tt--;\n \t\tint n = S.nextInt();\n \t\tint m = S.nextInt();\n \t\tString[] a = new String[n];\n \t\tfor(int i =0;i<n;i++) {\n \t\t\ta[i] = S.next();\n \t\t}\n \t\tint ans = 0;\n \t\tint sum = 0;\n \t\twhile(sum<=m && ans <= n-1) {\n \t\t\tsum = sum + a[ans].length();\n \t\t\tans = ans + 1;\n \t\t}\n \t\tif(ans !=n) {\n \t\t\tSystem.out.println(ans-1);\n \t\t}\n \t\telse {\n \t\t\tSystem.out.println(n);\n \t\t}\n \t}\n }\n }\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
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]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 295008644,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\npublic class A_Line_Breaks{\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n sc.nextLine();\n while(t-->0){\n int n = sc.nextInt();\n int m = sc.nextInt();\n sc.nextLine();\n \n int len = 0;\n int count = 0;String str[] = new String[n];\n for(int i=0;i<n;i++){\n str[i] = sc.next();\n }\n \n for(int i=0;i<n;i++){\n len += str[i].length();\n count++;\n if(len == m){\n System.out.println(count);\n break;\n }else if(len>m){\n System.out.println(count-1);\n break;\n \n }else if(i==0 && len>m){\n System.out.println(0);\n break;\n }\n \n \n }\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
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]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 295143638,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define f first\n#define s second\n#define pb push_back\n#define lb lower_bound\n#define ub upper_bound\n#define pi pair<int, int>\n#define bg begin()\n#define en end()\n#define sz(x) (int)x.size()\n#define all(x) x.begin(), x.end()\nusing namespace std;\n \nsigned main(){\n ios_base::sync_with_stdio(false);\n cin.tie(nullptr);\n int n;\n cin >> n;\n while (n--) {\n int n2, m, t = 0, total_s = 0, err = 0, first = 1;\n cin >> n2 >> m;\n while (n2--) {\n string s;\n cin >> s;\n int size = sz(s);\n if (first && size > m) err = 1;\n if (size <= m && total_s+size <= m) {\n t++;\n total_s += size;\n }\n first = 0;\n }\n cout << (err ? 0 : t) << '\\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": "2050A",
"submission_id": 295078545,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n #define int long long\n #define pb push_back\n using namespace std;\n signed main() {\n //freopen(\"rect.in\", \"r\", stdin);\n //freopen(\"rect.out\", \"w\", stdout);\n int T;\n cin >> T;\n while(T--){\n int n,m;\n cin >> n >> m;\n vector <int> a(n);\n for(int i = 0 ; i < n ; i++){\n string s;\n cin >> s;\n a[i] = s.size();\n }\n vector <int> pref(n);\n pref[0] = a[0];\n for(int i = 1;i < n; i++){\n pref[i] = pref[i - 1] + a[i];\n }\n int pos = 0;\n for(int i = 0 ; i < pref.size() ; i++){\n if(pref[i] == m){\n pos = i + 1;\n }\n else if(pref[i] > m){\n pos = i;\n break;\n }\n else if(pref[i] < m){\n continue;\n }\n }\n cout << pos << '\\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": "2050A",
"submission_id": 295235223,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\ncases = int(input())\n \nfor i in range(cases):\n nm = input().split(\" \")\n n = int(nm[0])\n m = int(nm[1])\n count = 0\n while count < n and m >= 0:\n m -= len(input())\n count += 1\n print(count if m == 0 else (count - 1))\n while count < n:\n input()\n count += 1\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 295331721,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\ndef code(num):\n for i in range(num):\n inp = input().split(\" \")\n lst = [None]*int(inp[0])\n for j in range(int(inp[0])):\n lst[j] = input()\n \n limit = int(inp[1])\n count = 0\n while True:\n \n limit -= len(lst[count])\n if limit == 0:\n print(count+1)\n break\n elif limit < 0:\n print(count)\n break\n count += 1\n \n \n \n \n \n \n \ncode(int(input()))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 302236672,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \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 int t = scanner.nextInt();\n \n while (t-- >0) {\n int ans=0;\n int current=0;\n int n=scanner.nextInt();\n int m = scanner.nextInt();\n int [] qais =new int[n];\n for (int i=0;i<n;i++){\n String g=scanner.next();\n qais[i]=g.length();\n }\n //Arrays.sort(qais);\n int p=0;\n while(m>current && p<n){\n current+=qais[p++];\n ans++;\n \n }\n if(current==m) System.out.println(ans);\n \n else System.out.println(--ans);\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": "2050A",
"submission_id": 296281229,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\nimport java.util.ArrayList;\nimport java.util.Scanner;\n \npublic class Solutions {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n ArrayList<String> words = new ArrayList<>();\n String minLengthString = null;\n int x;\n \n int t = scanner.nextInt();//t\n for(int i =0; i<t;i++){ //t\n x = 0;\n int count =0;\n \n words.clear();\n int n = scanner.nextInt();//n\n int m = scanner.nextInt();//m\n for(int j =0; j<n;j++){\n String s = scanner.next();\n words.add(s);\n }\n if(m>0 && n>0){\n while(true){\n if(m>0){\n minLengthString = words.get(count);\n }else{\n break;\n }\n \n if(minLengthString.length()<=m){\n x+=1;\n m-=minLengthString.length();\n words.remove(minLengthString);\n }else{\n break;\n }\n }\n System.out.println(x);\n \n }else {\n System.out.println(0);\n }\n \n }\n \n }\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 295054016,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n \nint main (){\n int t;\n cin >> t;\n while(t--) {\n int n, m;\n cin >> n >> m;\n string s[n];\n for (int i = 0; i < n; i++) {\n cin >> s[i];\n }\n int sum = 0;\n for (int i = 0; i < n; i++) {\n sum += s[i].length();\n if(sum > m) {\n cout << i << endl;\n break;\n }\n else if(sum == m) {\n cout << i+1 << endl;\n break;\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": "2050A",
"submission_id": 295014781,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n \ntypedef long long ll;\n \nint main() \n{\n ios::sync_with_stdio(0);\n cin.tie(0);\n \n int t;\n cin >> t;\n while (t--) \n {\n int n, m;\n cin>> n >> m;\n vector<string> v(n);\n for(int i=0; i<n; i++) cin>> v[i];\n int ct=0;\n int ind;\n for(int i=0; i<n && m>=0; i++)\n {\n m=m-v[i].size();\n ct++;\n ind=i;\n }\n if(ind==n-1 && m==0) cout<< ct <<\"\\n\";\n else cout<< --ct <<\"\\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": "2050A",
"submission_id": 299141731,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\nsint = lambda: int(input())\nmint = lambda: map(int, input().split())\naint = lambda: list(map(int, input().split()))\n \nfor _ in range(int(input())):\n n, m = mint()\n words = []\n for _ in range(n):\n words.append(input())\n words_fit = 0\n total_count = 0\n i = 0\n while total_count < m:\n total_count += len(words[i])\n if total_count > m:\n break\n words_fit += 1\n i += 1\n print(words_fit)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 296432629,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\nt=int(input())\nfor _ in range(t):\n n,m=map(int,input().split())\n a=0\n b=0\n k=n\n e=0\n o=True\n for i in range(k):\n s=input()\n a+=len(s)\n b+=1\n if a>m and o:\n e=b-1\n o=False\n if a==m and o:\n e = b\n o = False\n print(e)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050A",
"submission_id": 295010554,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.BufferedWriter;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.io.OutputStreamWriter;\nimport java.io.PrintWriter;\nimport java.util.StringTokenizer;\n \npublic class strip {\n public static StringTokenizer st = new StringTokenizer(\"\");\n public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n public static int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n public static String next() throws IOException {\n while (!st.hasMoreTokens()) {\n st = new StringTokenizer(br.readLine());\n }\n return st.nextToken();\n }\n \n public static void main(String args[]) throws IOException {\n \n PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n \n int num = nextInt();\n for (int i = 0; i < num; i++) {\n int wn = nextInt();\n int maxStrip = nextInt();\n int totalLen = 0;\n boolean done = false;\n for (int j = 0; j < wn; j++) {\n String word = next();\n int len = word.length();\n if (done) {\n continue;\n }\n if (len + totalLen > maxStrip) {\n out.println(j);\n done = true;\n } else if (len + totalLen >= maxStrip) {\n out.println(j + 1);\n done = true;\n }\n totalLen += len;\n }\n }\n out.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": "2050A",
"submission_id": 295022457,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Line Breaks\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Kostya has a text $$$s$$$ consisting of $$$n$$$ words made up of Latin alphabet letters. He also has two strips on which he must write the text. The first strip can hold $$$m$$$ characters, while the second can hold as many as needed.\n\nKostya must choose a number $$$x$$$ and write the first $$$x$$$ words from $$$s$$$ on the first strip, while all the remaining words are written on the second strip. To save space, the words are written without gaps, but each word must be entirely on one strip.\n\nSince space on the second strip is very valuable, Kostya asks you to choose the maximum possible number $$$x$$$ such that all words $$$s_1, s_2, \\dots, s_x$$$ fit on the first strip of length $$$m$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le 1000$$$) — the number of test cases.\n\nThe first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \\le n \\le 50$$$; $$$1 \\le m \\le 500$$$) — the number of words in the list and the maximum number of characters that can be on the first strip.\n\nThe next $$$n$$$ lines contain one word $$$s_i$$$ of lowercase Latin letters, where the length of $$$s_i$$$ does not exceed $$$10$$$.\n\nOutput: For each test case, output the maximum number of words $$$x$$$ such that the first $$$x$$$ words have a total length of no more than $$$m$$$.\n\nExamples:\ninput:\n5\n3 1\na\nb\nc\n2 9\nalpha\nbeta\n4 12\nhello\nworld\nand\ncodeforces\n3 2\nab\nc\nd\n3 2\nabc\nab\na\noutput:\n1\n2\n2\n1\n0\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\npublic class Sol{\n public static void main(String[] args){\n Scanner sc = new Scanner(System.in);\n int i,n,m,t = sc.nextInt(),sum;\n int[] arr;\n while(t-- != 0)\n {\n n = sc.nextInt();\n m = sc.nextInt();\n arr = new int[n];\n sum = 0;\n for( i = 0 ; i < n ; i++ )\n arr[i] = sc.next().length();\n for( i = 0 ; i < n ; i++ )\n {\n sum += arr[i];\n if(sum >= m)\n break;\n }\n if(sum > m)\n System.out.println(i);\n else\n System.out.println(i+1);\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": "2050A",
"submission_id": 295006006,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define ll long long\nusing namespace std;\ninline int read(){\n\tint a=0,b=1;char c=getchar();\n\tfor(;c<'0'||c>'9';c=getchar())if(c=='-')b=-1;\n\tfor(;c>='0'&&c<='9';c=getchar())a=a*10+c-'0';\n\treturn a*b;\n}\nint a[100001],b[100001];\nint main()\n{\n\tint T=read();\n\twhile(T--)\n\t{\n\t\tint n=read();\n\t\tint Max=0,Min=n+1;\n\t\tfor(int i=1;i<=n;i++)b[i]=0;\n\t\tfor(int i=1;i<=n;i++)\n\t\t{\n\t\t\ta[i]=read();\n\t\t\tif(a[i]!=i)Max=max(Max,a[i]),Min=min(Min,a[i]),b[a[i]]=i;\n\t\t}\n\t\tint can=0;\n\t\tfor(int i=1;i<=n;i++)\n\t\t{\n\t\t\tif(b[i]==a[i])\n\t\t\t{\n\t\t\t\tif(can==0)can=i+(a[i]);\n\t\t\t\telse if(can!=i+a[i])\n\t\t\t\t{\n\t\t\t\t\tcan=-1;\n\t\t\t\t\tbreak;\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t\tll ans=0;\n\t\tif(Max==0&&Min==n+1)\n\t\t{\n\t\t\tcout<<(1+2*n)*n<<endl;\n\t\t\tcontinue;\n\t\t}\n\t\tfor(int i=1;i<=Min+n;i++)\n\t\t{\n\t\t\tans+=2*n-max(i,Max);\n//\t\t\tcout<<i<<' '<<2*n-max(i,Max)<<endl;\n\t\t}\n\t\tif(can!=0&&can!=-1)ans++;\n\t\tcout<<ans<<endl;\n\t}\n\treturn 0;\n}\n/*\n1\n4\n3 2 1 4\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": "1973E",
"submission_id": 262031892,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n// typedef is used to create an alias for a data type\n \ntypedef long long ll;\ntypedef long long int lli;\ntypedef unsigned long long int ull;\ntypedef vector<int> vi;\ntypedef vector<ll> vll;\ntypedef vector<lli> vlli;\ntypedef pair<int, int> pi;\ntypedef pair<ll, ll> pll;\ntypedef vector<pi> vpi;\ntypedef vector<pll> vpll;\ntypedef long double ld;\ntypedef tuple<int, int, int> tiii;\n \n// #define is used to define a constant value\n#define int ll\n#define sz(x) (int)x.size()\n#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);\n#define lfastio static const auto speedup = []() { std::ios::sync_with_stdio(false); std::cin.tie(nullptr); return 0;}();\n#define pb(a) push_back(a)\n#define pbr(a, b) pb(make_pair(a, b))\n#define F first\n#define S second\n#define all(a) (a).begin(), (a).end()\n#define rall(a) (a).rbegin(), (a).rend()\n#define fr(i, a, b) for (int i = a; i <= b; i++)\n#define rfr(i, a, b) for (int i = a; i >= b; i--) \n#define forn(i, n) for (int i = 0; i < int(n); i++)\n#define pVec(vecName) for(auto a:vec){cout<<a<<\" \";}\n \n \nvoid solve(){\n int n;cin>>n;\n vector<int>per(n);\n forn(i,n)cin>>per[i];\n int mn=n+1,mx=0;\n int count=0;\n forn(i,n){\n if(per[i]!=i+1){\n count++;\n if(per[i]<mn)mn=per[i];\n if(per[i]>mx)mx=per[i];\n }\n }\n if(mn>mx){cout<<(2*n*(2*n+1))/2<<endl;return;}\n int ans=0;\n for(int i=1;i<=n+mn;i++){\n if(i<mx+1)ans+=2*n-mx;\n else{ans+=2*n-i;}\n // cout<<ans<<endl;\n }\n if(count==2)ans+=1;\n cout<<ans<<endl;\n \n \n}\nsigned main()\n{\n ios_base::sync_with_stdio(false); cin.tie(NULL);\n // cout<<\"Hello World\"<<endl;\n int t;cin>>t;\n while (t--)\n {\n solve();\n }\n \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": "1973E",
"submission_id": 262349158,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#include \"bits/stdc++.h\"\nusing namespace std;\n#pragma GCC optimize(\"Ofast\",\"inline\",\"-ffast-math\")\n#pragma GCC target(\"avx,sse2,sse3,sse4,mmx\")\n \n#define int long long\nint a[100001],n;\nvoid Delta() {\n cin >> n;\n for(int i=1;i<=n;++i) cin >> a[i];\n int cnt=0,mx=0,mn=2147483647,s=0;\n for(int i=1;i<=n;++i)\n if(a[i]!=i) {\n cnt++;\n mx=max(mx,a[i]);\n mn=min(mn,a[i]);\n }\n for(int i=1;i<=n;++i)\n s+=n*2-max(mx,i);\n for(int i=1;i<=n;++i)\n if(i<=mn)\n s+=n-i;\n cout << (cnt==0?(n*2)*(n*2+1)/2:s+(cnt==2)) << endl;\n}\nsigned main() {\n ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);\n int T=1; cin >> T;\n while(T--) Delta();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261356564,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\nimport heapq\nimport os, sys\nfrom io import BytesIO, IOBase\n \nfrom collections import defaultdict, deque, Counter\nfrom bisect import bisect_left, bisect_right\nfrom heapq import heappush, heappop\nfrom functools import lru_cache, reduce\nfrom itertools import accumulate\nfrom operator import xor\nimport random\nimport math\nimport sys\n \n# sys.setrecursionlimit(10000)\n# Fast IO Region\nBUFSIZE = 8192\n \n \nclass FastIO(IOBase):\n newlines = 0\n \n def __init__(self, file):\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n \n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n \n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n \n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\n \n \nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\n \n \nsys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\ninput = lambda: sys.stdin.readline().rstrip(\"\\r\\n\")\nsys.setrecursionlimit(10000)\n \nmag = random.getrandbits(20)\n \nlsum = lambda l, r: (l + r) * (r - l + 1) // 2\n \nclass D:\n def __init__(self, type=int):\n self.d = defaultdict(type)\n self.mag = random.getrandbits(20)\n \n def __setitem__(self, key, value):\n self.d[key + self.mag] = value\n \n def __getitem__(self, key):\n return self.d[key + self.mag]\n \n def __delitem__(self, key):\n del self.d[key + self.mag]\n \n def __contains__(self, key):\n return key + self.mag in self.d\n \n def __iter__(self):\n self.keys = iter(self.d.keys())\n return self\n \n def __next__(self):\n return next(self.keys) - self.mag\n \n def __len__(self):\n return len(self.d)\n \n \nii = lambda: int(input())\nmii = lambda: map(int, input().split())\nfii = lambda: map(float, input().split())\nlmii = lambda: list(map(int, input().split()))\ni2c = lambda n: chr(ord('a') + n)\nc2i = lambda c: ord(c) - ord('a')\n \n# mod = 998244353\n \nmod = 10 ** 9 + 7\n \n \ndef solve():\n n = ii()\n l = lmii()\n sl, sr = 0, n\n for i in range(n):\n if l[i] != i + 1:\n break\n sl += 1\n for i in range(n - 1, -1, -1):\n if l[i] != i + 1:\n break\n sr -= 1\n if sl == n:\n print(n * (2 * n + 1))\n return\n res = n * (n + 1) - 1\n if sl > 0:\n res += lsum(n - 2, n - sl - 1)\n if sr < n:\n res += lsum(sr, n - 1)\n S = set()\n for i in range(n):\n if l[i] != i + 1:\n S.add((l[i] + i + 1) ^ mag)\n if len(S) == 1:\n res += 1\n print(res)\n \nfor _ in range(ii()):\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": "1973E",
"submission_id": 292504999,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\nfrom sys import stdout\nfrom sys import stdin\nfrom math import ceil\nI=lambda:list(map(int,input().split()))\nn=int(input())\nfor i in range(n):\n N=int(input())\n P=I()\n dis=0\n dmi=999999\n dma=0\n for i in range(N):\n if P[i]!=i+1:\n dis+=1\n dmi=min(dmi,i)\n dma=max(dma,i)\n if dis==0:print(N*(2*N+1));continue\n else:\n dmi,dma=dma+1,dmi+N+2\n ans=N*(2*N+1)\n ans-=dmi*(dmi+1)//2\n tmp=N*2-dma+1\n ans-=(tmp+1)*tmp//2\n ans-=dma-dmi-1\n if dis==2:ans+=1\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261408496,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\nimport sys\nimport heapq\ninput = sys.stdin.readline\ndef fg():\n return int(input())\ndef fgh():\n return [int(xx) for xx in input().split()]\ndef tt():\n sys.stdout.flush()\ndef sd():\n print('Yes')\ndef df():\n print('No')\nfor __ in range(fg()):\n n = fg()\n p = fgh()\n if p == sorted(p):\n print(2 * n * (2 * n + 1) // 2)\n else:\n c = 0\n while p[c] == c + 1:\n c += 1\n d = 0\n while p[n - d - 1] == n - d:\n d += 1\n L = c + 1\n R = n - d\n ans = 0\n fl = 1\n for i in range(L, R + 1):\n if p[i - 1] != R - i or p[R - (i - L) - 1] != i:\n fl = 1\n for l in range(1, 2 * n + 1):\n minr = R + 1\n k = l - L\n if l <= L + n:\n if R - L <= 2 and l % 2 and 2 * L + 1 <= l <= 2 * R - 1:\n ans += max(0, 2 * n - max(minr, l) + 1)\n else:\n ans += max(0, 2 * n - max(minr, l + 1) + 1)\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261406213,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\nimport static java.lang.Math.*;\nimport java.util.*;\nimport java.io.*;\n \npublic class x1973\n{\n static final int INF = Integer.MAX_VALUE/2;\n public static void main(String[] args) throws Exception\n {\n FastScanner infile = new FastScanner();\n int T = infile.nextInt();\n while(T-->0)\n {\n int N = infile.nextInt();\n int[] p = new int[N+1];\n for(int i=1; i <= N; i++)\n p[i] = infile.nextInt();\n boolean identity = true;\n for(int i=1; i <= N; i++)\n identity &= p[i] == i;\n if(identity)\n {\n System.out.println((long)N*(2*N+1));\n continue;\n }\n long res = 1;\n int val = -1;\n for(int a=1; a <= N; a++)\n {\n if(a != p[p[a]])\n res = 0;\n else if(a != p[a])\n {\n if(val == -1)\n val = abs(a-p[a]);\n else if(val != abs(a-p[a]))\n res = 0;\n }\n }\n int[] pref = new int[N+1];\n for(int i=1; i <= N; i++)\n pref[i] = pref[i-1] + (p[i] != i ? 1 : 0);\n int needFix = pref[N];\n for(int R=3; R <= 2*N; R++)\n {\n int upper = min(N, R-1);\n if(pref[upper] != needFix)\n continue;\n int l = 1;\n int h = R-1;\n while(l != h)\n {\n int mid = (l+h+1)>>1;\n int lower = max(1, mid-N);\n if(pref[lower-1] == 0)\n l = mid;\n else\n h = mid-1;\n }\n //System.out.println(R+\": \"+l);\n res += l;\n }\n System.out.println(res);\n }\n }\n}\n/*\nE:\nL = N, R = N+1\n \nboolean[] seen = new boolean[N];\n for(int a=0; a < N; a++)\n if(!seen[a])\n {\n seen[a] = true;\n int curr = p[a];\n ArrayList<Integer> path = new ArrayList<>();\n path.add(a);\n while(curr != a)\n {\n path.add(curr);\n seen[curr] = true;\n curr = p[curr];\n }\n Collections.sort(path);\n if(path.size() >= 2)\n {\n int diff = path.get(1)-path.get(0);\n for(int i=2; i < path.size(); i++)\n if(path.get(i)-path.get(i-1) != diff)\n res = 0;\n }\n }\n */\nclass FastScanner\n{\n private int BS = 1 << 16;\n private char NC = (char) 0;\n private byte[] buf = new byte[BS];\n private int bId = 0, size = 0;\n private char c = NC;\n private double cnt = 1;\n private BufferedInputStream in;\n \n public FastScanner() {\n in = new BufferedInputStream(System.in, BS);\n }\n \n public FastScanner(String s) {\n try {\n in = new BufferedInputStream(new FileInputStream(new File(s)), BS);\n } catch (Exception e) {\n in = new BufferedInputStream(System.in, BS);\n }\n }\n \n private char getChar() {\n while (bId == size) {\n try {\n size = in.read(buf);\n } catch (Exception e) {\n return NC;\n }\n if (size == -1) return NC;\n bId = 0;\n }\n return (char) buf[bId++];\n }\n \n public int nextInt() {\n return (int) nextLong();\n }\n \n public int[] nextInts(int N) {\n int[] res = new int[N];\n for (int i = 0; i < N; i++) {\n res[i] = (int) nextLong();\n }\n return res;\n }\n \n public long[] nextLongs(int N) {\n long[] res = new long[N];\n for (int i = 0; i < N; i++) {\n res[i] = nextLong();\n }\n return res;\n }\n \n public long nextLong() {\n cnt = 1;\n boolean neg = false;\n if (c == NC) c = getChar();\n for (; (c < '0' || c > '9'); c = getChar()) {\n if (c == '-') neg = true;\n }\n long res = 0;\n for (; c >= '0' && c <= '9'; c = getChar()) {\n res = (res << 3) + (res << 1) + c - '0';\n cnt *= 10;\n }\n return neg ? -res : res;\n }\n \n public double nextDouble() {\n double cur = nextLong();\n return c != '.' ? cur : cur + nextLong() / cnt;\n }\n \n public double[] nextDoubles(int N) {\n double[] res = new double[N];\n for (int i = 0; i < N; i++) {\n res[i] = nextDouble();\n }\n return res;\n }\n \n public String next() {\n StringBuilder res = new StringBuilder();\n while (c <= 32) c = getChar();\n while (c > 32) {\n res.append(c);\n c = getChar();\n }\n return res.toString();\n }\n \n public String nextLine() {\n StringBuilder res = new StringBuilder();\n while (c <= 32) c = getChar();\n while (c != '\\n') {\n res.append(c);\n c = getChar();\n }\n return res.toString();\n }\n \n public boolean hasNext() {\n if (c > 32) return true;\n while (true) {\n c = getChar();\n if (c == NC) return false;\n else if (c > 32) return true;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261391798,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\nimport static java.lang.Math.*;\nimport java.util.*;\nimport java.io.*;\n \npublic class x1973\n{\n static final int INF = Integer.MAX_VALUE/2;\n public static void main(String[] args) throws Exception\n {\n FastScanner infile = new FastScanner();\n int T = infile.nextInt();\n while(T-->0)\n {\n int N = infile.nextInt();\n int[] p = infile.nextInts(N);\n for(int i=0; i < N; i++)\n p[i]--;\n boolean identity = true;\n for(int i=0; i < N; i++)\n identity &= p[i] == i;\n if(identity)\n {\n System.out.println((long)N*(2*N+1));\n continue;\n }\n long res = 1;\n int val = -1;\n for(int a=0; a < N; a++)\n {\n if(a != p[p[a]])\n res = 0;\n else if(a != p[a])\n {\n if(val == -1)\n val = abs(a-p[a]);\n else if(val != abs(a-p[a]))\n res = 0;\n }\n }\n int[] pref = new int[N];\n if(p[0] != 0)\n pref[0]++;\n for(int i=1; i < N; i++)\n pref[i] = pref[i-1] + (p[i] != i ? 1 : 0);\n int needFix = pref[N-1];\n for(int R=2; R <= 2*N; R++)\n {\n int upper = min(N, R-1);\n if(pref[upper-1] != needFix)\n continue;\n int l = 1;\n int h = R-1;\n while(l != h)\n {\n int mid = (l+h+1)>>1;\n int covered = pref[upper-1];\n int lower = max(1, mid-N);\n if(lower >= 2)\n covered -= pref[lower-2];\n if(covered == needFix)\n l = mid;\n else\n h = mid-1;\n }\n //System.out.println(R+\": \"+l);\n res += l;\n }\n System.out.println(res);\n }\n }\n}\n/*\nE:\nL = N, R = N+1\n \nboolean[] seen = new boolean[N];\n for(int a=0; a < N; a++)\n if(!seen[a])\n {\n seen[a] = true;\n int curr = p[a];\n ArrayList<Integer> path = new ArrayList<>();\n path.add(a);\n while(curr != a)\n {\n path.add(curr);\n seen[curr] = true;\n curr = p[curr];\n }\n Collections.sort(path);\n if(path.size() >= 2)\n {\n int diff = path.get(1)-path.get(0);\n for(int i=2; i < path.size(); i++)\n if(path.get(i)-path.get(i-1) != diff)\n res = 0;\n }\n }\n */\nclass FastScanner\n{\n private int BS = 1 << 16;\n private char NC = (char) 0;\n private byte[] buf = new byte[BS];\n private int bId = 0, size = 0;\n private char c = NC;\n private double cnt = 1;\n private BufferedInputStream in;\n \n public FastScanner() {\n in = new BufferedInputStream(System.in, BS);\n }\n \n public FastScanner(String s) {\n try {\n in = new BufferedInputStream(new FileInputStream(new File(s)), BS);\n } catch (Exception e) {\n in = new BufferedInputStream(System.in, BS);\n }\n }\n \n private char getChar() {\n while (bId == size) {\n try {\n size = in.read(buf);\n } catch (Exception e) {\n return NC;\n }\n if (size == -1) return NC;\n bId = 0;\n }\n return (char) buf[bId++];\n }\n \n public int nextInt() {\n return (int) nextLong();\n }\n \n public int[] nextInts(int N) {\n int[] res = new int[N];\n for (int i = 0; i < N; i++) {\n res[i] = (int) nextLong();\n }\n return res;\n }\n \n public long[] nextLongs(int N) {\n long[] res = new long[N];\n for (int i = 0; i < N; i++) {\n res[i] = nextLong();\n }\n return res;\n }\n \n public long nextLong() {\n cnt = 1;\n boolean neg = false;\n if (c == NC) c = getChar();\n for (; (c < '0' || c > '9'); c = getChar()) {\n if (c == '-') neg = true;\n }\n long res = 0;\n for (; c >= '0' && c <= '9'; c = getChar()) {\n res = (res << 3) + (res << 1) + c - '0';\n cnt *= 10;\n }\n return neg ? -res : res;\n }\n \n public double nextDouble() {\n double cur = nextLong();\n return c != '.' ? cur : cur + nextLong() / cnt;\n }\n \n public double[] nextDoubles(int N) {\n double[] res = new double[N];\n for (int i = 0; i < N; i++) {\n res[i] = nextDouble();\n }\n return res;\n }\n \n public String next() {\n StringBuilder res = new StringBuilder();\n while (c <= 32) c = getChar();\n while (c > 32) {\n res.append(c);\n c = getChar();\n }\n return res.toString();\n }\n \n public String nextLine() {\n StringBuilder res = new StringBuilder();\n while (c <= 32) c = getChar();\n while (c != '\\n') {\n res.append(c);\n c = getChar();\n }\n return res.toString();\n }\n \n public boolean hasNext() {\n if (c > 32) return true;\n while (true) {\n c = getChar();\n if (c == NC) return false;\n else if (c > 32) return true;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261378208,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#ifndef LOCAL\n#define FAST_IO\n#endif\n \n// ============\n#include <bits/stdc++.h>\n#define OVERRIDE(a, b, c, d, ...) d\n#define REP2(i, n) for (i32 i = 0; i < (i32)(n); ++i)\n#define REP3(i, m, n) for (i32 i = (i32)(m); i < (i32)(n); ++i)\n#define REP(...) OVERRIDE(__VA_ARGS__, REP3, REP2)(__VA_ARGS__)\n#define PER2(i, n) for (i32 i = (i32)(n)-1; i >= 0; --i)\n#define PER3(i, m, n) for (i32 i = (i32)(n)-1; i >= (i32)(m); --i)\n#define PER(...) OVERRIDE(__VA_ARGS__, PER3, PER2)(__VA_ARGS__)\n#define ALL(x) begin(x), end(x)\n#define LEN(x) (i32)(x.size())\nusing namespace std;\nusing u32 = unsigned int;\nusing u64 = unsigned long long;\nusing i32 = signed int;\nusing i64 = signed long long;\nusing f64 = double;\nusing f80 = long double;\nusing pi = pair<i32, i32>;\nusing pl = pair<i64, i64>;\ntemplate <typename T>\nusing V = vector<T>;\ntemplate <typename T>\nusing VV = V<V<T>>;\ntemplate <typename T>\nusing VVV = V<V<V<T>>>;\ntemplate <typename T>\nusing VVVV = V<V<V<V<T>>>>;\ntemplate <typename T>\nusing PQR = priority_queue<T, V<T>, greater<T>>;\ntemplate <typename T>\nbool chmin(T &x, const T &y) {\n if (x > y) {\n x = y;\n return true;\n }\n return false;\n}\ntemplate <typename T>\nbool chmax(T &x, const T &y) {\n if (x < y) {\n x = y;\n return true;\n }\n return false;\n}\ntemplate <typename T>\ni32 lob(const V<T> &arr, const T &v) {\n return (i32)(lower_bound(ALL(arr), v) - arr.begin());\n}\ntemplate <typename T>\ni32 upb(const V<T> &arr, const T &v) {\n return (i32)(upper_bound(ALL(arr), v) - arr.begin());\n}\ntemplate <typename T>\nV<i32> argsort(const V<T> &arr) {\n V<i32> ret(arr.size());\n iota(ALL(ret), 0);\n sort(ALL(ret), [&](i32 i, i32 j) -> bool {\n if (arr[i] == arr[j]) {\n return i < j;\n } else {\n return arr[i] < arr[j];\n }\n });\n return ret;\n}\n#ifdef INT128\nusing u128 = __uint128_t;\nusing i128 = __int128_t;\n#endif\n[[maybe_unused]] constexpr i32 INF = 1000000100;\n[[maybe_unused]] constexpr i64 INF64 = 3000000000000000100;\nstruct SetUpIO {\n SetUpIO() {\n#ifdef FAST_IO\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n#endif\n cout << fixed << setprecision(15);\n }\n} set_up_io;\nvoid scan(char &x) { cin >> x; }\nvoid scan(u32 &x) { cin >> x; }\nvoid scan(u64 &x) { cin >> x; }\nvoid scan(i32 &x) { cin >> x; }\nvoid scan(i64 &x) { cin >> x; }\nvoid scan(string &x) { cin >> x; }\ntemplate <typename T>\nvoid scan(V<T> &x) {\n for (T &ele : x) {\n scan(ele);\n }\n}\nvoid read() {}\ntemplate <typename Head, typename... Tail>\nvoid read(Head &head, Tail &...tail) {\n scan(head);\n read(tail...);\n}\n#define CHAR(...) \\\n char __VA_ARGS__; \\\n read(__VA_ARGS__);\n#define U32(...) \\\n u32 __VA_ARGS__; \\\n read(__VA_ARGS__);\n#define U64(...) \\\n u64 __VA_ARGS__; \\\n read(__VA_ARGS__);\n#define I32(...) \\\n i32 __VA_ARGS__; \\\n read(__VA_ARGS__);\n#define I64(...) \\\n i64 __VA_ARGS__; \\\n read(__VA_ARGS__);\n#define STR(...) \\\n string __VA_ARGS__; \\\n read(__VA_ARGS__);\n#define VEC(type, name, size) \\\n V<type> name(size); \\\n read(name);\n#define VVEC(type, name, size1, size2) \\\n VV<type> name(size1, V<type>(size2)); \\\n read(name);\n// ============\n \n#ifdef DEBUGF\n#else\n#define DBG(...) (void)0\n#endif\n \nvoid solve() {\n I32(n);\n VEC(i32, a, n);\n REP(i, n) {\n --a[i];\n }\n // -1 <= l <= r <= 2 * n - 2\n i32 back_match = 0;\n while (back_match < n && a[n - 1 - back_match] == n - 1 - back_match) {\n ++back_match;\n }\n i32 front_match = 0;\n while (front_match < n && a[front_match] == front_match) {\n ++front_match;\n }\n if (front_match == n) {\n cout << (i64)n * (2 * n + 1) << '\\n';\n return;\n }\n V<pi> prs;\n REP(i, n) {\n if (a[i] > i && a[a[i]] == i) {\n prs.emplace_back(i, a[i]);\n }\n }\n bool all_same = !prs.empty();\n i32 tgt = (prs.empty() ? -1 : prs[0].first + prs[0].second);\n for (auto [p, q] : prs) {\n if (p + q != tgt) {\n all_same = false;\n }\n }\n DBG(front_match, back_match, all_same, tgt);\n i64 ans = 0;\n REP(r, 1, n - 1) {\n if (n - r - 1 <= back_match) {\n if (all_same && tgt == r) {\n ans += r + 2;\n } else {\n ans += r + 1;\n }\n }\n }\n REP(l, n, 2 * n - 2) {\n if (l - n + 1 <= front_match) {\n if (all_same && tgt == l) {\n ans += 2 * n - 1 - l;\n } else {\n ans += 2 * n - 2 - l;\n }\n }\n }\n if (all_same && tgt == n - 1) {\n ans += (i64)(n + 1) * n;\n } else {\n ans += (i64)(n + 1) * n - 1;\n }\n cout << ans << '\\n';\n}\n \nint main() {\n i32 t = 1;\n cin >> t;\n while (t--) {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261395538,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nusing ll = long long;\n \nconst int N = 100005;\n \nint n, a[N];\n \nvoid solve()\n{\n cin >> n;\n for (int i = 1; i <= n; ++i) cin >> a[i];\n int lef = n + 1, rig = 0, cnt = 0;\n for (int i = 1; i <= n; ++i)\n {\n if (a[i] != i)\n {\n lef = min(lef, i);\n lef = min(lef, a[i]);\n rig = max(rig, a[i]);\n rig = max(rig, i);\n cnt++;\n }\n }\n if (rig == 0)\n {\n cout << (2 * n + 1) * n << '\\n';\n return;\n }\n ll ans = cnt == 2;\n for (int i = 1; i <= lef + n; ++i)\n {\n ans += 2 * n + 1 - max(rig + 1, i + 1);\n }\n cout << ans << '\\n';\n return;\n}\n \nint main()\n{\n ios::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n int T = 1;\n cin >> T;\n while (T--) solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261610275,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\nimport sys\nimport heapq\ninput = sys.stdin.readline\ndef fg():\n return int(input())\ndef fgh():\n return [int(xx) for xx in input().split()]\ndef tt():\n sys.stdout.flush()\ndef sd():\n print('Yes')\ndef df():\n print('No')\nfor __ in range(fg()):\n n = fg()\n p = fgh()\n if p == sorted(p):\n print(2 * n * (2 * n + 1) // 2)\n else:\n c = 0\n while p[c] == c + 1:\n c += 1\n d = 0\n while p[n - d - 1] == n - d:\n d += 1\n L = c + 1\n R = n - d\n ans = 0\n fl = 1\n for i in range(L, R + 1):\n if p[i - 1] != R - (i - L) or p[R - (i - L) - 1] != i:\n fl = 0\n for l in range(1, 2 * n + 1):\n minr = R + 1\n k = l - L\n if l <= L + n:\n if fl and l == R + L:\n ans += max(0, 2 * n - max(minr, l) + 1)\n else:\n ans += max(0, 2 * n - max(minr, l + 1) + 1)\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261408761,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\nimport sys\nimport os\nfrom io import BytesIO, IOBase\nfrom _collections import defaultdict\n \n# region fastio\nBUFSIZE = 8192\n \nclass FastIO(IOBase):\n newlines = 0\n \n def __init__(self, file):\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n \n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n \n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n \n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\n \nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\n \nif sys.version_info[0] < 3:\n sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))\nelse:\n sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\n \nfile = sys.stdin\nif os.environ.get('USER') == \"loic\":\n file = open(\"data.in\")\n \nline = lambda: file.readline().split()\nui = lambda: int(line()[0])\nmi = lambda: map(int,line())\nli = lambda: list(mi())\nlsi = lambda: list(map(int,line()[0]))\n#######################################################################\n \ndef solve():\n \n mn = 10**9\n mx = 0\n mp = defaultdict(int)\n for i,val in enumerate(A,1):\n if val != i:\n mn = min(mn,val)\n mx = max(mx,val)\n if A[val-1] == i:\n mp[val + i] += 1\n \n if mn > mx:\n return str(2 * N * (2 * N + 1) // 2)\n \n res = mx * (2 * N - mx)\n res += (2 * N - mx - 1) * (2 * N - mx) // 2\n rem = 2 * N - (mn + N + 1)\n res -= rem * (rem + 1) // 2\n if len(mp) == 1:\n res += 1\n \n return str(res)\n \n \nfor test in range(1,ui()+1):\n N = ui()\n A = li()\n \n if test == 61:\n print(\"_\".join(str(v) for v in A))\n \n print(solve())\n \nfile.close()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261667347,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\ntypedef long long ll;\n \nvoid solve(){\n int n;\n cin >> n;\n vector<int> q(n + 1);\n int mi = 1e9, ma = 0;\n for (int i = 1; i <= n; i++){\n cin >> q[i];\n if (q[i] != i) mi = min(mi, q[i]), ma = max(ma, q[i]);\n }\n if (ma < mi){\n cout << 1ll * (n * 2) * (n * 2 + 1) / 2 << endl;\n return;\n }\n int op = mi + ma;\n ll ans = 0;\n if (ma - mi == 1){\n ans = 1ll * op * (n * 2 - op + 1);\n for (int i = 1; i < op; i++){\n if (i - ma >= 1){\n ans += 1ll * (i - 1);\n }\n }\n for (int i = op + 2; i <= n * 2; i++){\n if (i - mi <= n + 1){\n ans += 1ll * (n * 2 - i + 1);\n }\n }\n }\n else {\n ans = 1ll * (op - 1) * (n * 2 - op + 1);\n for (int i = 1; i < op; i++){\n if (i - ma >= 1){\n ans += 1ll * (i - 1);\n }\n }\n for (int i = op + 1; i <= n * 2; i++){\n if (i - mi <= n + 1){\n ans += 1ll * (n * 2 - i + 1);\n }\n }\n int flag = 1;\n for (int i = 1; i <= n; i++){\n if (q[i] != i && q[i] != mi && q[i] != ma) flag = 0;\n }\n ans += flag;\n }\n cout << ans << endl;\n}\nint main(){\n ios::sync_with_stdio(false);\n cin.tie(0);\n int t;\n cin >> t;\n while(t--)\n solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261444932,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Main {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n \n int t = scanner.nextInt();\n while (t-- > 0) {\n int n = scanner.nextInt();\n int[] p = new int[n];\n for (int i = 0; i < n; i++) {\n p[i] = scanner.nextInt();\n }\n \n System.out.println(countValidPairs(n, p));\n }\n \n scanner.close();\n }\n \n private static int countValidPairs(int n, int[] p) {\n int totalPairs = 0;\n \n for (int l = 1; l <= 2 * n; l++) {\n for (int r = l; r <= 2 * n; r++) {\n if (canBeSorted(l, r, p)) {\n totalPairs++;\n }\n }\n }\n \n return totalPairs;\n }\n \n private static boolean canBeSorted(int l, int r, int[] p) {\n int n = p.length;\n UnionFind uf = new UnionFind(n + 1);\n \n for (int i = 0; i < n; i++) {\n for (int j = i + 1; j < n; j++) {\n int x = p[i];\n int y = p[j];\n if (l <= x + y && x + y <= r) {\n uf.union(x, y);\n }\n }\n }\n \n for (int i = 1; i <= n; i++) {\n if (uf.find(i) != uf.find(p[i - 1])) {\n return false;\n }\n }\n \n return true;\n }\n}\n \nclass UnionFind {\n private int[] parent;\n \n public UnionFind(int size) {\n parent = new int[size];\n for (int i = 0; i < size; i++) {\n parent[i] = i;\n }\n }\n \n public int find(int x) {\n if (parent[x] != x) {\n parent[x] = find(parent[x]);\n }\n return parent[x];\n }\n \n public void union(int x, int y) {\n int rootX = find(x);\n int rootY = find(y);\n if (rootX != rootY) {\n parent[rootX] = rootY;\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": "1973E",
"submission_id": 261376384,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nconst bool local = false;\n \nvector<int> v;\n \nvoid solve() {\n int n, k;\n cin >> n;\n v.resize(n + 1);\n int L = n + 1;\n int R = 0;\n int cnt = 0;\n for (int i = 1; i <= n; ++i) {\n cin >> k;\n v[k] = i;\n if (i != k) {\n L = min(k, L);\n R = max(k, R);\n ++cnt;\n }\n }\n int good = 0;\n for (int i = 1; i <= n; ++i) \n if (v[i] != i) {\n if (v[v[i]] != i or (i + v[i] != good and good != 0)) {\n good = -1;\n } else {\n good = i + v[i];\n }\n }\n int coef = L + n - R;\n int ans = 2 * n * (2 * n + 1) / 2 - R * (R + 1) / 2 - (n - L) * (n - L + 1) / 2 - coef + (good > 0);\n if (R == 0) {\n ans = 2 * n * (2 * n + 1) / 2;\n }\n cout << ans << endl;\n}\n \nsigned main() {\n //freopen(\"D:/code/codeforces/input.txt\", \"r\", stdin);\n // freopen(\"D:/code/codeforces/output.txt\", \"w\", stdout);\n int t = 1;\n cin >> t;\n while (t--) {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261483195,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n// LUOGU_RID: 160029361\n//Man always remember love because of romance only!\n#include<bits/stdc++.h>\nusing namespace std;\ninline int read(){\n int n=0,w=0; char ch=0;\n while(!isdigit(ch)) {w|=ch=='-';ch=getchar();}\n while(isdigit(ch)) n=(n<<3)+(n<<1)+(ch^48),ch=getchar();\n return w?-n:n;\n}\ninline void write(int x){\n\tif(x<0) putchar('-'),x=-x;\n if(x>9) write(x/10);\n putchar(x%10+'0');\n}\nint a[200001];\nint main(){\n\tint T=read();\n\twhile(T--){\n\t\tint n=read();\n\t\tset<int> s;\n\t\tfor(int i=1;i<=n;i++){\n\t\t\ta[i]=read();\n\t\t\tif(a[i]!=i) s.insert(a[i]+i);\n\t\t}\n\t\tif(s.size()==0){\n\t\t\twrite((2*n+1)*2*n/2);\n\t\t\tputs(\"\");\n\t\t\tcontinue;\n\t\t}\n\t\tint l=0,r=2*n;\n\t\tint ans=(s.size()==1);\n\t\tfor(int i=1;i<=n;i++){\n\t\t\tif(a[i]!=i) l=max(l,a[i]+1),r=min(r,a[i]+n);\n\t\t}\n\t\tans+=2*n*(2*n-1)/2-(l-1)*(l-2)/2-(2*n-r)*(2*n-r-1)/2;\n\t\twrite(ans);\n\t\tputs(\"\");\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": "1973E",
"submission_id": 262348610,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nusing ll = long long;\nconst int N = 1e5 + 10;\n \nint n, a[N];\nvoid mian() {\n cin >> n;\n for (int i = 1; i <= n; i++) cin >> a[i];\n if (is_sorted(a + 1, a + n + 1))\n return cout << 2 * n * (2 * n + 1) / 2 << '\\n', void();\n int v = -1;\n for (int i = 1; i <= n; i++)\n if (a[i] != i) {\n if (v == -1) v = a[i] + i;\n else if (v != a[i] + i) { v = -2; break; }\n }\n ll ans = 0;\n if (v != -2) ans += 1;\n int L = 1, R = n;\n while (a[L] == L) L++;\n while (a[R] == R) R--;\n L += n, R += 1;\n for (int l = 1; l <= L; l++)\n ans += (2 * n - max(R, l + 1)) + 1;\n cout << ans << '\\n';\n}\nint main() {\n cin.tie(0)->sync_with_stdio(0);\n int t;\n cin >> t;\n while (t--) mian();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261448935,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define MAXN 100005\nusing namespace std;\nint n;\nint a[MAXN];\nvoid work(){\n\tcin>>n;\n\tfor(int i=1;n>=i;i++)cin>>a[i];\n\tint mx=0,mn=1e9,ans=0,t=0,flg=0;\n\tfor(int i=1;n>=i;i++)\n\t\tif(a[i]!=i)mx=max(mx,a[i]),mn=min(mn,a[i]),flg=1,t=a[i]+i;\n\tfor(int i=1;n>=i;i++)\n\t\tif(a[i]!=i&&t!=a[i]+i)flg=-1;\n\tif(!flg){\n\t\tcout<<(1+2*n)*n<<endl;\n\t\treturn ;\n\t}\n\tif(flg!=-1)ans=1;\n\tfor(int i=mx+1;2*n>=i;i++)ans+=min(i-1,mn+n);\n\tcout<<ans<<endl;\n}\nsigned main(){\n\tint _T;\n\tcin>>_T;\n\twhile(_T--)work();\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": "1973E",
"submission_id": 265420163,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#include <bits/extc++.h>\nusing namespace std;\nconst int MAXN = 1e5 + 25;\ntypedef long long ll;\nll p[MAXN], n;\nbool vis[MAXN];\nvoid solve () {\n\tcin >> n;\n\tfor (int i = 1; i <= 2 * n; i++) {\n\t\tvis[i] = 0;\n\t}\n\tfor (int i = 1; i <= n; i++) {\n\t\tcin >> p[i];\n\t}\n\tll cnt = 0;\n\tif (is_sorted(p + 1, p + n + 1)) {\n\t\tcout << 2 * n * n + n << '\\n';\n\t\treturn;\n\t}\n\tll ans = 0;\n\tll l = n + 1, r = 0;\n\tfor (int i = 1; i <= n; i++) {\n\t\tif (p[i] != i) {\n\t\t\tif (!vis[p[i] + i]) {\n\t\t\t\tvis[p[i] + i] = 1;\n\t\t\t\tcnt++;\n\t\t\t}\n\t\t\tif (l == n + 1) l = i;\n\t\t\tr = i;\n\t\t}\t\n\t}\n\tfor (ll i = r + 1; i <= 2 * n; i++) {\n\t\tans += min(l + n, i - 1);\n\t}\n\tif (cnt == 1) ans++;\n\tcout << ans << '\\n';\n}\t\t\nsigned main () {\n\t #ifndef ONLINE_JUDGE \n\t\tfreopen(\"input_file\", \"r\", stdin);\n\t\tfreopen(\"output_file\", \"w\", stdout);\n\t#endif\n\tios::sync_with_stdio(0); cin.tie(0);\n\tint tc = 1; cin >> tc;\n\twhile (tc--) 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": "1973E",
"submission_id": 261436423,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Cat, Fox and Swaps\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Fox has found an array $p_1, p_2, \\ldots, p_n$, that is a permutation of length $n^\\dagger$ of the numbers $1, 2, \\ldots, n$. She wants to sort the elements in increasing order. Cat wants to help her — he is able to swap any two numbers $x$ and $y$ in the array, but only if $l \\leq x + y \\leq r$ (note that the constraint is imposed on the values of the elements, not their positions). He can make such swaps any number of times.\n\nThey don't know the numbers $l$, $r$ yet, they only know that it's true that $1 \\leq l \\leq r \\leq 2 \\cdot n$.\n\nYou are given the number $n$ and the array $p_1, p_2, \\ldots, p_n$. Determine how many pairs $(l, r)$ satisfying the conditions are there such that you can sort the permutation if you can only swap two number $(x, y)$ such that $l \\leq x + y \\leq r$ (arbitrary number of times, possibly $0$).\n\n$^\\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).\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\nDescription of each test case consists of two lines. The first line contains one integer $n$ ($1 \\leq n \\leq 10^5$).\n\nThe second line contains $n$ integers: the array $p_1, p_2, \\ldots, p_n$ ($1 \\le p_i \\le n$). It is guaranteed that this array is a permutation of length $n$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, print the number of pairs of integers $(l, r)$ such that $1 \\leq l \\leq r \\leq 2 \\cdot n$, and you can sort the array under the constraints.\n\nExamples:\ninput:\n7\n2\n2 1\n3\n3 1 2\n4\n3 2 1 4\n5\n5 3 1 2 4\n5\n1 2 3 4 5\n6\n3 2 1 5 4 6\n6\n1 3 2 4 5 6\noutput:\n6\n11\n23\n29\n55\n46\n58\n\nNote: In the first example, we need to be able to swap $1$ and $2$, so we must be able to swap numbers with sum $3$. There are exactly $6$ pairs satisfying the condition: $(1, 3), (2, 3), (3, 3), (1, 4), (2, 4)$ and $(3, 4)$, so the answer is $6$.\n\nIn the second example, the $11$ pairs satisfying the condition are $(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5)$ and $(4, 6)$. For example, if we pick the pair $(3, 4)$ we can first swap the numbers $1$ and $2$ and then the numbers $1$ and $3$, after this, the permutation is sorted.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define pb push_back\n#define lc (o<<1)\n#define rc (o<<1|1)\n#define Mid ((tr[o].l+tr[o].r)>>1)\n#define rint register int\n#define inv inline void\n#define ini inline int\n#define pair mypair\nusing namespace std;\ntypedef long long ll;\nstruct pair{\n\tll x,y;\n\tbool operator == (const pair &u)const{\n\t\tif(x==u.x&&y==u.y) return true;\n\t\treturn false;\n\t}\n\tbool operator < (const pair &u)const{\n\t\tif(x!=u.x) return x<u.x;\n\t\treturn y<u.y;\n\t}\n};\nconst int inf=2100000000;\nconst ll linf=8e18;\nconst ll N=500010; \nint a[N];\ninv solve(){\n\tint n;cin>>n;\n\tint mx=0,mn=inf;\n\tset <int> st;\n\tint res=0;\n\tfor(rint i=1;i<=n;i++){\n\t\tcin>>a[i];\n\t\tif(a[i]!=i){\n\t\t\tif(res==0) res=a[i]+i;\n\t\t\telse{\n\t\t\t\tif(res!=a[i]+i){\n\t\t\t\t\tres=-1;\n\t\t\t\t}\n\t\t\t}\n\t\t\tst.insert(a[i]);\n\t\t\tmx=max(mx,a[i]);\n\t\t\tmn=min(mn,a[i]);\n\t\t}\n\t}\n\tif(mx==0){\n\t\tcout<<2*n*(2*n+1)/2<<'\\n';\n\t\treturn ;\n\t}\n\tll ans=0;\n\tll l=mx+1,r=mn+n;\n\tfor(ll i=1;i<=2*n;i++){\n\t\tif(i>r) break;\n\t\tll num=l;\n\t\tnum=max(i,num);\n\t\tans+=2ll*(ll)n-(ll)(num)+1ll;\n\t\tif(num==i){\n\t\t\tif(i!=res){\n\t\t\t\tans--;\n\t\t\t}\n\t\t}\n\t}\n\tcout<<ans<<'\\n';\n\treturn ;\n}\nsigned main(){\n\tios::sync_with_stdio(false), cin.tie(0), cout.tie(0);\n\tint t;cin>>t;\n\twhile(t--){\n\t\tsolve();\n\t}\n\treturn 0;\n}\n//二分、dp、构造算法、数据结构、深搜广搜、排序\n//启发式合并、莫队分块 \n//分治、并查集、贪心、哈希、字典树 、双指针、拓扑 \n//最短路、字符串、树、图论 、数学、矩阵、数论\n//位掩码、暴力、组合数学、三分、博弈论、概率论\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1973E",
"submission_id": 261444591,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ll long long\n#define fr(i, a, b) for (ll i = a; i < b; i++)\n#define mmst(i,a) memset(i,a,sizeof(i))\n#define all(i) i.begin(),i.end()\n#define allr(i) i.rbegin(),i.rend()\n#define fast \\\n\tios_base::sync_with_stdio(false); \\\n\tcin.tie(NULL); \\\n\tcout.tie(NULL);\n#define SS \" \"\n#define ddd cout<<\"D\"<<endl;\n#define dd(i) cout<<i<<endl;\n#define yes cout << \"Yes\" << endl\n#define no cout << \"No\" << endl\nconst char nl = '\\n';\nusing namespace std;\n \nconst ll mod = 1e9 + 7;\nconst ll sz = 2e5 + 5;\nconst ll inf = 1e18;\n \nvoid solve()\n{\n\tll n;cin>>n;\n\tvector<ll> adj[n + 1];\n\tfr(i,0,n - 1){\n\t\tll u,v;cin>>u>>v;\n\t\tadj[u].push_back(v);\n\t\tadj[v].push_back(u);\n\t}\n\tstring s;cin>>s;\n\ts = '$' + s;\n\tll ans = 0, iris = 0, dora = 0, f = 0;\n\tfunction<ll(ll, ll, ll, ll, ll)> dfs = [&](ll node, ll par, ll onHp, ll Zhp, ll qs) -> ll{\n\t\tif(s[node] == '?')qs++;\n\t\telse if(s[node] == '1')onHp++;\n\t\telse Zhp++;\n\t\tif(adj[node].size() == 1 && node != 1){\n\t\t\tif(s[node] == '?'){\n\t\t\t\t// cout<<SS<<node<<nl;\n\t\t\t\t// ddd\n\t\t\t\t// if(qs >= 2)return 1;\n\t\t\t\t// // if(onHp ^ Zhp){iris++;return 0;}\n\t\t\t\t// if(onHp == 0 && Zhp == 0){iris++;return 0;}\t// only possible for 2 nodes otherwise it will go to first if\n\t\t\t\treturn 1;\n\t\t\t}\n\t\t\telse{\n\t\t\t\t// ddd\n\t\t\t\tif(s[1] != s[node]){iris++;return 0;}\n\t\t\t\telse{dora++;return 0;}\n\t\t\t\t// if(qs){\n\t\t\t\t// \tdora++;\n\t\t\t\t// \treturn 0;\n\t\t\t\t// }\n\t\t\t\t// else{\n\t\t\t\t// \tif(onHp ^ Zhp)iris++;\n\t\t\t\t// \telse dora++;\n\t\t\t\t// \treturn 0;\n\t\t\t\t// }\n\t\t\t}\n\t\t}\n\t\telse{\n\t\t\tif(s[node] == '?')f = 1;\n\t\t}\n\t\tll chinta = 0;\n\t\tfor(auto it: adj[node]){\n\t\t\tif(it != par){\n\t\t\t\tchinta += dfs(it, node, onHp, Zhp, qs);\n\t\t\t}\n\t\t}\n\t\treturn chinta;\n\t};\n\tif(s[1] != '?'){\n\t\tll totchinta = dfs(1, -1, 0,0,0);\n\t\tans = iris + (totchinta + 1) / 2;\n\t}\n\telse{\n\t\tll frDr = 0, frIr = 0;\n\t\ts[1] = '1';\n\t\tll totChinta = dfs(1,-1,0,0,0);\n\t\tfrDr = iris + (totChinta + f) / 2;\n\t\tfrIr = iris + (totChinta) / 2;\n\t\t// cout<<iris<<nl;\n\t\tiris = 0, s[1] = '0';\n\t\ttotChinta = dfs(1,-1,0,0,0);\n\t\tfrDr = min(frDr, iris + (totChinta + f) / 2);\n\t\tfrIr = max(frIr, iris + (totChinta) / 2);\n\t\tans = max(frIr, frDr);\n\t}\n\tcout<<ans<<nl;\n}\n \nint main()\n{\n\tfast ll T = 1;\n\tcin >> T;\n\tfor (int tc = 1; tc <= T; tc++)\n\t{\n\t\t// cout<<\"Case \"<<tc<<\": \";\n\t\tsolve();\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": "2006A",
"submission_id": 280567361,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nconst int N=1e5+5;\nint tl,n;vector<int> to[N];\nstring s;int tp[N],wl,ans,anl,wx;bool vl;\n \nvoid get(int x,int f,int w){int tl;\n\tfor(auto j:to[x]){\n\t\tif(j==f)continue;tl=w;\n\t\tif(tp[x]&&!tp[j])tl--;\n\t\telse if(tp[j]&&!tp[x])tl++;\t\n\t\tget(j,x,tl);\n\t}if(x>1&&to[x].size()==1){\n\t\tif(s[x-1]!='?')wl+=(w!=0);\n\t\telse{//cout<<w<<\"\\n\";\n\t\t\tif(w==0||w+1==0)wx++;\n\t\t\telse wl++;\n\t\t}\n\t}\n}\n \nvoid solve(){cin>>n;int x,y;\n\tfor(int i=1;i<n;i++){cin>>x>>y;\n\t\tto[x].push_back(y),to[y].push_back(x);\n\t}cin>>s;ans=anl=wx=vl=0;\n\tfor(int i=2;i<=n;i++)if(s[i-1]=='?'&&to[i].size()!=1)vl=1;\n\tfor(int i=1;i<n;i++)tp[i+1]=(s[i]=='1'?1:0);\n\tif(s[0]!='?'){\n\t\twl=wx=0;tp[1]=(s[0]=='1');get(1,0,0);ans=wl+wx/2;//cout<<wl<<\" \"<<wx<<\"\\n\";\n\t}else{\n\t\twl=wx=0;tp[1]=0;get(1,0,0);ans=wl+wx/2;anl=(wl+(wx+1)/2);\n\t\twl=wx=0;tp[1]=1;get(1,0,0);ans=max(ans,wl+wx/2);anl=min(anl,wl+(wx+1)/2);\n\t}cout<<max(ans,anl*vl)<<\"\\n\";\n\tfor(int i=1;i<=n;i++)to[i].clear(),tp[i]=0;\n}\n \nint main(){\n\tios::sync_with_stdio(false);\n\tcin.tie(0); cout.tie(0); \n\tcin>>tl;\n\tfor(int i=1;i<=tl;i++)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": "2006A",
"submission_id": 278826515,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define pii pair<int,int>\n#define pll pair<long long,long long>\n#define lowbit(x) x&(-x)\n#define ls rt<<1\n#define rs rt<<1|1\n#define all(x) x.begin(),x.end()\n#define rall(x) x.rbegin(),x.rend()\n// #define x first\n// #define y second\nusing namespace std;\nusing i64 = long long;\nusing ll = long long;\ni64 n, m, u, v, k, d, q, l, r, c, w, a, b, x, num1, num2;\nconst int mod = 1e9 + 7;\n \ni64 dfs(int now, int fa, i64 &out_minn, vector<vector<int>>&g, vector<i64>&a)\n{\n\ti64 minn = 1e18;\n\tfor (auto x : g[now])\n\t{\n\t\tif (x == fa) continue;\n\t\tminn = min(minn, dfs(x,now,out_minn,g,a));\n\t}\t\t\n\tif (now == 1)\n\t{\n\t\tout_minn = min(out_minn, minn);\n\t\treturn 1e18;\n\t}\n\tif (a[now] < minn && minn != 1e18)\n\t{\n\t\ti64 temp = (minn - a[now]) / 2;\n\t\ta[now] += temp;\n\t\treturn a[now];\n\t}\n\treturn min(minn, a[now]);\n};\n \n \nvector<int>le;\nvoid dfs(int u, int fa,vector<vector<int>> &g)\n{\n\tif (g[u].size() == 1 && g[u][0] == fa)\n\t{\n\t\tle.push_back(u);\n\t\treturn;\n\t}\n\tfor (auto v : g[u])\n\t{\n\t\tif (v == fa) continue;\n\t\tdfs(v,u,g);\n\t}\n}\nvoid solve() {\n\tint n, m, u, v;\n\tstring s;\n\tcin >> n;\n\tle.clear();\n\tvector<vector<int>> g(n + 1);\n\tfor (int i = 1; i <= n - 1; i++)\n\t{\n\t\tcin >> u >> v;\n\t\tg[u].push_back(v);\n\t\tg[v].push_back(u);\n\t}\n\tcin >> s;\n\ts = '=' + s;\n\tint cnt = 0;\n\tfor (int i = 2; i <= n; i++) if (s[i] == '?') cnt++;\n\tdfs(1,0,g);\n\tint yi = 0, li = 0, we = 0;\n\tfor (auto x : le) \n\t{\n\t\tif (s[x] == '1') yi++;\n\t\telse if (s[x] == '0') li++;\n\t\telse we++;\n\t}\n\tint ans = 0;\n\tif (s[1] == '?')\n\t{\n\t \tans = max(li,yi);\n\t\tans = max((we + (cnt - we) % 2) / 2, ans);\n\t}\n\telse\n\t{\n\t\tif (s[1] == '1') \n\t\t{\n\t\t\tans = li;\n\t\t\tans += (we + 1) / 2;\n\t\t}\n\t\telse \n\t\t{\n\t\t\tans = -yi; \n\t\t\tans -= (we + 1) / 2;\n\t\t}\n \n\t}\n\tcout << abs(ans) << '\\n';\n}\n \nint main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n int tt = 1;\n cin >> tt;\n while (tt--) {\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": "2006A",
"submission_id": 282109857,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class A_Iris_and_Game_on_the_Tree {\n \n int mod = (int) 1e9 + 7;\n static int a = 0;\n static int b = 0;\n static int c = 0;\n static int nl = 0;\n \n public static void main(String[] args) {\n int t = sc.nextInt();\n while (t-- > 0) {\n a = 0;\n b = 0;\n c = 0;\n nl = 0;\n int n = sc.nextInt();\n List<Integer>[] graph = new ArrayList[n];\n for(int i=0;i<n;i++) graph[i] = new ArrayList<>();\n for(int i=0;i<n-1;i++){\n int u = sc.nextInt()-1;\n int v = sc.nextInt()-1;\n graph[u].add(v);\n graph[v].add(u);\n }\n String s = sc.next();\n dfs(graph,0,-1,s);\n // System.out.println(a + \" \" + b + \" \" + c +\" \"+ nl);\n \n if(s.charAt(0)!='?'){\n if(s.charAt(0)=='1'){\n System.out.println(b+c/2+(c%2==0? 0:1));\n }\n else System.out.println(a+c/2+(c%2==0? 0 : 1));\n }\n else{\n System.out.println(Math.max(a,b) + c/2 + (a==b? nl%2 : 0));\n }\n }\n }\n \n public static void dfs(List<Integer>[] graph, int node, int par, String s){\n if(graph[node].size()==1 && par!=-1){\n if(s.charAt(node)=='0') b++;\n else if(s.charAt(node)=='1') a++;\n else c++;\n }\n else{\n if(node!=0 && s.charAt(node)=='?') nl++;\n }\n \n for(int x:graph[node]){\n if(x!=par) dfs(graph,x,node,s);\n }\n }\n \n public static void input(Object array) {\n if (array instanceof int[])\n for (int i = 0; i < ((int[]) array).length; i++)\n ((int[]) array)[i] = sc.nextInt();\n else if (array instanceof long[])\n for (int i = 0; i < ((long[]) array).length; i++)\n ((long[]) array)[i] = sc.nextLong();\n }\n \n public static void printArr(Object array) {\n if (array instanceof int[])\n for (int x : (int[]) array)\n System.out.print(x + \" \");\n else if (array instanceof long[])\n for (long x : (long[]) array)\n System.out.print(x + \" \");\n System.out.println();\n }\n \n public static <T extends Number> long gcd(T a, T b) {\n return b.longValue() == 0 ? a.longValue() : gcd(b, a.longValue() % b.longValue());\n }\n \n static FastReader sc = new FastReader();\n \n static class FastReader {\n BufferedReader br;\n StringTokenizer st;\n \n public FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() {\n while (st == null || !st.hasMoreTokens()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine() {\n String str = \"\";\n try {\n str = br.readLine().trim();\n } catch (Exception e) {\n e.printStackTrace();\n }\n return str;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006A",
"submission_id": 297097139,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class A_Iris_and_Game_on_the_Tree {\n \n int mod = (int) 1e9 + 7;\n static int a = 0;\n static int b = 0;\n static int c = 0;\n static int nl = 0;\n \n public static void main(String[] args) {\n int t = sc.nextInt();\n while (t-- > 0) {\n a = 0;\n b = 0;\n c = 0;\n nl = 0;\n int n = sc.nextInt();\n List<Integer>[] graph = new ArrayList[n];\n for(int i=0;i<n;i++) graph[i] = new ArrayList<>();\n for(int i=0;i<n-1;i++){\n int u = sc.nextInt()-1;\n int v = sc.nextInt()-1;\n graph[u].add(v);\n graph[v].add(u);\n }\n String s = sc.next();\n dfs(graph,0,-1,s);\n // System.out.println(a + \" \" + b + \" \" + c +\" \"+ nl);\n \n if(s.charAt(0)!='?'){\n if(s.charAt(0)=='1'){\n System.out.println(b+c/2+(c%2==0? 0:1));\n }\n else System.out.println(a+c/2+(c%2==0? 0 : 1));\n }\n else{\n if(a>0 || b>0) System.out.println(Math.max(a,b) + c/2 );\n else if(nl%2==1) System.out.println(Math.max(a,b) + (c+1)/2);\n else System.out.println(Math.max(a,b) + c/2);\n }\n }\n }\n \n public static void dfs(List<Integer>[] graph, int node, int par, String s){\n if(graph[node].size()==1 && par!=-1){\n if(s.charAt(node)=='0') b++;\n else if(s.charAt(node)=='1') a++;\n else c++;\n }\n else{\n if(node!=0 && s.charAt(node)=='?') nl++;\n }\n \n for(int x:graph[node]){\n if(x!=par) dfs(graph,x,node,s);\n }\n }\n \n public static void input(Object array) {\n if (array instanceof int[])\n for (int i = 0; i < ((int[]) array).length; i++)\n ((int[]) array)[i] = sc.nextInt();\n else if (array instanceof long[])\n for (int i = 0; i < ((long[]) array).length; i++)\n ((long[]) array)[i] = sc.nextLong();\n }\n \n public static void printArr(Object array) {\n if (array instanceof int[])\n for (int x : (int[]) array)\n System.out.print(x + \" \");\n else if (array instanceof long[])\n for (long x : (long[]) array)\n System.out.print(x + \" \");\n System.out.println();\n }\n \n public static <T extends Number> long gcd(T a, T b) {\n return b.longValue() == 0 ? a.longValue() : gcd(b, a.longValue() % b.longValue());\n }\n \n static FastReader sc = new FastReader();\n \n static class FastReader {\n BufferedReader br;\n StringTokenizer st;\n \n public FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() {\n while (st == null || !st.hasMoreTokens()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine() {\n String str = \"\";\n try {\n str = br.readLine().trim();\n } catch (Exception e) {\n e.printStackTrace();\n }\n return str;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006A",
"submission_id": 297095910,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define ll long long\n#define ss second\n#define ff first\n#define all(v) v.begin(),v.end()\n#define pb push_back\n#define display(a) { for (int i=0;i<a.size();i++) cout<<a[i]<<\" \"; cout<<endl; }\n#define IOS {ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);}\n \n#ifndef ONLINE_JUDGE\n#include \"debug.cpp\"\n#else\n#define debug(x) 1\n#endif\n \nvoid solve(){\n int n;\n cin>>n;\n vector<int> adj[n+1],deg(n+1,0);\n for (int i=0;i<n-1;i++){\n int u,v;\n cin>>u>>v;\n adj[u].pb(v);\n adj[v].pb(u);\n deg[u]++;\n deg[v]++;\n }\n string s;\n cin>>s;\n int root=1;\n int zero=0,one=0,mark=0;\n for(int i=2;i<=n;i++){\n if(!deg[i]) continue;\n if(s[i-1]=='1') one++;\n else if (s[i-1]=='0') zero++;\n else mark++;\n }\n int ans=0;\n if(s[root-1]=='1'){\n ans+=zero;\n ans+=(mark+1)/2;\n }else if(s[root-1]=='0'){\n ans+=one;\n ans+=(mark+1)/2;\n }else{\n ans+=min(one,zero)+mark/2;\n }\n cout<<ans<<endl;\n}\n \nint main(){\n IOS\n int t=1;\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": "2006A",
"submission_id": 279048230,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \nvoid solve(){\n int n;\n cin>>n;\n \n vector<vector<int>> g(n);\n for(int i=1;i<n;i++){\n int u,v;\n cin>>u>>v;\n u--,v--;\n g[u].push_back(v);\n g[v].push_back(u);\n }\n \n string s;\n cin>>s;\n \n if(s[0]!='?'){\n int ct=0,res=0;\n for(int i=1;i<n;i++)\n if(g[i].size()==1){\n if(s[i]==(s[0]^1)) res++;\n else if(s[i]=='?') ct++;\n }\n cout<<res+(ct+1)/2<<'\\n';\n }else{\n int ct1=0,ct2=0,ct3=0,ct4=0;\n for(int i=1;i<n;i++)\n if(g[i].size()==1){\n if(s[i]=='0') ct1++;\n else if(s[i]=='1') ct2++;\n else ct3++;\n }else if(s[i]=='?') ct4++;\n if(ct4&1) cout<<min(ct1,ct2)+(ct3+1)/2<<'\\n';\n else cout<<max(ct1,ct2)+ct3/2<<'\\n';\n }\n}\n \nint main(){\n ios::sync_with_stdio(false);\n cin.tie(nullptr),cout.tie(nullptr);\n \n int t=1;\n cin>>t;\n while(t--) solve();\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006A",
"submission_id": 281962892,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nint t,n,ans,tot;\nstring s;\nvector<int> g[100100];\nvector<int> rec;\nvoid dfs(int pos,int from){\n\tbool isl=1;\n\tfor(int i:g[pos]){\n\t\tif(i==from) continue;\n\t\tisl=0;\n\t\tdfs(i,pos);\n\t}\n\tif(isl){\n\t\trec.push_back(pos);\n\t}\n\telse if(s[pos]=='?'&&pos!=1){\n\t\ttot++;\n\t}\n\treturn;\n}\nint main(){\n\tios::sync_with_stdio(0);\n cin.tie(0),cout.tie(0);\n\tcin>>t;\n\tfor(int o=1;o<=t;o++){\n\t\tans=0; tot=0;\n\t\tcin>>n;\n for(int i=1;i<=n-1;i++){\n int u,v;\n cin>>u>>v;\n g[u].push_back(v);\n g[v].push_back(u);\n }\n cin>>s;\n s=\" \"+s;\n dfs(1,0);\n int rol=1;//0->max,1->min\n int muc[2]={0,0},siz=rec.size();\n if(s[1]=='?'){\n \tfor(int i:rec){\n \t\tif(s[i]!='?'){\n \t\t\tmuc[s[i]-'0']++;\n\t\t\t\t}\n\t\t\t}\n \trol^=1;\n \tif(muc[0]>muc[1]) s[1]='1';\n \telse s[1]='0';\n\t\t}\n if(s[1]=='0'){\n \tfor(int i:rec){\n \t\tif(s[i]=='?'){\n \t\t\trol^=1;\n \t\t\tif(rol==0) ans++;\n\t\t\t\t}\n\t\t\t\telse if(s[i]=='1') ans++;\n\t\t\t}\n\t\t}\n\t\telse if(s[1]=='1'){\n\t\t\tfor(int i:rec){\n \t\tif(s[i]=='?'){\n \t\t\trol^=1;\n \t\t\tif(rol==0) ans++;\n\t\t\t\t}\n\t\t\t\telse if(s[i]=='0') ans++;\n\t\t\t}\n\t\t}\n\t\tif(muc[0]==muc[1]&&muc[0]==0&&tot%2){\n\t\t\tans=max(ans,(siz+1)/2);\n\t\t}\n\t\tcout<<ans<<\"\\n\";\n for(int i=1;i<=n;i++){\n \tg[i].clear();\n\t\t}\n\t\trec.clear();\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": "2006A",
"submission_id": 278799719,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\nfrom sys import stdin,stdout\ninput = stdin.readline\nfrom bisect import bisect,bisect_right,bisect_left\nfrom math import inf,ceil,log2,gcd,lcm,sqrt,log\nfrom heapq import heapify,heappop,heappush\nfrom collections import Counter,deque\nmod = 10**9 + 7\n \n \nfor _ in range(int(input())):\n n = int(input()) \n adj = [[] for i in range(n+1)]\n for i in range(n-1):\n u,v = map(int,input().split())\n adj[u].append(v)\n adj[v].append(u)\n s = list(input().rstrip('\\n'))\n s = [''] + s\n tot2 = 0\n for i in range(1,n+1):\n if (s[i] == '?'):\n tot2 += 1\n \n l0 = 0\n l1 = 0\n l2 = 0 \n # print(s)\n for i in range(2,n+1):\n if (len(adj[i]) == 1):\n if (s[i] == '0'):\n l0 += 1\n elif (s[i] == '1'):\n l1 += 1\n else:\n l2 += 1\n if (s[1] == '1'):\n ans = l0 + ceil(l2/2) \n elif (s[1] == '0'):\n ans = l1 + ceil(l2/2)\n else:\n if (tot2 > (l2 + 1)):\n if (l0 == l1):\n ans = l0 + int((l2+1)/2)\n else:\n ans = max(l0,l1) + int(l2/2)\n else:\n ans = max(l0,l1) + int(l2/2)\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": "2006A",
"submission_id": 278832107,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\nimport os\nimport sys\nfrom io import BytesIO, IOBase\nfrom collections import *\nfrom math import *\nfrom functools import *\nfrom types import *\nfrom heapq import *\nfrom itertools import *\nfrom operator import *\nfrom bisect import *\nBUFSIZE = 8192\nclass FastIO(IOBase):\n newlines = 0\n def __init__(self, file):\n self._file = file\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\nsys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\ninput = lambda: sys.stdin.readline().rstrip(\"\\r\\n\")\n \ndef sol(n,edges,s):\n adj=defaultdict(set)\n for a,b in edges:\n adj[a].add(b)\n adj[b].add(a)\n queue=deque([0])\n d={'0':0, '1':0, '?':0}\n count=0\n for node in range(1,n):\n if len(adj[node])==1:\n d[s[node]]+=1\n else:\n count+=s[node]=='?'\n score=0\n if s[0]!='?':\n score=d['1'] if s[0]=='0' else d['0']\n score+=(d['?']+1)//2\n else:\n score=max(d['1'],d['0'])\n score+=(d['?'])//2\n score+=d['1']==d['0'] and count%2\n return score\n \nt=int(input().strip())\nfor _ in range(t):\n n=int(input().strip())\n edges=[]\n for _ in range(n-1):\n a,b=map(int,input().strip().split())\n edges.append((a-1,b-1))\n s=(input().strip())\n print(sol(n,edges,s))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006A",
"submission_id": 278820648,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\nimport java.util.*;\n \npublic class cf156 {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int T = sc.nextInt();\n \n for (int t = 0; t < T; t++) {\n int n = sc.nextInt();\n int x = 0, y = 0, z = 0, w = 0;\n int[] deg = new int[n + 1];\n \n // Reading the edges and calculating the degree of each node\n for (int i = 0; i < n - 1; i++) {\n int u = sc.nextInt();\n int v = sc.nextInt();\n deg[u]++;\n deg[v]++;\n }\n \n // Reading the string (s) and processing\n String s = \" \" + sc.next(); // Adding a space at the beginning to match 1-based indexing\n \n for (int i = 2; i <= n; i++) {\n if (deg[i] == 1) { // Leaf node\n if (s.charAt(i) == '?') {\n z++;\n } else if (s.charAt(i) == '0') {\n x++;\n } else {\n y++;\n }\n } else if (s.charAt(i) == '?') {\n w++;\n }\n }\n \n if (s.charAt(1) == '0') {\n System.out.println(y + (z + 1) / 2);\n } else if (s.charAt(1) == '1') {\n System.out.println(x + (z + 1) / 2);\n } else {\n System.out.println(Math.max(x, y) + (z + (w % 2 == (x == y ? 1 : 0) ? 1 : 0)) / 2);\n }\n }\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": "2006A",
"submission_id": 281909979,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\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 n = nextInt();\n int[][] edges = inEdges(n - 1);\n d = edges2List(edges, n);\n a = next().toCharArray();\n leafCnt = new int[200];\n dive(0, -1);\n if (a[0] != '?') sl(leafCnt[1 - (a[0] - '0') + '0'] + (leafCnt['?'] + 1) / 2);\n else {\n int allQ = 0;\n for (char x : a) if (x == '?') allQ++;\n int existMax = Math.max(leafCnt['0'], leafCnt['1']);\n if (existMax > 0) {\n sl(existMax + leafCnt['?'] / 2);\n } else {\n if ((allQ - 1 - leafCnt['?']) % 2 == 0) {\n sl((leafCnt['?'] - 1 + 1) / 2);\n } else {\n sl((leafCnt['?'] + 1) / 2);\n }\n }\n }\n}\n \nchar[] a;\nint[][] d;\nint[] leafCnt;\n \nvoid dive(int cur, int fa) {\n boolean isLeaf = true;\n for (int x : d[cur]) {\n if (x == fa) continue;\n isLeaf = false;\n dive(x, cur);\n }\n if (fa != -1 && isLeaf) {\n leafCnt[a[cur]]++;\n }\n}\n \nint[][] inEdges(int n) {\n int[][] edges = new int[n][2];\n for (int i = 0; i < n; ++i)\n for (int j = 0; j < 2; ++j)\n edges[i][j] = nextInt() - 1;\n return edges;\n}\n \nint[][] edges2List(int[][] edges, int n) {\n int[] g = new int[n];\n for (int[] en : edges) {\n g[en[0]]++;\n g[en[1]]++;\n }\n int[][] r = new int[n][0];\n for (int i = 0; i < n; ++i) {\n r[i] = new int[g[i]];\n }\n int[] ri = new int[n];\n for (int[] en : edges) {\n r[en[0]][ri[en[0]]++] = en[1];\n r[en[1]][ri[en[1]]++] = en[0];\n }\n return r;\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": "2006A",
"submission_id": 279830166,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\n// Working program with FastReader \nimport java.io.BufferedReader; \nimport java.io.IOException; \nimport java.io.InputStreamReader; \nimport java.util.*; \n \npublic class d {\n static class FastReader { \n BufferedReader br; \n StringTokenizer st; \n \n public FastReader() \n { \n br = new BufferedReader( \n new InputStreamReader(System.in)); \n } \n \n String next() \n { \n while (st == null || !st.hasMoreElements()) { \n try { \n st = new StringTokenizer(br.readLine()); \n } \n catch (IOException e) { \n e.printStackTrace(); \n } \n } \n return st.nextToken(); \n } \n \n int nextInt() { return Integer.parseInt(next()); } \n \n long nextLong() { return Long.parseLong(next()); } \n \n double nextDouble() \n { \n return Double.parseDouble(next()); \n } \n \n String nextLine() \n { \n String str = \"\"; \n try { \n if(st.hasMoreTokens()){ \n str = st.nextToken(\"\\n\"); \n } \n else{ \n str = br.readLine(); \n } \n } \n catch (IOException e) { \n e.printStackTrace(); \n } \n return str; \n } \n } \n // gcd method returns the GCD of a and b \n static long gcd(long a, long b) { \n \n // if b=0, a is the GCD \n if (b == 0) \n return a; \n \n // call the gcd() method recursively by \n // replacing a with b and b with \n // modulus(a,b) as long as b != 0 \n else\n return gcd(b, a % b); \n } \n \n // lcm() method returns the LCM of a and b \n static long lcm(long a, long b, long gcdValue) \n { \n return Math.abs(a * b) / gcdValue; \n } \n static FastReader sc = new FastReader();\n public static void main(String[] args) {\n int t= sc.nextInt();\n while (t-->0) {\n solve();\n }\n \n \n }\n private static void solve() {\n int n = sc.nextInt();\n HashMap<Integer , List<Integer>> map = new HashMap<>();\n while (n-->1) {\n int u = sc.nextInt();\n int v = sc.nextInt();\n if(!map.containsKey(u)){\n map.put(u, new ArrayList<>());\n }\n if(!map.containsKey(v)){\n map.put(v, new ArrayList<>());\n }\n map.get(u).add(v);\n map.get(v).add(u);\n }\n char[] arr = sc.next().toCharArray();\n HashSet<Integer> set = new HashSet<>();\n for(int x : map.keySet()){\n if(map.get(x).size() == 1 && x!=1){\n set.add(x);\n }\n }\n // System.out.println(set);\n int ans = 0 ;\n int c = 0 ;\n if(arr[0]!='?'){\n for(int x : set){\n if(arr[x-1]!='?' && arr[x-1]!=arr[0]){\n ans ++;\n }else if(arr[x-1]=='?'){\n c++;\n }\n }\n ans += (c+1)/2;\n System.out.println(ans);\n }else{\n int ones = 0 ;\n int zeros = 0 ;\n for(int x : set){\n if(arr[x-1]=='?'){\n c++;\n }else if(arr[x-1] == '0') {\n zeros++;\n }else ones++;\n }\n int nonleaf = 0 ;\n for(int i = 2 ; i<=arr.length ; i++){\n if(!set.contains(i) && arr[i-1] == '?'){\n nonleaf ++;\n }\n }\n int leafs = c;\n if(leafs%2==0)\n System.out.println(leafs/2 +Math.max(zeros, ones));\n else{\n if(nonleaf%2==1)\n System.out.println((leafs+1)/2+Math.min(zeros, ones));\n else{\n System.out.println(leafs/2 +Math.max(zeros, ones));\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": "2006A",
"submission_id": 278895344,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\n// Working program with FastReader \nimport java.io.BufferedReader; \nimport java.io.IOException; \nimport java.io.InputStreamReader; \nimport java.util.*; \n \npublic class d {\n static class FastReader { \n BufferedReader br; \n StringTokenizer st; \n \n public FastReader() \n { \n br = new BufferedReader( \n new InputStreamReader(System.in)); \n } \n \n String next() \n { \n while (st == null || !st.hasMoreElements()) { \n try { \n st = new StringTokenizer(br.readLine()); \n } \n catch (IOException e) { \n e.printStackTrace(); \n } \n } \n return st.nextToken(); \n } \n \n int nextInt() { return Integer.parseInt(next()); } \n \n long nextLong() { return Long.parseLong(next()); } \n \n double nextDouble() \n { \n return Double.parseDouble(next()); \n } \n \n String nextLine() \n { \n String str = \"\"; \n try { \n if(st.hasMoreTokens()){ \n str = st.nextToken(\"\\n\"); \n } \n else{ \n str = br.readLine(); \n } \n } \n catch (IOException e) { \n e.printStackTrace(); \n } \n return str; \n } \n } \n // gcd method returns the GCD of a and b \n static long gcd(long a, long b) { \n \n // if b=0, a is the GCD \n if (b == 0) \n return a; \n \n // call the gcd() method recursively by \n // replacing a with b and b with \n // modulus(a,b) as long as b != 0 \n else\n return gcd(b, a % b); \n } \n \n // lcm() method returns the LCM of a and b \n static long lcm(long a, long b, long gcdValue) \n { \n return Math.abs(a * b) / gcdValue; \n } \n static FastReader sc = new FastReader();\n public static void main(String[] args) {\n int t= sc.nextInt();\n while (t-->0) {\n solve();\n }\n \n \n }\n private static void solve() {\n int n = sc.nextInt();\n HashMap<Integer , List<Integer>> map = new HashMap<>();\n while (n-->1) {\n int u = sc.nextInt();\n int v = sc.nextInt();\n if(!map.containsKey(u)){\n map.put(u, new ArrayList<>());\n }\n if(!map.containsKey(v)){\n map.put(v, new ArrayList<>());\n }\n map.get(u).add(v);\n map.get(v).add(u);\n }\n char[] arr = sc.next().toCharArray();\n HashSet<Integer> set = new HashSet<>();\n for(int x : map.keySet()){\n if(map.get(x).size() == 1 && x!=1){\n set.add(x);\n }\n }\n // System.out.println(set);\n int ans = 0 ;\n int c = 0 ;\n if(arr[0]!='?'){\n for(int x : set){\n if(arr[x-1]!='?' && arr[x-1]!=arr[0]){\n ans ++;\n }else if(arr[x-1]=='?'){\n c++;\n }\n }\n ans += (c+1)/2;\n System.out.println(ans);\n }else{\n int ones = 0 ;\n int zeros = 0 ;\n for(int x : set){\n if(arr[x-1]=='?'){\n c++;\n }else if(arr[x-1] == '0') {\n zeros++;\n }else ones++;\n }\n int nonleaf = 0 ;\n for(int i = 2 ; i<arr.length ; i++){\n if(!set.contains(i) && arr[i-1] == '?'){\n nonleaf ++;\n }\n }\n int leafs = c;\n if(leafs%2==0)\n System.out.println(leafs/2 +Math.max(zeros, ones));\n else{\n if(nonleaf%2==1)\n System.out.println((leafs+1)/2+Math.min(zeros, ones));\n else{\n System.out.println(leafs/2 +Math.max(zeros, ones));\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": "2006A",
"submission_id": 278894398,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\nfor _ in range(int(input())):\n\tn = int(input())\n\tfreq = [0]*n\n\tfor i in range(n-1):\n\t\tu,v = map(int,input().split())\n\t\tfreq[u-1]+=1\n\t\tfreq[v-1]+=1\n\tweight = list(input())\n\tleafs = [i for i in range(1,n) if freq[i]==1]\n\t\n\tgame = sum(weight[leaf]=='?' for leaf in leafs)\n\tif weight[0]!='?':\n\t\tscore = sum(weight[leaf]!='?' and weight[leaf]!=weight[0] for leaf in leafs)\n\t\tprint(score + (game+1)//2)\n\telse:\n\t\tscore0 = sum(weight[leaf]!='?' and weight[leaf]!='0' for leaf in leafs)\n\t\tscore1 = sum(weight[leaf]!='?' and weight[leaf]!='1' for leaf in leafs)\n\t\tif score1!=score0:\n\t\t\tprint(max(score0,score1) + game//2)\n\t\telse:\n\t\t\tif sum(weight[i]=='?' for i in range(1,n)) - game == 0 :\n\t\t\t\tprint(score0+ game//2)\n\t\t\telse:\n\t\t\t\tprint(score0+ game//2 + game%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": "2006A",
"submission_id": 282803552,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\nfrom sys import stdin, stdout\nfrom collections import defaultdict,deque\ninput = stdin.readline\ninp = lambda : list(map(int,input().split()))\ninps= lambda :\tinput().strip()\ninpsl= lambda :\tlist(input().strip().split())\nimport math,random\nMOD = 1000000007\nclass CustomSet:\n def __init__(self, arr = []):\n self.RANDOM = random.randrange(2**62)\n self.set = set(arr)\n def wrapper(self,num):\n return num ^ self.RANDOM\n def add(self, key):\n self.set.add(self.wrapper(key))\n def remove(self, key):\n self.set.remove(self.wrapper(key))\n def __contains__(self, key):\n return self.wrapper(key) in self.set\n def __iter__(self):\n return iter(self.wrapper(i) for i in self.set)\n def __len__(self):\n return len(self.set)\n def __repr__(self):\n return repr({self.wrapper(i) for i in self.set})\nfor _ in range(*inp()):\n n,=inp()\n d={}\n \n for i in range(n-1):\n x,y=inp()\n if x in d:\n d[x]+=1\n else:\n d[x]=1\n if y in d:\n d[y]+=1\n else:\n d[y]=1\n le=[]\n for i in d:\n if i!=1 and d[i]==1: le.append(i)\n s=inps()\n ro=s[0]\n c0,c1,c2=0,0,0\n for i in le:\n if s[i-1]=='1': c1+=1\n elif s[i-1]=='0': c0+=1\n else: c2+=1\n if ro=='0':\n co=s.count('?')\n od=co-c2\n if c0==c1 and c2%2==0 and od%2:\n print(c1-(-c2//2)+1)\n else:\n print(c1-(-c2//2))\n elif ro=='1':\n co=s.count('?')\n od=co-c2\n if c0==c1 and c2%2==0 and od%2:\n print(c0-(-c2/2)+1)\n else:\n print(c0-(-c2//2))\n else:\n co=s.count('?')\n od=co-1-c2\n if c0!=c1 or od%2==0: \n print(max(c0,c1)+(c2//2))\n elif od%2:\n print(max(c0,c1)+(c2//2)+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": "2006A",
"submission_id": 278842015,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\nfor _ in range(int(input())):\n n=int(input())\n tr=[[] for i in range(n)]\n for i in range(n-1):\n u,v=map(int,input().split())\n tr[u-1].append(v-1)\n tr[v-1].append(u-1)\n s=input()\n li=[]\n co=s.count(\"?\")\n for i in range(1,n):\n if len(tr[i])==1:\n li.append(i)\n p=0\n q=0\n r=0\n for i in li:\n if s[i]==\"0\":\n p+=1\n elif s[i]==\"1\":\n q+=1\n else:\n r+=1\n if s[0]==\"0\":\n print(len(li)-p-r//2)\n elif s[0]==\"1\":\n print(len(li)-q-r//2)\n else:\n if (len(li)-co)%2==0:\n print(max(p,q)+(r+1)//2)\n else:\n print(max(p,q)+r//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": "2006A",
"submission_id": 278840847,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\nimport math\nt = int(input())\nfor _ in range(t):\n n = int(input())\n adj = [[]for _ in range(n)]\n for _ in range(n-1):\n u,v = map(int,input().split())\n u,v = min(u,v),max(u,v)\n adj[u-1].append(v-1)\n s = input()\n c0 = 0\n c1 = 0\n c2 = 0\n cnt = s.count('?') \n for node in range(n):\n if len(adj[node])==0:\n if s[node]=='0':\n c0 += 1\n elif s[node]=='1':\n c1 += 1\n else:\n c2 += 1\n res = 0\n if s[0]=='1':\n res = c0+math.ceil(c2/2)\n elif s[0]=='0':\n res = c1+math.ceil(c2/2)\n elif s[0]=='?':\n if c1==c0 and (cnt-c2-1)>0 and (cnt-c2-1)%2==1:\n res = c1+math.ceil(c2/2)\n else:\n res = max(c0,c1)+c2//2\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": "2006A",
"submission_id": 279018400,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \n \npublic class Main {\n static PrintWriter pw = new PrintWriter(System.out);\n static Scanner sc = new Scanner(System.in) ;\n static ArrayList<Integer> adjl [];\n \n public static void solve() throws IOException {\n int n = sc.nextInt();\n adjl = new ArrayList[n + 1];\n for(int i = 0; i < n ; i++) {\n adjl[i] = new ArrayList<>();\n }\n for(int i = 1; i < n ; i++){\n int u = sc.nextInt() - 1, v = sc.nextInt() - 1;\n adjl[u].add(v);\n adjl[v].add(u);\n }\n char a [] = sc.next().toCharArray();\n if(n == 100000){\n pw.println(0);\n return;\n }\n int ones = 0, zeros = 0, maybe = 0, unknown = 0;\n for(int i = 1; i < n ; i++){\n if(adjl[i].size() == 1){\n if(a[i] == '1') {\n ones++;\n }\n else if (a[i] == '0'){\n zeros++;\n }\n else{\n maybe++;\n }\n }\n else{\n if(a[i] == '?') {\n unknown++;\n }\n }\n }\n if(a[0] == '?'){\n if(ones == zeros){\n if(unknown % 2 == 1){\n pw.println(ones + (maybe + 1) / 2);\n }\n else{\n pw.println(ones + maybe / 2);\n }\n }\n else {\n pw.println(Math.max(ones, zeros) + maybe / 2);\n }\n }\n else{\n pw.println((a[0] == '1' ? zeros : ones) + (maybe + 1) / 2);\n }\n }\n \n public static void main(String[] args) throws IOException {\n int t = sc.nextInt();\n while(t -- > 0)\n solve();\n pw.flush();\n }\n \n \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 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": "2006A",
"submission_id": 278866357,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris and Game on the Tree\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Iris has a tree rooted at vertex $1$. Each vertex has a value of $\\mathtt 0$ or $\\mathtt 1$.\n\nLet's consider a leaf of the tree (the vertex $1$ is never considered a leaf) and define its weight. Construct a string formed by the values of the vertices on the path starting at the root and ending in this leaf. Then the weight of the leaf is the difference between the number of occurrences of $\\mathtt{10}$ and $\\mathtt{01}$ substrings in it.\n\nTake the following tree as an example. Green vertices have a value of $\\mathtt 1$ while white vertices have a value of $\\mathtt 0$.\n\n\n\n * Let's calculate the weight of the leaf $5$: the formed string is $\\mathtt{10110}$. The number of occurrences of substring $\\mathtt{10}$ is $2$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $2 - 1 = 1$. * Let's calculate the weight of the leaf $6$: the formed string is $\\mathtt{101}$. The number of occurrences of substring $\\mathtt{10}$ is $1$, the number of occurrences of substring $\\mathtt{01}$ is $1$, so the difference is $1 - 1 = 0$.\n\nThe score of a tree is defined as the number of leaves with non-zero weight in the tree.\n\nBut the values of some vertices haven't been decided and will be given to you as $\\texttt{?}$. Filling the blanks would be so boring, so Iris is going to invite Dora to play a game. On each turn, one of the girls chooses any of the remaining vertices with value $\\texttt{?}$ and changes its value to $\\mathtt{0}$ or $\\mathtt{1}$, with Iris going first. The game continues until there are no vertices with value $\\mathtt{?}$ left in the tree. Iris aims to maximize the score of the tree, while Dora aims to minimize that.\n\nAssuming that both girls play optimally, please determine the final score of the tree.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 5\\cdot 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 10^5$) — the number of vertices in the tree.\n\nThe following $n - 1$ lines each contain two integers $u$ and $v$ ($1 \\leq u, v \\leq n$) — denoting an edge between vertices $u$ and $v$.\n\nIt's guaranteed that the given edges form a tree.\n\nThe last line contains a string $s$ of length $n$. The $i$-th character of $s$ represents the value of vertex $i$. It's guaranteed that $s$ only contains characters $\\mathtt{0}$, $\\mathtt{1}$ and $\\mathtt{?}$.\n\nIt is guaranteed that the sum of $n$ over all test cases doesn't exceed $2\\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the final score of the tree.\n\nExamples:\ninput:\n6\n4\n1 2\n1 3\n4 1\n0101\n4\n1 2\n3 2\n2 4\n???0\n5\n1 2\n1 3\n2 4\n2 5\n?1?01\n6\n1 2\n2 3\n3 4\n5 3\n3 6\n?0????\n5\n1 2\n1 3\n1 4\n1 5\n11?1?\n2\n2 1\n??\noutput:\n2\n1\n1\n2\n1\n0\n\nNote: In the first test case, all the values of the vertices have been determined. There are three different paths from the root to a leaf:\n\n * From vertex $1$ to vertex $2$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. * From vertex $1$ to vertex $3$. The string formed by the path is $\\mathtt{00}$, so the weight of the leaf is $0-0=0$. * From vertex $1$ to vertex $4$. The string formed by the path is $\\mathtt{01}$, so the weight of the leaf is $0-1=-1$. \n\nThus, there are two leaves with non-zero weight, so the score of the tree is $2$.\n\nIn the second test case, one of the sequences of optimal choices for the two players can be:\n\n * Iris chooses to change the value of the vertex $3$ to $\\mathtt 1$. * Dora chooses to change the value of the vertex $1$ to $\\mathtt 0$. * Iris chooses to change the value of the vertex $2$ to $\\mathtt 0$. \n\nThe final tree is as follows:\n\n\n\nThe only leaf with non-zero weight is $3$, so the score of the tree is $1$. Note that this may not be the only sequence of optimal choices for Iris and Dora.\n\nBuggy Code:\n#include<iostream>\n#include<bits/stdc++.h>\nusing namespace std;\n \n#define IN(i,l,r) (l<i&&i<r)\n#define FOR(i,L,R) for (int i = L; i < R; i++)\n#define FORE(i,L,R) for (int i = L; i <= R; i++)\n#define FFOR(i,L,R) for (int i = L; i > R; i--)\n#define FFORE(i,L,R) for (int i = L; i >= R; i--)\n#define ll long long\n \nint c0=0;\nint c1=0;\nint c2=0;\n \n \nvoid dfs(int i,string val,vector<vector<int>>&adj,vector<bool>&vis){\n bool isLeaf=true;\n vis[i]=true;\n for(auto it:adj[i]){\n if(!vis[it]){\n isLeaf=false;\n dfs(it,val,adj,vis);\n }\n }\n if(isLeaf){\n if(val[i]=='1'){c1++;}\n if(val[i]=='0')c0++;\n if(val[i]=='?') c2++;\n }\n}\n \n \nint solve(){\n int n;cin>>n;\n vector<vector<int>>adj(n);\n vector<bool>vis(n,0);\n string val;\n c1=0;c2=0;c0=0;\n \n for(int i=0;i<n-1;i++){\n int a, b;cin>>a>>b;\n a--;b--;\n adj[a].push_back(b);\n adj[b].push_back(a);\n }\n cin>>val;\n \n dfs(0,val,adj,vis);\n \n int ans=0;\n int k=0;\n for(int i=0;i<n;i++){\n if(val[i]=='?'){\n k++;\n }\n }\n \n \n k=k-c2;\n if(val[0]=='1'){\n ans+=c0;\n ans=ans+(c2+1)/2;\n }\n else if(val[0]=='0'){\n ans=ans+c1;\n ans=ans+(c2+1)/2;\n }\n else{\n k--;\n \n int p1=c1+c2/2;\n int p0=c0+c2/2;\n \n if(c0==c1&&k%2){\n p1=c1+(c2+1)/2;\n }\n \n ans=max(p1,p0);\n }\n return ans;\n \n}\n \nint main()\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;cin>>t;\n \n while(t--){\n cout<<solve()<<endl;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006A",
"submission_id": 278849915,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\n#include \"bits/stdc++.h\"\n \nusing namespace std;\n \ntypedef long long ll;\ntypedef pair<ll, ll> pii;\n \n#define all(x) x.begin(), x.end()\n#define sz(x) ((int) ((x).size()))\n#define pb push_back\n#define F first\n#define S second\n#define FIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)\n \nint n;\nint s[1000010];\nmap<int, int> p;\n \nvoid solve() {\n\tcin >> n;\n \n\tint x = 0;\n \n\tfor (int i = 0; i < n; i++) {\n\t\tint t;\n\t\tcin >> t;\n\t\tx ^= p[s[t]];\n\t}\n \n\tif (x != 0) {\n\t\tcout << \"Bob\" << endl;\n\t} else \n\t\n\t\tcout << \"Alice\" << endl;\n}\n \nint main() {\n\tint cnt = 0;\n\tfor (int i = 2; i < 1000010; i++) {\n\t\tif (!s[i]) {\n\t\t\tp[i] = cnt;\n\t\t\tcnt++;\n\t\t\tfor (int j = 2; j < 1000010; j += i) {\n\t\t\t\ts[j] = i;\n\t\t\t}\n\t\t}\n\t}\n \n\tFIO;\n\tint t = 1;\n\tcin >> t;\n\twhile(t--) {\t\n\t\tsolve();\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": "2004E",
"submission_id": 276711904,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n#define ll long long\n#define ul unsigned long long\n#define ld long double\n#define pll pair<ll,ll>\n#define pii pair<int,int>\n#define vi vector<int>\n#define vl vector<ll>\n#define vpl vector<pll>\n#define vpi vector<pii>\n#define vvi vector<vi>\n#define vvl vector<vl>\n#define vb vector<bool>\n#define vs vector<string>\n#define rep(i,a,b) for(ll i=a;i<b;i++)\n#define rrep(i,a,b) for(ll i=a;i>=b;i--)\n#define frs(i,n) for(ll i=1;i*i<=n;i++)\n#define fr(i,a,b,c) for(ll i=a;i<b;i+=c)\n#define ff first\n#define ss second\n#define pb push_back\n#define be() begin()\n#define all(a) a.be(),a.end()\n#define lb(a,x) lower_bound(all(a),x)-a.be()\n#define nl cout<<\"\\n\";\n#define ub(a,x) upper_bound(all(a),x)-a.be()\n#define me max_element\n#define mpl map<ll,ll>\n#define umpl unordered_map<ll,ll>\n#define stl stack<ll>\n#define qul queue<ll>\n#define sz(a) a.size()\n#define rev(a) reverse(a.be(),a.end())\n#define zbit(x) __builtin_ctzll(x)\n#define sbit(x) __builtin_popcountll(x)\n#define lzbit(x) __builtin_clzll(x)\n#define sl set<ll>\n#define py cout<<\"YES\";\n#define pm cout<<\"-1\";\n#define pn cout<<\"NO\";\n#define fbo find_by_order\n#define ook order_of_key\n#define pnt(x) cout<<x\n#pragma GCC target(\"popcnt\")\n \nusing namespace std;\nusing u64 = uint64_t;\nusing u128 = __uint128_t;\nusing namespace __gnu_pbds;\n \n \ntemplate<typename T> \nusing oset= tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\n \nu64 binpower(u64 base, u64 e, u64 mod) {\n u64 result = 1;\n base %= mod;\n while (e) {\n if (e & 1)\n result = (u128)result * base % mod;\n base = (u128)base * base % mod;\n e >>= 1;\n }\n return result;\n}\n \nbool check_composite(u64 n, u64 a, u64 d, int s) {\n u64 x = binpower(a, d, n);\n if (x == 1 || x == n - 1)\n return false;\n for (int r = 1; r < s; r++) {\n x = (u128)x * x % n;\n if (x == n - 1)\n return false;\n }\n return true;\n};\nbool iPrime(u64 n) { \n if (n < 2)\n return false;\n \n int r = 0;\n u64 d = n - 1;\n while ((d & 1) == 0) {\n d >>= 1;\n r++;\n }\n \n for (int a : {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37}) {\n if (n == a)\n return true;\n if (check_composite(n, a, d, r))\n return false;\n }\n return true;\n}\n \n \n \n/*------------------------------------------------------------------------------------------------------------------------------------------------------------------*/\nll sm(vl a,ll s=0){for(auto i:a) s+=i;return s;}\nvoid pre(vl &a){rep(i,1,sz(a)) a[i]+=a[i-1];}\nvoid sieve(vb &x,ll n){x[0]=x[1]=0;for(ll i=2;i*i<=n;i++)if(x[i])for(ll j=i*i;j<=n;j+=i)x[j]=0;}\nll isprime(ll n){if(n==1||n==0)return 0;if(n==2||n==3)return 1;if(n%2==0||n%3==0)return 0;for(ll i=5;i<=n/2;i+=6)if(n%i==0||n%(i+2)==0)return 0;return 1;}\nvoid primef(ll n,vl &x){if(!(n&1)){x.pb(2);while(!(n&1))n=n>>1;}for(ll i=3;i*i<=n;i+=2){if(n%i==0){x.pb(i);while(n%i==0)n/=i;}}if(n>2)x.pb(n);}\nll freq(ll n,ll d){if(n%d!=0)return 0;return 1+freq(n/d,d);}\nll lcm(ll a,ll b){return (a*b)/gcd(a,b);}\nvoid pow2(vl &x){x.pb(1);while(sz(x)<=60)x.pb(2*x[sz(x)-1]);}\nbool sorta( pll a, pll b){return (a.ss < b.ss);}\nbool sortd( pll a, pll b){return (a.ss > b.ss);}\nvoid tobin(ll n,string &s){while(n>0){if(n&1)s.pb('1');else s.pb('0');n=n>>1;}}\nvoid printv(vl v){for(auto i: v)cout<<i<<\" \";}\nchar toc(ll n){char c='0';c+=n;return c;}\nll toi(char c){ll x=c-'0';return x;}\nll ain(char c){ll in=c-'a';return in;}\nll mi(ll a,ll b){return a+(b-a)/2;} \n/*------------------------------------------------------------------------------------------------------------------------------------------------------------------*/\n \n \n \n \n \n \n//-----SEGMENT TREE-----//\n \n/*ll st(vl a,vl& s,ll e,ll b=0,ll i=0)\n{\n if(b==e)\n {\n s[i]=a[b];\n return a[b];\n }\n ll m=b+(e-b)/2;\n s[i]=st(a,s,m,b,2*i+1)+st(a,s,e,m+1,2*i+2);\n return s[i];\n}\nll gsm(vl s,ll l,ll r,ll e,ll b=0,ll i=0)\n{\n if(l<=b&&r>=e)\n return s[i];\n if(l>e||r<b)\n return 0;\n ll m=b+(e-b)/2;\n return gsm(s,l,r,m,b,2*i+1)+gsm(s,l,r,e,m+1,2*i+2);\n}\nvoid update(vl& s,ll d,ll j,ll e,ll b=0,ll i=0)\n{\n if(j<b||j>e)\n return;\n s[i]+=d;\n if(b!=e)\n {\n ll m=b+(e-b)/2;\n update(s,d,j,m,b,2*i+1);\n update(s,d,j,e,m+1,2*i+2);\n }\n return;\n}\n*/\n \n \n \n/*------------------------------------------------------------------------------------------------------------------------------------------------------------------*/\n//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////\n//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////\n/*------------------------------------------------------------------------------------------------------------------------------------------------------------------*/\n \n \nconst ll hashp=1e18+7; //790738119649411319;\nconst ll mod=1e9+7; //998244353;\nconst ld pi= 3.141592653589793238;\nconst ll inf=2e18;\n \nll ext_gcd(ll a, ll b, ll &x, ll &y)\n{\n if(!b)\n {\n x=1,y=0;\n return a;\n }\n ll x1,y1;\n ll g=ext_gcd(b,a%b,x1,y1);\n x=y1,y=x1-(a/b)*y1;\n return g;\n}\nll powe(ll a,ll p,ll m=inf)\n{\n if(p==0)\n return 1;\n ll result=powe(a,p>>1,m);\n if(p&1)\n return ((((a%m)*result)%m)*result)%m;\n return (result*result)%m;\n}\nll add(ll a,ll b)\n{\n return ((a%mod)+(b%mod))%mod;\n}\nll mul(ll a,ll b)\n{\n return ((a%mod)*(b%mod))%mod;\n}\nll sub(ll a,ll b)\n{\n return ((a%mod)-(b%mod)+mod)%mod;\n}\nll divi(ll a,ll b)\n{\n b=powe(b,mod-2,mod);\n return mul(a,b);\n}\nvoid facto(vl &f, ll n)\n{\n f[0]=1,f[1]=1;\n rep(i,2,n)\n f[i]=mul(i,f[i-1]);\n return;\n}\n \n \n \n/*------------------------------------------------------------------------------------------------------------------------------------------------------------------*/\n//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////\n//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////\n/*------------------------------------------------------------------------------------------------------------------------------------------------------------------*/\n \n \n \n \n \nint main() \n{\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n //cout<<fixed<<setprecision(9);\n ll t;\n cin>>t;\n \n ll mxn=1e7+5;\n \n \n \n while(t--)\n {\n ll n,k;\n cin>>n;\n \n ll c1=0;\n rep(i,0,n){\n ll x;\n cin>>x;\n c1+=(x&1);\n }\n n-=c1;\n \n ((n&1)==(c1&1))?pnt(\"Alice\"):pnt(\"Bob\");\n \n \n \n nl \n }\n \n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2004E",
"submission_id": 279365235,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \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*/\n \nusing namespace std;\n// using namespace __gnu_pbds;\n \n// typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;\n \ntypedef long long ll;\ntypedef unsigned long long ull;\ntypedef long double lld;\ntypedef pair<int, int> pii;\ntypedef pair<long long, long long> pll;\ntypedef vector<int> vi;\ntypedef vector<ll> vll;\ntypedef vector<vi> vvi;\ntypedef vector<vll> vvll;\ntypedef vector<pii> vpii;\ntypedef vector<pll> vpll;\n \n#define forn(i, n) for(int i = 0; i < n; i++)\n#define ford(i, n) for(int i = n - 1; i >= 0; i--)\n#define for1(i, n) for(int i = 1; i <= n; i++)\n#define forr(i, l, r) for(int i = l; i <= r; i++)\n#define forrd(i, l, r) for(int i = r; i >= l; i--)\n#define all(a) (a).begin(), (a).end()\n#define rall(a) (a).rbegin(), (a).rend()\n#define len(a) (int)(a).size()\n \n#ifdef SARTHAK_LOCAL\n#include \"/Users/sarthakkumar/cpp/templates/CompCoding/debug.cpp\"\n#else \n#define debug(...) 69\n#endif\n \ntemplate<typename T> bool cmin(T& a, T b) { return b < a ? a = b, 1 : 0; }\ntemplate<typename T> bool cmax(T& a, T b) { return b > a ? a = b, 1 : 0; }\n \ntemplate<typename T>\nT gcd(T a, T b) {\n while (b) { a %= b; swap(a, b); }\n return a;\n}\n \nmt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());\n// mt19937_64 rng(696969);\n \nvoid Main() {\n int n;\n cin >> n;\n vi a(n);\n forn(i, n) cin >> a[i];\n \n vi type;\n for(int x: a) {\n type.push_back(x&1);\n }\n sort(rall(type));\n if(type.back() == 1) {\n cout << (len(type)%2 == 0 ? \"Bob\" : \"Alice\") << '\\n';\n } else {\n cout << (len(type)%2 == 1 ? \"Bob\" : \"Alice\") << '\\n';\n }\n \n}\n \n#define MULTI // comment for single test case\n \nint main() {\n\tios::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);\n\tcout << setprecision(12) << fixed;\n#ifdef SARTHAK_LOCAL\n clock_t start = clock();\n#endif\n \n\tint tt = 1;\n #ifdef MULTI\n cin >> tt;\n #endif\n \n\tfor(int test_case = 1; test_case <= tt; test_case++) {\n\t\tdebug(test_case);\n \n\t\tMain();\n // cout << Main() << '\\n';\n // cout << (Main() ? \"YES\" : \"NO\") << '\\n';\n \n #ifdef SARTHAK_LOCAL\n cerr << \"----------------------------------------------------------------------------------------------------------------------------------------\" << '\\n';\n #endif\n\t}\n \n#ifdef SARTHAK_LOCAL\n clock_t end = clock();\n cerr << \"Total Time: \" << (double)(end - start) * 1e3 / CLOCKS_PER_SEC << \"ms\" << '\\n';\n#endif\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": "2004E",
"submission_id": 276651695,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport math\n \ndef is_prime(n):\n if n < 2:\n return False\n for i in range(2, int(math.sqrt(n)) + 1):\n if n % i == 0:\n return False\n return True\n \ndef solve():\n n = int(input())\n a = list(map(int, input().split()))\n primes = 0\n for x in a:\n if is_prime(x):\n primes += 1\n if n == 1 or n % 2 == 1 and primes != n:\n print(\"Bob\")\n else:\n print(\"Alice\")\n \nt = int(input())\nfor _ in range(t):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2004E",
"submission_id": 279978829,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport math\nfrom typing import List\n \ndef gcd(a: int, b: int) -> int:\n \"\"\"Compute the Greatest Common Divisor of a and b.\"\"\"\n while b != 0:\n a, b = b, a % b\n return a\n \ndef primeFactors(n: int) -> List[int]:\n \"\"\"Return list of prime factors of n, excluding 'n' itself.\"\"\"\n factors = []\n # Check for the already checked factor\n i = 2\n while i * i <= n:\n if n % i:\n i += 1\n else:\n n //= i\n factors.append(i)\n if n > 1:\n factors.append(n)\n return factors\n \ndef nimWinner(a: List[int]) -> str:\n max_factor = -1\n # Find the maximum unique factor of all elements\n for i in range(len(a)):\n f = set()\n e = primeFactors(a[i])\n for j in e:\n f.add(j)\n if len(f) > max_factor:\n max_factor = len(f)\n \n # If the number of unique factors is odd, Bob wins; otherwise, Alice wins\n return \"Bob\" if max_factor % 2 == 1 else \"Alice\"\n \n# Main function to handle inputs and outputs\ndef main():\n t = int(input().strip())\n for _ in range(t):\n n = int(input().strip())\n a = list(map(int, input().split()))\n print(nimWinner(a))\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": "2004E",
"submission_id": 294898501,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nt = int(input())\nfor _ in range(t):\n n = int(input())\n input()\n print(\"Alice\" if n % 2 == 0 else \"Bob\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2004E",
"submission_id": 279012966,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class Codechef {\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(), a[] = new int[n];\n HashSet<Integer> hs = new HashSet<>();\n for (int i = 0; i < n; i++) {\n a[i] = sc.nextInt();\n if (hs.contains(a[i])) {\n hs.remove(a[i]);\n } else {\n hs.add(a[i]);\n }\n }\n long sum = 0;\n for (int x : hs) {\n sum += x;\n }\n \n if (sum % 2 != 0) {\n if (!hs.contains(3))\n System.out.println(\"Alice\");\n else\n System.out.println(\"Bob\");\n } else {\n System.out.println(\"Bob\");\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": "2004E",
"submission_id": 276679006,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport java.io.BufferedOutputStream;\nimport java.io.BufferedReader;\nimport java.io.BufferedWriter;\nimport java.io.File;\nimport java.io.FileNotFoundException;\nimport java.io.FileReader;\nimport java.io.IOException;\nimport java.io.InputStream;\nimport java.io.InputStreamReader;\nimport java.io.OutputStream;\nimport java.io.OutputStreamWriter;\t\nimport java.io.PrintWriter;\nimport java.lang.reflect.Array;\nimport java.math.BigInteger; \nimport java.util.ArrayList;\nimport java.util.Arrays;\nimport java.util.Collection;\nimport java.util.Collections;\nimport java.util.Comparator;\nimport java.util.HashMap;\nimport java.util.HashSet;\nimport java.util.Iterator;\nimport java.util.List;\nimport java.util.Map;\nimport java.util.PriorityQueue;\nimport java.util.Queue;\nimport java.util.Scanner;\nimport java.util.Set;\nimport java.util.Stack;\nimport java.util.StringTokenizer;\nimport java.util.TreeSet;\n \n \n \n \n \npublic class Main {\n\tstatic kattio sc = new kattio();\n\tstatic PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n \n\tpublic static void main(String[] args) {\n\t\t//stack=new Stack<Integer>();\n\t\t//set=new HashSet<List<Integer>>();\n\t\t//f(0, new int[3]);System.out.println(set.size());\n\t\t//p1=false;\n\t\tss=new int[(int) (1e7+1)];Arrays.fill(ss, 1);\n\t\tsize=new int[(int) (1e7+1)];size[1]=1;\n\t\tList<Integer>list=new ArrayList<Integer>();\n\t\tfor (int i = 2; i < ss.length; i++) {\n\t\t\tif (ss[i]==1) {\n\t\t\t\tint j=2;\n\t\t\t\twhile (i*j<ss.length) {\n\t\t\t\t\tss[i*j]=0;j++;\n\t\t\t\t}\n\t\t\t}\n\t\t\tsize[i]=size[i-1]+ss[i];\n\t\t}\n\t\t\n\t\t int t=sc.nextint(); while (t-->0) {\n\t\t\t\n\t\t \n\t\t sovle2(); } \n\t\t \n\t\t out.close();\n\t\t \n \n\t\t\n \n\t}\n\tstatic long[] ram,v;\n\tstatic int []size,ss;\n\t\n\tstatic void sovle2() {\n\t\tint n=sc.nextint();\n\t\tint []arr=new int[n];\n\t\tlong ans=0;\n\t\tfor (int i = 0; i < arr.length; i++) {\n\t\t\tarr[i]=sc.nextint();\n\t\t\tif (arr[i]%2!=0) {\n\t\t\t\tans+=size[arr[i]];\n\t\t\t\t//System.out.println(size[arr[i]]);\n\t\t\t}\n\t\t\t\n\t\t}\n\t\t//System.out.println(ans);\n\t\tif (ans%2==0) {\n\t\t\tout.println(\"Bob\");\n\t\t}else {\n\t\t\tout.println(\"Alice\");\n\t\t}\n\t}\n\t\n\tstatic boolean p1;\n\tstatic int[][] va,l,r;\n\tstatic int n,m;\n\tstatic void sovle() {\n\t\tint l=sc.nextint();\n\t\tint r=sc.nextint();\n\t\tint L=sc.nextint();\n\t\tint R=sc.nextint();\n\t\tif (R<l||r<L) {\n\t\t\tout.println(1);\n\t\t}else if((l>=L&&r<=R)||(l<=L&&r>=R)){\n\t\t\t\n\t\t\tif (l==L&&r==R) {\n\t\t\t\t//System.out.println(\"aaa\");\n\t\t\t\tout.println(r-l);\n\t\t\t}else if(l!=L&&R!=r){\n\t\t\t\tout.println(Math.min(r-l, R-L)+2);\n\t\t\t}else {\n\t\t\t\tout.println(Math.min(r-l, R-L)+1);\n\t\t\t}\n\t\t\t\n\t\t}else {\n\t\t\t\n\t\t\tout.println(Math.min(r-L, R-l)+2);\n\t\t}\n\t}\n \n \n \n\tstatic long pow(long x,long a) {\n\t\tif (a==0) {\n\t\t\treturn 1;\n\t\t}else {\n\t\t\tlong c=pow(x, a/2);\n\t\t\tif (a%2==0) {\n\t\t\t\treturn c*c%mod;\n\t\t\t}else {\n\t\t\t\treturn c*c%mod*x%mod;\n\t\t\t}\n\t\t}\n\t}\n\tstatic long mod = (long) (1e9+7);\n \n\tstatic void sovle1() {\n\t\t\n\t}\n \n\tstatic long []c;\n\tstatic int lowbit(int x) {\n\t\treturn x&-x;\n\t}\n\tstatic void add(int x,long v) {\n\t\twhile (x<c.length) {\n\t\t\t//System.out.println(x);\n\t\t\tc[x]^=v;\n\t\t\tx+=lowbit(x);\n\t\t}\n\t}\n\tstatic long sum(int x) {\n\t\tlong ans=0;\n\t\t//System.out.println(x+\"dasdsafadf\");\n\t\twhile (x>0) {\n\t\t\t//System.out.println(x);\n\t\t\tans^=c[x];\n\t\t\tx-=lowbit(x);\n\t\t}\n\t\treturn ans;\n\t}\n\tstatic Map<Integer, List<Integer>>map;\n\tstatic void mapinit(int chu,int n) {\n\t\tmap=new HashMap<Integer, List<Integer>>();\n\t\tfor (int i = chu; i < n+chu; i++) {\n\t\t\tmap.put(i, new ArrayList<Integer>());\n\t\t}\n\t}\n}\n \nclass kattio extends PrintWriter {\n \n\tBufferedReader r;\n\tStringTokenizer st;\n \n\tpublic kattio() {\n\t\tthis(System.in, System.out);\n\t}\n \n\tpublic kattio(InputStream i, OutputStream o) {\n\t\tsuper(o);\n\t\tr = new BufferedReader(new InputStreamReader(i));\n\t}\n \n\tpublic kattio(String in, String out) throws IOException {\n\t\tsuper(out);\n\t\tr = new BufferedReader(new FileReader(in));\n\t}\n \n\tpublic String next() {\n\t\ttry {\n\t\t\twhile (st == null || !st.hasMoreTokens()) {\n\t\t\t\tst = new StringTokenizer(r.readLine());\n\t\t\t}\n\t\t\treturn st.nextToken();\n\t\t} catch (Exception e) {\n\t\t\t// TODO: handle exception\n\t\t\treturn null;\n\t\t}\n\t}\n \n\tpublic int nextint() {\n\t\treturn Integer.parseInt(next());\n\t}\n \n\tpublic long nextlong() {\n\t\treturn Long.parseLong(next());\n\t}\n \n\tpublic double nextdouble() {\n\t\treturn Double.parseDouble(next());\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": "2004E",
"submission_id": 277308652,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \n#define fastio std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n#define nl \"\\n\"\ntypedef long long ll;\n#define all(a) (a).begin(), (a).end()\n#define show(C) for (auto J: C) cout << J << \" \";cout << endl;\nconst ll MAX = 1e7+3;\n \nvector<ll>grundy(MAX, -1);\nvector<ll>prime(MAX,-1);\n \nvoid sieve()\n{\n grundy[0] = 0;\n grundy[1] = 1;\n prime[1] = 1;\n for(ll i=2; i<MAX; i+=2){\n grundy[i] = 0;\n prime[i] = 2;\n }\n \n ll cnt = 2;\n for(ll i=3; i<MAX; i+=2)\n {\n if(prime[i]==-1)\n {\n prime[i] = i;\n grundy[i] = cnt++;\n for(ll j=i*i; j<MAX; j+=i)\n {\n prime[j] = i;\n if(j%2)\n grundy[j] = grundy[prime[j]];\n }\n }\n }\n}\n \nint main()\n{\n fastio\n ll t; cin>>t;\n while(t--)\n {\n int n; cin>>n;\n vector<int>v(n);\n for(int i=0; i<n; i++) cin>>v[i];\n ll ans = 0;\n for(auto x : v) ans ^= grundy[x];\n if(ans){\n cout<<\"Bob\"<<nl;\n }else cout<<\"Alice\"<<nl;\n }\n \n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2004E",
"submission_id": 276831256,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\n#ifndef VLSMB_VS\n#include <bits/stdc++.h>\n#else\n#include <iostream>\n#include <iomanip>\n#include <set>\n#include <algorithm>\n#include <map>\n#include <vector>\n#include <stack>\n#include <queue>\n#include <deque>\n#include <string>\n#include <cstdio>\n#include <cmath>\n#endif\n \n#define int long long\n#define pii pair<int,int>\n#define endl '\\n'\n#define veci vector<int>\n#define vecpi vector<pii >\n#define mpi map<int,int>\n#define mpc map<char,int>\n#define mppi map<pii,int>\n#define inf 0x7fffffffffffffff\n#define ninf (long long)(0x8000000000000000)\n#define eps 1e-8\n#define yes cout<<\"YES\"<<endl\n#define no cout<<\"NO\"<<endl\n#define seti set<int>\n#define mulseti multiset<int>\n#define setpi set<pii >\n#define priqli priority_queue<int,veci,less<int> >\n#define priqgi priority_queue<int,veci,greater<int> >\n#define priqgpi priority_queue<pii,vecpi,greater<pii > >\n#define ull unsigned long long\nusing namespace std;\n//const int N = 1000000;\nvoid solve();\nvoid init();\nint gcd(int a,int b)\n{\n return b?gcd(b,a%b):a;\n}\nint extgcd(int a,int b,int&x,int&y)\n{\n if(!b)\n {\n x=1,y=0;\n return a;\n }\n int d=extgcd(b,a%b,y,x);\n y-=a/b*x;\n return d;\n}\nint remod(int a,int m)\n{\n int x,y;\n extgcd(a,m,x,y);\n return (x%m+m)%m;\n}\nint mul(int a,int b,int m)\n{\n int ans=0;\n while(b)\n {\n if(b&1)ans=(ans+a)%m;\n a=(a+a)%m;\n b>>=1;\n }\n return ans;\n}\nint extcrt(veci&a,veci&m)\n{\n int x=0,y=0;\n int m1=m[0],a1=a[0];\n int ans=0;\n for(int i=1;i<min(a.size(),m.size());i++)\n {\n int a2=a[i],m2=m[i];\n int a=m1,b=m2,c=(a2-a1%m2+m2)%m2;\n int d=extgcd(a,b,x,y);\n if(c%d!=0)return -1;\n x=mul(x,c/d,b/d);\n ans=a1+x*m1;\n m1=m2/d*m1;\n ans=(ans%m1+m1)%m1;\n a1=ans;\n }\n return ans;\n}\n \nsigned main()\n{\n ios::sync_with_stdio(false);\n cin.tie(0);\n int t=1;\n init();\n cin>>t;\n while(t--)solve();\n cout.flush();\n return 0;\n}\nconst int N=2e7;\nbool notpri[N];\nvoid solve()\n{\n int n;\n cin>>n;\n veci a(n);\n int ans=0;\n for(auto&x:a)cin>>x;\n for(const auto&x:a)ans^=x;\n cout<<(notpri[ans]?\"Bob\":\"Alice\")<<endl;\n}\nvoid init()\n{\n notpri[1]=true;\n for(int i=2;i<N;i++)\n {\n if(!notpri[i])\n {\n for(int j=2*i;j<N;j+=i)notpri[j]=true;\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": "2004E",
"submission_id": 276619280,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport sys\ninput = sys.stdin.readline\n \ndef is_prime(num):\n if num == 1:\n return False\n \n for i in range(2, int(num**0.5)+1):\n if num%i == 0:\n return False\n \n return True\n \nt = int(input())\n \nfor _ in range(t):\n n = int(input())\n a = list(map(int, input().split()))\n \n xor = 0\n \n for i in a:\n xor ^= i\n \n print(\"Bob\" if xor == 1 or xor%2 == 0 else \"Alice\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2004E",
"submission_id": 276673597,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport sys \ninput = sys.stdin.buffer.readline\n \nfrom collections import Counter\n \n \ndef gcd(x, y):\n \"\"\"greatest common divisor of x and y\"\"\"\n while y:\n x, y = y, x % y\n return x\n \n \ndef memodict(f):\n \"\"\"memoization decorator for a function taking a single argument\"\"\"\n class memodict(dict):\n def __missing__(self, key):\n ret = self[key] = f(key)\n return ret\n \n return memodict().__getitem__\n \n \ndef pollard_rho(n):\n \"\"\"returns a random factor of n\"\"\"\n if n & 1 == 0:\n return 2\n if n % 3 == 0:\n return 3\n \n s = ((n - 1) & (1 - n)).bit_length() - 1\n d = n >> s\n for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:\n p = pow(a, d, n)\n if p == 1 or p == n - 1 or a % n == 0:\n continue\n for _ in range(s):\n prev = p\n p = (p * p) % n\n if p == 1:\n return gcd(prev - 1, n)\n if p == n - 1:\n break\n else:\n for i in range(2, n):\n x, y = i, (i * i + 1) % n\n f = gcd(abs(x - y), n)\n while f == 1:\n x, y = (x * x + 1) % n, (y * y + 1) % n\n y = (y * y + 1) % n\n f = gcd(abs(x - y), n)\n if f != n:\n return f\n return n\n \n \n@memodict\ndef prime_factors(n):\n \"\"\"returns a Counter of the prime factorization of n\"\"\"\n if n <= 1:\n return Counter()\n f = pollard_rho(n)\n return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)\n \n \ndef distinct_factors(n):\n \"\"\"returns a list of all distinct factors of n\"\"\"\n factors = [1]\n for p, exp in prime_factors(n).items():\n factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]\n return factors\n \n \ndef all_factors(n):\n \"\"\"returns a sorted list of all distinct factors of n\"\"\"\n small, large = [], []\n for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):\n if not n % i:\n small.append(i)\n large.append(n // i)\n if small[-1] == large[-1]:\n large.pop()\n large.reverse()\n small.extend(large)\n return small\n \n \ndef process(A):\n A = [x for x in A if x > 0]\n d = {}\n for x in A:\n pf = prime_factors(x)\n if len(pf)==0:\n p = 1 \n else:\n p = min(pf)\n if p not in d:\n d[p] = 0 \n d[p]+=1\n L = [p for p in d if d[p] % 2==1 and p != 2]\n if len(L)==0:\n print(\"Bob\")\n elif len(L)==1:\n print(\"Alice\")\n elif len(L) % 2==0:\n print(\"Alice\")\n else:\n print(\"Bob\")\nt = int(input())\nfor i in range(t):\n n = int(input())\n A = [int(x) for x in input().split()]\n process(A)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2004E",
"submission_id": 299827742,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n static HashSet<Integer> primeSet;\n static int[] firstPrimeArr;\n static int maxNum = (int) (1e7);\n \n static void findAllprimes(int maxNum) {\n boolean[] arr = new boolean[maxNum + 1];\n for (long i = 2; i <= maxNum; i++) {\n if (arr[(int)(i)])\n continue;\n for (long j = i * i; j <= maxNum; j += i) {\n arr[(int)(j)] = true;\n }\n }\n for (int i = 2; i <= maxNum; i++) {\n if (!arr[i]) {\n primeSet.add(i);\n }\n }\n }\n \n public static void fillFirstPrimeArr(int maxNum) {\n firstPrimeArr = new int[maxNum + 1];\n for (int i = 1; i <= maxNum; i++)\n firstPrimeArr[i] = -1;\n for (int i = 2; i <= maxNum; i++) {\n if (firstPrimeArr[i] != -1)\n continue;\n for (int j = i; j <= maxNum; j += i) {\n if (firstPrimeArr[j] == -1)\n firstPrimeArr[j] = i;\n }\n }\n }\n \n public static String fun(int n, int[] arr) {\n int xor = 0;\n for(int x: arr) xor ^= x;\n if(xor == 0 || xor == 2) return \"Bob\";\n if(xor == 1) return \"Bob\";\n if(primeSet.contains(xor)) {\n return \"Alice\";\n } \n int myFirstPrimeDivisor = firstPrimeArr[xor];\n if(myFirstPrimeDivisor == 2) return \"Bob\";\n return \"Alice\";\n }\n \n public static void main(String[] args) throws IOException {\n primeSet = new HashSet<>();\n fillFirstPrimeArr(maxNum);\n findAllprimes(maxNum);\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));\n StringTokenizer st = new StringTokenizer(br.readLine());\n int t = Integer.parseInt(st.nextToken());\n \n while (t-- > 0) {\n st = new StringTokenizer(br.readLine());\n int n = Integer.parseInt(st.nextToken());\n st = new StringTokenizer(br.readLine());\n int[] arr = new int[n];\n for (int i = 0; i < n; i++) {\n arr[i] = Integer.parseInt(st.nextToken());\n }\n bw.write(fun(n, arr) + \"\\n\");\n }\n \n br.close();\n bw.flush();\n bw.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": "2004E",
"submission_id": 287908672,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class Codechef {\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(), a[] = new int[n];\n int r[] = new int[10000001];\n for (int i = 0; i < n; i++) {\n a[i] = sc.nextInt();\n r[a[i]] = (r[a[i]] + 1) % 2;\n }\n long sum = 0;\n for (int i = 1; i < 10000001; i++) {\n if (r[i] == 1) {\n sum += i;\n }\n }\n if (sum % 2 != 0) {\n if (r[3] == 0)\n System.out.println(\"Alice\");\n else\n System.out.println(\"Bob\");\n } else {\n System.out.println(\"Bob\");\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": "2004E",
"submission_id": 276676016,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nconst int N = 1e7 + 10, M = 1e7;\nint t, n, primes[N], sg[N];\n \nint main() {\n sg[0] = 0, sg[1] = 1, sg[2] = 0;\n int cnt = 0, tot = 1;\n for (int i = 3; i <= M; i += 2 ) {\n if (!sg[i]) {\n sg[i] = ++ tot;\n primes[cnt ++ ] = i;\n for (int j = 0; j < cnt && primes[j] <= M / i; j ++ ) {\n sg[primes[j] * i] = sg[primes[j]];\n if (i % primes[j] == 0) break;\n }\n }\n }\n cin >> t;\n while (t -- ) {\n cin >> n;\n int res = 0;\n for (int i = 0; i < n; i ++ ) {\n int x;\n cin >> x;\n res ^= sg[x];\n }\n if (res) puts(\"Alice\");\n else puts(\"Bob\");\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": "2004E",
"submission_id": 276757319,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nusing LL = long long;\n \nconst int N = 1E3 + 5;\nint sg[N];\n \nvoid init() {\n\tfor (int i = 1 ; i < N ; ++i) {\n\t\tset<int> P;\n\t\tfor (int j = 1 ; j <= i ; ++j) {\n\t\t\tif (__gcd(i , j) == 1) {\n\t\t\t\tP.insert(sg[i - j]);\n\t\t\t}\n\t\t}\n\t\twhile (P.count(sg[i])) sg[i] += 1;\n\t}\n}\n \nvoid solve() {\n \n\tint n;\n\tcin >> n;\n\t\n\tint ans = 0;\n\tfor (int i = 1 ; i <= n ; ++i) {\n\t\tint x;\n\t\tcin >> x;\n\t\tans ^= sg[x];\n\t}\n\t\n\tif (ans) {\n\t\tcout << \"Alice\\n\";\n\t} else {\n\t\tcout << \"Bob\\n\";\n\t}\n \n}\n \nint main() {\n \n\tios::sync_with_stdio(false);\n\tcin.tie(nullptr);\n \n\tinit();\n\t\n\tint t = 1;\n\tcin >> t;\n\t\n\twhile (t--) {\n\t\tsolve();\n\t}\n \n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2004E",
"submission_id": 284775117,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport math\nfrom functools import lru_cache\nimport sys\n \n@lru_cache(None)\ndef gcd(a, b):\n return math.gcd(a, b)\n \n@lru_cache(None)\ndef valid_moves(x):\n moves = []\n for y in range(1, x + 1):\n if gcd(x, y) == 1:\n moves.append(y)\n return moves\n \n@lru_cache(None)\ndef calculate_grundy(x):\n if x == 0:\n return 0\n \n grundy_set = set()\n for move in valid_moves(x):\n grundy_set.add(calculate_grundy(x - move))\n \n grundy = 0\n while grundy in grundy_set:\n grundy += 1\n \n return grundy\n \ndef determine_winner(piles):\n xor_sum = 0\n for pile in piles:\n xor_sum ^= calculate_grundy(pile)\n \n return \"Alice\" if xor_sum != 0 else \"Bob\"\n \nfor tc in range(int(input())):\n n = int(input())\n stones = [int(x) for x in sys.stdin.readline().split()]\n print(determine_winner(stones))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2004E",
"submission_id": 276671480,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\ndef gcd(a, b):\n while b:\n a, b = b, a % b\n return a\n \ndef compute_dp(N):\n dp = [0] * N\n \n for i in range(1, N):\n moves = set()\n for j in range(1, i + 1):\n if gcd(i, j) == 1:\n moves.add(dp[i - j])\n \n mex = 0\n while mex in moves:\n mex += 1\n dp[i] = mex\n \n return dp\n \ndef find_winner(dp, piles):\n xor_sum = 0\n for pile in piles:\n xor_sum ^= dp[pile]\n return \"Alice\" if xor_sum != 0 else \"Bob\"\n \nt = int(input())\ntest_cases = []\n \nmax_value = 0\nfor _ in range(t):\n n = int(input())\n piles = list(map(int, input().split()))\n max_value = max(max_value, max(piles))\n test_cases.append(piles)\n \ndp = compute_dp(max_value + 1)\n \nfor piles in test_cases:\n print(find_winner(dp, piles))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2004E",
"submission_id": 277255247,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.StringTokenizer;\n \npublic class E {\n String filename = null;\n InputReader sc;\n \n int MAX = (int)1e7 + 2;\n int[] nb = new int[MAX + 1];\n void precompute() {\n nb[0] = 0;\n nb[1] = 1;\n \n int mex = 2;\n for (int i = 3; i < MAX; i += 2) {\n if (nb[i] == 0) {\n nb[i] = mex;\n mex++;\n }\n \n for (long j = (long)i * i; j < MAX; j += i) {\n int jj = (int)j;\n if (nb[jj] == 0) {\n nb[jj] = nb[i];\n }\n }\n }\n// for (int i = 0; i < 10; i++) System.out.println(i + \" -> \" + nb[i]);\n }\n \n void solve() {\n int n = sc.nextInt();\n int[] a = sc.nextArray(n);\n \n int sum = 0;\n for (int i = 0; i < n; i++) {\n sum ^= nb[a[i]];\n }\n \n if (sum == 0) {\n System.out.println(\"Bob\");\n } else {\n System.out.println(\"Alice\");\n }\n }\n \n public void run() throws FileNotFoundException {\n if (filename == null) {\n sc = new InputReader(System.in);\n } else {\n sc = new InputReader(new FileInputStream(filename));\n }\n \n int nTests = sc.nextInt();\n precompute();\n for (int test = 0; test < nTests; test++) {\n solve();\n }\n }\n \n public static void main(String[] args) {\n E sol = new E();\n try {\n sol.run();\n } catch (FileNotFoundException e) {\n e.printStackTrace();\n }\n }\n \n static class InputReader {\n public BufferedReader reader;\n public StringTokenizer tokenizer;\n \n public InputReader(InputStream stream) {\n reader = new BufferedReader(new InputStreamReader(stream));\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 float nextFloat() {\n return Float.parseFloat(next());\n }\n \n public double nextDouble() {\n return Float.parseFloat(next());\n }\n \n public long nextLong() {\n return Long.parseLong(next());\n }\n \n public int[] nextArray(int n) {\n int[] a = new int[n];\n for (int i = 0; i < n; i++) {\n a[i] = nextInt();\n }\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": "2004E",
"submission_id": 276672572,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Not a Nim Problem\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: Two players, Alice and Bob, are playing a game. They have $n$ piles of stones, with the $i$-th pile initially containing $a_i$ stones.\n\nOn their turn, a player can choose any pile of stones and take any positive number of stones from it, with one condition:\n\n * let the current number of stones in the pile be $x$. It is not allowed to take from the pile a number of stones $y$ such that the greatest common divisor of $x$ and $y$ is not equal to $1$. \n\nThe player who cannot make a move loses. Both players play optimally (that is, if a player has a strategy that allows them to win, no matter how the opponent responds, they will win). Alice goes first.\n\nDetermine who will win.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases.\n\nEach test case consists of two lines:\n\n * the first line contains a single integer $n$ ($1 \\le n \\le 3 \\cdot 10^5$); * the second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^7$). \n\nAdditional constraint on the input: the sum of $n$ across all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output Alice if Alice wins, or Bob if Bob wins.\n\nExamples:\ninput:\n3\n3\n3 2 9\n4\n3 3 6 1\n5\n1 2 3 4 5\noutput:\nBob\nAlice\nBob\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \n// sprague grundy theorem\n// if xor of grundy values of all elements is not zero then alice (first chance) wins.\n// grundy value of 1 is 1.\n// grundy value of a number is MEX (Minimum Excludant or smallest non-negative integer not present) of greedy values of those elements where we could reach from current number (in this case, those numbers with which gcd==1 && obviously smaller number than current number because we cannot remove number of piles more than its remaining existing quantity).\n// grundy value of 1->1 , 2->0 (mex=2 but gcd(2,mex)!=1), 3->2, 5->3,......so on.\n// grundy values of even numbers is 0.\n \npublic class Codechef {\n public static void main(String args[]) {\n Scanner sc = new Scanner(System.in);\n int c = 2;\n int pre[] = new int[10000001];\n pre[1] = 1;\n for (int i = 3; i < 10000001; i += 2) {// pre computing grundy values of all elements\n if (pre[i] == 0) {\n pre[i] = c;\n for (int j = i * i; j < 10000001 && j > 0; j += i) {\n if (pre[j] == 0) {\n pre[j] = c;\n }\n }\n c++;\n }\n }\n int t = sc.nextInt();\n while (t-- > 0) {\n int n = sc.nextInt(), a[] = new int[n], b[] = new int[n];\n int xor = 0;\n for (int i = 0; i < n; i++) {\n a[i] = sc.nextInt();\n b[i] = pre[a[i]];\n xor ^= b[i];\n }\n if (xor != 0) {\n System.out.println(\"Alice\");\n } else {\n System.out.println(\"Bob\");\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": "2004E",
"submission_id": 276716751,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define pb push_back\n#define F first\n#define S second\n#define fast ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)\ntypedef long long ll;\ntypedef long double ld;\nusing namespace std;\n \nconst int MAX_N = 5e5 + 10;\nconst int LOG = 25;\n \nvector<int> G[MAX_N];\nint par[LOG][MAX_N];\nint dep[MAX_N];\nint seen[MAX_N];\n \nint getDist(int u, int v) {\n int ret = 0;\n \n if (dep[u] > dep[v]) swap(u, v);\n \n int diff = dep[v] - dep[u];\n for (int i = 0; i < LOG; i++) {\n if (diff & (1 << i)) {\n v = par[i][v];\n ret += (1 << i);\n }\n }\n \n if (u == v) return ret;\n \n for (int i = LOG - 1; i >= 0; i--) {\n if (par[i][v] == par[i][u]) continue;\n v = par[i][v];\n u = par[i][u];\n \n ret += (1 << (i + 1));\n }\n \n ret += 2;\n return ret;\n}\n \nint bfs2(int i, vector<int> st) {\n queue<pair<int, int>> bat;\n for (auto i : st) bat.push({i, 0});\n while (!bat.empty()) {\n auto v = bat.front();\n bat.pop();\n \n if (G[v.F].size() < 3) {\n return v.S;\n }\n \n if (seen[v.F] == i) continue;\n seen[v.F] = i;\n \n for (auto i : G[v.F]) {\n if (seen[v.F] != i)\n bat.push({i, v.S + 1});\n }\n }\n \n return -1000;\n}\n \nvoid solve(int iter, int iter_count)\n{\n int n;\n cin >> n;\n \n for (int i = 0; i < n + 10; i++) {\n G[i].clear();\n dep[i] = 0;\n seen[i] = 0;\n for (int j = 0; j < LOG; j++) par[j][i] = 0;\n }\n \n vector<int> current_mid;\n current_mid.pb(1);\n int current_d = 1;\n bool fail = false;\n \n cout << 1 << ' ';\n for (int i = 2; i <= n; i++) {\n int p;\n cin >> p;\n \n G[p].pb(i);\n G[i].pb(p);\n \n dep[i] = dep[p] + 1;\n \n if (G[p].size() > 3)\n fail = true;\n \n if (fail) {\n cout << -1 << ' ';\n continue;\n }\n \n par[0][i] = p;\n for (int j = 1; j < LOG; j++)\n par[j][i] = par[j - 1][par[j - 1][i]];\n \n vector<int> finals;\n for (auto j : current_mid) {\n int dist = getDist(i, j) + 1;\n if (dist <= current_d) finals.pb(j);\n }\n \n if (finals.size() == 0) {\n current_d++;\n finals.pb(current_mid[0]);\n for (auto j : G[current_mid[0]]) {\n int dist = getDist(i, j) + 1;\n if (dist == current_d - 1) finals.pb(j);\n }\n }\n current_mid = finals;\n \n cout << current_d + bfs2(i, current_mid) << ' ';\n }\n \n cout << '\\n';\n}\n \nint main()\n{\n fast;\n int iter_count = 1;\n cin >> iter_count;\n for (int iter = 0; iter < iter_count; iter++)\n {\n solve(iter, iter_count);\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": "2006E",
"submission_id": 280543367,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \nconst int N = 510000;\nconst int LOGN = 20;\n \nint jmp[N][LOGN], dep[N], d[N], p[N], sub[N];\nvector<pair<int, int>> cent[N];\nmultiset<int> dists[N];\nbool bad, used[N];\nvector<int> g[N];\n \nvoid dfs(int v, int p = -1) {\n\tdep[v] = (p == -1 ? 0 : dep[p] + 1);\n\tjmp[v][0] = (p == -1 ? v : p);\n\tfor (int k = 1; k < LOGN; k++)\n\t\tjmp[v][k] = jmp[jmp[v][k - 1]][k - 1];\n\tfor (int u : g[v]) if (u != p)\n\t\tdfs(u, v);\n}\n \nint lca(int u, int v) {\n\tif (dep[u] < dep[v])\n\t\tswap(u, v);\n\tfor (int k = LOGN - 1; k >= 0; k--)\n\t\tif (dep[u] - (1 << k) >= dep[v])\n\t\t\tu = jmp[u][k];\n\tif (u == v) return u;\n\tfor (int k = LOGN - 1; k >= 0; k--)\n\t\tif (jmp[u][k] != jmp[v][k]) {\n\t\t\tu = jmp[u][k];\n\t\t\tv = jmp[v][k];\n\t\t}\n\treturn jmp[u][0];\n}\n \nint goup(int v, int k) {\n\tfor (int t = LOGN - 1; t >= 0; t--)\n\t\tif (k >= (1 << t)) {\n\t\t\tk -= (1 << t);\n\t\t\tv = jmp[v][t];\n\t\t}\n\treturn v;\n}\n \nint vert(int u, int v, int k) {\n\tint l = lca(u, v);\n\tint d1 = dep[u] - dep[l];\n\tint d2 = dep[v] - dep[l];\n\tif (k <= d1) return goup(u, k);\n\tk -= d1;\n\tk = d2 - k;\n\treturn goup(v, k);\n}\n \nint dist(int u, int v) {\n\tint l = lca(u, v);\n\treturn dep[u] + dep[v] - 2 * dep[l];\n}\n \nvoid add_set(int v) {\n\tfor (auto [u, w] : cent[v])\n\t\tdists[u].insert(w);\n}\n \nvoid rem_set(int v) {\n\tfor (auto [u, w] : cent[v])\n\t\tdists[u].extract(w);\n}\n \nint query(int v) {\n\tint ans = (int)(1e9);\n\tfor (auto [u, w] : cent[v]) {\n\t\tif (dists[u].empty())\n\t\t\tcontinue;\n\t\tans = min(ans, *dists[u].begin() + w);\n\t}\n\treturn ans;\n}\n \nvoid add(int v) {\n\td[v]++;\n\tif (d[v] == 1 && v != 1) add_set(v);\n\tif (d[v] == 3) rem_set(v);\n\tif (d[v] > 3) bad = true;\n}\n \nvoid dfs2(int v, int p = -1) {\n\tsub[v] = 1;\n\tfor (int u : g[v]) if (p != u && !used[u]) {\n\t\tdfs(u, v);\n\t\tsub[v] += sub[u];\n\t}\n}\n \nint centroid(int v, int t, int p = -1) {\n\tfor (int u : g[v]) if (p != u && !used[u] && sub[u] > t / 2)\n\t\treturn centroid(u, t, v);\n\treturn v;\n}\n \nvoid push(int v, int x, int p = -1, int d = 0) {\n\tcent[v].push_back({x, d});\n\tfor (int u : g[v]) if (p != u && !used[u])\n\t\tpush(u, x, v, d + 1);\n}\n \nvoid build(int v) {\n\tdfs2(v);\n\tint x = centroid(v, sub[v]);\n\tpush(x, x);\n\tused[x] = true;\n\tfor (int u : g[x]) if (!used[u])\n\t\tbuild(u);\n}\n \nvoid test_case() {\n\tint n; cin >> n;\n\tfor (int i = 1; i <= n; i++) {\n\t\tcent[i].clear();\n\t\tdists[i].clear();\n\t\tg[i].clear();\n\t}\n\tfor (int i = 2; i <= n; i++) {\n\t\tcin >> p[i];\n\t\tg[p[i]].push_back(i);\n\t\tg[i].push_back(p[i]);\n\t}\n\tfor (int i = 1; i <= n; i++)\n\t\td[i] = 0;\n\tdfs(1);\n\tfor (int i = 1; i <= n; i++)\n\t\tused[i] = false;\n\tbuild(1);\n\tbad = false;\n\tint l = 1, r = 1;\n\tadd_set(1);\n\tfor (int i = 1; i <= n; i++) {\n\t\tint cur = dist(l, r);\n\t\tif (i > 1) {\n\t\t\tif (dist(l, i) > cur) r = i;\n\t\t\telse if (dist(i, r) > cur) l = i;\n\t\t\tcur = dist(l, r);\n\t\t\tadd(i);\n\t\t\tadd(p[i]);\n\t\t}\n\t\tif (bad) {\n\t\t\tcout << -1 << ' ';\n\t\t\tcontinue;\n\t\t}\n\t\tint a = vert(l, r, cur / 2);\n\t\tint b = vert(r, l, cur / 2);\n\t\tcout << min(query(a), query(b)) + (cur + 1) / 2 + 1 << ' ';\n\t}\n\tcout << '\\n';\n}\n \nint32_t main() {\n\tios::sync_with_stdio(0); cin.tie(0);\n\tint t; cin >> t;\n\twhile (t--) test_case();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 278928841,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \ntypedef long long ll;\nconst ll mod = 998244353;\nconst int maxn = 5e5 + 5;\nconst int inf = 0x3f3f3f3f;\n \n#define ls k << 1\n#define rs k << 1 | 1\n \nstruct segtree\n{\n struct segnode\n {\n int l, r;\n int lz, mn;\n }t[maxn << 2];\n \n inline void pushup(int k) { t[k].mn = min(t[ls].mn, t[rs].mn); }\n inline void upd(int k, int d) { t[k].mn += d; t[k].lz += d; }\n inline void pushdown(int k) { if (t[k].lz) upd(ls, t[k].lz), upd(rs, t[k].lz), t[k].lz = 0; }\n \n void build(int k, int l, int r)\n {\n t[k].l = l, t[k].r = r, t[k].mn = inf, t[k].lz = 0;\n if (l == r) return; int mid = (l + r) >> 1;\n build(ls, l, mid); build(rs, mid + 1, r);\n }\n \n void update(int k, int l, int r, int d)\n {\n if (l <= t[k].l && t[k].r <= r) { upd(k, d); return; }\n pushdown(k); int mid = (t[k].l + t[k].r) >> 1;\n if (l <= mid) update(ls, l, r, d);\n if (r > mid) update(rs, l, r, d);\n pushup(k);\n }\n \n void change(int k, int x, int d)\n {\n if (t[k].l == t[k].r) { t[k].mn = d; return; }\n pushdown(k); int mid = (t[k].l + t[k].r) >> 1;\n if (x <= mid) change(ls, x, d); else change(rs, x, d);\n pushup(k);\n }\n}T;\n \nvector<int> g[maxn];\nint deg[maxn], fa[maxn], dfn[maxn], siz[maxn], dep[maxn], cnt;\nint f[maxn][19];\n \nvoid dfs(int u)\n{\n dfn[u] = ++cnt, siz[u] = 1;\n f[u][0] = fa[u]; for (int i = 1; i <= 18; i++) f[u][i] = f[f[u][i - 1]][i - 1];\n for (int v : g[u]) dep[v] = dep[u] + 1, dfs(v), siz[u] += siz[v];\n}\n \nint lca(int u, int v)\n{\n if (dep[u] < dep[v]) swap(u, v);\n for (int i = 18; ~i; i--) if (dep[f[u][i]] >= dep[v]) u = f[u][i];\n if (u == v) return u;\n for (int i = 18; ~i; i--) if (f[u][i] != f[v][i]) u = f[u][i], v = f[v][i];\n return f[u][0];\n}\nint dist(int u, int v) { return dep[u] + dep[v] - 2 * dep[lca(u, v)]; }\n \nint jump(int u, int x)\n{\n for (int i = 18; ~i; i--)\n if ((x >> i) & 1) u = f[u][i];\n return u;\n}\n \nvoid work()\n{\n int n; cin >> n; cnt = 0;\n for (int i = 1; i <= n; i++) g[i].clear(), deg[i] = 0;\n T.build(1, 1, n);\n for (int i = 2; i <= n; i++) cin >> fa[i], g[fa[i]].emplace_back(i);\n dep[0] = -1; dfs(1);\n \n bool Fl = 0; cout << \"1 \";\n int pl = 1, pr = 1, len = 0;\n for (int i = 2; i <= n; i++)\n {\n deg[i]++, deg[fa[i]]++;\n if (deg[i] >= 4 || deg[fa[i]] >= 4) Fl = 1;\n if (Fl) { cout << \"-1 \"; continue; }\n if (deg[i] >= 3) T.change(1, dfn[i], inf);\n if (deg[fa[i]] >= 3) T.change(1, dfn[fa[i]], inf);\n \n int dx = dist(i, pl), dy = dist(i, pr), nlen = max(dx, dy);\n T.change(1, dfn[i], nlen);\n if (nlen > len)\n {\n if (dx < dy) swap(pl, pr), swap(dx, dy);\n if (len == 0) T.change(1, dfn[pl], 1);\n else\n {\n int nx = (len - 1) / 2, lc = lca(pl, pr), dt;\n if (dep[pr] - dep[lc] > nx)\n {\n dt = jump(pr, nx);\n T.update(1, 1, n, 1); T.update(1, dfn[dt], dfn[dt] + siz[dt] - 1, -1);\n }\n else\n {\n dt = jump(pl, len - nx - 1);\n T.update(1, dfn[dt], dfn[dt] + siz[dt] - 1, 1);\n }\n }\n len = nlen; pr = i;\n }\n cout << T.t[1].mn + 1 << \" \";\n }\n cout << \"\\n\";\n}\n \nint main()\n{\n ios::sync_with_stdio(false);\n cin.tie(0);\n \n int T; cin >> T;\n while (T--) work();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 278833556,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\nimport heapq\n \npar = []\ndepth = []\ndef build(u, p):\n par[u][0] = p\n i = 0\n while par[u][i]:\n par[u][i + 1] = par[par[u][i]][i]\n i += 1\n depth[u] = depth[p] + 1\n \ndef lca(u, v):\n d = depth[u] + depth[v]\n if depth[v] > depth[u]:\n u, v = v, u\n for i in range(19, -1, -1):\n if depth[par[u][i]] >= depth[v]:\n u = par[u][i]\n for i in range(19, -1, -1):\n if par[u][i] != par[v][i]:\n u, v = par[u][i], par[v][i]\n p = u if u == v else par[u][0]\n return p, d - 2 * depth[p]\n \ndef solve():\n n = int(input())\n \n global par, depth\n par = [[0] * 20 for _ in range(n + 1)]\n depth = [0] * (n + 1)\n depth[1] = 1\n \n u, v = 1, 1\n dist = 0\n heap = [(0, 1)]\n deg = [0] * (n + 1)\n max_deg = 1\n \n print(1, end = ' ')\n i = 1\n for p in list(map(int, input().split())):\n i += 1\n build(i, p)\n deg[i] += 1\n deg[p] += 1\n max_deg = max(max_deg, deg[p])\n if max_deg > 3:\n print(-1, end = ' ')\n continue\n \n dist1 = lca(i, u)[1]\n dist2 = lca(i, v)[1]\n if dist1 > dist2:\n if dist1 > dist:\n v, dist = i, dist1\n else:\n if dist2 > dist:\n u, dist = i, dist2\n \n heapq.heappush(heap, (max(lca(i, u)[1], lca(i, v)[1]), i))\n while True:\n d, t = heap[0]\n if deg[t] > 2:\n heapq.heappop(heap)\n continue\n mx = max(lca(t, u)[1], lca(t, v)[1])\n if mx != d:\n heapq.heappop(heap)\n heapq.heappush(heap, (mx, t))\n continue\n print(d + 1, end = ' ')\n break\n print()\n \n \nt = int(input())\nfor _ in range(t):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 281710910,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nusing ll = long long;\n#include <iostream>\n#include <vector>\n#include <queue>\n#include <algorithm>\n \nusing namespace std;\n \nclass Tree {\npublic:\n int n; // number of nodes\n vector<vector<int>> adj; // adjacency list\n vector<int> depth, parent, subtree_size;\n vector<vector<int>> up; // for binary lifting in LCA\n \n Tree(int nodes) : n(nodes), adj(nodes), depth(nodes), parent(nodes), subtree_size(nodes) {\n int log = 1;\n while ((1 << log) <= n) log++;\n up.assign(n, vector<int>(log));\n }\n \n void add_edge(int u, int v) {\n adj[u].push_back(v);\n adj[v].push_back(u);\n }\n \n void dfs(int v, int p) {\n parent[v] = p;\n subtree_size[v] = 1;\n up[v][0] = p;\n for (int i = 1; i < up[v].size(); i++) {\n up[v][i] = up[up[v][i - 1]][i - 1];\n }\n for (int u : adj[v]) {\n if (u != p) {\n depth[u] = depth[v] + 1;\n dfs(u, v);\n subtree_size[v] += subtree_size[u];\n }\n }\n }\n \n void prepare(int root = 0) {\n depth[root] = 0;\n dfs(root, root);\n }\n \n int lca(int u, int v) {\n if (depth[u] < depth[v]) swap(u, v);\n int log = up[u].size();\n for (int i = log - 1; i >= 0; i--) {\n if (depth[u] - (1 << i) >= depth[v]) {\n u = up[u][i];\n }\n }\n if (u == v) return u;\n for (int i = log - 1; i >= 0; i--) {\n if (up[u][i] != up[v][i]) {\n u = up[u][i];\n v = up[v][i];\n }\n }\n return parent[u];\n }\n \n int dist(int u, int v) {\n int lca_uv = lca(u, v);\n return depth[u] + depth[v] - 2 * depth[lca_uv];\n }\n \n pair<int, int> find_midpoints(int u, int v) {\n int lca_uv = lca(u, v); // Get the LCA of u and v\n int dist_uv = depth[u] + depth[v] - 2 * depth[lca_uv]; // Calculate the distance between u and v\n \n // Ensure that `u` is the deeper node\n if (depth[u] < depth[v]) {\n swap(u, v);\n }\n \n // Calculate the first midpoint\n int midpoint1 = move_up(u, dist_uv / 2);\n \n // Calculate the second midpoint (if it exists)\n int midpoint2 = (dist_uv % 2 == 0) ? move_up(u, dist_uv / 2) : move_up(u, dist_uv / 2 + 1);\n \n return {midpoint1, midpoint2};\n }\n \n int move_up(int node, int steps) {\n for (int i = up[node].size() - 1; i >= 0; i--) {\n if (steps >= (1 << i)) {\n node = up[node][i];\n steps -= (1 << i);\n }\n }\n return node;\n }\n \n \n};\n \nvoid solve(){\n int n;\n cin>>n;\n Tree tree(n);\n vector<int> p(n);\n for(int i=1;i<n;i++) cin>>p[i];\n for(int i=1;i<n;i++) --p[i];\n for(int i=1;i<n;i++) tree.add_edge(i, p[i]);\n tree.prepare();\n auto diam = pair(0, 0);\n int diam_length = 0;\n vector<int> deg(n);\n cout<<\"1 \";\n bool failed = 0;\n for(int u=1;u<n;u++){\n deg[u]++;\n deg[p[u]]++;\n if(deg[p[u]] > 3) failed = true;\n if(failed){\n cout<<\"-1 \";\n continue;\n }\n if(tree.dist(u, diam.first) > diam_length){\n diam.second = u;\n }\n else if(tree.dist(u, diam.second) > diam_length){\n diam.first = u;\n }\n diam_length = tree.dist(diam.first, diam.second);\n auto centres = tree.find_midpoints(diam.first, diam.second);\n int ans = (diam_length + 1) / 2 + 2;\n if(deg[centres.first] < 3 || deg[centres.second] < 3){\n ans--;\n }\n cout<<ans<<\" \";\n }\n cout<<\"\\n\";\n}\nint main(){\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n int T = 1;\n cin>>T;\n for(int t=1;t<=T;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": "2006E",
"submission_id": 278851862,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nint T,N,dg[500009],p[500009],anc[500009][19],hd[500009],to[1000009],nxt[1000009],k,vv[500009],\nd[500009],as[500009][22],dep[500009],r[500009];\nvoid l(int u,int v) {\n\tto[++k]=v;nxt[k]=hd[u];hd[u]=k;\n}\ninline int lca(int a,int b) {\n if(dep[a]<dep[b]) std::swap(a,b);\n for(int i=18;i>=0;i--)\n if(dep[a]-dep[b]>=(1<<i))\n a=anc[a][i];\n for(int i=18;i>=0;i--)\n if(anc[a][i]!=anc[b][i]) \n \ta=anc[a][i],b=anc[b][i];\n if(a!=b) a=anc[a][0];return a;\n} \ninline int jp(int a,int x) {\n\tfor(int i=18;i>=0;i--) if(x>=(1<<i)) {\n\t\tx-=(1<<i);a=anc[a][i];\n\t}\n\tassert(a);\n\treturn a;\n}\ninline int qiu(int x,int v) {\n\tfor(int i=0;i<=21;i++) {\n\t\tif(as[x][i]>=v) return i;\n\t}\n\treturn 22;\n}\ninline int dis(int x,int y) {\n\treturn dep[x]+dep[y]-2*dep[lca(x,y)];\n}\nsigned main(void) {\n\tscanf(\"%d\",&T);\n\twhile(T--) {\n\t\tscanf(\"%d\",&N);\n\t\tfor(int i=1;i<=N;i++) dg[i]=0,hd[i]=0;\n\t\tprintf(\"1 \");\n\t\tk=0;\n\t\tbool ff=0;\n\t\tint ll=N+1;\n\t\tfor(int i=1;i<=N;i++) r[i]=N,vv[i]=0;\n\t\tfor(int i=2;i<=N;i++) {\n\t\t\tscanf(\"%d\",&p[i]);\n\t\t\tanc[i][0]=p[i];\n\t\t\tdep[i]=dep[p[i]]+1;\n\t\t\tl(p[i],i);l(i,p[i]);\n\t\t\tdg[p[i]]++;\n\t\t\tdg[i]++;\n\t\t\tif(dg[p[i]]>3) ff=1;\n\t\t\tif(dg[i]>3) ff=1;\n\t\t\tif(dg[p[i]]==3) {\n\t\t\t\tr[p[i]]=i-1;\n\t\t\t}\n\t\t\tif(ff&&ll==N+1) {\n\t\t\t\tll=i;\n\t\t\t}\n\t\t}\n\t\tfor(int i=1;i<=18;i++) {\n\t\t\tfor(int j=1;j<=N;j++) {\n\t\t\t\tanc[j][i]=anc[anc[j][i-1]][i-1];\n\t\t\t}\n\t\t}\n\t\tfor(int i=1;i<ll;i++) {\n\t\t\tstd::queue<int> q;\n\t\t\tq.push(i);\n\t\t\tvv[i]=i;\n\t\t\td[i]=0;\n\t\t\tmemset(as[i],0,sizeof(as[i]));\n\t\t\twhile(!q.empty()) {\n\t\t\t\tint a=q.front();\n\t\t\t\tq.pop();\n\t\t\t\tas[i][d[a]]=std::max(as[i][d[a]],r[a]);\n\t\t\t\tassert(d[a]<=21);\n\t\t\t\tif(r[a]==N) break;\n\t\t\t\tfor(int j=hd[a];j;j=nxt[j]) {\n\t\t\t\t\tif(vv[to[j]]!=i&&to[j]<ll) {\n\t\t\t\t\t\td[to[j]]=d[i]+1;\n\t\t\t\t\t\tvv[to[j]]=i;\n\t\t\t\t\t\tq.push(to[j]);\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t\tint a=1,b=1,dd=0;\n\t\tfor(int i=2;i<ll;i++) {\n\t\t\tint aa=a,bb=b;\n\t\t\tint t1=dis(i,aa);\n\t\t\tif(t1>=dd) {\n\t\t\t\tdd=t1;\n\t\t\t\ta=i;b=aa;\n\t\t\t}\n\t\t\tint t2=dis(bb,i);\n\t\t\tif(t2>=dd) {\n\t\t\t\tdd=t2;\n\t\t\t\ta=bb;\n\t\t\t\tb=i;\n\t\t\t}\n\t\t//\tprintf(\"%d %d %d\\n\",a,b,dd);\n\t\t\tint ans=0;\n\t\t\tif(dd%2==0) {\n\t\t\t\tif(dep[a]<=dep[b]) {\n\t\t\t\t\tans=qiu(jp(b,dd/2),i)+dd/2;\n\t\t\t\t} else {\n\t\t\t\t\tans=qiu(jp(a,dd/2),i)+dd/2;\t\t\t\t\t\n\t\t\t\t}\n\t\t\t} else {\n\t\t\t\tif(dep[a]<=dep[b]) {\n\t\t\t\t\tint x=jp(b,dd/2);\n\t\t\t\t\tans=std::min(qiu(x,i),qiu(p[x],i))+dd/2+1;\n\t\t\t\t} else {\n\t\t\t\t\tint x=jp(a,dd/2);\n\t\t\t\t\tans=std::min(qiu(x,i),qiu(p[x],i))+dd/2+1;\n\t\t\t\t}\t\t\t\t\n\t\t\t}\n\t\t\tprintf(\"%d \",ans+1);\n\t\t}\n\t\tfor(int i=ll;i<=N;i++) printf(\"-1 \");\n\t\tprintf(\"\\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": "2006E",
"submission_id": 278842540,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include<map>\n#include<set>\n#include<ctime>\n#include<cmath>\n#include<queue>\n#include<bitset>\n#include<cstdio>\n#include<vector>\n#include<random>\n#include<cstdlib>\n#include<cstring>\n#include<iostream>\n#include<algorithm>\n#define ll long long\nusing namespace std;\n#define I ll\n#define her1 20081214\n#define IV void\n#define cht 998244353\n#define ld long double\n#define Aestas16 392699\n#define ull unsigned long long\n#define cp(x,y)memcpy(x,y,sizeof y)\n#define mem(x,val)memset(x,val,sizeof x)\n#define D(i,j,n)for(register int i=j;i>=n;i--)\n#define E(i,now)for(register int i=first[now];i;i=e[i].nxt)\n#define F(i,j,n)for(register int i=j;i<=n;i++)\n#define DL(i,j,n)for(register i64 i=j;i>=n;i--)\n#define EL(i,now)for(register i64 i=first[now];i;i=e[i].nxt)\n#define FL(i,j,n)for(register i64 i=j;i<=n;i++)\n//#define D(i,j,n)for(int i=j;i>=n;i--)\n//#define E(i,now)for(int i=first[now];i;i=e[i].nxt)\n//#define F(i,j,n)for(int i=j;i<=n;i++)\n//#define DL(i,j,n)for(register ll i=j;i>=n;i--)\n//#define EL(i,now)for(register ll i=first[now];i;i=e[i].nxt)\n//#define FL(i,j,n)for(register ll i=j;i<=n;i++)\nll read(){\n\tll ans=0,f=1;\n\tchar c=getchar();\n\twhile(c<'0'||c>'9'){\n\t\tif(c=='-')f=-1;\n\t\tc=getchar();\n\t}\n\twhile(c>='0'&&c<='9')ans=ans*10+c-'0',c=getchar();\n\treturn ans*f;\n}\n#undef ll\n#include \"assert.h\"\nmt19937_64 rnd(her1);\n#include \"functional\"\nusing i64 = long long;\nconst int maxn = 5e5+5;\ni64 n,f[maxn][22],p[maxn],s[maxn][22],d[maxn],zk[maxn];\ni64 lca(i64 x,i64 y){\n if(d[x]<d[y])swap(x,y);i64 a=d[x]-d[y];D(i,20,0)if(a>>i&1)x=f[x][i];\n\tD(i,20,0)if(f[x][i]!=f[y][i])x=f[x][i],y=f[y][i];return x==y?x:f[x][0];\n}\ni64 sp(i64 x){i64 e=0;while(!s[x][e])e++;return e;}\ni64 dis(i64 x,i64 y){return d[x]+d[y]-2*d[lca(x,y)];}\ni64 kth(i64 x,i64 k){F(i,0,20)if(k>>i&1)x=f[x][i];return x;}\nIV doi(i64 x,i64 z){i64 o=x;F(i,0,20){if(i)o=p[o];if(!o)break;s[o][i]+=z;}}\ni64 ask2(i64 x){\n\ti64 fk=1e9,o=x;\n\tF(i,0,20){\n\t\tif(i)o=p[o];if(!o)break;\n\t\tfk=min(fk,i+sp(o));\n\t}\n\treturn fk;\n}\ni64 ask1(i64 x){\n\ti64 fk=1e9,o=x;\n\tF(i,0,20){\n\t\tif(i)o=p[o];if(!o)break;\n\t\tfk=min(fk,max(0,i-1)+sp(o));\n\t}\n\treturn fk;\n}\nIV solve(){\n\tn=read();F(i,2,n)f[i][0]=p[i]=read(),d[i]=d[p[i]]+1;\n\tF(o,1,20)F(i,1,n)f[i][o]=f[f[i][o-1]][o-1];F(i,1,n)F(o,0,20)s[i][0]=0;\n\ti64 a=1,b=1;F(i,1,n)zk[i]=0;zk[1]=-1;bool gg=0;\n\tF(i,1,n){\n if(gg){printf(\"-1 \");continue;}\n if(dis(a,b)<dis(b,i))a=i;if(dis(a,b)<dis(a,i))b=i;\n if(d[a]<d[b])swap(a,b);doi(i,1);\n if(i>1){\n i64 f=p[i];zk[f]++;\n if(zk[f]>2){gg=1,printf(\"-1 \");continue;}\n if(zk[f]==2)doi(f,-1);\n }\n i64 lc=lca(a,b),di=d[a]+d[b]-d[lc]*2;\n if((d[a]^d[b])&1)printf(\"%lld \",1+ask1(kth(a,di/2))+(di+1)/2);\n else printf(\"%lld \",1+ask2(kth(a,di/2))+di/2);\n }\n puts(\"\");\n}\nint main(){\n\t// freopen(\"1.in\",\"r\",stdin);\n\t// freopen(\"1.out\",\"w\",stdout);\n\ti64 T=read();while(T--)solve();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": "2006E",
"submission_id": 280284222,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nint T,N,dg[500009],p[500009],anc[500009][19],hd[500009],to[1000009],nxt[1000009],k,vv[500009],\nd[500009],as[500009][22],dep[500009],r[500009];\nvoid l(int u,int v) {\n\tto[++k]=v;nxt[k]=hd[u];hd[u]=k;\n}\ninline int lca(int a,int b) {\n if(dep[a]<dep[b]) std::swap(a,b);\n for(int i=18;i>=0;i--)\n if(dep[a]-dep[b]>=(1<<i))\n a=anc[a][i];\n for(int i=18;i>=0;i--)\n if(anc[a][i]!=anc[b][i]) \n \ta=anc[a][i],b=anc[b][i];\n if(a!=b) a=anc[a][0];return a;\n} \ninline int jp(int a,int x) {\n\tfor(int i=18;i>=0;i--) if(x>=(1<<i)) {\n\t\tx-=(1<<i);a=anc[a][i];\n\t}\n\tassert(a);\n\treturn a;\n}\ninline int qiu(int x,int v) {\n\tfor(int i=0;i<=21;i++) {\n\t\tif(as[x][i]>=v) return i;\n\t}\n\treturn 22;\n}\ninline int dis(int x,int y) {\n\treturn dep[x]+dep[y]-2*dep[lca(x,y)];\n}\nsigned main(void) {\n\tscanf(\"%d\",&T);\n\twhile(T--) {\n\t\tscanf(\"%d\",&N);\n\t\tfor(int i=1;i<=N;i++) dg[i]=0,hd[i]=0;\n\t\tprintf(\"1 \");\n\t\tk=0;\n\t\tbool ff=0;\n\t\tint ll=N+1;\n\t\tfor(int i=1;i<=N;i++) r[i]=N,vv[i]=0;\n\t\tfor(int i=2;i<=N;i++) {\n\t\t\tscanf(\"%d\",&p[i]);\n\t\t\tanc[i][0]=p[i];\n\t\t\tdep[i]=dep[p[i]]+1;\n\t\t\tl(p[i],i);l(i,p[i]);\n\t\t\tdg[p[i]]++;\n\t\t\tdg[i]++;\n\t\t\tif(dg[p[i]]>3) ff=1;\n\t\t\tif(dg[i]>3) ff=1;\n\t\t\tif(dg[p[i]]==3) {\n\t\t\t\tr[p[i]]=i-1;\n\t\t\t}\n\t\t\tif(ff&&ll==N+1) {\n\t\t\t\tll=i;\n\t\t\t}\n\t\t}\n\t\tfor(int i=1;i<=18;i++) {\n\t\t\tfor(int j=1;j<=N;j++) {\n\t\t\t\tanc[j][i]=anc[anc[j][i-1]][i-1];\n\t\t\t}\n\t\t}\n\t\tfor(int i=1;i<ll;i++) {\n\t\t\tstd::queue<int> q;\n\t\t\tq.push(i);\n\t\t\tvv[i]=i;\n\t\t\td[i]=0;\n\t\t\tmemset(as[i],0,sizeof(as[i]));\n\t\t\twhile(!q.empty()) {\n\t\t\t\tint a=q.front();\n\t\t\t\tq.pop();\n\t\t\t\tas[i][d[a]]=std::max(as[i][d[a]],r[a]);\n\t\t\t\tassert(d[a]<=21);\n\t\t\t\tif(r[a]==N) break;\n\t\t\t\tfor(int j=hd[a];j;j=nxt[j]) {\n\t\t\t\t\tif(vv[to[j]]!=i&&to[j]<ll) {\n\t\t\t\t\t\td[to[j]]=d[i]+1;\n\t\t\t\t\t\tvv[to[j]]=i;\n\t\t\t\t\t\tq.push(to[j]);\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t\tint a=1,b=1,dd=0;\n\t\tfor(int i=2;i<ll;i++) {\n\t\t\tint t1=dis(i,a);\n\t\t\tif(t1>dd) {\n\t\t\t\tdd=t1;\n\t\t\t\tb=i;\n\t\t\t}\n\t\t\tint t2=dis(i,b);\n\t\t\tif(t2>dd) {\n\t\t\t\tdd=t2;\n\t\t\t\ta=i;\n\t\t\t}\n\t\t//\tprintf(\"%d %d %d\\n\",a,b,dd);\n\t\t\tint ans=0;\n\t\t\tif(dd%2==0) {\n\t\t\t\tif(dep[a]<=dep[b]) {\n\t\t\t\t\tans=qiu(jp(b,dd/2),i)+dd/2;\n\t\t\t\t} else {\n\t\t\t\t\tans=qiu(jp(a,dd/2),i)+dd/2;\t\t\t\t\t\n\t\t\t\t}\n\t\t\t} else {\n\t\t\t\tif(dep[a]<=dep[b]) {\n\t\t\t\t\tint x=jp(b,dd/2);\n\t\t\t\t\tans=std::min(qiu(x,i),qiu(p[x],i))+dd/2+1;\n\t\t\t\t} else {\n\t\t\t\t\tint x=jp(a,dd/2);\n\t\t\t\t\tans=std::min(qiu(x,i),qiu(p[x],i))+dd/2+1;\n\t\t\t\t}\t\t\t\t\n\t\t\t}\n\t\t\tprintf(\"%d \",ans+1);\n\t\t}\n\t\tfor(int i=ll;i<=N;i++) printf(\"-1 \");\n\t\tprintf(\"\\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": "2006E",
"submission_id": 278841125,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \n// #define int long long\ntypedef long long ll;\ntypedef long double ld;\n \n#define all(x) (x).begin(), (x).end()\n#define rall(x) (x).rbegin(), (x).rend()\n#define pb push_back\n#define ar(x) array<int, x>\nconst int INF = (int) 2e9;\nconst int L = 20;\n \nvoid solve() {\n int n;\n cin >> n;\n vector<vector<int>> g(n);\n cout << \"1 \";\n bool ban = false;\n vector<int> xx(n - 1);\n vector<vector<int>> g0(n);\n for (int i = 0; i < n - 1; i++) {\n cin >> xx[i];\n xx[i]--;\n g0[xx[i]].pb(i + 1);\n g0[i + 1].pb(xx[i]);\n }\n \n vector<int> d(n);\n vector<vector<int>> up(n, vector<int>(L));\n \n auto dfs = [&](auto dfs, int v, int p = -1) -> void {\n for (auto u: g0[v]) {\n if (u == p)continue;\n up[u][0] = v;\n d[u] = 1 + d[v];\n dfs(dfs, u, v);\n }\n };\n \n dfs(dfs, 0);\n for (int l = 1; l < L; l++) {\n for (int v = 0; v < n; v++) {\n up[v][l] = up[up[v][l - 1]][l - 1];\n }\n }\n \n auto lca = [&](int v, int u) {\n if (d[v] > d[u]) {\n swap(v, u);\n }\n int diff = d[u] - d[v];\n for (int i = 0; i < L; i++) {\n if ((diff >> i) & 1) {\n u = up[u][i];\n }\n }\n if (v == u)return v;\n for (int i = L - 1; i >= 0; i--) {\n if (up[u][i] != up[v][i]) {\n v = up[v][i];\n u = up[u][i];\n }\n }\n return up[u][0];\n };\n \n auto dist = [&](int v, int u) {\n return d[v] + d[u] - 2 * d[lca(v, u)];\n };\n \n int v1 = 0, v2 = 0, curr = 0;\n int ans = INF;\n \n set<int> good = {0};\n \n for (int i = 0; i < n - 1; i++) {\n int x = xx[i];\n g[x].pb(i + 1);\n g[i + 1].pb(x);\n good.insert(i + 1);\n if (g[x].size() >= 4 || g[i + 1].size() >= 4) {\n ban = true;\n }\n if (g[x].size() >= 3) {\n good.erase(x);\n }\n if (ban) {\n cout << \"-1 \";\n continue;\n }\n \n {\n int w1 = dist(v1, i + 1);\n int w2 = dist(v2, i + 1);\n if (w1 == max({w1, w2, curr})) {\n curr = w1;\n v2 = i + 1;\n } else if (w2 == max({w1, w2, curr})) {\n curr = w2;\n v1 = i + 1;\n }\n }\n \n ans = INF;\n \n int tick=0;\n \n for (auto v: good) {\n ++tick;\n ans = min(ans, 1 + max(dist(v, v2), dist(v, v1)));\n if(tick==90)break;\n }\n \n \n cout << ans << ' ';\n }\n \n \n cout << '\\n';\n \n}\n \nsigned main() {\n ios_base::sync_with_stdio(false);\n cin.tie(nullptr);\n int t;\n cin >> t;\n while (t--) {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 278843121,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n//A tree without skin will surely die.\n//A man without face will be alive.\n#include<bits/stdc++.h>\nusing namespace std;\n#define rep(i,j,k) for (int i=j;i<=k;++i)\n#define per(i,j,k) for (int i=j;i>=k;--i)\nint const N=5e5+10;\nint n,p[N],siz[N],du[N],mx,maxn[N],rt;\nvector<int>a[N];\ninline void dfs(int x,int fa,int dep){\n mx=max(mx,dep);\n int cnt=0;\n for (auto v:a[x]) if (v^fa) dfs(v,x,dep+1),++cnt;\n if (cnt>2) mx=1e9;\n}\ninline void solve(){\n cin>>n;\n rep(i,1,n) a[i].clear();\n rep(i,2,n) cin>>p[i];\n cout<<\"1 \";\n rep(i,2,n){\n a[p[i]].push_back(i);\n a[i].push_back(p[i]);\n int ans=1e9;\n rep(j,1,i)\n mx=0,dfs(j,0,1),ans=min(ans,mx);\n if (ans==1e9) ans=-1;\n cout<<ans<<' ';\n }\n cout<<'\\n';\n}\nsigned main(){\n ios::sync_with_stdio(false);\n cin.tie(0),cout.tie(0);\n int t=1;\n cin>>t;\n while (t--) solve();\n // cerr<<\"Time: \"<<(double)clock()/CLOCKS_PER_SEC<<\" s\\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": "2006E",
"submission_id": 278930416,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n \nvoid dfs(int i, int depth, vector<vector<int>> &adj, vector<bool> &visited, vector<int> &order, vector<pair<int,int>> &idx){\n\t\tif(visited[i]) return;\n\t\t// cerr << i << \" \";\n\t\tvisited[i] = true;\n\t\tidx[i].first = order.size();\n\t\torder.push_back(depth);\n\t\tfor(int c : adj[i]) dfs(c, depth + 1, adj, visited, order, idx);\n\t\tidx[i].second = order.size();\n}\n \nvoid update(vector<int> &order, int l, int r){\n\tfor(int i = l; i < r; i++) order[i]++;\n}\n \nvoid u1(set<int> &m, int i, vector<vector<int>> &adj, vector<pair<int,int>> &idx, vector<int> &order){\n\tint mm = *m.begin();\n\t// cerr << mm << endl;\n\tif(idx[i].first < idx[mm].first || idx[mm].second < idx[i].first + 1){\n\t\tint to = adj[mm][0];\n\t\tm.insert(to);\n\t\tupdate(order, idx[mm].first, idx[mm].second);\n\t\treturn;\n\t}\n\tfor(int j = 0; j < adj[mm].size(); j++){\n\t\tint cur = adj[mm][j];\n\t\tif(cur < mm) continue;\n\t\tif(idx[cur].first - 1 < idx[i].first && idx[i].first < idx[cur].second) m.insert(cur);\n\t\telse update(order, idx[cur].first, idx[cur].second);\n\t}\n\tupdate(order, 0, idx[mm].first + 1);\n\tupdate(order, idx[mm].second, order.size());\n}\n \nvoid u2(set<int> &m, int i, vector<vector<int>> &adj, vector<pair<int,int>> &idx, vector<int> &order){\n\tint m1 = *m.begin(), m2 = *m.rbegin();\n\t// cerr << m1 << \" \" << m2 << endl;\n\tif(idx[m2].first - 1 < idx[i].first && idx[i].first < idx[m2].second){\n\t\tm.erase(m1);\n\t\tfor(int j = 0; j < adj[m1].size(); j++){\n\t\t\tint cur = adj[m1][j];\n\t\t\tif(cur < m1) continue;\n\t\t\tif(cur == m2) continue;\n\t\t\tupdate(order, idx[cur].first, idx[cur].second);\n\t\t}\n\t\tupdate(order, 0, idx[m1].first + 1);\n\t\tupdate(order, idx[m1].second, order.size());\n\t\treturn;\n\t}\n\tm.erase(m2);\n\tupdate(order, idx[m2].first, idx[m2].second);\n\treturn;\n}\n \nvoid solve(){\n\t// Basic Approach\n\t// Once every vertex is added, run dfs from that vertex.\n\t// Update max distance based dfs. O(n)\n\t// Compute minimum. O(n)\n\t// Total complexity O(n^2)\n\t// Optimizing - All the nodes which get +1 are in a subtree.\n\t// One of the diameter endpoints must get a +1\n\t// If diameter root gets updated, then everything either on left or right gets updated. \n\t// How do we update the diameter connections - point towards furthest node?\n\t// I can dfs on the whole tree initially. \n\t// start - distance to node 1 from every node.\n\t// adding node 2 - all tree apart from subtree with node 2 get +1\n\t// if I create a new diameter, the midpoints don't change (one or two)\n\t// adding a vertex which increases the diameter\n\t// two midpoints to one - subtrees of vanishing midpoint increases.\n\t// one midpoint to two - the one midpoint and it's subtree increases. \n\tint n; cin >> n;\n\tvector<vector<int>> adj(n, vector<int>());\n\tfor(int i = 0; i < n - 1; i++){\n\t\tint p; cin >> p;\n\t\tadj[p-1].push_back(i+1);\n\t\tadj[i+1].push_back(p-1);\n\t}\n \n\tvector<pair<int,int>> idx(n);\n\tvector<int> order;\n\tvector<bool> visited(n,false);\n\tdfs(0, 1, adj, visited, order, idx);\n\t// cerr << endl << endl;\n\t\n\t// for(int i : order) cerr << i << \" \";\n\t// cerr << endl;\n \n\tint diam = order[0];\n\tbool can = true;\n\tvector<int> ans(1, order[0]), deg(n, 0);\n\tset<int> m({0});\n\tfor(int i = 1; i < n; i++){\n\t\tdeg[i] = 1;\n\t\tint p = adj[i][0];\n\t\tdeg[p]++;\n\t\tif(deg[p] > 2) order[idx[p].first] = n + 1;\n\t\tif(deg[p] > 3) can = false;\n\t\tif(order[idx[i].first] > diam){\n\t\t\tdiam++;\n\t\t\tif(m.size() == 1) u1(m, i, adj, idx, order);\n\t\t\telse u2(m, i, adj, idx, order);\n\t\t}\n\t\t// for(int j : order) cerr << j << \" \";\n\t\t// cerr << endl;\n\t\tans.push_back(n+1);\n\t\tfor(int j = 0; j < i + 1; j++) ans.back() = min(ans.back(), order[idx[j].first]);\n\t\tif(n < ans.back() || !can) ans.back() = -1;\n\t}\n\t// cerr << endl;\n\tfor(int &i : ans) cout << i << \" \";\n\tcout << endl;\n\t// cerr << endl;\n}\n \nvoid testcases(){\n\tint t = 0;\n\tcin >> t;\n\twhile(t--) solve();\n}\n \nint main() {\n\tcin.tie(nullptr);\n\tcout.tie(nullptr);\n\tios::sync_with_stdio(false);\n\ttestcases();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 280183318,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define Pr pair<int,int>\n#define fi first\n#define nd second\n#define psb push_back\nusing namespace std;\ntypedef long long LL;\ninline int read(){\n\tchar ch=getchar();\n\twhile(!isdigit(ch) && ch!='-') ch=getchar();\n\tint x=0,ff=1; if(ch=='-') ff=-1,ch=getchar();\n\twhile(isdigit(ch)) x=(x<<3) + (x<<1) + (ch^48),ch=getchar();\n\treturn x*ff;\n}\nconst int N=5e5+5,inf=1e9;\nint n,p[N],b[N],ld[N],tot,son[N],fa[N],top[N],dep[N],sz[N],ru,rv,rds,ans,ind[N],gfa[N],rT; bool vis[N]; vector<int> e[N],kf[N],kd[N]; set<Pr> st[N];\nvoid dfs1(int u,int Fa){\n\tsz[u]=1; fa[u]=Fa; dep[u]=dep[Fa]+1;\n\tfor(int v:e[u]) if(v!=Fa){\n\t\tdfs1(v,u); sz[u]+=sz[v];\n\t\tif(sz[v]>sz[son[u]]) son[u]=v;\n\t}\n}\nvoid dfs2(int u,int tpp){\n\ttop[u]=tpp; b[ld[u]=++tot]=u;\n\tif(son[u]){\n\t\tdfs2(son[u],tpp);\n\t\tfor(int v:e[u]) if(v!=son[u] && v!=fa[u]) dfs2(v,v);\n\t}\n}\nint LCA(int u,int v){\n\twhile(top[u]!=top[v]){\n\t\tif(dep[top[u]]<dep[top[v]]) swap(u,v);\n\t\tu=fa[top[u]];\n\t} return dep[u]<dep[v]?u:v;\n}\nint K_fa(int u,int kx){\n\tkx=dep[u]-kx; while(dep[top[u]]>kx) u=fa[top[u]];\n\treturn b[ld[u]-dep[u]+kx];\n}\nint gdis(int x,int y) {return dep[x]+dep[y]-2*dep[LCA(x,y)];}\nvoid getrt(int u,int Fa,int S){\n\tsz[u]=1; int mx=0;\n\tfor(int v:e[u]) if(v!=Fa && !vis[v]) {\n\t\tgetrt(v,u,S); sz[u]+=sz[v];\n\t\tmx=max(mx,sz[v]);\n\t} mx=max(mx,S-sz[u]); rT=mx*2<=S?u:rT;\n}\nvoid rdfs(int u,int Fa,int d,int gf){\n kf[u].psb(gf); kd[u].psb(d);\n for(int v:e[u]) if(v!=Fa && v!=Fa) rdfs(v,u,d+1,gf);\n}\nvoid dfz(int u,int Fa,int S){\n\tgetrt(u,0,S); u=rT; gfa[u]=Fa; vis[u]=1; rdfs(u,0,0,u); getrt(u,0,S);\n\tfor(int v:e[u]) if(!vis[v]) dfz(v,u,sz[v]);\n}\nvoid addx(int x,int o){\n int len=kf[x].size(),u;\n for(int i=len-1;i>=0;i--){\n u=kf[x][i]; Pr bf={kd[x][i],x};\n\t\tif(o) st[u].insert(bf);\n\t\telse st[u].erase(bf);\n }\n}\nint qry(int x){\n\tint len=kf[x].size(),u; Pr z={inf,0};\n\tfor(int i=len-1;i>=0;i--){ u=kf[x][i];\n\t\tif(st[u].size()){\n\t\t\tPr bf=*st[u].begin(); bf.fi+=kd[x][i];\n\t\t\tz=min(z,bf);\n\t\t}\n\t} return z.nd;\n}\nvoid vmain(){\n\tn=read(); tot=0; for(int i=1;i<=n;i++) son[i]=ind[i]=gfa[i]=vis[i]=0,e[i].clear(),st[i].clear(),kd[i].clear(),kf[i].clear();\n\tfor(int i=2;i<=n;i++) p[i]=read(),e[p[i]].push_back(i),e[i].push_back(p[i]);\n\tdfs1(1,0); dfs2(1,1); dfz(1,0,n); ans=0;\n\tfor(int i=1;i<=n;i++){\n\t\tif(~ans){\n\t\t\taddx(i,1);\n\t\t\tif(i==1) ans=1,ru=rv=1,rds=0;\n\t\t\telse { int x=p[i]; ++ind[x]; ind[i]=1;\n\t\t\t\tif(ind[x]<=3){\n\t\t\t\t\tif(ind[x]==3) addx(x,0);\n\t\t\t\t\tif(gdis(ru,i)>rds) ++rds,rv=i;\n\t\t\t\t\tif(gdis(rv,i)>rds) ++rds,ru=i;\n\t\t\t\t\tint lc=LCA(ru,rv),px=0,z=rds/2; px=(z<=dep[ru]-dep[lc])?K_fa(ru,z):K_fa(rv,rds-z);\n\t\t\t\t\tpx=qry(px); ans=max(gdis(ru,px),gdis(rv,px))+1;\n\t\t\t\t\tif(rds&1){\n\t\t\t\t\t\t++z; px=(z<=dep[ru]-dep[lc])?K_fa(ru,z):K_fa(rv,rds-z); px=qry(px);\n\t\t\t\t\t\tans=min(ans,max(gdis(ru,px),gdis(rv,px))+1);\n\t\t\t\t\t}\n\t\t\t\t} else ans=-1;\n\t\t\t}\n\t\t}\n\t\tprintf(\"%d \",ans);\n\t} puts(\"\");\n}\nint main(){\n\tint T=read(); while(T--) vmain();\n\treturn 0;\n}\n/*\ng++ 1.cpp -o 1 -std=c++20 -Wall -Wl,-stack=999999999\n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 279475310,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n// Author: YE Minghan\n#include<bits/stdc++.h>\nusing namespace std;\n#ifdef DEBUG\n#include \"templates/debug.h\"\n#else\n#define dbg(...) (void)0\n#define msg(...) (void)0\n#endif\n#define ll long long\n#define endl '\\n'\n#define PB emplace_back\n#define PPB pop_back\n#define MP make_pair\n#define ALL(Name) Name.begin(),Name.end()\n#define PII pair<int,int>\n#define VI vector<int>\n#define GI greater<int>\n#define fi first\n#define se second\n \nconst int N=500005,L=20,inf=1e9;\nint n,f[N<<1][L],dep[N],dfn[N],ed[N],df,deg[N];\nVI g[N<<1];\n \nvoid dfs(int x)\n{\n\tdfn[x]=++df,dep[x]=dep[f[x][0]]+1;\n\tfor(int i=1;i<L;i++)f[x][i]=f[f[x][i-1]][i-1];\n\tfor(int y:g[x])dfs(y);\n\ted[x]=df;\n}\nint anc(int x,int k)\n{\n\twhile(k){int t=__builtin_ctz(k);k^=1<<t,x=f[x][t];}\n\treturn x;\n}\nint lca(int x,int y)\n{\n\tif(dep[x]>dep[y])swap(x,y);\n\ty=anc(y,dep[y]-dep[x]);\n\tif(x==y)return x;\n\tfor(int i=L-1;~i;i--)\n\t\tif(f[x][i]!=f[y][i])x=f[x][i],y=f[y][i];\n\treturn f[x][0];\n}\nint dis(int x,int y){return dep[x]+dep[y]-(dep[lca(x,y)]<<1);}\n \nstruct info\n{\n\tint add,mn;\n}tr[1<<21];\n#define lid id<<1\n#define rid lid|1\n#define mid (l+r>>1)\n#define rmd (mid+1)\nvoid atg(int id,int v){tr[id].add+=v,tr[id].mn+=v;}\nvoid push_down(int id)\n{\n\tatg(lid,tr[id].add),atg(rid,tr[id].add);\n\ttr[id].add=0;\n}\nvoid add(int L,int R,int v,int l=1,int r=(n<<1)-1,int id=1)\n{\n\tif(L==l&&R==r)return atg(id,v);\n\tpush_down(id);\n\tif(L<=mid)add(L,min(R,mid),v,l,mid,lid);\n\tif(R>=rmd)add(max(L,rmd),R,v,rmd,r,rid);\n\ttr[id].mn=min(tr[lid].mn,tr[rid].mn);\n}\nint mn(){return tr[1].mn;}\n \nint main()\n{\n\tios::sync_with_stdio(false),cin.tie(nullptr);\n\tint _;cin>>_;while(_--)\n{\n\tcin>>n;\n\tfor(int i=2;i<=n;i++)cin>>f[n+i-1][0],f[i][0]=n+i-1,g[f[n+i-1][0]].PB(n+i-1),g[n+i-1].PB(i);\n\tdfs(1);\n\tadd(2,(n<<1)-1,inf);\n\tcout<<\"1 \";\n\tint x=1,y=1,z=1,d=0;\n\tbool bang=0;\n\tfor(int i=2;i<=n;i++)\n\t{\n\t\tdeg[i]++,add(dfn[i],dfn[i],dis(i,1)-inf);\n\t\tif(++deg[f[i][1]]==3)add(dfn[f[i][1]],dfn[f[i][1]],inf);\n\t\tif(deg[f[i][1]]>3)bang=1;\n\t\tif(bang){cout<<-1<<\" \\n\"[i==n];continue;}\n\t\tint b=dis(i,x),c=dis(i,y);\n\t\td=max({b,c,d});\n\t\tif(b==d)y=i;\n\t\telse if(c==d)x=i;\n\t\tint m=anc(dep[x]>dep[y]?x:y,d>>1);\n\t\tif(dep[z]>dep[m])add(1,(n<<1)-1,-1),add(dfn[z],ed[z],2);\n\t\tif(dep[z]<dep[m])add(1,(n<<1)-1,1),add(dfn[m],ed[m],-2);\n\t\tz=m;\n\t\tcout<<((d>>1)+mn()>>1)+1<<\" \\n\"[i==n];\n\t}\n\tfill(deg+1,deg+n+1,0),df=0,fill(g+1,g+(n<<1),VI());\n\tmemset(tr+1,0,sizeof(info)<<33-__builtin_clz(n-1<<1));\n}\n \n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 292461425,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\ntypedef long long ll;\n \n#define N 500000\n \nint i,j,k,n,m,t,dfn[N+50],it,fa[N+50],in[N+50],mx;\nbasic_string<int> v[N+50];\nint dep[N+50];\n \nstruct SB{\n\tint i,j,k,n;\n\tint f[21][N+50];\n\tint _get(int l,int r){\n\t\treturn (dfn[l]<dfn[r]?l:r);\n\t}\n\tvoid init(int nn){\n\t\tn=nn;\n\t\tfor(i=1;i<=20;i++)for(j=1;j+(1<<i)-1<=n;j++)f[i][j]=_get(f[i-1][j],f[i-1][j+(1<<(i-1))]);\n\t}\n\tint lca(int l,int r){\n\t\tif(l==r)return l;\n\t\tl=dfn[l]; r=dfn[r]; if(l>r)swap(l,r);\n\t\tl++; int k=__lg(r-l+1);\n\t\treturn _get(f[k][l],f[k][r-(1<<k)+1]);\n\t}\n\tint dis(int x,int y){\n\t\tint z=lca(x,y);\n\t\treturn dep[x]+dep[y]-2*dep[z];\n\t}\n}sb;\n \nvoid dfs0(int x){\n\tdfn[x]=++it; sb.f[0][it]=fa[x];\n\tfor(auto i:v[x])dfs0(i);\n}\n \nstruct JB{\n\tint x,y,w;\n}f,tmp;\n \nJB hb(JB x,JB y){\n\tauto [x1,y1,w1]=x;\n\tauto [x2,y2,w2]=y;\n\tvector<int> v={x1,y1,x2,y2};\n\tint rx=0,ry=0,rw=-1,k;\n\tfor(int i=0;i<4;i++)for(int j=i+1;j<4;j++){\n\t\tk=sb.dis(v[i],v[j]);\n\t\tif(k>rw){rw=k; rx=v[i]; ry=v[j];}\n\t}\n\treturn {rx,ry,rw};\n}\n \nint main(){\n\tios::sync_with_stdio(0); cin.tie(0);\n\tcin>>t;\n\twhile(t--){\n\t\tcin>>n; it=0; mx=0;\n\t\tfor(i=2;i<=n;i++){\n\t\t\tcin>>fa[i];\n\t\t\tv[fa[i]]+=i; dep[i]=dep[fa[i]]+1;\n\t\t}\n\t\tdfs0(1); sb.init(n);\n\t\t\n\t\tf={1,1,0};\n\t\tcout<<\"1 \";\n\t\tfor(i=2;i<=n;i++){\n\t\t\tin[fa[i]]++; in[i]++;\n\t\t\tmx=max({mx,in[fa[i]],in[i]});\n\t\t\tif(mx>=4){cout<<-1<<' '; continue;}\n\t\t\ttmp={i,i,0};\n\t\t\tf=hb(f,tmp);\n\t\t\t\n\t\t\tint res=1e9;\n\t\t\tfor(j=1;j<=i;j++)if(in[j]<=2){\n\t\t\t\tres=min(res,max(sb.dis(f.x,j),sb.dis(f.y,j))+1);\n\t\t\t}\n\t\t\tif(res>1e8)res=-1;\n\t\t\tcout<<res<<' ';\n\t\t}\n\t\tcout<<'\\n';\n\t\t\n\t\tfor(i=1;i<=n;i++){\n\t\t\tv[i]={}; in[i]=0;\n\t\t}\n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 278847305,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long\n#define pii pair<int,int>\n#define all(v) v.begin(),v.end()\n#define pb push_back\n#define REP(i,b,e) for(int i=(b);i<(int)(e);++i)\n#define over(x) {cout<<(x)<<endl;return;}\nstruct ds{\n\tint seg[2000005],tag[2000005];\n\tvoid build(int l,int r,int p){\n\t\tseg[p]=(l==0? 0:1e18),tag[p]=0;\n\t\tif(l==r)return;\n\t\tint m=(l+r)>>1;\n\t\tbuild(l,m,p*2+1);build(m+1,r,p*2+2);\n\t}\n\tvoid pushdown(int p){\n\t\tif(!tag[p])return;\n\t\ttag[p*2+1]+=tag[p];tag[p*2+2]+=tag[p];\n\t\tseg[p*2+1]+=tag[p];seg[p*2+2]+=tag[p];\n\t\ttag[p]=0;\n\t}\n\tvoid add(int l,int r,int s,int t,int p,int val){\n\t\tif(l<=s&&t<=r){\n\t\t\tseg[p]+=val;tag[p]+=val;\n\t\t\treturn;\n\t\t}\n\t\tint m=(s+t)>>1;pushdown(p);\n\t\tif(m>=l)add(l,r,s,m,p*2+1,val);\n\t\tif(m<r)add(l,r,m+1,t,p*2+2,val);\n\t\tseg[p]=min(seg[p*2+1],seg[p*2+2]);\n\t}\n\tvoid update(int pos,int l,int r,int p,int val){\n\t\tif(l==r){\n\t\t\tseg[p]=val;\n\t\t\treturn;\n\t\t}\n\t\tint m=(l+r)>>1;pushdown(p);\n\t\tif(m>=pos)update(pos,l,m,p*2+1,val);\n\t\telse update(pos,m+1,r,p*2+2,val);\n\t\tseg[p]=min(seg[p*2+1],seg[p*2+2]);\n\t}\n}seg;\nint n;\nvector<int>v[500005];\nint fa[500005],an[500005][21];\nint dep[500005],dfn[500005],rev[500005];\nint tot,deg[500005],ls[500005];\nvoid dfs(int x,int pre,int d){\n\tdep[x]=d;fa[x]=pre;an[x][0]=fa[x];ls[x]=1;\n\tREP(i,0,__lg(n+1))if(an[x][i]==-1)an[x][i+1]=-1;else an[x][i+1]=an[an[x][i]][i];\n\trev[tot]=x;dfn[x]=tot++;\n\tfor(auto i:v[x])dfs(i,x,d+1),ls[x]+=ls[i];\n}\nint getlca(int x,int y){\n\tif(dep[x]<dep[y])swap(x,y);\n\tint d=dep[x]-dep[y];\n\tfor(int i=__lg(d);i>=0;--i)if((d>>i)&1)x=an[x][i];\n\tif(x==y)return x;\n\td=dep[x];\n\tfor(int i=__lg(d);i>=0;--i)if((1<<i)<=dep[x]&&an[x][i]!=an[y][i])x=an[x][i],y=an[y][i];\n\treturn fa[x];\n}\nint getan(int x,int y){\n\tif(dfn[y]>=dfn[x]&&dfn[y]<=ls[x]){\n\t\tint d=dep[y]-dep[x]-1;\n\t\tfor(int i=0;(1<<i)<=d;++i)if((d>>i)&1)y=an[y][i];\n\t\treturn y;\n\t}else return fa[x];\n}\nvoid updside(int x,int y){\n\tif(dfn[y]>=dfn[x]&&dfn[y]<=ls[x]){\n\t\tseg.add(0,n-1,0,n-1,0,1);\n\t\tx=getan(x,y);\n\t\tseg.add(dfn[x],ls[x],0,n-1,0,-1);\n\t}else seg.add(dfn[x],ls[x],0,n-1,0,1);\n}\nint getdist(int x,int y){return dep[x]+dep[y]-2*dep[getlca(x,y)];}\nstruct diameter{\n\tint x,y,len;\n\tint update(int z){\n\t\tif(x==y){\n\t\t\tint d=getdist(x,z);\n\t\t\tif(d<=len)return d+len;\n\t\t\tupdside(y,z);\n\t\t\ty=getan(y,z);\n\t\t\treturn d+len;\n\t\t}else{\n\t\t\tint d1=getdist(x,z),d2=getdist(y,z),d3=min(d1,d2);\n\t\t\tif(d3<=len)return d3+len+1;\n\t\t\tif(d1>d2)swap(x,y);\n\t\t\tupdside(y,z);\n\t\t\ty=x;++len;return len*2;\n\t\t}\n\t}\n\tint query(int z){\n\t\tif(x==y)return len+getdist(x,z);\n\t\telse return len+min(getdist(x,z),getdist(y,z))+1;\n\t}\n};\nvoid Main() {\n\tcin>>n;\n\tREP(i,0,n)v[i].clear();\n\tREP(i,1,n){\n\t\tcin>>fa[i];--fa[i];\n\t\tv[fa[i]].pb(i);\n\t}\n\ttot=0;dfs(0,-1,0);seg.build(0,n-1,0);\n\tREP(i,0,n)deg[i]=0,ls[i]+=dfn[i]-1;\n\tdiameter d={0,0,0};\n\tcout<<\"1 \";\n\tREP(i,1,n){\n\t\t++deg[fa[i]];++deg[i];\n\t\tif(deg[fa[i]]==4){\n\t\t\tREP(j,i,n)cout<<-1<<' ';\n\t\t\tcout<<endl;return;\n\t\t}\n\t\tif(deg[fa[i]]==3)seg.update(dfn[fa[i]],0,n-1,0,1e18);\n\t\tseg.update(dfn[i],0,n-1,0,d.update(i));\n\t\tcout<<seg.seg[0]+1<<' ';\n\t}\n\tcout<<endl;\n}\nvoid TC() {\n int tc=1;\n cin>>tc;\n\twhile(tc--){\n\t\tMain();\n\t\tcout.flush();\n\t}\n}\nsigned main() {\n\treturn cin.tie(0),cout.tie(0),ios::sync_with_stdio(0),TC(),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": "2006E",
"submission_id": 278943352,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nusing i64=long long;\nconst int N=5e5+5;\ni64 n;\nvector<i64>v[N];\ni64 fa[N],an[N][21],dep[N],dfn[N],rev[N],tot,deg[N],ls[N];\nstruct STree{\n\ti64 mn[N<<2],lazy[N<<2];\n\tvoid init(i64 l,i64 r,i64 p){\n\t\tmn[p]=(l==0? 0:1e18),lazy[p]=0;\n\t\tif(l==r)return;\n\t\ti64 m=(l+r)>>1;\n\t\tinit(l,m,p*2+1);init(m+1,r,p*2+2);\n\t}\n\tvoid pushdown(i64 p){\n\t\tif(!lazy[p])return;\n\t\tlazy[p*2+1]+=lazy[p];lazy[p*2+2]+=lazy[p];\n\t\tmn[p*2+1]+=lazy[p];mn[p*2+2]+=lazy[p];\n\t\tlazy[p]=0;\n\t}\n\tvoid change(i64 l,i64 r,i64 s,i64 t,i64 p,i64 val){\n\t\tif(l<=s&&t<=r){\n\t\t\tmn[p]+=val;lazy[p]+=val;\n\t\t\treturn;\n\t\t}\n\t\ti64 m=(s+t)>>1;pushdown(p);\n\t\tif(m>=l)change(l,r,s,m,p*2+1,val);\n\t\tif(m<r)change(l,r,m+1,t,p*2+2,val);\n\t\tmn[p]=min(mn[p*2+1],mn[p*2+2]);\n\t}\n\tvoid change2(i64 pos,i64 l,i64 r,i64 p,i64 val){\n\t\tif(l==r){\n\t\t\tmn[p]=val;\n\t\t\treturn;\n\t\t}\n\t\ti64 m=(l+r)>>1;pushdown(p);\n\t\tif(m>=pos)change2(pos,l,m,p*2+1,val);\n\t\telse change2(pos,m+1,r,p*2+2,val);\n\t\tmn[p]=min(mn[p*2+1],mn[p*2+2]);\n\t}\n}seg;\nvoid dfs(i64 x,i64 pre,i64 d){\n\tdep[x]=d;fa[x]=pre;an[x][0]=fa[x];ls[x]=1;\n\tfor(int i=0;i<__lg(n+1);i++)\tif(an[x][i]==-1)an[x][i+1]=-1;else an[x][i+1]=an[an[x][i]][i];\n\trev[tot]=x;dfn[x]=tot++;\n\tfor(auto i:v[x])dfs(i,x,d+1),ls[x]+=ls[i];\n}\ni64 getlca(i64 x,i64 y){\n\tif(dep[x]<dep[y])swap(x,y);\n\ti64 d=dep[x]-dep[y];\n\tfor(i64 i=__lg(d);i>=0;--i)if((d>>i)&1)x=an[x][i];\n\tif(x==y)return x;\n\td=dep[x];\n\tfor(i64 i=__lg(d);i>=0;--i)if((1<<i)<=dep[x]&&an[x][i]!=an[y][i])x=an[x][i],y=an[y][i];\n\treturn fa[x];\n}\ni64 getan(i64 x,i64 y){\n\tif(dfn[y]>=dfn[x]&&dfn[y]<=ls[x]){\n\t\ti64 d=dep[y]-dep[x]-1;\n\t\tfor(i64 i=0;(1<<i)<=d;++i)if((d>>i)&1)y=an[y][i];\n\t\treturn y;\n\t}else return fa[x];\n}\nvoid updside(i64 x,i64 y){\n\tif(dfn[y]>=dfn[x]&&dfn[y]<=ls[x]){\n\t\tseg.change(0,n-1,0,n-1,0,1);\n\t\tx=getan(x,y);\n\t\tseg.change(dfn[x],ls[x],0,n-1,0,-1);\n\t}else seg.change(dfn[x],ls[x],0,n-1,0,1);\n}\ni64 getdist(i64 x,i64 y){return dep[x]+dep[y]-2*dep[getlca(x,y)];}\nstruct Node{\n\ti64 x,y,len;\n\ti64 change2(i64 z){\n\t\tif(x==y){\n\t\t\ti64 d=getdist(x,z);\n\t\t\tif(d<=len)return d+len;\n\t\t\tupdside(y,z);\n\t\t\ty=getan(y,z);\n\t\t\treturn d+len;\n\t\t}else{\n\t\t\ti64 d1=getdist(x,z),d2=getdist(y,z),d3=min(d1,d2);\n\t\t\tif(d3<=len)return d3+len+1;\n\t\t\tif(d1>d2)swap(x,y);\n\t\t\tupdside(y,z);\n\t\t\ty=x;++len;return len*2;\n\t\t}\n\t}\n\ti64 query(i64 z){\n\t\tif(x==y)return len+getdist(x,z);\n\t\telse return len+min(getdist(x,z),getdist(y,z))+1;\n\t}\n};\nint main(){\n\ti64 T;\n\tcin>>T;\n\twhile(T--){\n\t\tcin>>n;\n\t\tfor(int i=0;i<n;i++)\tv[i].clear();\n\t\tfor(int i=1;i<n;i++){\n\t\t\tcin>>fa[i];--fa[i];\n\t\t\tv[fa[i]].push_back(i);\n\t\t}\n\t\ttot=0;dfs(0,-1,0);seg.init(0,n-1,0);\n\t\tfor(int i=0;i<n;i++)\tdeg[i]=0,ls[i]+=dfn[i]-1;\n\t\tNode d={0,0,0};\n\t\tcout<<\"1 \";\n\t\tfor(int i=1;i<n;i++){\n\t\t\t++deg[fa[i]];++deg[i];\n\t\t\tif(deg[fa[i]]==4){\n\t\t\t\tfor(int j=i;j<n;j++)\tcout<<-1<<' ';\n\t\t\t\tbreak;\n\t\t\t}\n\t\t\tif(deg[fa[i]]==3)seg.change2(dfn[fa[i]],0,n-1,0,1e18);\n\t\t\tseg.change2(dfn[i],0,n-1,0,d.change2(i));\n\t\t\tcout<<seg.mn[0]+1<<' ';\n\t\t}\n\t\tcout<<endl;\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 292592720,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\nimport heapq\n \npar = []\ndepth = []\ndef build(u, p):\n par[u][0] = p\n i = 0\n while par[u][i]:\n par[u][i + 1] = par[par[u][i]][i]\n i += 1\n depth[u] = depth[p] + 1\n \ndef lca(u, v):\n d = depth[u] + depth[v]\n if depth[v] > depth[u]:\n u, v = v, u\n for i in range(19, -1, -1):\n if depth[par[u][i]] >= depth[v]:\n u = par[u][i]\n for i in range(19, -1, -1):\n if par[u][i] != par[v][i]:\n u, v = par[u][i], par[v][i]\n p = u if u == v else par[u][0]\n return p, d - 2 * depth[p]\n \ndef solve():\n n = int(input())\n \n global par, depth\n par = [[0] * 20 for _ in range(n + 1)]\n depth = [0] * (n + 1)\n depth[1] = 1\n \n u, v = 1, 1\n dist = 0\n heap = [(0, 1)]\n deg = [0] * (n + 1)\n max_deg = 1\n \n print(1, end = ' ')\n i = 1\n for p in list(map(int, input().split())):\n i += 1\n build(i, p)\n deg[i] += 1\n deg[p] += 1\n max_deg = max(max_deg, deg[p])\n if max_deg > 3:\n print(-1, end = ' ')\n continue\n \n dist1 = lca(i, u)[1]\n dist2 = lca(i, v)[1]\n if dist1 > dist2:\n if dist1 > dist:\n v, dist = i, dist1\n else:\n if dist2 > dist:\n u, dist = i, dist2\n \n heapq.heappush(heap, (max(lca(i, u)[1], lca(i, v)[1]), i))\n while True:\n d, t = heap[0]\n if deg[t] > 2:\n heapq.heappop(heap)\n continue\n mx = max(lca(t, u)[1], lca(t, v)[1])\n if mx != d:\n heapq.heappop(heap)\n heapq.heappush(heap, (mx, t))\n continue\n print(d + 1, end = ' ')\n break\n print()\n \n \nt = int(input())\nfor _ in range(t):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 281711081,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include \"bits/stdc++.h\"\nusing namespace std;\n#define all(x) x.begin(),x.end()\ntemplate<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << p.first << \" \" << p.second; }\ntemplate<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = \" \"; return os; }\n#ifdef LOCAL\n#include \"debug.h\"\n#else\n#define debug(...) 42\n#define ASSERT(...) 42\n#endif\ntypedef long long ll;\ntypedef vector<int> vi;\ntypedef vector<vi> vvi;\ntypedef pair<int,int> pi;\nconst int oo = 1e9;\nstruct node {\n int mn;\n node() {\n // *this = node(0);\n mn=oo;\n }\n node(int val) {\n mn=val;\n }\n void merge(const node& o) {\n mn=min(mn,o.mn);\n }\n friend node merge(node a, const node& b) {\n a.merge(b);\n return a;\n }\n};\nstruct segtree {\n int ptwo;\n vector<node> seg;\n segtree(){}\n node& operator[](int i) {\n return seg[i+ptwo];\n }\n segtree(int nn) {\n ptwo=nn;\n seg.assign(ptwo*2,{});\n }\n auto query(int l, int r) {\n assert(l>=0 and l<ptwo and r >=0 and r<ptwo);\n l+=ptwo; r+=ptwo;\n node resl, resr;\n while(l and l<=r) {\n if(l%2==1) resl = merge(resl,seg[l++]);\n if(r%2==0) resr = merge(seg[r--],resr);\n l/=2; r/=2;\n }\n return merge(resl,resr);\n }\n void update(int i, int val) {\n assert(i>=0 and i<ptwo);\n i+=ptwo;\n seg[i] = node(val);\n for(i/=2;i>=1;i/=2) {\n seg[i] = merge(seg[2*i],seg[2*i+1]);\n }\n }\n void build() {\n for(int i=ptwo-1;i>=1;--i) {\n seg[i] = merge(seg[2*i],seg[2*i+1]);\n }\n }\n};\nsegtree seg;\nstruct LCA {\n vi par,d,jmp,in,out;\n vector<basic_string<int>> adj;\n int root;\n vector<set<int>> h;\n LCA(int n,int root=0) : par(n,root),d(n),jmp(n,root),in(n),out(n),adj(n),root(root),h(n) {}\n \n void add(int i) {\n int p = par[i];\n d[i]=1+d[p];\n if(d[p] - d[jmp[p]] == d[jmp[p]] - d[jmp[jmp[p]]]) jmp[i] = jmp[jmp[p]];\n else jmp[i] = p;\n }\n int jump(int a, int k) {\n int D = max(0,d[a]-k);\n while(d[a]>D) {\n if(d[jmp[a]]>=D) a = jmp[a];\n else a = par[a];\n }\n return a;\n }\n template<typename F> pi find(int a, F f) {\n int steps=d[a];\n while(!f(a)) {\n if(!f(jmp[a])) a = jmp[a];\n else a = par[a];\n }\n steps-=d[a]-1;\n return {a,steps};\n }\n int lca(int a, int b) {\n if(d[a]<d[b]) swap(a,b);\n a = jump(a,d[a]-d[b]);\n while(a!=b) {\n if(jmp[a]!=jmp[b]) a=jmp[a],b=jmp[b];\n else a=par[a],b=par[b];\n }\n return a;\n }\n int dist(int u,int v) {\n return d[u]+d[v]-2*d[lca(u,v)];\n }\n void addE(int u, int v) {\n adj[u].push_back(v);\n adj[v].push_back(u);\n }\n int cnt=0;\n void dfs(int at) {\n in[at]=cnt++;\n for(int to: adj[at]) if(to!=par[at]) {\n par[to]=at;\n add(to);\n dfs(to);\n }\n out[at]=cnt;\n }\n void init() {\n dfs(root);\n }\n const int B = 21;\n \n int get(int a, int b, int w, int di) {\n int l = lca(a,b);\n if(d[a]-d[l]>=w) {\n return jump(a,w);\n } else {\n return jump(b,di-w);\n }\n }\n int a,b;\n bool insub(int u, int v) {\n return in[u]<=in[v] and in[v]<out[u];\n }\n int midpoint(int u, int v) {\n a=u,b=v;\n int di = dist(a,b);\n int at = get(a,b,di/2,di);\n // get your answer immediately\n int ans=oo;\n for(int v=at,rep=0;rep<B;v=par[v],rep++) {\n // do range min query\n int cur = max(dist(v,a),dist(v,b));\n ans=min(ans, cur + seg.query(in[v],out[v]-1).mn - d[v]);\n }\n for(int rep2=0;rep2<2;++rep2) {\n for(int v=at,rep=0;rep<B;rep++) {\n int cur = max(dist(v,a),dist(v,b));\n ans=min(ans, cur + seg.query(in[v],out[v]-1).mn - d[v]);\n int nxt=-1;\n for(int to : adj[at]) {\n if(to!=par[at] and insub(to,a)) {\n nxt=to;\n }\n }\n if(nxt==-1) break;\n v=nxt;\n }\n swap(a,b);\n }\n return ans;\n \n \n }\n};\nvoid solve() {\n int n; cin >> n;\n LCA lca(n);\n vi par(n);\n for(int i=1;i<n;++i) {\n int p; cin >> p;\n --p;\n par[i]=p;;\n lca.addE(p,i); \n }\n lca.init();\n int a=0,b=0;\n vi deg(n);\n bool bad=0;\n seg = segtree(n);\n seg.build();\n for(int i=0;i<n;++i) {\n int diam = lca.dist(a,b);\n if(lca.dist(a,i)>diam) {\n b=i;\n } else if(lca.dist(i,b)>diam) {\n a=i;\n }\n \n if(i) {\n deg[par[i]]++,deg[i]++;\n if(deg[par[i]]==3) {\n seg.update(lca.in[par[i]],oo);\n }\n \n }\n seg.update(lca.in[i],lca.d[i]);\n if(deg[i]>=4) bad=1;\n if(deg[par[i]]>=4) bad=1;\n int ans = lca.midpoint(a,b);\n if(bad) ans=-2;\n cout << ans+1 << ' ';\n \n }\n cout << '\\n';\n \n \n \n}\nint main() {\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n int t; 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": "2006E",
"submission_id": 278846141,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define Pr pair<int,int>\n#define fi first\n#define nd second\n#define psb push_back\nusing namespace std;\ntypedef long long LL;\ninline int read(){\n\tchar ch=getchar();\n\twhile(!isdigit(ch) && ch!='-') ch=getchar();\n\tint x=0,ff=1; if(ch=='-') ff=-1,ch=getchar();\n\twhile(isdigit(ch)) x=(x<<3) + (x<<1) + (ch^48),ch=getchar();\n\treturn x*ff;\n}\nconst int N=5e5+5,inf=1e9;\nint n,p[N],b[N],ld[N],tot,son[N],fa[N],top[N],dep[N],sz[N],ru,rv,rds,ans,ind[N],gfa[N],rT; bool vis[N]; vector<int> e[N]; set<Pr> st[N];\nvoid dfs1(int u,int Fa){\n\tsz[u]=1; fa[u]=Fa; dep[u]=dep[Fa]+1;\n\tfor(int v:e[u]) if(v!=Fa){\n\t\tdfs1(v,u); sz[u]+=sz[v];\n\t\tif(sz[v]>sz[son[u]]) son[u]=v;\n\t}\n}\nvoid dfs2(int u,int tpp){\n\ttop[u]=tpp; b[ld[u]=++tot]=u;\n\tif(son[u]){\n\t\tdfs2(son[u],tpp);\n\t\tfor(int v:e[u]) if(v!=son[u] && v!=fa[u]) dfs2(v,v);\n\t}\n}\nint LCA(int u,int v){\n\twhile(top[u]!=top[v]){\n\t\tif(dep[top[u]]<dep[top[v]]) swap(u,v);\n\t\tu=fa[top[u]];\n\t} return dep[u]<dep[v]?u:v;\n}\nint K_fa(int u,int kx){\n\tkx=dep[u]-kx; while(dep[top[u]]>kx) u=fa[top[u]];\n\treturn b[ld[u]-dep[u]+kx];\n}\nint gdis(int x,int y) {return dep[x]+dep[y]-2*dep[LCA(x,y)];}\nvoid getrt(int u,int Fa,int S){\n\tsz[u]=1; int mx=0;\n\tfor(int v:e[u]) if(v!=Fa && !vis[v]) {\n\t\tgetrt(v,u,S); sz[u]+=sz[v];\n\t\tmx=max(mx,sz[v]);\n\t} mx=max(mx,S-sz[u]); rT=mx*2<=S?u:rT;\n}\nvoid dfz(int u,int Fa,int S){\n\tgetrt(u,0,S); u=rT; gfa[u]=Fa; vis[u]=1; getrt(u,0,S);\n\tfor(int v:e[u]) if(!vis[v]) dfz(v,u,sz[v]);\n}\nvoid addx(int x,int o){ int u=x;\n\twhile(u){\n\t\tPr bf={gdis(x,u),x};\n\t\tif(o) st[u].insert(bf);\n\t\telse st[u].erase(bf);\n\t\tu=gfa[u];\n\t}\n}\nint qry(int x){\n\tint u=x; Pr z={inf,0};\n\twhile(u){\n\t\tif(st[u].size()){\n\t\t\tPr bf=*st[u].begin(); bf.fi+=gdis(x,u);\n\t\t\tz=min(z,bf);\n\t\t}\n\t\tu=gfa[u];\n\t} return z.nd;\n}\nvoid vmain(){\n\tn=read(); tot=0; for(int i=1;i<=n;i++) son[i]=ind[i]=gfa[i]=vis[i]=0,e[i].clear(),st[i].clear();\n\tfor(int i=2;i<=n;i++) p[i]=read(),e[p[i]].push_back(i),e[i].push_back(p[i]);\n\tdfs1(1,0); dfs2(1,1); dfz(1,0,n); ans=0;\n\tfor(int i=1;i<=n;i++){\n\t\tif(~ans){\n\t\t\taddx(i,1);\n\t\t\tif(i==1) ans=1,ru=rv=1,rds=0;\n\t\t\telse { int x=p[i]; ++ind[x]; ind[i]=1;\n\t\t\t\tif(ind[x]<=3){\n\t\t\t\t\tif(ind[x]==3) addx(x,0);\n\t\t\t\t\tif(gdis(ru,i)>rds) ++rds,rv=i;\n\t\t\t\t\tif(gdis(rv,i)>rds) ++rds,ru=i;\n\t\t\t\t\tint lc=LCA(ru,rv),px=0,z=rds/2; px=(z<=dep[ru]-dep[lc])?K_fa(ru,z):K_fa(rv,rds-z);\n\t\t\t\t\tpx=qry(px); ans=max(gdis(ru,px),gdis(rv,px))+1;\n\t\t\t\t\tif(rds&1){\n\t\t\t\t\t\t++z; px=(z<=dep[ru]-dep[lc])?K_fa(ru,z):K_fa(rv,rds-z); px=qry(px);\n\t\t\t\t\t\tans=min(ans,max(gdis(ru,px),gdis(rv,px))+1);\n\t\t\t\t\t}\n\t\t\t\t} else ans=-1;\n\t\t\t}\n\t\t}\n\t\tprintf(\"%d \",ans);\n\t} puts(\"\");\n}\nint main(){\n\tint T=read(); while(T--) vmain();\n\treturn 0;\n}\n/*\ng++ 1.cpp -o 1 -std=c++20 -Wall -Wl,-stack=999999999\n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 279340157,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Iris's Full Binary Tree\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: Iris likes full binary trees.\n\nLet's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $d$ is a binary tree of depth $d$ with exactly $2^d - 1$ vertices.\n\nIris calls a tree a $d$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $d$. Note that any vertex can be chosen as the root of a full binary tree.\n\nSince performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $d$ satisfying that the tree is $d$-binary. Specifically, if there is no integer $d \\ge 1$ such that the tree is $d$-binary, the binary depth of the tree is $-1$.\n\nIris now has a tree consisting of only vertex $1$. She wants to add $n - 1$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $i$ ($2 \\leq i \\leq n$), she'll give you an integer $p_i$ ($1 \\leq p_i < i$), and add a new edge connecting vertices $i$ and $p_i$.\n\nIris wants to ask you the binary depth of the tree formed by the first $i$ vertices for each $1 \\le i \\le n$. Can you tell her the answer?\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 5 \\cdot 10^5$) — the final size of the tree.\n\nThe second line of each test case contains $n - 1$ integers $p_2, p_3, \\ldots, p_n$ ($1 \\leq p_i < i$) — descriptions of all edges of the tree.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $5 \\cdot 10^5$.\n\nOutput: For each test case output $n$ integers, $i$-th of them representing the binary depth of the tree formed by the first $i$ vertices.\n\nExamples:\ninput:\n7\n3\n1 1\n6\n1 2 3 4 5\n7\n1 1 3 2 5 1\n10\n1 1 2 1 4 2 4 5 8\n10\n1 1 3 1 3 2 2 2 6\n20\n1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12\n25\n1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23\noutput:\n1 2 2 \n1 2 2 3 3 4 \n1 2 2 3 3 4 4 \n1 2 2 3 3 3 4 4 5 5 \n1 2 2 3 3 4 4 4 -1 -1 \n1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 \n1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8\n\nNote: In the first test case, the final tree is shown below:\n\n\n\n * The tree consisting of the vertex $1$ has the binary depth $1$ (the tree itself is a full binary tree of depth $1$). * The tree consisting of the vertices $1$ and $2$ has the binary depth $2$ (we can add the vertex $3$ to make it a full binary tree of depth $2$). * The tree consisting of the vertices $1$, $2$ and $3$ has the binary depth $2$ (the tree itself is a full binary tree of depth $2$).\n\nIn the second test case, the formed full binary tree after adding some vertices to the tree consisting of $n$ vertices is shown below (bolded vertices are added):\n\n\n\nThe depth of the formed full binary tree is $4$.\n\nIn the fifth test case, the final tree is shown below:\n\n\n\nIt can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $-1$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define Pr pair<int,int>\n#define fi first\n#define nd second\n#define psb push_back\nusing namespace std;\ntypedef long long LL;\ninline int read(){\n\tchar ch=getchar();\n\twhile(!isdigit(ch) && ch!='-') ch=getchar();\n\tint x=0,ff=1; if(ch=='-') ff=-1,ch=getchar();\n\twhile(isdigit(ch)) x=(x<<3) + (x<<1) + (ch^48),ch=getchar();\n\treturn x*ff;\n}\nconst int N=5e5+5,inf=1e9;\nint n,p[N],b[N],ld[N],tot,son[N],fa[N],top[N],dep[N],sz[N],ru,rv,rds,ans,ind[N],gfa[N],rT; bool vis[N]; vector<int> e[N]; set<Pr> st[N];\nvoid dfs1(int u,int Fa){\n\tsz[u]=1; fa[u]=Fa; dep[u]=dep[Fa]+1;\n\tfor(int v:e[u]) if(v!=Fa){\n\t\tdfs1(v,u); sz[u]+=sz[v];\n\t\tif(sz[v]>sz[son[u]]) son[u]=v;\n\t}\n}\nvoid dfs2(int u,int tpp){\n\ttop[u]=tpp; b[ld[u]=++tot]=u;\n\tif(son[u]){\n\t\tdfs2(son[u],tpp);\n\t\tfor(int v:e[u]) if(v!=son[u] && v!=fa[u]) dfs2(v,v);\n\t}\n}\nint LCA(int u,int v){\n\twhile(top[u]!=top[v]){\n\t\tif(dep[top[u]]<dep[top[v]]) swap(u,v);\n\t\tu=fa[top[u]];\n\t} return dep[u]<dep[v]?u:v;\n}\nint K_fa(int u,int kx){\n\tkx=dep[u]-kx; while(dep[top[u]]>kx) u=fa[top[u]];\n\treturn b[ld[u]-dep[u]+kx];\n}\nint gdis(int x,int y) {return dep[x]+dep[y]-2*dep[LCA(x,y)];}\nvoid getrt(int u,int Fa,int S){\n\tsz[u]=1; int mx=0;\n\tfor(int v:e[u]) if(v!=Fa && !vis[v]) {\n\t\tgetrt(v,u,S); sz[u]+=sz[v];\n\t\tmx=max(mx,sz[v]);\n\t} mx=max(mx,S-sz[u]); rT=mx*2<=S?u:rT;\n}\nvoid dfz(int u,int Fa,int S){\n\tgetrt(u,0,S); u=rT; gfa[u]=Fa; vis[u]=1; getrt(u,0,S);\n\tfor(int v:e[u]) if(!vis[v]) dfz(v,u,sz[v]);\n}\nvoid addx(int x,int o){ int u=x;\n\twhile(u){\n\t\tPr bf={gdis(x,u),x};\n\t\tif(o) st[u].insert(bf);\n\t\telse st[u].erase(bf);\n\t\tu=gfa[u];\n\t}\n}\nint qry(int x){\n\tint u=x; Pr z={inf,0};\n\twhile(u){\n\t\tif(st[u].size()){\n\t\t\tPr bf=*st[u].begin(); bf.fi+=gdis(x,u);\n\t\t\tz=min(z,bf);\n\t\t}\n\t\tu=gfa[u];\n\t} return z.nd;\n}\nvoid vmain(){\n\tn=read(); tot=0; for(int i=1;i<=n;i++) son[i]=ind[i]=gfa[i]=vis[i]=0,e[i].clear(),st[i].clear();\n\tfor(int i=2;i<=n;i++) p[i]=read(),e[p[i]].push_back(i),e[i].push_back(p[i]);\n\tdfs1(1,0); dfs2(1,1); dfz(1,0,n); ans=0;\n\tfor(int i=1;i<=n;i++){\n\t\tif(~ans){\n\t\t\taddx(i,1);\n\t\t\tif(i==1) ans=1,ru=rv=1,rds=0;\n\t\t\telse { int x=p[i]; ++ind[x]; ind[i]=1;\n\t\t\t\tif(ind[x]<=3){\n\t\t\t\t\tif(ind[x]==3) addx(x,0);\n\t\t\t\t\tif(gdis(ru,i)>rds) ++rds,rv=i;\n\t\t\t\t\tif(gdis(rv,i)>rds) ++rds,ru=i;\n\t\t\t\t\tint lc=LCA(ru,rv),px=0,z=rds/2; px=(z<=dep[ru]-dep[lc])?K_fa(ru,z):K_fa(rv,rds-z);\n\t\t\t\t\tpx=qry(px); ans=max(gdis(ru,px),gdis(rv,px))+1;\n\t\t\t\t\tif(rds&1){\n\t\t\t\t\t\t++z; px=(z<=dep[ru]-dep[lc])?K_fa(ru,z):K_fa(rv,rds-z); px=qry(px);\n\t\t\t\t\t\tans=min(ans,max(gdis(ru,px),gdis(rv,px))+1);\n\t\t\t\t\t}\n\t\t\t\t} else ans=-1;\n\t\t\t}\n\t\t}\n\t\tprintf(\"%d \",ans);\n\t} puts(\"\");\n}\nint main(){\n\tint T=read(); while(T--) vmain();\n\treturn 0;\n}\n/*\ng++ 1.cpp -o 1 -std=c++20 -Wall -Wl,-stack=999999999\n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2006E",
"submission_id": 279473857,
"task_type": "TestCase-Eval-Task2"
} |
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