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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: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\n#include<stdio.h>\nint main()\n{\n\tint t, i, j, flag = 0;\n\tchar a[100] = {0};\n\tscanf(\"%d\", &t);\n\tif (t < 1 || t>100)\n\t\treturn 0;\n\tfor (i = 0; i < t; i++)\n\t{\n\t\tfor (j = 0; j < 100; j++)\n\t\t{\n\t\t\tscanf(\"%c\", &a[j]);\n\t\t\tif (j!=0 && a[j] == '\\n')\n\t\t\t{\n\t\t\t\tflag = j ;\n\t\t\t\tbreak;\n\t\t\t}\n\t\t}\n\t\tfor (j = flag-1 ; j>=0; j--)\n\t\t{\n\t\t\tif ((a[j] == 'p') || (a[j] == 'q') || (a[j] == 'w'))\n\t\t\t{\n\t\t\t\tif (a[j] == 'p')\n\t\t\t\t\tprintf(\"q\");\n\t\t\t\telse if (a[j] == 'q')\n\t\t\t\t\tprintf(\"p\");\n\t\t\t\telse\n\t\t\t\t\tprintf(\"w\");\n\t\t\t}\n\t\t}\n\t\tprintf(\"\\n\");\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
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]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 296698671,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\n#include<stdio.h>\n#include<stdlib.h>\n#include<math.h>\n#include<string.h>\nint main()\n{\n int test;\n scanf(\"%d\", &test);\n for(int t=1; t<=test; t++)\n {\n int i;\n char a[100], b[100];\n scanf(\"%s\", a);\n for(int i=0; a[i]; i++)\n {\n if(a[i]=='q') a[i]='p';\n else if(a[i]=='p') a[i]='q';\n }\n for(i=0; i<strlen(a); i++)\n {\n b[strlen(a)-i-1] = a[i];\n }\n b[i] = '\\0';\n printf(\"%s\\n\", b);\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": "2044B",
"submission_id": 296612561,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\n#include<iostream>\nusing namespace std;\n \nint main(){\n int n;\n cin>>n;\n string store[n];\n string rev[n];\n for (int i = 0; i < n; i++)\n {\n cin>>store[i];\n \n }\n// cout<<store[0][store[0].length()-1];\n \nfor (int i = 0; i < n; i++)\n{\n for (int j = 0; j < store[i].length(); j++)\n {\n rev[i][j]=store[i][store[i].length()-1-j];\n }\n \n}\n \n \n \nfor (int i = 0; i < n; i++)\n{\n for (int j = 0; j < store[i].length(); j++)\n { \n if (rev[i][j]=='p')\n {\n cout<<\"q\";\n }\n else if(rev[i][j]=='q'){\n cout<<\"p\";\n }\n else{\n cout<<rev[i][j];\n }\n \n \n }\n cout<<endl;\n}\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 296827063,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\ndef mirror_character(ch):\n if ch == 'p':\n return 'q'\n elif ch == 'q':\n return 'p'\n else: # ch == 'w'\n return 'w'\n \ndef transform_string(s):\n # Reverse the string\n reversed_s = s[::-1]\n # Replace each character with its mirrored character\n transformed_s = ''.join(mirror_character(ch) for ch in reversed_s)\n return transformed_s\n \ndef main():\n import sys\n input = sys.stdin.read\n data = input().split()\n \n t = int(data[0])\n results = []\n \n for i in range(1, t + 1):\n a = data[i]\n b = transform_string(a)\n results.append(b)\n \n for result in results:\n print(result)\n \n# Example usage with provided sample input\nif __name__ == \"__main__\":\n import sys\n from io import StringIO\n \n sample_input = \"\"\"5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\"\"\"\n \n sys.stdin = StringIO(sample_input)\n main()\n \n \n# aNHMA7qwEjWbiK065O1\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 304202175,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\nt = int(input())\nwhile t != 0:\n a = str(input())\n l = list(a)\n l1=[]\n n1=l.count('p')\n n2=l.count('q')\n if n1==n2:\n print(a)\n else:\n for i in range(len(l)):\n if l[i]=='p':\n l1.append('q')\n elif l[i]=='q':\n l1.append('p')\n else:\n l1.append('w')\n l1.reverse()\n \n for i in l1:\n print(i,end='')\n print()\n \n t -= 1\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 301125784,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\nfor _ in range(int(input())):\n a = list(input())\n for i in range(len(a)):\n if a[i] == 'p':\n a[i] = 'q'\n elif a[i] == 'q':\n a[i] = 'p'\n a.reverse()\n print(''.join(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": "2044B",
"submission_id": 298853785,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\n#include <iostream>\n \nusing namespace std;\n \nint main()\n{\n int t;\n cin>>t;\n while(t--){\n string s;\n cin>>s;\n int sl=s.size();\n if((s[0]=='q' && s[sl-1]=='q') || (s[0]=='p' && s[sl-1]=='p')){\n for(int i=0;i<sl;i++){\n if(s[i]=='q') s[i]='p';\n else if(s[i]=='p') s[i]='q';\n }\n }\n else if((s[0]=='w' && s[sl-1]=='p' || s[sl-1]=='q') ||\n (s[sl-1]=='w' && s[0]=='p' || s[0]=='q')){\n for(int i=0;i<(sl/2);i++){\n char temp=s[i];\n s[i]=s[sl-i-1];\n s[sl-i-1]=temp;\n if(s[i]=='p') s[i]='q';\n else if(s[i]=='q') s[i]='p';\n if(sl%2!=0 && s[(sl/2)]=='p') s[(sl/2)]='q';\n else if(sl%2!=0 && s[(sl/2)]=='q') s[(sl/2)]='p';\n if(s[sl-i-1]=='p') s[sl-i-1]='q';\n else if(s[sl-i-1]=='q') s[sl-i-1]='p';\n }\n }\n \n cout<<s<<endl;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 300428363,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <string>\n#include <algorithm>\nusing namespace std;\nint main(){\n int t;\n cin>>t;\n vector<string>b;\n while(t--){\n char p,q;\n string a;\n cin>>a;\n for(int i =0;i<a.size();i++){\n reverse(a.begin(),a.end());\n if(a[i]=='p'){\n a[i]='q';\n }\n else if(a[i]=='q'){\n a[i]='p';\n }\n else{\n continue;\n }\n }\n b.push_back(a);\n }\n for(string i:b){\n cout<<i<<endl;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 301648036,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \nint main() {\n int t;\n cin>>t;\n while(t--){\n string a,b;\n cin>>a;\n int k=a.size();\n for(int i=0;i<k;i++){\n if(a[k-1-i]=='p'){\n b[i]='q';\n }else if(a[k-1-i]=='q'){\n b[i]='p';\n }else{\n b[i]='w';\n }\n }\n for(int i=0;i<k;i++){\n cout<<b[i];\n }\n cout<<endl;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 296704326,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nint main(){\n int t;\n cin>>t;\n while(t--){\n string s;\n cin>>s;\n string m;\n for(int i=0;i<s.size();i++){\n m[i]=s[s.size()-1-i];\n }\n for(int i=0;i<s.size();i++){\n if(m[i]=='p') m[i]='q';\n else if(m[i]=='q') m[i]='p';\n }\n for(int i=0;i<s.size();i++){\n cout<<m[i];\n }\n cout<<endl;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 296634002,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\ndef ship_string(a):\n b=a[::-1]\n c=''\n for i in b:\n if i=='p':\n c+='q'\n elif i=='q':\n c+='p'\n elif i=='w':\n c+='w'\n return c \n \n \nt=int(input())\nl=[] \nif t>=1 and t<=99:\n for i in range(t):\n n=input()\n if 1<=len(n)<=100:\n l.append(ship_string(n))\nfor j in l: \n print(j)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 296708613,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\nt = int(input())\n \nans = []\n \nfor _ in range(t):\n s = list(input())\n for i in range(len(s)):\n s = s[::-1]\n if s[i] == 'q':\n s[i] = 'p'\n elif s[i] == 'p':\n s[i] = 'q'\n \n ans.append(s)\n \nfor line in ans:\n print(*line,sep='')\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 303287453,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class cpcode {\n public static void main(String[] args) {\n \n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt(); \n \n for (int j = 0; j < t; j++) {\n String s = sc.next();\n \n char[] arr = s.toCharArray();\n \n for(int i=0;i<=arr.length/2;i++){\n char te;\n if(arr[i]=='p'){\n te='q';\n }\n else if(arr[i]=='q'){\n te='p';\n }\n else{\n te=arr[i];\n }\n if(arr[arr.length-1-i]=='p'){\n arr[i]='q';\n }\n else if(arr[arr.length-1-i]=='q'){\n arr[i]='p';\n }\n else{\n arr[i]=arr[arr.length-1-i];\n }\n arr[arr.length-1-i]=te;\n }\n \n String string = new String(arr); \n \n \n System.out.println(string);\n }\n \n sc.close();\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": "2044B",
"submission_id": 301548592,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class Normal_Prob {\n public static void main(String[] args) {\n Scanner sc=new Scanner(System.in);\n int t=sc.nextInt();\n \n while(t>0){\n \n String str = sc.next();\n int n=str.length();\n StringBuilder stri = new StringBuilder(str);\n for(int i=0;i<n;i++){\n if(stri.charAt(i)=='p')\n stri.setCharAt(i,'q');\n else if(stri.charAt(i)=='q')\n stri.setCharAt(i,'p');\n stri.reverse();\n \n \n }\n System.out.println(stri);\n \n \n t--;\n } sc.close();\n }}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 296653756,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <string>\nusing namespace std;\nvoid vol(){\n string s,a;\n cin>>s;\n for(int i=0;i<(int)s.size();i++){\n a[i]=s[(int)s.size()-1-i];\n if(a[i]=='p') a[i]='q';\n else if(a[i]=='q') a[i]='p';\n }\n for(int i=0;i<(int)s.size();i++) cout<<a[i];\n cout<<endl;\n}\nint main(){\n int t;\n cin>>t;\n while(t--) vol();\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": "2044B",
"submission_id": 296678358,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nint main() {\n int t;\n cin >> t;\n \n while (t--) {\n string a;\n cin >> a;\n int c1 = 0, c2 = 0;\n for (char c : a) {\n if (c == 'p') c1++;\n if (c == 'q') c2++;\n }\n if (c1 == c2) {\n cout << a << endl;\n }\n else {\n reverse(a.begin(), a.end());\n if (c1 == 0) {\n for (char &c : a) {\n if (c == 'q') {\n c = 'p';\n }\n }\n }\n else if (c2 == 0) {\n for (char &c : a) {\n if (c == 'p') {\n c = 'q';\n }\n }\n }\n else {\n for (char &c : a) {\n if (c == 'p') {\n c = 'q';\n }\n else if (c == 'q') {\n c = 'p';\n }\n }\n }\n cout << a << endl;\n }\n }}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2044B",
"submission_id": 296734294,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\nimport java.util.*;\npublic class Normal\n{\n public static String solution(Scanner sc)\n {\n String s=sc.next();\n ArrayList<Character> res = new ArrayList<>();\n \n int pcount=0,qcount=0;\n for(int i=0;i<s.length();i++)\n {\n if(s.charAt(i)== 'p') pcount++;\n else if(s.charAt(i)=='q') qcount++; \n }\n \n if(pcount==qcount) return s;\n else\n {\n for(int i=s.length()-1;i>=0;i--)\n {\n if(s.charAt(i)=='p') res.add('q');\n else if(s.charAt(i)=='q') res.add('p');\n else res.add(s.charAt(i));\n }\n \n StringBuilder sb = new StringBuilder();\n for (char c : res) {\n sb.append(c);\n }\n return sb.toString();\n \n }\n }\n public static void main(String args[])\n {\n Scanner sc=new Scanner(System.in);\n int t=sc.nextInt();\n while(t-- >0)\n {\n System.out.println(solution(sc));\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": "2044B",
"submission_id": 300812079,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\nimport java.util.*;\npublic class Normal\n{\n public static String solution(Scanner sc)\n {\n String s=sc.nextLine();\n ArrayList<Character> res = new ArrayList<>();\n \n int pcount=0,qcount=0;\n for(int i=0;i<s.length();i++)\n {\n if(s.charAt(i)== 'p') pcount++;\n else if(s.charAt(i)=='q') qcount++; \n }\n \n if(pcount==qcount) return s;\n else\n {\n for(int i=s.length()-1;i>=0;i--)\n {\n if(s.charAt(i)=='p') res.add('q');\n else if(s.charAt(i)=='q') res.add('p');\n else res.add(s.charAt(i));\n }\n \n StringBuilder sb = new StringBuilder();\n for (char c : res) {\n sb.append(c);\n }\n return sb.toString();\n \n }\n }\n public static void main(String args[])\n {\n Scanner sc=new Scanner(System.in);\n int t=sc.nextInt();\n System.out.println();\n while(t-- >=0)\n {\n System.out.println(solution(sc));\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": "2044B",
"submission_id": 300813895,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nint main()\n{\n int n;\n char a[110][110];\n char b[110][110];\n cin>>n;\n for(int i = 0; i < n; i ++)\n {\n cin>>a[i];\n }\n for(int i = 0; i < n; i ++)\n {\n int temp = strlen(a[i]);\n for(int j = temp-1; j >= 0; j --)\n {\n if(a[i][j] == 'w') b[i][temp - j - 1] = 'w';\n else if(a[i][j] == 'p') b[i][temp - j - 1] = 'q';\n else if(a[i][j] == 'q') b[i][temp - j - 1] = 'p';\n }\n }\n for(int i = 0; i < n; i ++)\n {\n cout<<b[i]<<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": "2044B",
"submission_id": 296630994,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Normal Problem\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.\n\nShip gives you string $$$a$$$. Your job is to find and output $$$b$$$.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\leq t \\leq 100$$$) — the number of test cases.\n\nThe only line of each test case contains a string $$$a$$$ ($$$1 \\leq |a| \\leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.\n\nOutput: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.\n\nExamples:\ninput:\n5\nqwq\nppppp\npppwwwqqq\nwqpqwpqwwqp\npqpqpqpq\noutput:\npwp\nqqqqq\npppwwwqqq\nqpwwpqwpqpw\npqpqpqpq\n\nNote: \n\nBuggy Code:\nimport java.util.*;\npublic class Normal\n{\n public static String solution(Scanner sc)\n {\n String s=sc.nextLine();\n ArrayList<Character> res = new ArrayList<>();\n \n int pcount=0,qcount=0;\n for(int i=0;i<s.length();i++)\n {\n if(s.charAt(i)== 'p') pcount++;\n else if(s.charAt(i)=='q') qcount++; \n }\n \n if(pcount==qcount) return s;\n else\n {\n for(int i=s.length()-1;i>=0;i--)\n {\n if(s.charAt(i)=='p') res.add('q');\n else if(s.charAt(i)=='q') res.add('p');\n else res.add(s.charAt(i));\n }\n \n StringBuilder sb = new StringBuilder();\n for (char c : res) {\n sb.append(c);\n }\n return sb.toString();\n \n }\n }\n public static void main(String args[])\n {\n Scanner sc=new Scanner(System.in);\n int t=sc.nextInt();\n while(t-- >=0)\n {\n System.out.println(solution(sc));\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": "2044B",
"submission_id": 300809683,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\n// semenInRussia 2024\n#include <cstdint>\n#include <iostream>\n#include <vector>\nusing namespace std;\n \n#pragma GCC optimize(\"O3,unroll-loops\")\n#pragma GCC target(\"avx2,popcnt,lzcnt,abm,bmi,fma,bmi2,tune=native\")\n \n#define isz(x) int(x.size())\n \nconst int inf = 1e9 + 5;\n \nvoid solve() {\n string a, b, c;\n cin >> a >> b >> c;\n \n // f[i][j] - minimum errors that a[0..i) located on c[0..j]\n // answer = f[isz(a)][isz(c)]\n \n vector<vector<int>> f(isz(a) + 3, vector<int>(isz(c) + 3, 0));\n \n // f[i][j] - is minimum errors to do\n // that on c[0..j] j of a\n \n f[0][0] = 0;\n int t = 0;\n for (int j = 0; j < isz(c); j++) {\n bool bad = (t >= isz(b)) || (c[t] != b[j]);\n f[0][j + 1] = f[0][j] + bad;\n if (!bad) {\n t++;\n }\n }\n \n for (int i = 0; i < isz(a); i++) {\n for (int j = 0; j < isz(c); j++) {\n auto &ans = f[i + 1][j + 1];\n ans = +inf;\n \n ans = min(ans, f[i][j] + (a[i] != c[j]));\n \n int k = j - i - 1;\n if (k < 0) {\n continue;\n }\n \n ans = min(ans, f[i + 1][j] + (b[k] != c[j]));\n \n // i = 1, a = 2\n // j = 2, c = 3\n }\n }\n \n cout << f[isz(a)][isz(c)] << \"\\n\";\n}\n \nint main() {\n cin.tie(nullptr)->sync_with_stdio(0);\n int tt;\n cin >> tt;\n while (tt--) {\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": "2050E",
"submission_id": 295084619,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\ntypedef long long int ll;\n#define int long long\n \nvoid solve()\n{\n string a, b, c;\n cin >> a >> b >> c;\n vector<vector<int> > dp(a.size()+2, vector<int>(b.size()+2, INT_MAX));\n dp[0][1] = 0;\n dp[1][0] = 0;\n for(int i=1; i<=a.size()+1; i++){\n for(int j=1; j<=b.size()+1; j++){\n \n if(c[i+j-2]==a[i-1]){\n dp[i][j] = min(dp[i][j],dp[i-1][j]);\n }\n else{\n dp[i][j] = min(dp[i][j],dp[i-1][j]+1);\n }\n if(c[i+j-2]==b[j-2]){\n dp[i][j] = min(dp[i][j],dp[i][j-1]);\n }\n else{\n dp[i][j] = min(dp[i][j],dp[i][j-1]+1);\n }\n \n // cout << dp[i][j] << \" \";\n \n }\n // cout << endl;\n \n }\n cout << dp[a.size()+1][b.size()+1];\n return;\n}\n \nint32_t main()\n{\n int t;\n cin >> t;\n while (t--){\n solve();\n cout << endl;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050E",
"submission_id": 295055031,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nint T,dp[2000][2000],n,m,x1,x2,ans[5000];\nstring a,b,c;\nint main()\n{\n\tscanf(\"%d\",&T);\n\tgetchar();\n\twhile(T--)\n\t{\n\t\tgetline(cin,a);\n\t\tgetline(cin,b);\n\t\tgetline(cin,c);\n\t\tn=a.length();\n\t\tm=b.length();\n\t\tdp[0][0]=0;\n\t\tfor (int i=1;i<=n;i++)\n\t\t{\n\t\t\tdp[i][0]=dp[i-1][0];\n\t\t\tif(a[i-1]!=c[i-1])\n\t\t\tdp[i][0]++;\n\t\t}\n\t\tfor (int i=1;i<=m;i++)\n\t\t{\n\t\t\tdp[0][i]=dp[0][i-1];\n\t\t\tif(b[i-1]!=c[i-1])\n\t\t\tdp[0][i]++;\n\t\t}\n\t\tfor (int i=1;i<=n;i++)\n\t\t{\n\t\t\tfor (int j=1;j<=m;j++)\n\t\t\tdp[i][j]=n+m;\n\t\t}\n\t\tfor (int i=1;i<=n+m;i++)\n\t\t{\n\t\t\tfor (int j=1;j<=min(n,i);j++)\n\t\t\t{\n\t\t\t\tx1=dp[j-1][i-j];\n\t\t\t\tif(a[j-1]!=c[i-1])\n\t\t\t\tx1++;\n\t\t\t\tx2=dp[j][i-j-1];\n\t\t\t\tif(b[i-j-1]!=c[i-1])\n\t\t\t\tx2++;\n\t\t\t\tdp[j][i-j]=min(x1,x2);\n\t\t\t}\n\t\t}\n\t\tcout<<dp[n][m]<<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": "2050E",
"submission_id": 295058650,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\ndef f(a, b, c, i, j, dp):\n if i == 0 and j == 0:\n return 0\n if i == 0:\n return sum(1 for k in range(j) if b[k] != c[k])\n if j == 0:\n return sum(1 for k in range(i) if a[k] != c[k])\n if dp[i][j] != -1:\n return dp[i][j]\n indx = i + j - 1\n ans = float(\"inf\")\n if i > 0 and c[indx] == a[i - 1]:\n ans = min(ans, f(a, b, c, i - 1, j, dp))\n if j > 0 and c[indx] == b[j - 1]:\n ans = min(ans, f(a, b, c, i, j - 1, dp))\n if i > 0 and c[indx] != a[i - 1]:\n ans = min(ans, f(a, b, c, i - 1, j, dp) + 1)\n if j > 0 and c[indx] != b[j - 1]:\n ans = min(ans, f(a, b, c, i, j - 1, dp) + 1)\n dp[i][j] = ans\n return ans\nn = int(input())\nfor _ in range(n):\n a = input()\n b = input()\n c = input()\n i = len(a)\n j = len(b)\n dp = [[-1] * (1000) for _ in range(1000)]\n res = f(a, b, c, i, j, dp)\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": "2050E",
"submission_id": 297773015,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\nfor _ in range(int(input())):\n a = input()\n b = input()\n c = input()\n changes = 1\n while changes:\n changes = 0\n if len(a) and c[0]==a[0]:\n a = a[1:]\n c = c[1:]\n changes += 1\n elif len(b) and c[0]==b[0]:\n b = b[1:]\n c = c[1:]\n changes += 1\n elif len(a) and c[-1]==a[-1]:\n c = c[:-1]\n a = a[:-1]\n changes += 1\n elif len(b) and c[-1]==b[-1]:\n b = b[:-1]\n c = c[:-1]\n changes += 1\n k = min(len(a),len(b))\n ans = k\n for i in range(k,len(c)):\n if len(b) > len(a):\n if c[i]!=b[i-k]:\n ans += 1\n else:\n if c[i]!=a[i-k]:\n 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": "2050E",
"submission_id": 295068261,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\ndef solve(a,b,c,ind1,ind2,dp):\n if(ind1>=len(a) and ind2>=len(b)):\n return 0\n takea=float(\"inf\")\n takeb=float(\"inf\")\n if(dp[ind1][ind2]!=-1):\n return dp[ind1][ind2]\n if(ind1<len(a)):\n takea=int(c[ind1+ind2]!=a[ind1])+solve(a,b,c,ind1+1,ind2,dp)\n if(ind2<len(b)):\n takeb=int(c[ind1+ind2]!=b[ind2])+solve(a,b,c,ind1,ind2+1,dp)\n dp[ind1][ind2]=min(takea,takeb)\n return min(takea,takeb)\nfor _ in range(int(input())):\n a=input()\n b=input()\n c=input()\n dp=[[-1 for _ in range(len(b)+1343)] for _ in range(len(a)+1128)]\n ans=solve(a,b,c,0,0,dp)\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": "2050E",
"submission_id": 295117930,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class three {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n sc.nextLine(); \n \n while (t-- > 0) {\n String a = sc.nextLine();\n String b = sc.nextLine();\n String c = sc.nextLine();\n \n int n = a.length();\n int m = c.length();\n int dp[][] = new int[n + 1][m + 1];\n for(int i=0;i<=n;i++){\n Arrays.fill(dp[i],100000000);\n }\n for (int i = 0; i <= m; i++) {\n dp[0][i] = i; \n }\n \n for (int i = 1; i <= n; i++) {\n for (int k = 1; k <= m; k++) {\n int g1 = Integer.MAX_VALUE; \n int g2 = Integer.MAX_VALUE; \n \n if (a.charAt(i - 1) == c.charAt(k - 1)) {\n g1 = dp[i - 1][k - 1];\n } else {\n g1 = 1 + dp[i - 1][k - 1]; \n }\n \n int j = k - i; \n if (j >= 1 && j <= b.length()) { \n if (b.charAt(j - 1) == c.charAt(k - 1)) {\n g2 = dp[i][k - 1]; \n } else {\n g2 = 1 + dp[i][k - 1]; \n }\n } \n \n dp[i][k] = Math.min(dp[i][k],Math.min(g1, g2));\n }\n }\n \n System.out.println(dp[n][m]);\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": "2050E",
"submission_id": 296951801,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\nimport static java.lang.Math.max;\nimport static java.lang.Math.min;\nimport static java.lang.Math.abs;\n \nimport java.io.BufferedInputStream;\nimport java.io.File;\nimport java.io.PrintWriter;\nimport java.io.FileInputStream;\nimport java.io.InputStream;\nimport java.util.ArrayList;\nimport java.util.Arrays;\nimport java.util.List;\nimport java.util.Objects;\nimport java.util.Map;\nimport java.util.Random;\nimport java.util.TreeMap;\nimport java.util.TreeSet;\n \npublic class E_Three_Strings {\n static final int MOD = (int) 1e9 + 7;\n static final FastScanner in = new FastScanner();\n static final PrintWriter out = new PrintWriter(System.out);\n static int n;\n static int m;\n static char[] a, b, c;\n static final int INF = (int)1e9;\n \n public static void main(String[] args) {\n \n int T = in.nextInt();\n \n while (T-- > 0) {\n \n a = in.next().toCharArray();\n b = in.next().toCharArray();\n c = in.next().toCharArray();\n \n n = a.length;\n m = b.length;\n \n int[][] dp = new int[n + m + 1][n + 1];\n \n for(int i = 0; i <= n + m; ++i) {\n Arrays.fill(dp[i], INF);\n }\n \n dp[0][0] = 0;\n \n for(int t = 1; t <= n; ++t) {\n dp[t][0] = dp[t - 1][0] + (b[t - 1] != c[t - 1] ? 1 : 0);\n }\n \n for(int t = 1; t <= n + m; ++t) {\n for(int j = 1; j <= n; ++j) {\n dp[t][j] = min(dp[t][j], dp[t - 1][j - 1] + (a[j - 1] != c[t - 1] ? 1 : 0));\n if(t - j - 1 >= 0 && t - j - 1 < m) {\n dp[t][j] = min(dp[t][j], dp[t - 1][j] + (b[t - j - 1] != c[t - 1] ? 1 : 0));\n }\n }\n }\n \n \n out.println(dp[n + m][n]);\n \n \n }\n \n out.close();\n \n }\n \n // Remember to initialize the whole adj list\n // with empty arraylist.\n // static void dfs(int node) {\n // visited[node] = true;\n // Just after entering the node.\n \n // for (int child : g[node]) {\n // Take action on each child\n // if (!visited[child])\n // dfs(child);\n // }\n \n // Take action after exiting all the\n // childs\n \n // }\n \n static long binPow(long a, long n, long mod) {\n long res = 1L;\n \n while (n > 0L) {\n if ((n & 1L) == 1L)\n res = ((a % mod) * (res % mod)) % mod;\n \n a = ((a % mod) * (a % mod)) % mod;\n n >>= 1;\n }\n \n return res;\n }\n \n // Strictly lesser than target if less than equal to target < := <=\n static int lowerBound(long[] a, int l, int r, long target) {\n while (l <= r) {\n int mid = (r - l) / 2 + l;\n \n if (a[mid] < target) {\n l = mid + 1;\n } else {\n r = mid - 1;\n }\n }\n return r;\n }\n \n // Strictly greater than target if greater than equal to target <= := <\n static int upperBound(long[] a, int l, int r, long target) {\n while (l <= r) {\n int mid = (r - l) / 2 + l;\n \n if (a[mid] <= target) {\n l = mid + 1;\n } else {\n r = mid - 1;\n }\n }\n return l;\n }\n \n // Strictly greater than target if greater than equal to target <= := <\n static int upperBound(int[] a, int l, int r, int target) {\n while (l <= r) {\n int mid = (r - l) / 2 + l;\n \n if (a[mid] <= target) {\n l = mid + 1;\n } else {\n r = mid - 1;\n }\n }\n return l;\n }\n \n // Strictly lesser than target if less than equal to target < := <=\n static int lowerBound(int[] a, int l, int r, int target) {\n while (l <= r) {\n int mid = (r - l) / 2 + l;\n \n if (a[mid] < target) {\n l = mid + 1;\n } else {\n r = mid - 1;\n }\n }\n return r;\n }\n \n static void push(Map<Long, Long> mp, long k, long v) {\n if (mp.containsKey(k)) {\n mp.put(k, mp.get(k) + v);\n } else {\n mp.put(k, v);\n }\n }\n \n static void pull(Map<Long, Long> mp, long k, long v) {\n long x = mp.get(k);\n \n if (v >= x)\n mp.remove(k);\n else\n mp.put(k, x - v);\n }\n \n static boolean isPrime(long n) {\n if (n < 2)\n return false;\n if (n == 2 || n == 3)\n return true;\n if (n % 2 == 0 || n % 3 == 0)\n return false;\n \n for (long i = 6; i * i <= n; i += 6) {\n if ((n % (i - 1) == 0) || (n % (i + 1) == 0)) {\n return false;\n }\n }\n \n return true;\n }\n \n static long gcd(long a, long b) {\n return b == 0L ? a : gcd(b, a % b);\n }\n \n static void reverse(int[] a, int l, int r) {\n while (l < r)\n swap(a, l++, r--);\n }\n \n static void reverse(long[] a, int l, int r) {\n while (l < r)\n swap(a, l++, r--);\n }\n \n static void swap(long[] a, int l, int r) {\n long temp = a[l];\n a[l] = a[r];\n a[r] = temp;\n }\n \n static void swap(int[] a, int l, int r) {\n int temp = a[l];\n a[l] = a[r];\n a[r] = temp;\n }\n \n static void ruffleSort(long[] a, int l, int r) {\n shuffleArray(a, l, r);\n Arrays.sort(a, l, r + 1);\n }\n \n static void ruffleSort(int[] a, int l, int r) {\n shuffleArray(a, l, r);\n Arrays.sort(a, l, r + 1);\n }\n \n static void shuffleArray(int[] arr, int l, int r) {\n Random rnd = new Random();\n for (int i = l; i <= r; ++i) {\n int tmp = arr[i];\n int randomPos = i + rnd.nextInt(r - i + 1);\n arr[i] = arr[randomPos];\n arr[randomPos] = tmp;\n }\n }\n \n static void shuffleArray(long[] arr, int l, int r) {\n Random rnd = new Random();\n for (int i = l; i <= r; ++i) {\n long tmp = arr[i];\n int randomPos = i + rnd.nextInt(r - i + 1);\n arr[i] = arr[randomPos];\n arr[randomPos] = tmp;\n }\n }\n \n static long[] readArrayL(int n) {\n long[] a = new long[n + 1];\n \n for (int i = 1; i <= n; ++i) {\n a[i] = in.nextLong();\n }\n \n return a;\n }\n \n static ArrayList<Integer> readList(int n) {\n ArrayList<Integer> list = new ArrayList<>();\n \n for (int i = 0; i < n; i++) {\n list.add(in.nextInt());\n }\n \n return list;\n }\n \n static int[] readArray(int n) {\n int[] arr = new int[n + 1];\n \n for (int i = 1; i <= n; i++) {\n arr[i] = in.nextInt();\n }\n \n return arr;\n }\n \n static class FastScanner {\n // I don't understand how this works lmao\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 InputStream 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 nextChar() {\n while (bId == size) {\n try {\n size = in.read(buf);\n } catch (Exception e) {\n return NC;\n }\n if (size == -1)\n 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 long nextLong() {\n cnt = 1;\n boolean neg = false;\n if (c == NC)\n c = nextChar();\n for (; (c < '0' || c > '9'); c = nextChar()) {\n if (c == '-')\n neg = true;\n }\n long res = 0;\n for (; c >= '0' && c <= '9'; c = nextChar()) {\n res = (res << 3) + (res << 1) + c - '0';\n cnt *= 10;\n }\n return neg ? -res : res;\n }\n \n public double nextDouble() {\n double cur = nextLong();\n return c != '.' ? cur : cur + nextLong() / cnt;\n }\n \n public String next() {\n StringBuilder res = new StringBuilder();\n while (c <= 32)\n c = nextChar();\n while (c > 32) {\n res.append(c);\n c = nextChar();\n }\n return res.toString();\n }\n \n public String nextLine() {\n StringBuilder res = new StringBuilder();\n while (c <= 32)\n c = nextChar();\n while (c != '\\n') {\n res.append(c);\n c = nextChar();\n }\n return res.toString();\n }\n \n public boolean hasNext() {\n if (c > 32)\n return true;\n while (true) {\n c = nextChar();\n if (c == NC)\n return false;\n else if (c > 32)\n return true;\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": "2050E",
"submission_id": 295314279,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\n#include <algorithm>\n#include <ctime>\n \nusing namespace std;\n#define ull unsigned long long\n#define cy cout << \"YES\\n\"\n#define cn cout << \"NO\\n\"\n#define new cout << endl\n#define ll long long\n#define pb push_back\n#define vll vector<long long>\n#define INF (long long)1e9 \n#define MOD 1000000007\n#define mp make_pair\n#define F first\n#define S second\n#define f(i, a, b) for(long long i = a; i < b; i++) // Loop from a to b\nlong long gcd(long long a, long long b) {\n while (b != 0) {\n long long temp = b;\n b = a % b;\n a = temp;\n }\n return a;\n}\nlong long lcm(long long a, long long b) {\n return (a * b) /gcd(a, b);\n}\nlong long factorial(int n) {\n long long result = 1;\n for (int i = 1; i <= n; ++i) {\n result *= i;\n }\n return result;\n}\n \nlong long combination(int n, int r) {\n if (r > n) return 0;\n if (r == 0 || r == n) return 1; \n \n \n if (r > n - r) r = n - r;\n \n long long result = 1;\n \n for (int i = 0; i < r; ++i) {\n result *= (n - i);\n result /= (i + 1); \n }\n \n return result;\n}\n \nbool damn(const pair<int, int>& a, const pair<int, int>& b) { \n return a.first<b.first;\n \n};\n \n \n \n \nvector<long long> sieve_of_eratosthenes(long long limit) {\n vector<bool> prime(limit + 1, true);\n prime[0] = prime[1] = false;\n \n for (long long num = 2; num * num <= limit; ++num) {\n if (prime[num]) {\n for (long long multiple = num * num; multiple <= limit; multiple += num) {\n prime[multiple] = false;\n }\n }\n }\n \n vector<long long> primes;\n for (long long i = 2; i <= limit; ++i) {\n if (prime[i]) {\n primes.push_back(i);\n }\n }\n \n return primes;\n};\nvoid merge(int arr[], int left, int right) {\n if (left >= right) return;\n int mid = left + (right - left) / 2;\n merge(arr, left, mid);\n merge(arr, mid + 1, right);\n \n int n1 = mid - left + 1;\n int n2 = right - mid;\n \n int L[n1], R[n2];\n for (int i = 0; i < n1; i++) L[i] = arr[left + i];\n for (int i = 0; i < n2; i++) R[i] = arr[mid + 1 + i];\n \n int i = 0, j = 0, k = left;\n while (i < n1 && j < n2) {\n if (L[i] <= R[j]) arr[k++] = L[i++];\n else arr[k++] = R[j++];\n }\n \n while (i < n1) arr[k++] = L[i++];\n while (j < n2) arr[k++] = R[j++];\n};\n \nvoid mergeSort(int arr[], int left, int right) {\n if (left < right) {\n int mid = left + (right - left) / 2;\n mergeSort(arr, left, mid);\n mergeSort(arr, mid + 1, right);\n merge(arr, left, right);\n }\n};\nstruct item{\n ll m,c;\n};\nint count_trailing_zeros(ll n) {\n int count = 0;\n \n \n while ((n & 1) == 0 && n != 0) {\n count++;\n n >>= 1; \n }\n \n return count;\n}\n \n \nint popcount(ll n) {\n int count = 0;\n \n \n while (n) {\n \n n = n & (n - 1); \n count++; \n }\n \n return count;\n}\n \n \n \nstruct segTree {\n ll size;\n vll sums;\n \n void init(int n) {\n size = 1;\n while (size < n) {\n size *= 2;\n }\n sums.assign(2 * size, 0LL);\n }\n \n void set(ll i, ll v, ll x, ll lx, ll rx) {\n if (rx - lx == 1) {\n sums[x] = v;\n return;\n }\n ll m = (lx + rx) / 2;\n if (i < m) {\n set(i, v, 2 * x + 1, lx, m);\n } \n else {\n set(i, v, 2 * x + 2, m, rx);\n }\n sums[x] = gcd(sums[2 * x + 1] , sums[2 * x + 2]);\n }\n \n void set(ll i, ll v) {\n set(i, v, 0, 0, size);\n }\n \n ll sum(ll l, ll r, ll x, ll lx, ll rx) {\n if (lx >= r || l >= rx) return 1e9;\n if (lx >= l && rx <= r) return sums[x];\n ll m = (lx + rx) / 2;\n ll s1 = sum(l, r, 2 * x + 1, lx, m);\n ll s2 = sum(l, r, 2 * x + 2, m, rx);\n return gcd(s1 , s2);\n }\n \n ll sum(ll l, ll r) {\n return sum(l, r, 0, 0, size);\n }\n};\nbool isPerfectPowerOf3(int n) {\n if (n <= 0) {\n return false;\n }\n while (n % 3 == 0) {\n n /= 3;\n }\n \n return n == 1;\n}\n \nint modInverse(int a, int m) {\n a = a % m;\n for (int x = 1; x < m; x++) {\n if ((a * x) % m == 1) {\n return x;\n }\n }\n return -1;\n}\n \nint chineseRemainderTheorem(const vector<int>& a, const vector<int>& n) {\n int k = a.size();\n int N = 1;\n for (int i = 0; i < k; i++) {\n N *= n[i];\n }\n \n int result = 0;\n for (int i = 0; i < k; i++) {\n int Ni = N / n[i];\n int yi = modInverse(Ni, n[i]);\n if (yi == -1) {\n return -1;\n }\n result += a[i] * Ni * yi;\n }\n \n return result % N;\n}\nll dp[1005][1005];\nll sumi(string str1,string str2,string str3,ll ptr1,ll ptr2,ll count){\n if(ptr1==str1.length() && ptr2==str2.length()){\n return dp[ptr1][ptr2]=count;\n }\n if(count>=dp[ptr1][ptr2]) return dp[ptr1][ptr2];\n if(ptr1==str1.length()){\n if(str2[ptr2]!=str3[ptr2+ptr1]) return dp[ptr1][ptr2]=sumi(str1,str2,str3,ptr1,ptr2+1,count+1);\n else return dp[ptr1][ptr2]=sumi(str1,str2,str3,ptr1,ptr2+1,count);\n }\n if(ptr2==str2.length()){\n if(str1[ptr1]!=str3[ptr2+ptr1]) return dp[ptr1][ptr2]=sumi(str1,str2,str3,ptr1+1,ptr2,count+1);\n else return dp[ptr1][ptr2]=sumi(str1,str2,str3,ptr1+1,ptr2,count);\n }\n //if(dp[ptr1][ptr2]!=-1) return dp[ptr1][ptr2];\n if(str1[ptr1]==str3[ptr1+ptr2] && str2[ptr2]!=str3[ptr1+ptr2] ) return dp[ptr1][ptr2]=min(sumi(str1,str2,str3,ptr1+1,ptr2,count),sumi(str1,str2,str3,ptr1,ptr2+1,count+1));\n else if(str1[ptr1]!=str3[ptr1+ptr2] && str2[ptr2]==str3[ptr1+ptr2]) return dp[ptr1][ptr2]=min(sumi(str1,str2,str3,ptr1,ptr2+1,count),sumi(str1,str2,str3,ptr1+1,ptr2,count+1));\n else if(str1[ptr1]==str3[ptr1+ptr2] && str2[ptr2]==str3[ptr1+ptr2]) return dp[ptr1][ptr2]=min(sumi(str1,str2,str3,ptr1,ptr2+1,count),sumi(str1,str2,str3,ptr1+1,ptr2,count));\n else return dp[ptr1][ptr2]=min(sumi(str1,str2,str3,ptr1,ptr2+1,count+1),sumi(str1,str2,str3,ptr1+1,ptr2,count+1));\n}\n \n \nint main() {\n \n #ifndef ONLINE_JUDGE\n freopen(\"input.txt\", \"r\", stdin);\n freopen(\"output.txt\", \"w\", stdout);\n #endif\n ll test_cases;\n cin>>test_cases;\n auto start = chrono::high_resolution_clock::now();\n //vll vec=sieve_of_eratosthenes(10000);\n f(j,0,test_cases){\n memset(dp,10000,sizeof(dp));\n ll n,m,l1,l2,r1,r2,k,x=0,maxi2=0,maxi=0;ll maxi_pos=0;\n bool change=false, sorted=true;\n ll sum=0;ll sum1=0;string str;\n //cout<<dp[0][0]<<endl;\n string str1, str2,str3;\n cin>>str1;\n cin>>str2;\n cin>>str3;\n \n ll pointer_1=0;\n ll pointer_2=0;\n ll count=0;\n \n cout<<sumi(str1,str2,str3,0,0,0)<<endl;\n \n \n } \n \n \n \n auto end = chrono::high_resolution_clock::now();\n chrono::duration<double> duration = end - start;\n //cout << \"Time taken: \" << duration.count() << \" seconds\" << endl;\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050E",
"submission_id": 306536947,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ll long long\nconst ll mod = 1e9 + 7;\nusing namespace std;\n \nll f(ll ind1, ll ind2, ll &n, ll &m, string &s1, string &s2, string &s, vector<vector<ll>> &dp)\n{\n if (ind1 + ind2 == n + m)\n return 0;\n ll mn = 1e9;\n if (ind1 < n)\n mn = min(mn, (s1[ind1] != s[ind1 + ind2]) + f(ind1 + 1, ind2, n, m, s1, s2, s, dp));\n if (ind2 < m)\n mn = min(mn, (s2[ind2] != s[ind1 + ind2]) + f(ind1, ind2 + 1, n, m, s1, s2, s, dp));\n return dp[ind1][ind2] = mn;\n}\n \nvoid Ankur()\n{\n string s1, s2, s;\n cin >> s1 >> s2 >> s;\n ll n = s1.size(), m = s2.size();\n vector<vector<ll>> dp(n + m, vector<ll>(n + m, -1));\n cout << f(0, 0, n, m, s1, s2, s, dp) << \"\\n\";\n}\n \nint32_t main()\n{\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n ll t = 1;\n cin >> t;\n while (t--)\n {\n Ankur();\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": "2050E",
"submission_id": 295055595,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\nimport sys\nimport os\nimport time \nimport math\nimport random\n \ndef min_changes_required(a, b, c):\n n = len(a)\n m = len(b)\n total = n + m\n \n def get_possible_strings(a, b):\n if not a:\n return {b}\n if not b:\n return {a}\n \n # Try taking from a first\n result1 = set()\n for s in get_possible_strings(a[1:], b):\n result1.add(a[0] + s)\n \n # Try taking from b first\n result2 = set()\n for s in get_possible_strings(a, b[1:]):\n result2.add(b[0] + s)\n \n return result1.union(result2)\n \n # Get all possible strings\n possible = get_possible_strings(a, b)\n \n # Find minimum differences\n min_diff = total\n for p in possible:\n diff = sum(1 for x, y in zip(p, c) if x != y)\n min_diff = min(min_diff, diff)\n \n return min_diff\n \ndef solve():\n t = int(input())\n for _ in range(t):\n a = input()\n b = input()\n c = input()\n print(min_changes_required(a, b, c))\n \nsolve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050E",
"submission_id": 295042846,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\nimport sys\nsys.setrecursionlimit(2*(10**3))\nfor _ in range(int(input())):\n a = input()\n b = input()\n c = input()\n dic = {}\n dic = {}\n key = lambda i, j, k: i * len(b) * len(c) + j * len(c) + k\n def calculate(i,j,k,dic):\n key_ = key(i,j,k)\n if (i,j,k) in dic:\n return dic[key_]\n if i == len(a):\n count = 0\n while k <len(c):\n if c[k] != b[j]:\n count+=1\n k+=1\n j+=1\n dic[key_] = count\n return count\n if j == len(b):\n count = 0\n while k < len(c):\n if c[k] != a[i]:\n count+=1\n i+=1\n k+=1\n dic[key_] = count\n return count \n \n if c[k]!=a[i] and c[k]!=b[j]:\n count = 1 + min(calculate(i+1,j,k+1,dic),calculate(i,j+1,k+1,dic))\n elif c[k]!=a[i] and c[k] == b[j]:\n count = min(calculate(i,j+1,k+1,dic),calculate(i+1,j,k+1,dic)+1)\n elif c[k]==a[i] and c[k] != b[j]:\n count = min(calculate(i,j+1,k+1,dic)+1,calculate(i+1,j,k+1,dic))\n else:\n count = min(calculate(i+1,j,k+1,dic),calculate(i,j+1,k+1,dic))\n \n dic[key_] = count\n return count \n \n print(calculate(0,0,0,dic))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050E",
"submission_id": 298781251,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\npublic class Main\n{\n private static int helper(int i,int j,int n1,int n2,int k,String a,String b,String c)\n {\n if(i==n1)\n {\n int count=0;\n for(int idx=j;idx<n2;idx++)\n {\n if(b.charAt(idx)!=c.charAt(k))\n {\n count++;\n }\n k++;\n }\n return count;\n }\n \n if(j==n2)\n {\n int count=0;\n for(int idx=i;idx<n1;idx++)\n {\n if(a.charAt(idx)!=c.charAt(k))\n {\n count++;\n }\n k++;\n }\n return count;\n }\n \n int min = helper(i+1, j, n1, n2, k+1, a, b, c) + (a.charAt(i) != c.charAt(k) ? 1 : 0);\n min = Math.min(min, helper(i, j+1, n1, n2, k+1, a, b, c) + (b.charAt(j) != c.charAt(k) ? 1 : 0));\n \n return min;\n \n }\n\tpublic static void main(String[] args) {\n\t\tScanner sc=new Scanner(System.in);\n\t\tint t=sc.nextInt();\n\t\tsc.nextLine();\n\t\twhile(t-->0)\n\t\t{\n\t\t String a=sc.nextLine();\n\t\t String b=sc.nextLine();\n\t\t String c=sc.nextLine();\n\t\t int n1=a.length();\n\t\t int n2=b.length();\n\t\t int ans=helper(0,0,n1,n2,0,a,b,c);\n\t\t System.out.println(ans);\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": "2050E",
"submission_id": 302500570,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n\tstatic String a,b,c;\n\tstatic int[][][] dp;\n\tstatic int solve(int ind1, int ind2, int ind3) {\n\t\tif(ind3 >= c.length() || (ind1 >= a.length() && ind2 >= b.length())) {\n\t\t\treturn (c.length() - (ind3+1));\n\t\t}\n\t\tif(dp[ind1][ind2][ind3] != -1) return dp[ind1][ind2][ind3];\n\t\tint res = Integer.MAX_VALUE;\n\t\tif(ind1 < a.length() && a.charAt(ind1) == c.charAt(ind3)) {\n\t\t\tres = Math.min(res, 0+solve(ind1 + 1, ind2, ind3 + 1));\n\t\t}\n\t\tif(ind2 < b.length() && b.charAt(ind2) == c.charAt(ind3)) {\n\t\t\tres = Math.min(res, 0+solve(ind1, ind2 + 1, ind3 + 1));\n\t\t}\n\t\tif(ind1 < a.length()) res = Math.min(res, 1+solve(ind1+1, ind2, ind3+1));\n\t\tif(ind2 < b.length()) res = Math.min(res, 1+solve(ind1, ind2 + 1, ind3 + 1));\n\t\treturn dp[ind1][ind2][ind3] = res;\n\t}\n\tpublic static void main(String[] args) {\n\t\tFastIO sc = new FastIO();\n\t\tFastWriter out = new FastWriter();\n \n\t\tint t= sc.nextInt();\n\t\twhile(t-- > 0) {\n\t\t\ta = sc.next();\n\t\t\tb = sc.next();\n\t\t\tc = sc.next();\n\t\t\tdp = new int[a.length()+1][b.length()+1][c.length()+1];\n\t\t\tfor(int[][] it2 : dp) for(int[] it : it2) Arrays.fill(it,-1);\n\t\t\tint res = solve(0,0,0);\n\t\t\tout.println(res+1);\n\t\t}\n \n\t\ttry {\n\t\t\tout.close();\n\t\t} catch (IOException e) {\n\t\t\te.printStackTrace();\n\t\t}\n\t\tsc.close();\n\t}\n \n\t\t// #region FastIO and FastWriter\n\tstatic class FastIO extends PrintWriter {\n\t\tprivate InputStream stream;\n\t\tprivate byte[] buf = new byte[1 << 16];\n\t\tprivate int curChar, numChars;\n \n\t\tpublic FastIO() {\n\t\t\tthis(System.in, System.out);\n\t\t}\n \n\t\tpublic FastIO(InputStream i, OutputStream o) {\n\t\t\tsuper(o);\n\t\t\tstream = i;\n\t\t}\n \n\t\tpublic FastIO(String i, String o) throws IOException {\n\t\t\tsuper(new FileWriter(o));\n\t\t\tstream = new FileInputStream(i);\n\t\t}\n \n\t\tprivate int nextByte() {\n\t\t\tif (numChars == -1)\n\t\t\t\tthrow new InputMismatchException();\n\t\t\tif (curChar >= numChars) {\n\t\t\t\tcurChar = 0;\n\t\t\t\ttry {\n\t\t\t\t\tnumChars = stream.read(buf);\n\t\t\t\t} catch (IOException e) {\n\t\t\t\t\tthrow new InputMismatchException();\n\t\t\t\t}\n\t\t\t\tif (numChars == -1)\n\t\t\t\t\treturn -1; // End of file\n\t\t\t}\n\t\t\treturn buf[curChar++];\n\t\t}\n \n\t\tpublic String next() {\n\t\t\tint c;\n\t\t\tdo {\n\t\t\t\tc = nextByte();\n\t\t\t} while (c <= ' ');\n\t\t\tStringBuilder res = new StringBuilder();\n\t\t\tdo {\n\t\t\t\tres.appendCodePoint(c);\n\t\t\t\tc = nextByte();\n\t\t\t} while (c > ' ');\n\t\t\treturn res.toString();\n\t\t}\n \n\t\tpublic int nextInt() {\n\t\t\tint c;\n\t\t\tdo {\n\t\t\t\tc = nextByte();\n\t\t\t} while (c <= ' ');\n\t\t\tint sgn = 1;\n\t\t\tif (c == '-') {\n\t\t\t\tsgn = -1;\n\t\t\t\tc = nextByte();\n\t\t\t}\n\t\t\tint res = 0;\n\t\t\tdo {\n\t\t\t\tif (c < '0' || c > '9')\n\t\t\t\t\tthrow new InputMismatchException();\n\t\t\t\tres = 10 * res + c - '0';\n\t\t\t\tc = nextByte();\n\t\t\t} while (c > ' ');\n\t\t\treturn res * sgn;\n\t\t}\n \n\t\tpublic long nextLong() {\n\t\t\tint c;\n\t\t\tdo {\n\t\t\t\tc = nextByte();\n\t\t\t} while (c <= ' ');\n\t\t\tint sgn = 1;\n\t\t\tif (c == '-') {\n\t\t\t\tsgn = -1;\n\t\t\t\tc = nextByte();\n\t\t\t}\n\t\t\tlong res = 0;\n\t\t\tdo {\n\t\t\t\tif (c < '0' || c > '9')\n\t\t\t\t\tthrow new InputMismatchException();\n\t\t\t\tres = 10 * res + c - '0';\n\t\t\t\tc = nextByte();\n\t\t\t} while (c > ' ');\n\t\t\treturn res * sgn;\n\t\t}\n \n\t\tpublic double nextDouble() {\n\t\t\treturn Double.parseDouble(next());\n\t\t}\n \n\t\tpublic String nextLine() {\n\t\t\tStringBuilder res = new StringBuilder();\n\t\t\tint c = nextByte();\n\t\t\t\n\t\t\t// If the current character is a new line, read the next character\n\t\t\tif (c == '\\n') {\n\t\t\t\tc = nextByte();\n\t\t\t}\n \n\t\t\t// Read until the end of the line or the end of the stream\n\t\t\twhile (c != -1 && c != '\\n') {\n\t\t\t\tres.appendCodePoint(c);\n\t\t\t\tc = nextByte();\n\t\t\t}\n \n\t\t\treturn res.toString();\n\t\t}\n \n\t\tpublic char nextChar() {\n\t\t\tint c = nextByte();\n\t\t\twhile (c <= ' ')\n\t\t\t\tc = nextByte();\n\t\t\treturn (char) c;\n\t\t}\n \n\t\tpublic float nextFloat() {\n\t\t\treturn Float.parseFloat(next());\n\t\t}\n\t}\n \n\tstatic class FastWriter {\n\t\tprivate final StringBuilder sb;\n \n\t\tpublic FastWriter() {\n\t\t\tthis.sb = new StringBuilder();\n\t\t}\n \n\t\tpublic void print(Object object) {\n\t\t\tsb.append(object);\n\t\t}\n \n\t\tpublic void println(Object object) {\n\t\t\tsb.append(object).append(\"\\n\");\n\t\t}\n \n\t\tpublic void print(char c) {\n\t\t\tsb.append(c);\n\t\t}\n \n\t\tpublic void println(char c) {\n\t\t\tsb.append(c).append(\"\\n\");\n\t\t}\n \n\t\tpublic void print(int i) {\n\t\t\tsb.append(i);\n\t\t}\n \n\t\tpublic void println(int i) {\n\t\t\tsb.append(i).append(\"\\n\");\n\t\t}\n \n\t\tpublic void print(long l) {\n\t\t\tsb.append(l);\n\t\t}\n \n\t\tpublic void println(long l) {\n\t\t\tsb.append(l).append(\"\\n\");\n\t\t}\n \n\t\tpublic void print(double d) {\n\t\t\tsb.append(d);\n\t\t}\n \n\t\tpublic void println(double d) {\n\t\t\tsb.append(d).append(\"\\n\");\n\t\t}\n \n\t\tpublic void close() throws IOException {\n\t\t\tSystem.out.print(sb);\n\t\t}\n\t}\n\t\t// #endregion\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050E",
"submission_id": 295370662,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long\nint solve(int i, int j, string a, string b, string c, vector<vector<int>>& dp){\n if(i + j >= c.size()) return 0;\n if(dp[i][j] != 0) return dp[i][j];\n \n int l = (i < a.size()) ? (((a[i] == c[i+j]) ? 0 : 1) + solve(i+1, j, a, b, c, dp)) : 1e9;\n int r = (j < b.size()) ? ((b[j] == c[i+j] ? 0 : 1) + solve(i, j+1, a, b, c, dp)) : 1e9; \n \n dp[i][j] = min(l, r);\n \n return dp[i][j];\n}\n \nvoid solve(){\n int t;\n cin>>t;\n while(t--){\n string a, b, c;\n cin>>a>>b>>c;\n \n vector<vector<int>>dp(a.length()+1, vector<int>(b.length()+1, 0));\n \n int ans = solve(0, 0, a, b, c, dp);\n \n cout<<ans<<\"\\n\";\n }\n}\n \nint32_t main(){\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": "2050E",
"submission_id": 295539669,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\n# Author: rishabhxchoudhary\n# Created: Friday, 27.12.2024 01:20 PM (GMT+5:30)\nimport sys\nsys.setrecursionlimit(2147483647)\nMOD = 1000000007\ninput = sys.stdin.readline\n \ndef solve():\n a = input().strip()\n b = input().strip()\n c = input().strip()\n n = len(a)\n m = len(b)\n memo = [[0 for _ in range(m+1)] for _ in range(n+1)]\n def dp(j=0, k=0):\n nonlocal memo\n if memo[j][k]:\n return memo[j][k]\n i = j + k\n if i==len(c): return 0\n if (j<n and c[i]==a[j]) and (k<m and c[i]==b[k]):\n ans = min(dp(j,k+1), dp(j+1,k))\n elif (j<n and c[i]==a[j]):\n if k==m:\n ans = dp(j+1,k)\n else: \n ans = min(1+dp(j,k+1), dp(j+1,k))\n elif (k<m and c[i]==b[k]):\n if j==n:\n ans = dp(j,k+1)\n else:\n ans = min(1+dp(j+1,k), dp(j,k+1))\n else:\n if j<n and k<m:\n ans = 1 + min(dp(j,k+1), dp(j+1,k))\n elif j<n:\n ans = 1 + dp(j+1,k)\n else:\n ans = 1 + dp(j,k+1)\n memo[j][k] = ans\n return ans\n print(dp())\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": "2050E",
"submission_id": 299247540,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\n# ======= author: kuanhungchen.com | created: 05/02/23 08:33:06 CST ======= #\nfrom sys import stdin, stderr, stdout, setrecursionlimit\nfrom bisect import bisect_left, bisect_right\nfrom collections import defaultdict, deque, Counter\nfrom itertools import accumulate, combinations, permutations, product\nfrom functools import lru_cache, cmp_to_key, reduce\nfrom heapq import heapify, heappush, heappop, heappushpop, heapreplace\nfrom pypyjit import set_param\nset_param(\"max_unroll_recursion=-1\")\n# setrecursionlimit(300005)\norda = 97\nordA = 65\nINF = 1 << 60\nMOD = 1000000007\nMOD2 = 998244353\ninput = lambda: stdin.readline().rstrip(\"\\r\\n\")\nlinen = lambda cf_: \"[L\" + str(cf_.f_back.f_lineno) + \"] \"\ndbg = lambda *A, **M: stderr.write(\"\\033[91m\" + \\\n M.get(\"sep\", \" \").join(map(str, A)) + M.get(\"end\", \"\\n\") + \"\\033[0m\")\n# ============================ START OF MY CODE ============================ #\n \ndef solve(_tc):\n a = input()\n b = input()\n c = input()\n \n @lru_cache(None)\n def dp(i, j):\n ans = 0\n if i == len(a):\n for jj in range(j, len(b)):\n ans += int(c[i + jj] != b[jj])\n elif j == len(b):\n for ii in range(i, len(a)):\n ans += int(c[j + ii] != a[ii])\n else:\n ans = int(c[i + j] != a[i]) + dp(i + 1, j)\n ans = min(ans, int(c[i + j] != b[j]) + dp(i, j + 1))\n return ans\n \n print(dp(0, 0))\n \n \nif __name__ == \"__main__\":\n _tcs = int(input())\n for _tc in range(1, vars().get(\"_tcs\", 1) + 1):\n dbg(\"=== Case {} ===\".format(str(_tc).rjust(2)))\n solve(_tc)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2050E",
"submission_id": 297718948,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\n#include <stdio.h>\n#include <string.h>\n \n#define MIN(x, y) ((x)<=(y)?(x):(y))\n \nconst int N = 1e3+5, inf = N*3;\n \nint dp[N][N];\n \nint solver(const char* a, const char* b, const char* c){\n int n = strlen(a), m = strlen(b), k = strlen(c);\n for(int i=0; i<=k; ++i)\n for(int j=0; j<=n; ++j)\n dp[i][j] = inf;\n dp[0][0] = 0;\n for(int i=1; i<=k; ++i){\n for(int j=0; j<=MIN(i,n); ++j){\n if(i-j>m) continue;\n int val1 = (i-j-1>=0) ? (dp[i-1][j]+(b[i-j-1]!=c[i-1])) : inf;\n int val2 = (j-1>=0) ? dp[i-1][j-1]+(a[j-1]!=c[i-1]) : inf;\n dp[i][j] = MIN(val1, val2);\n }\n }\n return dp[k][n];\n}\n \nint main(void){\n int T;\n char a[N], b[N], c[N];\n scanf(\"%d\", &T);\n while(T--){\n scanf(\"%s%s%s\", a, b, c);\n printf(\"%d\\n\", solver(a, b, c));\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": "2050E",
"submission_id": 297558392,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\nimport java.util.*;\npublic class A {\n private static int isInterleaveHelper(String s1, String s2, String s3, int i, int j, Map<String, Integer> memo) {\n // Base case: if we have reached the end of s3, we successfully interleaved\n if (i==s1.length() && j == s2.length()) return 0;\n \n // Create a unique key for the current state\n String key = i + \",\" + j;\n if (memo.containsKey(key)) return memo.get(key);\n int takeFromS1 = 10000000 , takeFromS2 = 10000000;\n \n if(i<s1.length())\n takeFromS1 = isInterleaveHelper(s1, s2, s3, i + 1, j, memo) + (s1.charAt(i)==s3.charAt(i+j)? 0:1) ;\n \n if(j<s2.length())\n takeFromS2 = isInterleaveHelper(s1, s2, s3, i, j + 1, memo) + (s2.charAt(j)==s3.charAt(i+j)? 0:1);\n \n int result = Math.min(takeFromS1,takeFromS2);\n memo.put(key, result);\n return result;\n }\n public static void main(String[] args) {\n Scanner sc =new Scanner(System.in);\n int t = sc.nextInt();\n \n while(t-->0){\n String a = sc.next();\n String b = sc.next();\n String c = sc.next();\n \n int result = isInterleaveHelper(a, b, c, 0, 0, new HashMap<>());\n System.out.println(result);\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": "2050E",
"submission_id": 301017997,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Three Strings\ntime_limit_ms: 2500\nmemory_limit_mb: 256\nDescription: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:\n\n1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.\n2. Then, a certain number of characters in string $$$c$$$ were randomly changed.\n\nFor example, from the strings $$$a=\\color{red}{\\text{abra}}$$$ and $$$b=\\color{blue}{\\text{cada}}$$$, without character replacements, the strings $$$\\color{blue}{\\text{ca}}\\color{red}{\\text{ab}}\\color{blue}{\\text{d}}\\color{red}{\\text{ra}}\\color{blue}{\\text{a}}$$$, $$$\\color{red}{\\text{abra}}\\color{blue}{\\text{cada}}$$$, $$$\\color{red}{\\text{a}}\\color{blue}{\\text{cada}}\\color{red}{\\text{bra}}$$$ could be obtained.\n\nFind the minimum number of characters that could have been changed in string $$$c$$$.\n\nInput: The first line of the input contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^3$$$) — the number of test cases.\n\nThe first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \\leq |a| \\leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.\n\nThe second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \\leq |b| \\leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.\n\nThe third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.\n\nIt is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \\cdot 10^3$$$.\n\nOutput: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.\n\nExamples:\ninput:\n7\na\nb\ncb\nab\ncd\nacbd\nab\nba\naabb\nxxx\nyyy\nxyxyxy\na\nbcd\ndecf\ncodes\nhorse\ncodeforces\negg\nannie\negaegaeg\noutput:\n1\n0\n2\n0\n3\n2\n3\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.text.DateFormatSymbols;\nimport java.util.*;\n \n \n \n \n \n \n \npublic class Main {\n\tstatic long cr[][]=new long[1001][1001];\n\t//static double EPS = 1e-7;\n static long mod=998244353;\n static long val=0;\n static boolean flag=true;\n \npublic static void main(String[] args) \n{\n\tFScanner sc = new FScanner();\n\t//Arrays.fill(prime, true);\n\t//sieve();\t\n\t//ncr();\n\tint t=sc.nextInt();\n\tStringBuilder sb = new StringBuilder();\n \n while(t-->0) \n {\t\n \tchar a[]=sc.nextLine().toCharArray();\n \tchar b[]=sc.nextLine().toCharArray();\n \tchar c[]=sc.nextLine().toCharArray();\n \t HashMap<String, Integer> hMap=new HashMap<>();\n \t sb.append(Check(a,b,c,a.length-1,b.length-1,c.length-1,hMap));\n \n\t sb.append(\"\\n\");\n }\n \n System.out.println(sb.toString());\n} \npublic static int Check(char a[],char b[],char c[],int i,int j,int k,HashMap<String, Integer> hMap)\n{ \n\tif(i<0 && j<0 && k<0) {\n\t\treturn 0;\n\t}\n\tif(k<0) {\n\t\treturn 100000;\n\t}\n\tif(hMap.containsKey(i+\" \"+j+\" \"+k)) {\n\t\treturn hMap.get(i+\" \"+j+\" \"+k);\n\t}\n\t int res=100000;\n\tif(i>=0 && k>=0 && a[i]==c[k]) {\n\t\tres= Check(a, b, c, i-1, j, k-1,hMap);\n\t}\n\tif(j>=0 && k>=0 && b[j]==c[k]) {\n\t\tres=Math.min( res,Check(a, b, c, i, j-1, k-1,hMap));\n\t}\n\t\n\tres = Math.min(res,1+Math.min(Check(a, b, c, i-1, j, k-1,hMap), Check(a, b, c, i, j-1, k-1,hMap)));\n\thMap.put(i+\" \"+j+\" \"+k, res);\n\treturn res;\n}\n \n \npublic static long powerLL(long x, long n)\n{\n long result = 1;\n while (n > 0)\n {\n if (n % 2 == 1)\n {\n result = result * x % mod;\n }\n n = n / 2;\n x = x * x % mod;\n }\n return result;\n}\n \n \n \npublic static long gcd(long a,long b)\n{ \n\t\n return b==0 ? a:gcd(b,a%b);\t\n\t\n}\n \n /* public static void sieve()\n{ prime[0]=prime[1]=false;\nint n=1000000;\n\tfor(int p = 2; p*p <=n; p++)\n {\n if(prime[p] == true)\n {\n \n for(int i = p*p; i <= n; i += p)\n prime[i] = false;\n }\n }\n */\n \npublic static void ncr()\n{\n\tcr[0][0]=1;\n\tfor(int i=1;i<=1000;i++)\n\t{\n\t\tcr[i][0]=1;\n\t\tfor(int j=1;j<i;j++)\n\t\t{\n\t\t\tcr[i][j]=(cr[i-1][j-1]+cr[i-1][j])%mod;\n\t\t}\n\t\tcr[i][i]=1;\n\t}\n}\n \n}\n \n class pair\n //implements Comparable<pair> \n {\n\t long a;long b;\n\t pair(long a,long b)\n\t{ \n\t\tthis.b=b;\n\t\tthis.a=a;\n\t}\n\t\n }\n \n class FScanner {\n\tBufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n\tStringTokenizer sb = new StringTokenizer(\"\");\n \n\tString next(){\n\t\twhile(!sb.hasMoreTokens()){\n\t\t\ttry{\n\t\t\t\tsb = new StringTokenizer(br.readLine());\n\t\t\t} catch(IOException e){ }\n\t\t}\n\t\treturn sb.nextToken();\n\t}\n\tString nextLine(){\n\t\ttry{ return br.readLine(); } \n\t\tcatch(IOException e) { } return \"\";\n\t}\n \n\tint nextInt(){\n\t\treturn Integer.parseInt(next());\n\t}\n \n\tlong nextLong() {\n\t\treturn Long.parseLong(next());\n\t}\n \n\tint[] readArray(int n) {\n\t\tint a[] = new int[n];\n\t\tfor(int i=0;i<n;i++) a[i] = nextInt();\n\t\treturn a;\n\t}\n \n\tfloat nextFloat(){\n\treturn Float.parseFloat(next());\n\t}\n\t\n\tdouble nextDouble(){\n\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": "2050E",
"submission_id": 295282049,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#define long long int\nusing namespace std;\n \nsigned main(){\n int n_test;\n cin >> n_test;\n while(n_test--){\n int n;\n cin >> n;\n int remainder = n%4;\n if (remainder > 1 && remainder <3){\n cout<<\"Sakurako\\n\";\n }\n else{\n cout<<\"Kosuke\\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": "2033A",
"submission_id": 287696076,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <vector>\nusing namespace std;\n#define ll long long\nint n;\nint main()\n{\nint t;cin>>t;\nwhile(t--)\n{\n cin>>n;int i=0,z=1;\n while(i<=n&&i>=(-1*n))\n {\n if(z==1)\n {\n i+=(i*2+1);z=0;\n }\n else {i+=(i*2+1);z=1;}\n }\n if(z==0)cout<<\"Sakurako\\n\";\n else cout<<\"Kosuke\\n\";z=1;\n}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2033A",
"submission_id": 287713297,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\n// Problem: A. Sakurako and Kosuke\n// Contest: Codeforces - Codeforces Round 981 (Div. 3)\n// URL: https://codeforces.com/contest/2033/problem/0\n// Memory Limit: 256 MB\n// Time Limit: 1000 ms\n// \n// Powered by CP Editor (https://cpeditor.org)\n \n#include <bits/stdc++.h>\n#include <iostream>\nusing namespace std;\n#define utkarsh 1\n#define JAI ios_base::sync_with_stdio(false);\n#define SHREE cin.tie(NULL);\n#define RAM cout.tie(NULL);\nusing ll= long long;\nusing sg=string;\nusing lld= long double;\nusing ull= unsigned long long;\nconst lld pi= 3.141592653589793238;\nconst ll mod=1e9+7;\ntypedef pair<ll, ll> pll;\ntypedef vector<ll> vll;\ntypedef vector<pll> vpll;\ntypedef vector<vll> vvll;\ntypedef vector<string> vs;\ntypedef unordered_map<ll,ll> umll;\ntypedef map<ll,ll> mll;\n#define ff first\n#define ss second\n#define pb push_back\n#define mp make_pair\n#define fl(i,n) for(int i=0;i<n;i++)\n#define rl(i,m,n) for(int i=n;i>=m;i--)\n#define py cout<<\"YES\\n\";\n#define pm cout<<\"-1\\n\";\n#define pn cout<<\"NO\\n\";\n#define tl(n) ll n;cin>>n;\n#define ts(n) sg n;cin>>n;\n#define vr(v) v.begin(),v.end()\n#define rv(v) v.end(),v.begin()\n#define memz(a) memset(a, 0, sizeof(a))\n#define memn(a) memset(a, -1, sizeof(a)) \n#define p(x) cout<<x<<endl;\n#define flp(i,a,b) for(int i = a; i <= b; i++)\n \n/*------------------------------Graph Moves----------------------------*/\n// const int fx[]={+1,-1,+0,+0};\n// const int fy[]={+0,+0,+1,-1};\n// const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move\n// const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move\n// const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move\n// const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move\n/*---------------------------------------------------------------------*/\n \n//Debug\n#ifdef utkarsh\n#define debug(x) cerr << #x<<\" \"; cerr<<x<<\" \"; cerr << endl;\n#else\n#define debug(x);\n#endif\n \n// Operator overloads\ntemplate<typename T1, typename T2> // cin >> pair<T1, T2>\nistream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }\ntemplate<typename T> // cin >> vector<T>\nistream& operator>>(istream &istream, vector<T> &v){for (auto &it : v)cin >> it;return istream;}\ntemplate<typename T1, typename T2> // cout << pair<T1, T2>\nostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << \" \" << p.second); }\ntemplate<typename T> // cout << vector<T>\nostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << \" \"; return ostream; }\n \n// Utility functions\ntemplate <typename T>\nvoid print(T &&t) { cout << t << \"\\n\"; }\nvoid printarr(ll arr[], ll n){fl(i,n) cout << arr[i] << \" \";cout << \"\\n\";}\ntemplate<typename T>\nvoid printvec(vector<T>v){ll n=v.size();fl(i,n)cout<<v[i]<<\" \";cout<<\"\\n\";}\ntemplate<typename T>\nll sumvec(vector<T>v){ll n=v.size();ll s=0;fl(i,n)s+=v[i];return s;}\n \n \n \n \nvoid square(){\n //write code here\n tl(n);\n ll c=0;\n flp(i,1,100){\n \tll sp=2*i-1;\n \tif(i%2==1){\n \t\tc-=sp;\n \t}else{\n \t\tc+=sp;\n \t} \n \tif(abs(c)>n){\n \t\tif(i%2){\n \t\t\tcout<<\"Sakurako\"<<endl;\n \t\t}else{\n \t\t\tcout<<\"Kosuke\"<<endl;\n \t\t}\n \t\treturn;\n \t}\n }\n \n \n \n }\n \nint main()\n{\n JAI SHREE RAM\n int n = 1;\n cin>>n;\n while(n--){\n square();\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": "2033A",
"submission_id": 287693278,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nn = int(input())\n \nfor i in range(n):\n m = int(input())\n if (i + 1) % 2 == 0:\n print('Sakurako')\n else:\n print('Kosuke')\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2033A",
"submission_id": 287732082,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nx = 0\nflag = 0\nname = \"Sakurako\"\nt = int(input())\nfor k in range(t):\n n = int(input())\n #print(n)\n #for j in range (1,n-1):\n #l = 2 * j -1\n #if (flag%2 ==0):\n #x-=l\n #else:\n #x+=l\n #print(x)\n \n \n if (flag%2 ==0): #четный\n name = \"Kosuke\"\n else:\n name = \"Sakurako\"\n flag+=1\n print(name)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2033A",
"submission_id": 287750435,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\ndef solve(n):\n if(n==1):\n print(\"Kosuke\")\n return \n s=0\n i=0\n l=0\n while(abs(s)<=n):\n if(i%2==0):\n l=(2*l-1)\n s=s-abs(l)\n \n else:\n l=(2*l-1) \n s=s+abs(l)\n # print(s)\n i+=1\n if(i%2==1):\n print(\"Kosuke\")\n else:\n print(\"Sakurako\")\n \n \n \nfor _ in range(int(input())):\n n=int(input())\n solve(n)\n # print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2033A",
"submission_id": 287758770,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\npublic class SakurakoAndKosuke {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int p,x=0;\n int t= sc.nextInt();\n int n[]=new int[t+1];\n for(int i=1;i<=t;i++){\n n[i]=sc.nextInt();\n }\n for(int j=0;j<t;j++){\n if(j%2==0){\n System.out.println(\"Kosuke\");\n }\n else{\n System.out.println(\"Sakurako\");\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": "2033A",
"submission_id": 293059862,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Game{\n public static void main(String args[]) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n while (t-- > 0) {\n int n = sc.nextInt();\n if (n == 1) {\n System.out.println(\"Kosuke\");\n continue;\n }\n int rem = n % 2;\n n = n / 2 + rem;\n int s = n % 2;\n s = s / 2 + s;\n \n if (s % 2 == 0) {\n System.out.println(\"Kosuke\");\n } else {\n System.out.println(\"Sakurako\");\n }\n } \n sc.close();\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": "2033A",
"submission_id": 287738747,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\n// Code by \"sword\".\n// University of Science, VNU-HCM.\n//------------------------------------------------------//\n \n#include <iostream>\n#include <vector>\n#include <algorithm>\n#include <string>\n#include <cmath>\n#include <map>\n#include <set>\n#include <queue>\n#include <stdio.h>\n#include <bits/stdc++.h>\nusing namespace std;\n \ntypedef long long ll;\ntypedef long double ld;\n \n#define int long long\n \n#define pii pair<int,int>\n#define pll pair<ll, ll>\n \n#define mp(a,b) make_pair(a,b)\n#define fi first\n#define se second\n \n#define vi vector<int>\n#define vll vector<ll>\n#define vpii vector<pii>\n#define vs vector<string>\n \n#define For(i,a,n) for (int i = a; i <= n; i++)\n#define ForD(i,a,n) for (int i = a; i >= n; i--)\n#define trav(x,a) for (auto x: a)\n \n#define pb push_back\n#define eb emplace_back\n \n#define num1(x) __builtin_popcount(x) // count number of 1 bit\n#define num1ll(x) __builtin_popcountll(x)\n#define clz(x) __builtin_clz(x) // count leading zero\n#define clzll(x) __builtin_clzll(x)\n#define ctz(x) __builtin_ctz(x) // count trailing zero\n#define ctzll(x) __builtin_ctzll(x)\n \n#define all(x) (x).begin()+1, (x).end()\n#define ALL(x) (x).begin(), (x).end()\n \n#define sfmax(a,b) (a) = max((a), (b))\n#define sfmin(a,b) (a) = min((a), (b))\n \n#define endl \"\\n\"\n#define MULTI_TEST 1\n \nint bound2(int x) {\n // Compute the smallest power of 2 that is not smaller than x.\n // Example: x = 5 -> return 8, x = 1 -> return 1\n return 1 << (32 - __builtin_clz(x-1));\n}\n \nvoid yes() {cout << \"YES\\n\"; return;}\nvoid no() {cout << \"NO\\n\"; return;}\n \nconst int N = 2e5 + 5;\nconst int inf = INT_MAX;\nconst long long linf = 1e18;\nconst int mod = 1e9 + 7;\n \nvoid solve() {\n int n; cin >> n;\n int t = 1;\n int pos = -1;\n if (n == 1) {\n cout << \"Kosuke\" << endl;\n return;\n }\n else {\n while (pos <= n and pos >= -n) {\n t += 1;\n pos = 2*t - 1;\n if (t % 2 == 1) pos = -pos;\n }\n cout << (t % 2 != 0 ? \"Kosuke\" : \"Sakurako\") << endl;\n }\n \n}\n \n \nsigned main() {\n ios::sync_with_stdio(false);\n cin.tie(0); cout.tie(0);\n if (MULTI_TEST) {\n int tc; cin >> tc;\n while (tc--) solve();\n }\n else 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": "2033A",
"submission_id": 287684988,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nint main() \n{\n int t; cin >> t;\n while(t--){\n int n; cin >> n;\n int ans = 0;\n for(int i = 0; i < 100; i++){\n if(i & 1){\n ans += 2*(i+1)-1;\n }else{\n ans -= 2*(i+1)-1;\n }\n if(abs(ans) > n){\n if(i % 2 == 0){\n cout << \"Sakurako\";\n }else{\n cout << \"Kosuke\";\n }\n break;\n }\n }\n cout << endl;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2033A",
"submission_id": 289031044,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nt = int(input())\nfor i in range(t):\n n = int(input())\n a = 1\n i = 1\n ans = 0\n while a < n:\n if i % 2 != 0:\n ans -= a\n else:\n ans += a\n a = i * 2 - 1\n i += 1\n if ans == 0:\n print(\"Kosuke\")\n else:\n print(\"Sakurako\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2033A",
"submission_id": 287718704,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nt = int(input());\ndef ehe(x):\n if (x%2==0): return \"Kosuke\"\n else: return \"Sakurako\"\nfor i in range (t):\n n = int(input()); x = 0;\n for i in range (1, 100):\n x += ((-1)**i)*(2*i-1);\n if abs(x) > n:\n print (ehe(i));\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"
}
]
| TestCase-Eval | null | {
"problem_id": "2033A",
"submission_id": 287700061,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n \n public static void main(String[] args) throws IOException {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n while (t-- > 0) {\n int n = sc.nextInt();\n int x = 0;\n int count = 0;\n while (Math.abs(x) <= n) {\n x += 2 * x - 1;\n count++;\n }\n if (count % 2 == 0) {\n System.out.println(\"Kosuke\");\n }\n else {\n System.out.println(\"Sakurako\");\n }\n }\n sc.close();\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": "2033A",
"submission_id": 297993891,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class step {\n public static void main(String[] args) {\n Scanner scan = new Scanner(System.in);\n int t = scan.nextInt();\n int count = 0;\n for (int i = 1; i <= t; i++) {\n int n = scan.nextInt();\n int every = 2 * i - 1;\n if(every <= n + 2){\n count++;\n }\n if(count % 2 == 0){\n System.out.println(\"Sakurako\");\n }else{\n System.out.println(\"Kosuke\");\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": "2033A",
"submission_id": 288241643,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \n//#define int long long\n#define pii pair<int, int>\n#define fi first\n#define se second\n#define rep(i, n) for(int i = 0; i < (n); i++)\n \ntypedef long long ll;\ntypedef long double ld;\n \nconst int MAXN = 1e6 + 7;\nconst int INF = 1e9 + 7;\nconst int MAX = 3e3 + 7;\nconst int MOD = 1e9 + 7;\nconst int base = 1<<20;\nconst ld EPS = 1e-9;\n \n \n \n \n \nvoid solve(){\n\t\n\tint n;\n\tcin >> n;\n\t\n\tint ok = 1;\n\tint pos = 0;\n\t\n\tfor(int i = 1; i <= 100; i++){\n\t\t\n\t\tif(ok) pos -= (2 * i - 1);\n\t\telse pos += (2 * i - 1);\n\t\t\n\t\tif(abs(pos) > n){\n\t\t\t\n\t\t\tif(!ok) cout << \"Kosuke\\n\";\n\t\t\telse cout << \"Sakurako\\n\";\n\t\t\t\n\t\t\treturn;\n\t\t}\n\t\tok ^= 1;\n\t}\n\t\n\t\n\t\n}\n \nbool multi = 1;\n \nsigned main(){\n\t\n\tios_base::sync_with_stdio(0);\n\tcin.tie(0); cout.tie(0);\n\t\n\tint t = 1;\n\tif(multi) cin >> t;\t\n\twhile(t--) solve();\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": "2033A",
"submission_id": 287683570,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\n#define ll long long int\n#define nl '\\n'\n#define pi pair<int, int>\n#define fastIO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)\nusing namespace std;\n \n \nvoid solve()\n{\n int x=0, mov= 1, n;\n cin>>n;\n bool sakru= true;\n while (true)\n {\n \n \n if (sakru) x -=mov;\n else x +=mov;\n \n if (x>n || x< -n)\n {\n if (sakru) cout<<\"Sakurako\"<<nl;\n else cout<<\"Kosuke\"<<nl;\n break;\n \n \n }\n sakru= !sakru;\n x+=2;\n }\n \n \n \n}\nint main()\n{\n fastIO;\n \n \n int test;\n cin>>test;\n while (test--) solve();\n \n \n \n \n \n \n \n return 0;\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": "2033A",
"submission_id": 287779108,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nt=int(input())\nwhile t:\n n=int(input())\n if n==1:\n print(\"Kosuke\")\n elif ((n+1)/2)%2==0:\n print(\"Kosuke\")\n else:\n print(\"Sakurako\")\n t-=1\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2033A",
"submission_id": 287706193,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nimport math\nfor _ in range(int(input())):\n n = int(input())\n if round(math.log(n, 2)) % 2 == 0:\n print(\"Kosuke\")\n else:\n print(\"Sakurako\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2033A",
"submission_id": 287778508,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Main {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t=sc.nextInt();\n int n=t;\n for (int i=0;i<t;i++) {\n if (n%2==0)\n System.out.println(\"Kosuke\");\n else\n System.out.println(\"Sakurako\");\n n = sc.nextInt();\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": "2033A",
"submission_id": 294104965,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako and Kosuke\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.\n\nOn the $i$-th move, the current player will move the dot in some direction by $2\\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.\n\nIn other words, the following will happen:\n\n 1. Sakurako will change the position of the dot by $-1$, $x = -1$ now 2. Kosuke will change the position of the dot by $3$, $x = 2$ now 3. Sakurako will change the position of the dot by $-5$, $x = -3$ now 4. $\\cdots$ \n\nThey will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\\le x\\le n$. It can be proven that the game will always end.\n\nYour task is to determine who will be the one who makes the last turn.\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 100$) — the number of games that Sakurako and Kosuke played.\n\nEach game is described by one number $n$ ($1 \\le n\\le 100$) — the number that defines the condition when the game ends.\n\nOutput: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output \"Sakurako\" (without quotes); else output \"Kosuke\".\n\nExamples:\ninput:\n4\n1\n6\n3\n98\noutput:\nKosuke\nSakurako\nKosuke\nSakurako\n\nNote: \n\nBuggy Code:\nimport java.math.*;\nimport java.text.DecimalFormat;\nimport java.time.*;\nimport java.util.*;\nimport java.util.Map.Entry;\nimport java.util.function.Function;\nimport java.util.stream.*;\nimport java.util.regex.*;\n \npublic class App {\n \n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int tc = sc.nextInt();\n \n while (tc-- > 0) {\n int n = sc.nextInt();\n \n int sum = 0;\n for (int i = 1; i <= 100; i++) {\n sum += (i % 2 == 0) ? (2 * i - 1) : -(2 * i - 1);\n if (Math.abs(sum) > n) {\n System.out.println((i % 2 == 0) ? \"Kosuke\" : \"Sakurako\");\n break;\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": "2033A",
"submission_id": 294325101,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define LL long long\n#define int long long\n#define endl \"\\n\"\n#define Endl \"\\n\"\nconst LL mod = 1e9 + 7;\nconst LL sz = 1e7 + 5;\n \nvector<LL> prefix_function(string &s)\n{\n LL n = (LL)s.length();\n vector<LL> pi(n);\n for (LL i = 1; i < n; i++)\n {\n LL j = pi[i - 1];\n while (j > 0 && s[i] != s[j])\n j = pi[j - 1];\n if (s[i] == s[j])\n j++;\n pi[i] = j;\n }\n return pi;\n}\n \nLL bin(LL a, LL p)\n{\n if (p == 0)\n return 1;\n LL res = bin(a, p / 2ll);\n res = (res * res) % mod;\n if (p & 1ll)\n res = (res * a) % mod;\n return res;\n}\n \nLL mmi(LL a)\n{\n return bin(a, mod - 2);\n}\n \nvector<LL> primes;\nvoid sieve()\n{\n vector<bool> isp(sz, true);\n isp[0] = false, isp[1] = false;\n for (LL i = 2; i * i <= sz; i++)\n if (isp[i])\n for (LL j = i * i; j < sz; j += i)\n isp[j] = false;\n for (int i = 2; i < sz; i++)\n if (isp[i])\n primes.push_back(i);\n}\n \nbool cmp(const pair<int, int> &p1, const pair<int, int> &p2)\n{\n // ip / op\n return (p1.second * p2.first) >= (p1.first * p2.second);\n}\nvoid solve()\n{\n int n, p, x, pt = 0;\n cin >> n >> p;\n vector<pair<int, int>> vp(n);\n multiset<int> ms;\n for (int i = 0; i < n; i++)\n cin >> vp[i].first;\n for (int i = 0; i < n; i++)\n cin >> vp[i].second;\n for (int i = 0; i < p; i++)\n {\n cin >> x;\n ms.insert(x);\n }\n sort(vp.begin(), vp.end(), cmp);\n for (int i = 0; i < n;)\n {\n auto val = vp[i];\n if (ms.size() == 0)\n break;\n int cost = val.first;\n int last = *prev(ms.end());\n if (cost > last)\n {\n i++;\n continue;\n }\n ms.erase(prev(ms.end()));\n int ways = last / cost;\n ms.insert((last % cost) + (ways * val.second));\n pt += (2ll * ways);\n }\n cout << pt << endl;\n}\nsigned main()\n{\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n int t = 1;\n // cin >> t;\n while (t--)\n {\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1989D",
"submission_id": 268278702,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\n#include<iostream>\n#include<algorithm>\nusing namespace std;\nconst int N = 2e6 + 1003;\n#define ll long long int\nll n, m, c[N];\nstruct red {\n\tint a, b, c;\n}node[N];\nbool cmp(red x, red y) {\n\treturn x.c < y.c;\n}\nint main() {\n\tstd::ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);\n\tcin >> n >> m;\n\tfor (int i = 1; i <= n; i++) cin >> node[i].a;\n\tfor (int i = 1; i <= n; i++) {\n\t\tcin >> node[i].b;\n\t\tnode[i].c = node[i].a - node[i].b;\n\t}\n\tsort(node + 1, node + 1 + n, cmp);\n\tll sum = 0;\n\tfor (int i = 1; i <= m; i++) { cin >> c[i]; }\n\tfor (int i = 1; i <= n; i++) {\n\t\tfor (int j = 1; j <= m; j++) {\n\t\t\tif (c[j] >= node[i].a) { sum += 2; c[j] -= (node[i].c); }\n\t\t}\n\t}\n\tcout << sum;\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": "1989D",
"submission_id": 268899243,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\ndef solve():\n n, m = map(int, input().split())\n a = list(map(int, input().split()))\n b = list(map(int, input().split()))\n c = list(map(int, input().split()))\n \n mi = min(a)\n d = sorted([(x, x - y) for x, y in zip(a, b)])\n dp = [0] * 1000005\n dp[mi] = 1\n for i in range(mi + 1, 1000005):\n l, r = 0, n - 1\n idx = n\n while l <= r:\n mid = (l + r) // 2\n if d[mid][0] > i:\n idx = mid\n r = mid - 1\n else:\n l = mid + 1\n ai, diff = d[idx-1]\n pro = ((i - ai) // diff) + 1\n res = i - (pro * diff)\n dp[i] = dp[res] + pro\n \n ans = 0\n for i in c:\n ans += dp[i]\n print(2 * ans)\n \nsolve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1989D",
"submission_id": 267953400,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\ndef finnMinsteDiff(mengde, ordbok, differanser, start, slutt):\n mid = int((start + slutt) / 2)\n if mengde >= ordbok[differanser[mid]]:\n if mid == start or mengde < ordbok[differanser[mid + 1]]:\n return differanser[mid], ordbok[differanser[mid]]\n else:\n return finnMinsteDiff(mengde, ordbok, differanser, start, mid)\n else:\n return finnMinsteDiff(mengde, ordbok, differanser, mid + 1, slutt)\n \n \ninfo = input().split()\nantallVaapen = int(info[0])\nantallMetaller = int(info[1])\nvaapenKrav = [int(tall) for tall in input().split()]\nfaarTilbake = [int(tall) for tall in input().split()]\ninventoryMetaller = [int(tall) for tall in input().split()]\ndifferanse = []\nxp = 0\nminsteKrav = min(vaapenKrav)\n \nfor i in range(antallVaapen):\n differanse.append(vaapenKrav[i] - faarTilbake[i])\n \nsortedDifferanse = differanse.copy()\nsortedDifferanse.sort()\n \nordbok = {}\n \nfor i in range(len(vaapenKrav)):\n if differanse[i] in ordbok:\n if vaapenKrav[i] < ordbok[differanse[i]]:\n ordbok[differanse[i]] = vaapenKrav[i]\n else:\n ordbok[differanse[i]] = vaapenKrav[i]\n \n \ni = 0\nwhile i < antallVaapen:\n fjern = []\n for k in range(i+1, len(sortedDifferanse)):\n if ordbok[sortedDifferanse[i]] < ordbok[sortedDifferanse[k]]:\n fjern.append(sortedDifferanse[k])\n fjern = list(set(fjern))\n \n for objekt in fjern:\n ordbok.pop(objekt)\n antallVaapen -= 1\n i += 1\n \ndifferanser = list(ordbok.keys())\ndifferanser.sort()\n \n \nfor mengde in inventoryMetaller:\n igjen = mengde\n while igjen >= minsteKrav:\n minstediff, kravet = finnMinsteDiff(igjen, ordbok, differanser, 0, len(differanser))\n rest = int(kravet / minstediff)\n if rest == 0:\n nyxp = int(igjen / minstediff)\n igjen = igjen - nyxp * minstediff\n else:\n nyxp = int(igjen / minstediff) - rest\n igjen = igjen - nyxp * minstediff\n if igjen >= kravet and igjen >= minstediff:\n igjen -= minstediff\n nyxp += 1\n \n xp += nyxp * 2\n \nprint(xp)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1989D",
"submission_id": 268501651,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\ndef finnMinsteDiff(mengde, sortertDifferanse, ordbok):\n for diff in sortertDifferanse:\n if mengde >= ordbok[diff]:\n return diff, ordbok[diff]\n \n \ninfo = input().split()\nantallVaapen = int(info[0])\nantallMetaller = int(info[1])\nvaapenKrav = [int(tall) for tall in input().split()]\nfaarTilbake = [int(tall) for tall in input().split()]\ninventoryMetaller = [int(tall) for tall in input().split()]\ndifferanse = []\nxp = 0\nminsteKrav = min(vaapenKrav)\n \nfor i in range(antallVaapen):\n differanse.append(vaapenKrav[i] - faarTilbake[i])\n \nsortedDifferanse = differanse.copy()\nsortedDifferanse.sort()\n \nordbok = {}\n \nfor i in range(len(vaapenKrav)):\n if differanse[i] in ordbok:\n if vaapenKrav[i] < ordbok[differanse[i]]:\n ordbok[differanse[i]] = vaapenKrav[i]\n else:\n ordbok[differanse[i]] = vaapenKrav[i]\n \nfjern = []\nfor i in range(len(sortedDifferanse)):\n for k in range(i+1, len(sortedDifferanse)):\n if ordbok[sortedDifferanse[i]] < ordbok[sortedDifferanse[k]]:\n fjern.append(sortedDifferanse[k])\n \nfor objekt in fjern:\n ordbok.pop(objekt)\n \n \nfor mengde in inventoryMetaller:\n igjen = mengde\n while igjen >= minsteKrav:\n minstediff, kravet = finnMinsteDiff(igjen, sortedDifferanse, ordbok)\n rest = int(kravet / minstediff)\n if rest == 0:\n nyxp = int(igjen / minstediff)\n igjen = igjen - nyxp * minstediff\n else:\n nyxp = int(igjen / minstediff) - rest\n igjen = igjen - nyxp * minstediff\n if igjen >= kravet and igjen >= minstediff:\n igjen -= minstediff\n nyxp += 1\n \n xp += nyxp * 2\n \nprint(xp)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1989D",
"submission_id": 268480736,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\nimport java.text.DecimalFormat;\nimport java.util.*;\n \npublic class Main {\n public static void main(String[] args) {\n Scanner in = new Scanner(System.in);\n int N = in.nextInt();\n int M = in.nextInt();\n int[] a = new int[N];\n int[] b = new int[N];\n int[] c = new int[M];\n for(int i = 0; i< N; i++) a[i] = in.nextInt();\n for (int i = 0; i < N; i++) b[i] = in.nextInt();\n for (int i = 0; i < M; i++) c[i] = in.nextInt();\n in.close();\n \n List<int[]> list = new ArrayList<>();// (实际消耗， 制造消耗) \n for(int i = 0; i < N; i++){\n list.add(new int[]{a[i] - b[i], a[i]});\n }\n list.sort((x, y) -> { //排序，实际消耗升序，相同则制造消耗升序\n if (x[0] == y[0]) return x[1] - y[1];\n return x[0] - y[0];\n });\n \n double exe = 0;\n for (int ignot: c) {\n for(int[] item: list){\n if (ignot > item[1]) {\n int ans = (ignot - item[1]) / item[0];\n ignot -= ans*item[0];\n exe += ans;\n }\n if (ignot == item[1]) {\n ignot -= item[0];\n exe++;\n }\n }\n }\n System.out.println(new DecimalFormat(\"#\").format(exe*2));\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": "1989D",
"submission_id": 270222115,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <string>\n#include <string.h>\n#include <queue>\n#include <bitset>\n#include<numeric>\n#include <iomanip>\n#include <map>\n#include <stack>\n#include <set>\n#include <algorithm>\n#include <math.h>\nusing namespace std;\n#define ll long long\n#define INF 0x3f3f3f3f\n#define MOD 998244353\n#define N 2000007\nint f[N], g[N];\nint main()\n{\n int n,m;\n scanf(\"%d%d\",&n,&m);\n int a[n],b[n],c[m];\n for(int i=0;i<n;i++)\n scanf(\"%d\",&a[i]);\n int p = INF, id = -1;\n int minv = INF;\n for(int i=0;i<N;i++)\n {\n f[i] = INF;\n g[i] = -1;\n }\n for(int i=0;i<n;i++)\n {\n scanf(\"%d\",&b[i]);\n minv = min(minv, a[i]);\n if(a[i]-b[i]<p)\n {\n p=a[i]-b[i];\n id = i;\n }\n f[a[i]] = a[i]-b[i];\n g[a[i]] = i;\n }\n for(int i=1;i<N;i++)\n if(f[i]>f[i-1])\n {\n f[i]=f[i-1];\n g[i]=g[i-1];\n }\n ll ans = 0;\n priority_queue<int>que;\n for(int i=0;i<m;i++)\n {\n scanf(\"%d\",&c[i]);\n if(c[i] >= minv)\n {\n if(c[i] >= a[id])\n {\n int y = c[i] - a[id];\n if(y % p == 0)\n {\n int z = y / p + 1;\n ans += z * 2;\n que.push(c[i] - z * p);\n }\n else\n {\n int z = y / p + 1;\n ans += z * 2;\n que.push(c[i] - z * p);\n }\n }\n else\n que.push(c[i]);\n }\n }\n while(!que.empty())\n {\n int x = que.top();que.pop();\n if(x<minv) break;\n int y = x - a[g[x]];\n if(y % f[x] == 0)\n {\n int z = y / f[x] +1;\n ans += z * 2;\n que.push(x - z * f[x]);\n }\n else\n {\n int z = y / f[x] + 1;\n ans += z * 2;\n que.push(x - z * f[x]);\n }\n }\n printf(\"%ld\\n\",ans);\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": "1989D",
"submission_id": 270518159,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\nimport java.util.*;\npublic class smithing_skill {\n // public static int sort rowWise(int mtx[][], int n){\n // for(int i=0; i<n; i++){\n // Arrays.sort(mtx[1][i]);\n // }\n \n // }\n public static void main(String args[]){\n try(Scanner sc = new Scanner(System.in)){\n \n long n = sc.nextLong();\n long m = sc.nextLong();\n \n long mtx[][] = new long[(int)n][3];\n \n long min = Long.MAX_VALUE;\n for(long i=0; i<n; i++){\n mtx[(int)i][0] = sc.nextLong(); \n min = (long)Math.min(min, mtx[(int)i][0]);\n }\n \n for(long i=0; i<n; i++){\n mtx[(int)i][2] = sc.nextLong();\n mtx[(int)i][1] = (mtx[(int)i][0] - mtx[(int)i][2]);\n \n // mtx[1][-1] = Integer.MIN_VALUE; \n }\n \n // for(int i=0; i<n; i++){\n // System.out.print(mtx[1][i]+\" \");\n // }\n // System.out.println();\n \n Arrays.sort(mtx, Comparator.comparingInt(a -> (int)(a[0] - a[2])));\n \n // for(int i=0; i<n; i++){\n // System.out.print(mtx[i][0]+\" \");\n // }\n // System.out.println();\n // for(int i=0; i<n; i++){\n // System.out.print(mtx[i][2]+\" \");\n // }\n // System.out.println();\n // for(int i=0; i<n; i++){\n // System.out.print(mtx[i][1]+\" \");\n // }\n \n // for(long i=1; i<n; i++){\n \n // long temp[][] = new long[3][(int)n];\n \n // if(mtx[1][(int)i] < mtx[1][(int)(i-1)]){\n // for(int j=0; j<3; j++){\n // temp[j][(int)i] = mtx[j][(int)i];\n // mtx[j][(int)i] = mtx[j][(int)(i-1)];\n // mtx[j][(int)(i-1)] = temp[j][(int)i];\n // }\n \n // }\n \n // if(mtx[1][(int)i] == mtx[1][(int)(i-1)]){\n // if(mtx[0][(int)i] > mtx[0][(int)(i-1)])\n // for(int j=0; j<3; j++){\n // temp[j][(int)i] = mtx[j][(int)i];\n // mtx[j][(int)i] = mtx[j][(int)(i-1)];\n // mtx[j][(int)(i-1)] = temp[j][(int)i];\n // }\n \n // }\n // }\n \n // for(int i=0; i<n; i++){\n // System.out.print(mtx[0][i]+\" \");\n // }\n // System.out.println();\n // for(int i=0; i<n; i++){\n // System.out.print(mtx[2][i]+\" \");\n // }\n // System.out.println();\n // for(int i=0; i<n; i++){\n // System.out.print(mtx[1][i]+\" \");\n // }\n \n Long types[] = new Long[(int)m];\n for(long i=0; i<m; i++){\n types[(int)i] = sc.nextLong();\n }\n Arrays.sort(types, Collections.reverseOrder());\n long max_type = types[0];\n \n long exper = 0;\n if(n==1){\n if(types[0]>=mtx[0][0]){\n exper = 1 + ((types[0]-mtx[0][0])/mtx[0][1]); \n }\n }else{\n for(long i=0; i<n; i++){\n long excess = 0;\n long diff = 0;\n long use = 0;\n long type_1 = types[1];\n \n if(types[0]>=mtx[(int)i][0]){\n excess = (types[0] - mtx[(int)i][0]);\n diff = mtx[(int)i][1];\n use = 1 + (excess/diff);\n types[0] -= use*diff;\n // System.out.println(\"use:\"+use);\n // System.out.println(\"types:\"+types[0]);\n exper += use; \n // System.out.println(\"exper=\"+exper);\n }else if(types[0]==mtx[(int)i][0]){\n exper ++;\n types[0] = mtx[(int)i][1]; \n // System.out.println(\"types:\"+types[0]);\n // System.out.println(\"exper=\"+exper);\n }\n Arrays.sort(types, Collections.reverseOrder());\n // System.out.print(\"i:\"+i+\" \");\n // for(int j=0; j<m; j++){\n // System.out.print(types[j]+\" \");\n // }\n max_type = types[0];\n // System.out.println(\"type_1:\"+type_1);\n // System.out.println(\"max_type:\"+max_type);\n if(type_1==max_type){\n i = -1;\n }\n \n \n // System.out.println();\n if(types[0] < min){\n i = n; \n }\n \n // System.out.println(\"exp\"+exper);\n }\n }\n \n \n System.out.println(exper*2);\n \n \n \n \n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1989D",
"submission_id": 267898043,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\nn,m=list(map(int,input().split(' ')))\na=list(map(int,input().split(' ')))\nb=list(map(int,input().split(' ')))\nc=list(map(int,input().split(' ')))\nans=0\nReq=[]\nfor i in range(n):\n Req.append([a[i]-b[i],a[i]])\nReq.sort()\nreq=[Req[0]]\nfor i in range(1,n):\n if(Req[i][0]!=Req[i-1][0]):\n req.append(Req[i])\nbar=0\nc.sort(reverse=True)\nfor i in range(m):\n pre=c[i]\n while(bar<len(req) and pre<a[bar]):\n bar+=1\n for j in range(bar,len(req)):\n ans+=2*(max(0,(pre-req[j][1])//req[j][0]))\n pre-=req[j][0]*max(0,((pre-req[j][1])//req[j][0]))\n if(pre>=req[j][1]):\n pre-=req[j][0]\n ans+=2\nprint(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": "1989D",
"submission_id": 268471743,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\n# ------------------------------- ESSENTIALS -------------------------------\nimport sys\nimport platform\n \ndef is_online_judge():\n if platform.system()=='Windows' and platform.win32_ver()[0]=='10' and platform.win32_ver()[2]=='SP0' and platform.node()=='DESKTOP-07FV8E1':\n return False\n else:\n return True\nis_judge : bool = is_online_judge()\n \nif is_judge:\n input = sys.stdin.readline\nelse:\n inp = open('./InpOut/input.txt','r')\n out = open('./InpOut/output.txt','w')\n input = inp.readline\n if str(sys.stdout)[1:29] == 'ipykernel.iostream.OutStream':\n og_stdout = sys.stdout\n sys.stdout = out\n \ndef readList():\n return list(map(int, input().split()))\ndef readLists(n):\n return [readList() for _ in range(n)]\ndef readInt():\n return int(input())\ndef readInts():\n return map(int, input().split())\ndef readStr():\n return input().strip()\ndef print_array(arr):\n print(\" \".join(map(str, arr)))\ndef printer(is_True):\n print('YES') if is_True else print('NO')\n \n# ------------------------------- __MAIN__ -------------------------------\n \n \n \n \ndef solve(n, m, a, b, c):\n cnt = 0\n c.sort(reverse=True)\n duh = dict()\n zero = float('inf')\n \n for i in range(n):\n if b[i]==0:\n zero = min(zero, a[i])\n elif a[i] in duh:\n if duh[a[i]]>b[i]:\n duh[a[i]] = b[i]\n else:\n duh[a[i]] = b[i]\n \n cla = sorted([[x,duh[x]] for x in duh], key=lambda x:(x[0]/x[1], x[0]))\n f = len(cla)\n # print(cla)\n useless =(zero!=float('inf'))\n low_lim = min(a)\n \n if f==0 and useless:\n for met in c:\n cnt += met//zero\n print(2*cnt)\n return\n \n first = True\n noah = 0\n \n for i,met in enumerate(c):\n # print('for',met)\n k= noah\n \n while met>0 and k<f:\n # print('while', met, k, cla[k][0])\n if met<low_lim:\n break\n if met<cla[k][0]:\n k+=1\n continue\n \n if first:\n noah = k\n first = False\n di = cla[k][0]-cla[k][1]\n tim = (met-cla[k][0])//di + 1\n # print(tim)\n cnt += tim\n met -= tim*di\n k+=1\n first = True\n \n if useless: cnt += met//zero\n \n print(2*cnt)\n return\n \n \n \n \nt = 1\n# t = readInt()\nfor _ in range(t):\n n, m, = readInts()\n a = readList()\n b = readList()\n c = readList()\n solve(n, m, a, b, c)\n \n \nif not is_judge:\n sys.stdout = og_stdout\n out.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": "1989D",
"submission_id": 267755068,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\nimport java.util.*;\n \npublic class D {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int n = sc.nextInt(), m = sc.nextInt();\n int[][] a = new int[n][2];\n for (int i = 0; i < n; i++) {\n a[i][0] = sc.nextInt();\n }\n for (int i = 0; i < n; i++) {\n a[i][1] = sc.nextInt();\n }\n Arrays.sort(a, (o1, o2) -> {\n if ((o1[0] - o1[1]) != o2[0] - o2[1]) return (o1[0] - o1[1]) - (o2[0] - o2[1]);\n return o1[0] - o2[0];\n });\n int N = (int) 1e6 + 1;\n int[] cnt = new int[N];\n long ans = 0;\n for (int i = 0; i < m; i++) {\n int d = sc.nextInt();\n if (d >= a[0][0]) {\n int c = ((d - a[0][0]) / (a[0][0] - a[0][1])) + 1;\n ans += c * 2;\n cnt[d - c * (a[0][0] - a[0][1])]++;\n } else {\n cnt[d]++;\n }\n \n }\n int pt = 0;\n for (int i = N - 1; i > 0; i--) {\n if (cnt[i] > 0) {\n while (pt < n && a[pt][0] > i) pt++;\n if (pt == n) break;\n int c = ((i - a[pt][0]) / (a[pt][0] - a[pt][1])) + 1;\n ans += (long) c * 2 * cnt[i];\n cnt[i - c * (a[pt][0] - a[pt][1])] += cnt[i];\n }\n }\n \n \n System.out.println(ans);\n }\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1989D",
"submission_id": 267712059,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.io.PrintWriter;\nimport java.util.Arrays;\nimport java.util.StringTokenizer;\n \n/**\n * @Author congge\n * @Date 2024/6/28 0:09\n * @description https://codeforces.com/contest/1989/problem/D\n */\npublic class Main {\n static FastReader sc = new FastReader();\n static PrintWriter out = new PrintWriter(System.out);\n static int N = 1000010;\n static int[][] a = new int[N][2];\n static int[] c = new int[N];\n static int[] stack = new int[N], r = new int[N];\n static int top = 0;\n static int n, m;\n \n public static void main(String[] args) {\n n = sc.nextInt();\n m = sc.nextInt();\n int minV = N;\n for (int i = 1; i <= n; i++) {\n a[i][0] = sc.nextInt();\n minV = Math.min(minV, a[i][0]);\n }\n \n for (int i = 1; i <= n; i++) a[i][1] = sc.nextInt();\n for (int i = 1; i <= m; i++) c[i] = sc.nextInt();\n Arrays.sort(a, 1, n + 1, ((o1, o2) -> {\n if (o1[0] - o1[1] == o2[0] - o2[1]) {\n return o1[0] - o2[0];\n } else {\n return o1[0] - o1[1] - (o2[0] - o2[1]);\n }\n }));\n long res = 0;\n a[n + 1][0] = 0;\n // 单调栈\n stack[++top] = n + 1;\n for (int i = n; i >= 1; i--) {\n while (top >= 1 && a[stack[top]][0] >= a[i][0]) top--;\n r[i] = stack[top];\n stack[++top] = i;\n }\n \n for (int i = 1; i <= m; i++) {\n int j = 1;\n while (j <= n) {\n if (c[i] >= a[j][0]) {\n long x = (c[i] - a[j][0]) / (a[j][0] - a[j][1]) + 1;\n res += x * 2;\n c[i] -= x * (a[j][0] - a[j][1]);\n }\n if (c[i] < minV) break;\n j = r[j];\n }\n }\n out.println(res);\n out.flush();\n }\n}\n \nclass FastReader {\n StringTokenizer st;\n BufferedReader br;\n \n FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() {\n while (st == null || !st.hasMoreElements()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n String nextLine() {\n String s = \"\";\n try {\n s = br.readLine();\n } catch (IOException e) {\n e.printStackTrace();\n }\n return s;\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n // 是否由下一个\n boolean hasNext() {\n while (st == null || !st.hasMoreTokens()) {\n try {\n String line = br.readLine();\n if (line == null)\n return false;\n st = new StringTokenizer(line);\n } catch (IOException e) {\n throw new RuntimeException(e);\n }\n }\n return true;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1989D",
"submission_id": 267762388,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define debug(n1) cout << n1 << endl\n#define SORT(array, initial, final) sort(&array[initial], &array[final]);\n#define MOD 1000000007\n \ntypedef long long int ll;\ntypedef vector<int> vi;\ntypedef pair<int,int> pi;\n \n#define F first\n#define S second\n#define PB push_back\n#define MP make_pair\n \n/*\nHELP: \nMod of Division:\n \nMethod 1 - \n \n IF a Overflows OR/AND b overflows use ALWAYS (EVEN if 'a')\n \n For (a/b)%c = ((a%c)*((1/b)%c))%c = ((a%c)*(modInverse(b, MOD)%c))%c ;\n For b=x*y which overflows - take mod where required - \n WRONG : (a/(x*y))%c != ((a%c)*(modInverse(x, MOD)%c)*(modInverse(y, MOD)%c))%c - (NOT EQUAL)\n CORRECT: (a/(x*y))%c = ((a%c)*(modInverse(modMul(x,y,MOD), MOD)%c))%c\n \nMethod 2 - \n \n For (a/b)%c = ((a%c)*((1/b)%c))%c = ((a%c)*(modExpo(b, MOD-2, MOD)%c))%c ;\n \nHere modExpo with exponent = MOD-2 is same modInverse.\n \nHELP END;\n*/\n \n// To compute (a * b) % mod when a*b overflows (i.e when mod > INT_MAX so mod*mod > ll_MAX)\nll modMul(ll a, ll b, ll mod){\n ll res = 0; // Initialize result\n a = a % mod;\n b = b % mod;\n while (b > 0){\n // If b is odd, add 'a' to result\n if(b % 2 == 1){\n res = (res + a) % mod;\n }\n // Multiply 'a' with 2\n a = (a * 2) % mod;\n \n // Divide b by 2\n b /= 2;\n }\n \n // Return result\n return res % mod;\n}\n \nll modExpo(ll x, ll n, ll modulo){ \t\t//pass MOD for modulo usually\n//\tint modulo = 1000000007;\n ll product = 1;\n ll p = x%modulo;\n while (n != 0) {\n if ((n & 1) == 1) {\n product = product * p % modulo;\n }\n p = (p * p % modulo);\n n >>= 1;\n }\n return product;\n}\n \n// Function for extended Euclidean Algorithm\nint gcdExtended(ll a, ll b, int* x, int* y){\n // Base Case\n if (a == 0) {\n *x = 0, *y = 1;\n return b;\n }\n // To store results of recursive call\n int x1, y1;\n int gcd = gcdExtended(b % a, a, &x1, &y1);\n \n // Update x and y using results of recursive\n // call\n *x = y1 - (b / a) * x1;\n *y = x1; \n return gcd;\n}\n \nll modInverse(ll A, int M){ \t\t//pass MOD for M usually\n int x, y;\n ll g = gcdExtended(A, M, &x, &y);\n if (g != 1){\n//\t\tcout << \"Inverse doesn't exist\";\n return -1;\n }else {\n//\t\tm is added to handle negative x\n ll res = (x % M + M) % M;\n//\t\tcout << \"Modular multiplicative inverse is \" << res;\n return res;\n }\n}\n \nll gcd(ll a, ll b){\n if(b%a==0)return a;\n else return gcd(b%a,a);\n}\n \nll fac(ll n, ll p){\n if(n==0 || n==1)return 1;\n return (fac(n-1,p)*n)%p;\n}\n \nvector<bool> mysieve(ll n){\n vector<bool>arr(n+1,true);\n arr[1]=false;\n for(int i=2;i*i<=n;i++){\n if(arr[i]==true){\n for(int j=2*i;j<n;j+=i){\n arr[j]=false;\n }\n }\n }\n return arr;\n}\n \nvoid readArr(vector<ll> &array, ll n){\n for(ll i=0; i<n; i++){\n cin>>array[i];\n }\n}\n \nvoid printArr(vector<ll> &array, ll n){\n cout<<\"Array : \";\n for(ll i=0; i<n-1; i++){\n cout<<array[i]<<\", \";\n }\n cout << array[n-1] << endl;\n}\n \n \nvoid solve(){\n ll ans = 0;\n ll sum = 0;\n \n ll n, m;\n cin >> n >> m;\n vector<ll> a(n, 0);\n \n ll mn = 1000000000;\n \n for(int i=0; i<n; i++){\n cin>>a[i];\n mn = min(a[i], mn);\n }\n vector<ll> b(n, 0);\n \n for(int i=0; i<n; i++){\n cin>>b[i];\n }\n \n vector<ll> c(m, 0);\n \n priority_queue<ll> pq;\n \n map <ll, ll> cnt;\n for(int i=0; i<m; i++){\n cin>>c[i];\n cnt[c[i]]++;\n } \n \n for(auto x: cnt){\n pq.push(x.first);\n }\n vector<tuple<ll, ll, ll>> d(n);\n \n for(int i=0; i<n; i++){\n d[i] = {a[i]-b[i], a[i], b[i]};\n }\n sort(d.begin(), d.end());\n \n ll curr = 0;\n \n while(!pq.empty() && curr < n){\n \n ll tempAns = 0;\n ll mx;\n while(!pq.empty() && pq.top() >= get<1>(d[curr])){\n mx = pq.top();\n ll diff = get<0>(d[curr]);\n ll steps = (((mx - get<1>(d[curr]))/diff)+1);\n tempAns += 2*steps;\n pq.pop();\n ll left = mx - diff*steps;\n if(left >= mn){\n cnt[left] = cnt[mx];\n pq.push(left);\n } \n }\n tempAns *= cnt[mx];\n ans += tempAns;\n curr++;\n }\n cout << ans << \"\\n\";\n}\nint main(){\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n ll t;\n t=1;\n // cin >> t;\n \n while(t-->0){\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": "1989D",
"submission_id": 267744049,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#include<unordered_map>\nusing namespace std;\n#define The_Flash_mode ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)\n#define ll long long\n#define ull unsigned long long\n#define dd double\n#define all(v) ((v).begin()),((v).end())\n#define allr(v) ((v).rbegin()),((v).rend())\n#define pll pair<ll,ll>\n#define pii pair<int,int>\n#define dpp(arr,val) memset(arr,val,sizeof(arr))\n#define F first\n#define S second\nvoid File() {\n#ifndef ONLINE_JUDGE\n freopen(\"input.txt\", \"r\", stdin);\n freopen(\"output.txt\", \"w\", stdout);\n#endif\n}\nll gcd(ll x, ll y) { return ((!y) ? x : gcd(y, x % y)); }\nll lcm(ll x, ll y) { return ((x / gcd(x, y)) * y); }\nconst int dx[8] = {0,-1,0,1,1,-1,1,-1 };\nconst int dy[8] = {-1,0,1,0,1,1,-1,-1 };\nconst int N = 2e5 + 5, mod = 1e9 + 7, md = 998244353;\nll fpow (ll x, ll y) {\n ll res = 1;\n while (y) {\n if (y & 1)res *= x;\n y >>= 1, x *= x;\n }\n return res;\n}\nbool cmp(pll p, pll p2) {\n return (p.F - p.S) > (p2.F - p2.S);\n}\nvoid Solve() {\n int n, m;\n cin >> n >> m;\n vector<pll > v(n);\n for (auto &it: v)cin >> it.F;\n for (auto &it: v)cin >> it.S;\n set<ll> s;\n unordered_map<ll, ll> mp;\n while (m--) {\n int x;\n cin >> x;\n mp[x]++;\n s.insert(x);\n }\n ll res = 0;\n for (int i = 0; i < n; i++) {\n if (s.empty())break;\n auto it = s.end();\n it--;\n if (v[i].F > *it)continue;\n ll diff = v[i].F - v[i].S, num = *it;\n res += ((((num - v[i].F) / diff) + 1) * 2) * mp[num];\n s.erase(it);\n if (num - (((num - v[i].F) / diff) + 1)) {\n s.insert(num - (((num - v[i].F) / diff) + 1) * diff);\n mp[num - (((num - v[i].F) / diff) + 1) * diff] = mp[num];\n }\n i--;\n }\n cout << res << '\\n';\n}\nsigned main() {\n The_Flash_mode;\n File();\n int t = 1; //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": "1989D",
"submission_id": 267751323,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <algorithm>\n#include <climits>\n#include <tuple>\n#include <queue>\n#include <cmath>\n#define i64 long long\nusing namespace std;\ni64 INF = 4e6 + 3e5;\n \nvoid solve(){\n i64 n, m;\n cin >> n >> m;\n vector<pair<i64,i64>> X(n);\n for (int i = 0; i < n; i++) {\n cin >> X[i].first;\n } for (int i=0; i < n; i++){\n cin >> X[i].second;\n }\n sort(X.begin(),X.end());\n vector<i64> MAXI(INF,0);\n pair<i64,i64> DD = {-1,-1};\n vector<pair<i64,i64>> D(1e6+2,{-1,-1});\n for (i64 i = 0; i < n; i++){\n if (DD.first == -1 or (DD.first * X[i].first < DD.second * X[i].second)){\n D[X[i].first] = {X[i].second, X[i].first};\n DD = {X[i].second, X[i].first};\n }\n }\n pair<i64,i64> DEAL = {-1,-1};\n for (i64 i = 0; i < INF; i++){\n if (i <= 1e6 and D[i].first != -1){\n DEAL = D[i];\n } if (DEAL.first == -1) continue;\n i64 left = i - DEAL.second + DEAL.first;\n \n MAXI[i] = MAXI[left] + 2;\n }\n i64 minus = DEAL.second - DEAL.first;\n i64 SOL = 0;\n for (i64 i = 0; i<m;i ++){\n i64 k;\n cin >> k;\n if (k >= INF){\n i64 HAHA = 2e6;\n i64 xD = min((k-DEAL.second),k-HAHA);\n i64 z = (xD) / minus;\n if ((xD)%minus != 0){\n z++;\n }\n SOL += 2*z;\n k-=minus*z;\n }\n SOL += MAXI[k];\n }\n cout << SOL << endl;\n \n}\nint main() {\n ios::sync_with_stdio(false);\n cin.tie(0);\n int t;\n t = 1;\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": "1989D",
"submission_id": 267758983,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\n// LUOGU_RID: 168711980\n#include<bits/stdc++.h>\n#define int long long\nusing namespace std;\nconst int N=1e6+5;\nint n,m,tot,x[N],y[N],c,dp[N];\nstruct Node {\n\tint a,b;\n} a[N];\nsigned main() {\n\tios::sync_with_stdio(false);\n\tcin.tie(nullptr),cout.tie(nullptr);\n\tcin>>n>>m;\n\tfor(int i=1; i<=n; i++) cin>>a[i].a;\n\tfor(int i=1; i<=n; i++) cin>>a[i].b;\n\tsort(a+1,a+n+1,[](const Node&x,const Node&y) {\n\t\treturn x.a-x.b==y.a-y.b?x.a<y.a:x.a-x.b<y.a-y.b;\n\t});\n\tx[++tot]=a[1].a,y[tot]=a[1].b;\n\tfor(int i=2; i<=n; i++)\n\t\tif(a[i].a<a[i-1].a) x[++tot]=a[i].a,y[tot]=a[i].b;\n\tn=tot;\n//\tfor(int i=1; i<=n; i++) cout<<x[i]<<' '<<y[i]<<'\\n';\n\tint pos=n;\n\tfor(int i=x[n]; i<=x[1]; i++) {\n\t\twhile(i>=x[pos-1]&&pos>1) pos--;\n\t\tdp[i]=dp[i-x[pos]+y[pos]]+1;\n//\t\tcout<<pos<<' '<<i<<' '<<dp[i]<<'\\n';\n\t}\n\tint ans=0;\n\tfor(int i=1; i<=m; i++) {\n\t\tcin>>c;\n\t\tif(c>x[1]) {\n\t\t\tint p=(c-y[1])/(x[1]-y[1]);\n\t\t\tc-=p*(x[1]-y[1]),ans+=p;\n\t\t}\n\t\tans+=dp[c];\n\t}\n\tcout<<ans*2;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1989D",
"submission_id": 272673996,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\nimport sys; input=sys.stdin.readline\nfrom math import ceil\nn,m=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\nc=list(map(int,input().split()))\na=sorted([(a[j],a[j]-b[j]) for j in range(n)])\ndp=[0 for j in range(10**6+1)]\np,d=1,a[0][1]\nfor j in range(a[0][0],10**6+1):\n if p<n:\n if j==a[p][0]:\n d=min(d,a[p][1])\n p+=1\n dp[j]=dp[j-d]+1\nans=0\nfor j in range(m):\n if c[j]>10**6:\n ans+=dp[-1]+(c[j]-10**6)//d+(c[j]-10**6)%d\n else:\n ans+=dp[c[j]]\nprint(2*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": "1989D",
"submission_id": 267801789,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\nimport sys\ninput = sys.stdin.readline\n \nimport os\nimport sys\nfrom io import BytesIO, IOBase\n \nBUFSIZE = 8192\n \n \nclass FastIO(IOBase):\n newlines = 0\n \n def __init__(self, file):\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n \n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n \n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n \n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\n \n \nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\n \n \nsys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\ninput = lambda: sys.stdin.readline().rstrip(\"\\r\\n\")\n \ndef readList():\n return list(map(int, input().split()))\ndef readInt():\n return int(input())\ndef readInts():\n return map(int, input().split())\ndef readStr():\n return input().strip()\n \n \n# BF-OB5, EC, DB, CC, CL\ndef solve():\n n, m = readInts()\n a, b, c = readList(), readList(), readList()\n thr = sorted([i for i in range(n)], key=lambda x: (a[x]-b[x], a[x]))\n d, th = a[thr[0]] - b[thr[0]], a[thr[0]]\n ma = max(a)\n arr, ans = [0] * (ma + 1), 0\n for i in range(m):\n if c[i] > ma:\n cnt = (c[i] - th + d) // d\n ans += 2 * cnt\n c[i] -= cnt * d\n arr[c[i]] += 1\n \n j = 0\n for i in range(ma, 0, -1):\n if arr[i] == 0:\n continue\n while j < n and a[thr[j]] > i:\n j += 1\n if j < n:\n d, th = a[thr[j]] - b[thr[j]], a[thr[j]]\n cnt = (i - th + d) // d\n ans += 2 * arr[i] * cnt\n arr[i - cnt * d] += arr[i]\n return ans\n \n \nprint(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": "1989D",
"submission_id": 268111937,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\n\nBuggy Code:\nimport java.util.*;\nimport java.util.function.BinaryOperator;\nimport java.lang.*;\nimport java.lang.reflect.Array;\nimport java.io.*;\npublic class Main\n{\n static final PrintWriter out =new PrintWriter(System.out);\n static final FastReader sc = new FastReader();\n static class Pair{\n int f; int s; \n public Pair(int first,int second)\n {\n this.f=first;\n this.s=second;\n }\n }\n static class Triple{\n long f,s,t;\n public Triple(long first,long second,long third)\n {\n this.f=first;\n this.s=second;\n this.t=third;\n }\n }\n public static boolean prime(long n)\n {\n if(n==1L) return false;\n if(n==2L) return true;\n for(long i=2;i*i<=n;i++)\n if(n%i==0) return false;\n return true;\n } \n public static long mod=(long)1e9+7;\n public static int cnt[];\n public static boolean flag;\n\tpublic static void main (String[] args) throws java.lang.Exception\n\t{\n int tes=1;\n StringBuilder sb=new StringBuilder();\n\t lable:for(int fuck=1;fuck<=tes;fuck++)\n\t {\n int n=sc.nextInt(),m=sc.nextInt(),i,j;\n ArrayList<Triple> arr=new ArrayList<>();\n long a[]=new long[n];\n for(i=0;i<n;i++) a[i]=sc.nextLong();\n for(i=0;i<n;i++){\n long x=sc.nextLong();\n arr.add(new Triple(a[i],x,a[i]-x));\n }\n Collections.sort(arr,(x,y)->Long.compare(x.t,y.t));\n PriorityQueue<Long> pq=new PriorityQueue<>(Collections.reverseOrder());\n HashMap<Long,Integer> map=new HashMap<>();\n for(i=0;i<m;i++){\n long x=sc.nextLong();\n map.put(x,map.getOrDefault(x,0)+1);\n if(map.get(x)==1) pq.add(x);\n }\n long ans=0;\n for(i=0;i<n;i++){\n long x=arr.get(i).f,y=arr.get(i).s,z=arr.get(i).t;\n while(x<=pq.peek()){\n long top=pq.poll();\n long k=(top-x)/z+1;\n ans+=2*k*map.get(top);\n long tmp=top-k*z;\n if(!map.containsKey(tmp)) { map.put(tmp,0); pq.add(tmp); }\n map.put(tmp,map.get(top)+map.get(tmp));\n }\n }\n sb.append(ans+\"\\n\");\n }\n System.out.println(sb);\n } \n public static void dfs(ArrayList<ArrayList<Integer>> arr,int u,int dp[],int k){\n if(dp[u]!=-1){\n if(dp[u]!=k) flag=false;\n return;\n }\n dp[u]=k;\n cnt[k]++;\n for(int it:arr.get(u)) dfs(arr,it,dp,k^1);\n }\n public static long inv(long a){\n if(a<=1) return a;\n return mod-(long)(mod/a)*inv(mod%a)%mod;\n }\n public static boolean palin(String s)\n {\n String tmp=s;\n String rev=new StringBuilder(tmp).reverse().toString();\n return s.equals(rev);\n }\n public static long nCr(long n, long r, long p)\n {\n if (r > n - r) r = n - r;\n long C[] = new long[(int)r + 1];\n C[0] = 1;\n for (long i = 1; i <= n; i++) {\n for (long j = Math.min(i, r); j > 0; j--)\n C[(int)j] = (C[(int)j] + C[(int)j - 1]) % p;\n }\n return C[(int)r];\n }\n public static int upper(ArrayList<Integer> arr,int val)\n {\n int low=-1,high=arr.size(),id=-1;\n while(high>1+low)\n {\n int mid=(low+high)/2;\n if(arr.get(mid)>=val) high=mid;\n else low=mid;\n }\n return high;\n }\n public static int lower(long a[][],long val)\n {\n int low=-1,high=a.length,id=-1;\n while(high>1+low)\n {\n int mid=(low+high)/2;\n if(a[mid][0]<=val) low=mid;\n else high=mid;\n }\n return low;\n }\n\t public static long pow(long x, long y,long mod)\n {\n if(y==0)\n return 1;\n if(y==1)\n return x;\n long ans=pow(x,y/2,mod);\n ans=(ans*ans)%mod;\n if(y%2==1)\n ans=(ans*x)%mod;\n return (long)ans;\n }\n private static long gcd(long a, long b) {\n if (b == 0)\n return a;\n return gcd(b, a % b);\n }\n private static long lcm(long a, long b) {\n return (a / gcd(a, b)) * b;\n }\n\tstatic 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 str = br.readLine();\n }\n catch (IOException e) {\n e.printStackTrace();\n }\n return str;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1989D",
"submission_id": 267845049,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Smithing Skill\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the \"Smithing\" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.\n\nThere are $n$ classes of weapons you can forge and $m$ types of metal ingots.\n\nYou can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.\n\nYou have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.\n\nWhat is the maximum total amount of experience you can earn by crafting and melting weapons?\n\nInput: The first line contains two integers $n$ and $m$ ($1 \\le n, m \\le 10^6$) — the number of weapon classes and metal types.\n\nThe second line contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1 \\le a_i \\le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.\n\nThe third line contains $n$ integers $b_1, b_2, \\dots, b_n$ ($0 \\le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.\n\nThe fourth line contains $m$ integers $c_1, c_2, \\dots, c_m$ ($1 \\le c_j \\le 10^9$) — the number of ingots you have of the corresponding metal type.\n\nOutput: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.\n\nExamples:\ninput:\n5 3\n9 6 7 5 5\n8 4 5 1 2\n10 4 7\noutput:\n12\n\ninput:\n3 4\n10 20 20\n0 0 0\n9 10 19 20\noutput:\n8\n\ninput:\n1 5\n3\n1\n1000000000 1000000000 1000000000 1000000000 1000000000\noutput:\n4999999990\n\nNote: In the first example, you can do the following:\n\n 1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots; 2. melt that weapon, returning $8$ ingots of the $1$-st metal type; 3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type; 4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type; 5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type; 6. craft and melt one weapon of the $4$-th class from the $1$-st metal type; 7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type; \n\nIn the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.\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;\nimport java.io.PrintWriter;\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;\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//System.out.println(3|1 |1| 32| 2|12);\n\t\t//System.out.println(32 |16| 2| 1| 1| 1 );\n\t\t//long s=System.currentTimeMillis();\n\t\t//System.out.println((long)200000*(long)Math.sqrt(200000));\n\t\tp1=false;\n\t\t int t=1; while (t-->0) {\n\t\t\t\n\t\t \n\t\t sovle1(); } \n\t\t \n\t\t out.close();\n\t\t \n \n\t\t\n \n\t}\n\tstatic void sovle2() {\n\t\t\n\t}\n\tstatic boolean p1;\n\tstatic int[][] va,l,r;\n\tstatic void sovle() {\n\t\t\n\t}\n\tstatic long suan(long x) {\n\t\tif (x%2==0) {\n\t\t\treturn x*(x/2+1)/2;\n\t\t}else {\n\t\t\treturn (x+1)*(x/2+1)/2;\n\t\t}\n\t}\n\tstatic Map<Integer,List<Integer>>lpMap;\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) (998244353);\n \n\tstatic void sovle1() {\n\t\tint n=sc.nextint();\n\t\tint m=sc.nextint();\n\t\tint[][]arr=new int[n][2];\n\t\t\n\t\tfor (int i = 0; i < arr.length; i++) {\n\t\t\tarr[i][0]=sc.nextint();\n\t\t}\n\t\tfor (int i = 0; i < arr.length; i++) {\n\t\t\tarr[i][1]=sc.nextint();\n\t\t}\n\t\tint []b=new int[m];\n\t\tfor (int i = 0; i < m; i++) {\n\t\t\tb[i]=sc.nextint();\n\t\t}\n\t\tMap<Integer, Integer>map=new HashMap<Integer, Integer>();\n\t\tSet<Integer>set=new HashSet<Integer>();\n\t\tfor (int i = 0; i < arr.length; i++) {\n\t\t\tset.add(arr[i][0]-arr[i][1]);\n\t\t\tmap.put(arr[i][0]-arr[i][1], map.get(arr[i][0]-arr[i][1])==null?arr[i][0]:Math.min(arr[i][0], map.get(arr[i][0]-arr[i][1])));\n\t\t}\n\t\tlong ans=0;\n\t\tList<Integer>list=new ArrayList<Integer>();list.addAll(set);list.sort(null);\n\t\tList<Integer>list2=new ArrayList<Integer>();\n\t\tlist2.add(list.get(0));\n\t\tfor (int i = 1; i < list.size(); i++) {\n\t\t\tint ui=list.get(i);\n\t\t\tif (map.get(ui)<map.get(list2.get(i-1))) {\n\t\t\t\tlist2.add(ui);\n\t\t\t}\n\t\t}\n\t\tlong []geshu=new long[(int) (1e6+1)];\n\t\tfor (int i = 0; i < b.length; i++) {\n\t\t\tif (b[i]<map.get(list.get(0))) {\n\t\t\t\tgeshu[b[i]]++;continue;\n\t\t\t}\n\t\t\tlong cha=b[i]-map.get(list.get(0));\n\t\t\tans+=cha/list.get(0)+1;\n\t\t\tb[i]-=(cha/list.get(0)+1)*list.get(0);\n\t\t\t//System.out.println(b[i]);\n\t\t\tif (b[i]>0) {\n\t\t\t\tgeshu[b[i]]++;\n\t\t\t}\n\t\t\t\n\t\t}\n\t\t//System.out.println(Arrays.toString(geshu)+\" \"+ans);\n\t\tint di=1;\n\t\tfor (int i = map.get(list.get(0))-1; i >0&&di<list2.size(); i--) {\n\t\t\tif (geshu[i]==0) {\n\t\t\t\tcontinue;\n\t\t\t}\n\t\t\twhile (di<list2.size()&&i<map.get(list2.get(di))) {\n\t\t\t\tdi++;\n\t\t\t}\n\t\t\tif (di>=list2.size()||i<map.get(list2.get(di))) {\n\t\t\t\tbreak;\n\t\t\t}else {\n\t\t\t\tlong cha=i-map.get(list.get(di));ans+=1l*(cha/list.get(di)+1)*geshu[i];\n\t\t\t\tlong sheng=i-1l*(cha/list.get(di)+1)*list2.get(di);\n\t\t\t\tif (sheng>0) {\n\t\t\t\t\tgeshu[(int) sheng]+=geshu[i];\n\t\t\t\t}\n\t\t\t}\n\t\t\t//System.out.println(ans);\n\t\t}\n\t\tout.println(ans*2);\n\t}\n\tstatic boolean p(char[]a,char[]b) {\n\t\tint id=0;\n\t\tfor (int i = 0; i < a.length; i++) {\n\t\t\tif (a[i]==b[id]) {\n\t\t\t\tid++;\n\t\t\t}\n\t\t\tif (id>=b.length) {\n\t\t\t\treturn true;\n\t\t\t}\n\t\t}\n\t\treturn false;\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;c[x]%=mod;\n\t\t\tx+=lowbit(x);\n\t\t}\n\t}\n\tstatic long sum(int x) {\n\t\tlong ans=0;\n\t\twhile (x>0) {\n\t\t\t//System.out.println(x);\n\t\t\tans+=c[x];ans%=mod;\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": "1989D",
"submission_id": 267740971,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\nint main()\n{\n long long int t,f,s,y,d,a;\n cin >> t;\n for (int i = 0; i < t; i++)\n {\n cin >> f; \n vector<long long> h(f);\n vector<long long> g(f);\n for (long long j = 0; j < f; j++)\n {\n cin >> h[j];\n }\n sort(h.begin(),h.end());\n g = h;\n s = 0;\n y = 0;\n d = 0;\n a = f - 1;\n while (d < a)\n {\n if (h[f - 1] - h[a - 1] >= h[d])\n {\n h[d] = h[f - 1];\n d++;\n s++;\n }\n else\n {\n a--;\n } \n } \n d = 0;\n a = f - 1;\n while (d < a)\n {\n if (g[a] - g[0] >= g[a - 1])\n {\n g[a] = g[0];\n a--;\n y++;\n } \n else\n {\n a--;\n }\n \n } \n cout << min(s,y) << 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": "2032C",
"submission_id": 289275258,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\n/**\n Author: AmanSR\n**/\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long \n#define pb push_back\n#define nl \"\\n\"\nconst int mod = 1e9 + 7, inf = 1e18;\n#define all(a) a.begin(),a.end()\n#define forn(a, c) for (int (a) = 0; (a) < (c); (a)++) \n#define forl(a, b, c) for (int (a)= (b); (a) <= (c); (a)++) \n#define forr(a, b, c) for (int (a) = (b); (a) >= (c); (a)--)\n#define cin(v) for(auto &i:v){cin>>i;}\n#define cout(v) for(auto &i:v){cout<<i<<\" \";}cout<<nl;\n#define coutpr(v) for(auto &i:v){cout<<i.first<<\" -> \"<<i.second<<nl;}cout<<nl;\n#define coutmp(v) for(auto &x:v){cout<<x.first<<\" -> \"<<x.second<<nl;}cout<<nl;\n#define super() (ios_base::sync_with_stdio(false), cin.tie(NULL));\n#define dbg(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); pri(_it, args); cout<<nl;}\nvoid pri(istream_iterator<string> it) {}\ntemplate<typename T, typename... Args>\nvoid pri(istream_iterator<string> it, T a, Args... args) {\n cout << *it << \" \" << a << nl;\n pri(++it, args...);\n}\nstring dtobin(int n,int k = 0){string res = \"\";while(n>0 || k>0){k--;if(n&1){res += \"1\";}else{res += \"0\";}n>>= 1;}reverse(res.begin(),res.end());return res;}\nint binexp(int a,int b){if(b==0){return 1;}int res = 1;while(b){if(b&1){res*= a;}b>>=1; a *= a;}return res;}\nmt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());\nvoid setbit(int i,int &x){\n x |= ((int)1<<i);\n}\nbool bitv(int i,int x){\n return (((int)1<<i)&x);\n}\n \nvoid solve()\n{\n int n;\n cin>>n;\n std::vector<int> v(n);\n forn(i,n)cin>>v[i];\n map<int,int>mp;\n forn(i, n)mp[v[i]]++;\n vector<int>a;\n for(auto &x:mp){\n a.push_back(x.first);\n }\n n = a.size();\n vector<int>pref(n+1);\n forn(i, n){\n pref[i+1] = pref[i]+mp[a[i]];\n }\n int mini = inf, sum = 0;\n forn(i, n-1){\n int sm = a[i]+a[i+1];\n if(a[n-1] < sm){\n mini = min(mini, sum);break;\n }\n int ub = lower_bound(all(a), sm)-a.begin();\n mini = min(mini, sum + pref[n]-pref[ub]);\n sum += mp[a[i]];\n }\n if(mini == inf){\n mini = 0;\n }\n cout<<mini<<nl;\n}\n \nsigned main()\n{\n super();int tt;tt = 1;\n cin>>tt;\n while(tt--){ solve();}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2032C",
"submission_id": 289278978,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define fr(i, a, b) for (int i = a; i < b; i++)\n#define fl(i, a, b) for (int i = b - 1; i >= a; i--)\n \nusing ll = long long;\n \nint main(){\n int t;\n cin >> t;\n while(t--) {\n int n;\n cin >> n;\n vector<ll> a(n);\n vector<ll> prefixSum;\n \n fr(i, 0, n) cin >> a[i];\n \n // Determine the maximum element in the array\n int maxValue = *max_element(a.begin(), a.end());\n \n // Count frequencies of each element\n map<int, int> freqMap;\n fr(i, 0, n) {\n freqMap[a[i]]++; \n }\n \n // Calculate prefix sums of frequencies\n prefixSum.push_back(0);\n prefixSum.push_back(freqMap.begin()->second);\n for (auto it = next(freqMap.begin()); it != freqMap.end(); ++it) {\n prefixSum.push_back(prefixSum.back() + it->second);\n }\n \n ll minRemovals = LLONG_MAX;\n \n // Find the minimum elements to remove\n for (auto it = freqMap.begin(); it != freqMap.end(); ++it) {\n ll threshold = 2 * it->first;\n auto upper = freqMap.upper_bound(threshold);\n int upperIndex = distance(freqMap.begin(), upper);\n int currentIndex = distance(freqMap.begin(), it);\n \n minRemovals = min(minRemovals, n - (prefixSum[upperIndex] - prefixSum[currentIndex]));\n }\n \n cout << minRemovals << \"\\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": "2032C",
"submission_id": 289359657,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nfrom collections import Counter \nt=int(input())\nfor i in range(t):\n p=int(input())\n L=list(map(int, input().split()))\n a=Counter(L)\n k=sorted(a.keys())\n s=0\n h=1\n j=0\n while(h+j+2<=len(k) and k[j]+k[j+1]<=k[-h]):\n if a[k[j]]<=a[k[-h]]:\n s+=a[k[j]]\n j+=1\n if a[k[j]]>a[k[-h]]:\n s+=a[k[-h]]\n h+=1\n print(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": "2032C",
"submission_id": 289451502,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nimport bisect\nimport typing\nfrom collections import Counter, defaultdict, deque\nfrom copy import deepcopy\nfrom functools import cmp_to_key, lru_cache, reduce\nfrom heapq import merge, heapify, heappop, heappush, heappushpop, nlargest, nsmallest\nfrom itertools import accumulate, combinations, permutations, count, product\nfrom operator import add, iand, ior, itemgetter, mul, xor\nfrom string import ascii_lowercase, ascii_uppercase, ascii_letters\nfrom typing import *\n \ndef solve():\n n = int(input())\n arr = list(map(int, input().split()))\n arr.sort()\n ans = n+1\n for i in range(n-1):\n d = arr[n-1]-arr[i]\n idx1 = bisect.bisect_right(arr, d, 0, n-1)\n if idx1 <= i:\n idx1 -= 1\n idx2 = bisect.bisect_left(arr, arr[i], 0, n-1)\n idx = max(idx1+1, idx2)\n ans = min(ans, idx)\n print(ans)\nt = 1\nt = int(input())\nwhile t > 0:\n solve()\n t -= 1\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2032C",
"submission_id": 293162511,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nt = int(input())\nfor _ in range (t):\n n = int(input())\n a = list(map(int, input().split()))\n a.sort()\n d = {}\n for i in a:\n if i not in d:\n d[i] = 1\n else:\n d[i] += 1\n \n l = 0\n r = 1\n Max = 0\n bestl = -1\n bestr = -1\n while r != n:\n if a[l] * 2 < a[r]:\n l += 1\n elif a[l]*2 == a[r]:\n if r - l > Max and d[a[l]] == 1 and d[a[r]] == 1:\n Max = r - l\n bestl = l\n bestr = r\n r += 1\n else:\n if r - l > Max:\n Max = r - l\n bestl = l\n bestr = r\n r += 1\n \n k = 0\n for i in a:\n if i < a[bestl] or i > a[bestr]:\n k += 1\n \n print(k)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2032C",
"submission_id": 298705493,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nimport java.util.*;\n \npublic class Main {\n \n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int t = scanner.nextInt();\n \n for (int i = 0; i < t; i++) {\n int[] times = new int[200001];\n int n = scanner.nextInt();\n Set<Integer> set = new HashSet<Integer>();\n \n for (int j = 0; j < n; j++) {\n int k = scanner.nextInt();\n times[k]++;\n set.add(k);\n }\n \n ArrayList<Integer> list = new ArrayList<>(set);\n Collections.sort(list);\n \n if (list.get(0) + list.get(1) > list.get(list.size() - 1)) {\n System.out.println(0);\n continue;\n }\n \n int max = Integer.MIN_VALUE;\n int left = 0, right = 0;\n int twoside;\n int arrmax = list.get(list.size() - 1);\n int temp = 0;\n do {\n left++;\n int x = list.get(left - 1), y = list.get(left);\n twoside = x + y;\n if (left == 1) {\n temp = times[x] + times[y];\n right = left + 1;\n }\n while (right <= list.size() - 1 && list.get(right) <= twoside) {\n temp += times[list.get(right)];\n right++;\n }\n temp -= times[x];\n max = Math.max(max, temp);\n } while (right < list.size() - 1 && twoside <= arrmax);\n System.out.println(n - max);\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": "2032C",
"submission_id": 289353428,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\nimport java.math.*;\n \n// map.merge(key,newV,(oldV,newV)->oldV+newV); = if x is absent x will be inserted with newV.\n// map.computeIfAbsent(key, k -> v); // Only adds a value if the key is absent.\n// map.computeIfPresent(k, (k,oldV) -> newV); it will modify if k is present.\n// map.getOrDefault(k, v); If the key is absent, value will hold v, but this does not modify the map.\n \npublic class Task{\n // ..............code begins here..............\n \nstatic int mod=(int)1e9+7;\nstatic int timer = 0;\nstatic int n,x;\nstatic int[][] dp,pre;\n \n \nstatic void solve() throws IOException {\n n=int_v(read());\n int[] a=int_arr();\n sort(a);\n int res=n,i=0; \n while(i<n){\n while(i+1<n&&a[i]==a[i+1])i++;\n int tmp=n-(i+1);\n int l=0,r=n-1;\n //serch for the index j st: a[j]<=a[i]/2\n while(l<r){\n int m=(l+r+1)/2;\n if(a[m]<=a[i]/2){\n l=m;\n }else{\n r=m-1;\n }\n }\n if(a[i]%2==0)l--;\n if(a[i]%2==1&&l==0&&a[1]+a[0]>a[i])l--;\n if(l>=0&&a[l]>a[i]/2)l=-1;\n tmp+=l+1;\n res=Math.min(res,tmp);\n i++;\n // 3 4 5\n }\n out.write(res+\"\\n\");\n}\n \n \npublic static void main(String[] args) throws IOException{\n assign();\n int t=int_v(read());\n while(t--!=0) solve();\n out.flush();\n}\n \n// taking inputs\n \nstatic BufferedReader s1;\nstatic BufferedWriter out;\nstatic String read() throws IOException{String line=\"\";while(line.length()==0){line=s1.readLine();continue;}return line;}\nstatic int int_v (String s1){return Integer.parseInt(s1);}\nstatic long long_v(String s1){return Long.parseLong(s1);}\nstatic void sort(int[] a){List<Integer> l=new ArrayList<>();for(int x:a){l.add(x);}Collections.sort(l);for(int i=0;i<a.length;i++){a[i]=l.get(i);}}\nstatic int[] int_arr() throws IOException{String[] a=read().split(\" \");int[] b=new int[a.length];for(int i=1;i<=a.length;i++){b[i-1]=int_v(a[i-1]);}return b;}\nstatic long[] long_arr() throws IOException{String[] a=read().split(\" \");long[] b=new long[a.length];for(int i=1;i<=a.length;i++){b[i-1]=long_v(a[i-1]);}return b;}\nstatic void assign(){s1=new BufferedReader(new InputStreamReader(System.in));out=new BufferedWriter(new OutputStreamWriter(System.out));}\n//static String setpreciosion(double d,int k){BigDecimal d1 = new BigDecimal(Double.toString(d));return d1.setScale(k,RoundingMode.HALF_UP).toString();}//UP DOWN HALF_UP\nstatic long Modpow(long a,long p,long m){long res=1;while(p>0){if((p&1)!=0){res=(res*a)%m;}p >>=1;a=(a*a)%m;}return res%m;}\nstatic long Modmul(long a,long b,long m){return ((a%m)*(b%m))%m;}\nstatic long ModInv(long a,long m){return Modpow(a,m-2,m);}\n//static long nck(int n,int r,long m){if(r>n){return 0l;}return Modmul(f[n],ModInv(Modmul(f[n-r],f[r],m),m),m);}\n//static long[] f;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2032C",
"submission_id": 289298567,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\n#include <cassert>\n#include <cctype>\n#include <climits>\n#include <cmath>\n#include <cstdio>\n#include <cstdlib>\n#include <cstring>\n#include <iostream>\n#include <vector>\n#include <algorithm>\n#include <queue>\n#include <deque>\n#include <stack>\n#include <set>\n#include <map>\n#include <unordered_map>\n#include <unordered_set>\n#include <string>\n#include <bitset>\n#include <sstream>\n \nusing namespace std;\n \n#define INF 1e9\n#define F first\n#define S second\n#define mp make_pair\n#define int long long\n#define all(x) (x).begin(), (x).end()\nconst int mod = 1e9+7;\nconst int maxn = 200202;\nusing state = pair<int,int>;\n \nint check(multiset<int> &mt){\n auto it = mt.begin();\n int a = *it;\n int b = *(next(it));\n int c = *mt.rbegin();\n return (a + b) <= c;\n}\n \nvoid solve(){\n int n;\n cin >> n;\n int a[n];\n multiset<int> mt;\n for (int i = 0; i < n; i++) {\n cin >> a[i];\n mt.insert(a[i]);\n }\n \n int cnt = 0;\n \n while(check(mt)){\n mt.erase(mt.find(*mt.begin()));\n mt.insert(*mt.rbegin());\n cnt++;\n }\n cout << cnt << '\\n';\n}\n \nsigned main() {\n ios_base:: sync_with_stdio(0);cin.tie(0);cout.tie(0);\n int t; cin>>t;\n while(t--){\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2032C",
"submission_id": 289270435,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nusing ll = long long;\nusing vll = vector<long long>;\nusing vpll = vector<pair<long long, long long>>;\nusing pll = pair<long long, long long>;\n#define PB push_back\n#define MP make_pair\n#define F first\n#define S second\n#define REP(i, a, b) for(long long i = a; i < b; i++)\n#define PI 4*atan(1)\n \nvoid fileio(){\n #ifdef ONLINE_JUDGE\n freopen(\"input.txt\", \"r\", stdin);\n freopen(\"output.txt\", \"w\", stdout);\n #endif\n}\n \n/*------------------------------------------------------------*/\n \nvoid solve(){\n ll n; cin >> n; vll a(n); REP(i, 0, n) cin >> a[i];\n sort(a.begin(), a.end());\n ll counter = 0;\n REP(i, 1, n) {\n if (a[i] + a[i-1] <= a[n-1]) {\n counter++;\n } else break;\n }\n cout << counter << \"\\n\";\n return;\n}\n \n \nint main(){\n ios::sync_with_stdio(0);\n cin.tie(0);\n \n //fileio();\n ll T = 1;\n cin >> T;\n while(T--){\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2032C",
"submission_id": 289252263,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nt = int(input())\nfor _ in range(t):\n n = int(input())\n l = list(map(int,input().split()))\n l.sort()\n i = 0\n j = 1\n count = 0\n while(j<n-1):\n if(l[i]+l[j]>l[-1]):\n j+=1\n elif(l[i]+l[j]<=l[-1]):\n count+=1\n i+=1\n j+=1\n print(count)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2032C",
"submission_id": 289262406,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nfrom itertools import combinations\nt=int(input())\nl=[]\nfor i in range(0,t):\n n=int(input())\n l.append([int(x) for x in input().split(\" \")])\nfor i in range(0,t):\n ar=l[i]\n n=len(ar)\n w=[1]*n\n for j in range(0,n):\n w[j]=j\n comb=list(combinations(w,2))\n nn=len(comb)\n c=2\n countt=999999999\n s=0.0\n #sarr=[0]*99999999\n #for j in range(0,99999999):\n # sarr[j]=j\n #for x in ar:\n # s=s+x\n # sarr[x]=sarr[x]+1\n #s=s/n\n ar.sort(reverse=False)\n inc=0\n while inc<n-2 and ar[n-1]>=ar[inc]+ar[inc+1]:\n inc=inc+1\n c=c+1\n #print(ar)\n print(inc)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2032C",
"submission_id": 289576397,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class QUES1 {\n static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));\n static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));\n static StringTokenizer st;\n \n static long MOD = (long) (1e9 + 7);\n \n \n \n public static void solve() throws IOException {\n int n = Integer.parseInt(reader.readLine());\n int[] arr = new int[n];\n st = new StringTokenizer(reader.readLine());\n for (int i=0;i<n;i++) arr[i] = Integer.parseInt(st.nextToken());\n Arrays.sort(arr);\n// System.out.println(Arrays.toString(arr));\n int ans = 0;\n for (int i=0;i<n-1;i++){\n if (arr[i]+arr[i+1] > arr[n-1]) break;\n ans++;\n }\n writer.write(ans+\"\\n\");\n }\n \n \n \n \n \n \n \n public static void main(String[] args) throws IOException {\n \n int t = Integer.parseInt(reader.readLine());\n// int t = 1;\n while (t-- > 0) {\n solve();\n }\n \n writer.flush();\n writer.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": "2032C",
"submission_id": 290508608,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nimport java.util.*;\nimport java.lang.*;\nimport java.io.*;\n \npublic class Solution\n{\n\tpublic static void main (String[] args) throws java.lang.Exception\n\t{\n\t\tScanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n while (t--> 0) {\n int n = sc.nextInt();\n long arr[] = new long[n];\n for(int i = 0; i<n; i++) arr[i] = sc.nextLong();\n Arrays.sort(arr);\n int i = 0;\n for(; i<n-1; i++) {\n if(arr[i]+arr[i+1] > arr[n-1]) break;\n }\n System.out.println(i);\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": "2032C",
"submission_id": 289243102,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nfrom io import BytesIO, IOBase\nimport sys\nimport os\n \n# import time\nimport bisect\n# import functools\nimport math\nimport random\n# import re\nfrom collections import Counter, defaultdict, deque\nfrom copy import deepcopy\nfrom functools import cmp_to_key, lru_cache, reduce\nfrom heapq import heapify, heappop, heappush, heappushpop, nlargest, nsmallest\nfrom itertools import accumulate, combinations, permutations, count, product\nfrom operator import add, iand, ior, itemgetter, mul, xor\nfrom string import ascii_lowercase, ascii_uppercase\nfrom typing import *\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 \nBUFSIZE = 5096\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 \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 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 \nsys.stdin = IOWrapper(sys.stdin)\nsys.stdout = IOWrapper(sys.stdout)\ninput = lambda: sys.stdin.readline().rstrip(\"\\r\\n\")\n \ndef I():\n return input()\n \ndef II():\n return int(input())\n \ndef MII():\n return map(int, input().split())\n \ndef LI():\n return list(input().split())\n \ndef LII():\n return list(map(int, input().split()))\n \ndef GMI():\n return map(lambda x: int(x) - 1, input().split())\n \ndef LGMI():\n return list(map(lambda x: int(x) - 1, input().split()))\n \ninf = float('inf')\n \nT = II()\n'''\n1. 考虑最小的2个元素 和最大元素组成的(min1, min2, max)\nmin1 + min2 > max 即可使得所有都满足\n2. 考虑操作\n (1)\n'''\nxx = random.randint(64, 128)\n \ndef main():\n n = II()\n nums = LII()\n nums.sort()\n # print(nums)\n ans = inf\n \n pre = defaultdict(int)\n for i in range(n - 1):\n \n if nums[i] not in pre:\n pre[nums[i]] = i\n # print(i, pre)\n tmp = nums[i] + nums[i + 1]\n index = bisect.bisect_left(nums, tmp)\n \n if index < n:\n # if i == 7:\n # print(pre[nums[i]], index)\n ans = min(ans, pre[nums[i]] + n - index)\n \n print(ans if ans != inf else 0)\n \n \n \n \n \n \n \nfor _ in range(T):\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": "2032C",
"submission_id": 289241572,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nT = int(input())\nfor t in range(T):\n n = int(input())\n s = list(map(int, input().split()))\n s.sort()\n MOD = 10**9\n s[-1] %= MOD\n c = i = 0\n while i < len(s) - 2 and (s[i]%MOD) + (s[i + 1]%MOD) <= s[-1]:\n c += 1\n i += 1\n print(c)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2032C",
"submission_id": 289254503,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nimport java.util.*;\n \npublic class Codeforces {\n public static void main(String[] args) {\n Scanner in = new Scanner(System.in);\n int t = in.nextInt();\n while (t-- > 0) {\n int n = in.nextInt(), count = 0;\n long[] arr = new long[n];\n long max = -1;\n for (int i = 0; i < n; i++) {\n arr[i] = in.nextLong();\n max = Math.max(max, arr[i]);\n }\n Arrays.sort(arr);\n for (int i = 0; i < n - 1; i++) {\n if (arr[i] + arr[i + 1] <= max) {\n count++;\n arr[i] = max;\n }\n // System.out.println(Arrays.toString(arr));\n }\n System.out.println(count);\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": "2032C",
"submission_id": 289218503,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\nimport java.util.Arrays;\nimport java.util.Scanner;\n \npublic class any {\n public static void main(String[] args) {\n Scanner s=new Scanner(System.in);\n int t=s.nextInt();\n while(t-->0){\n int n=s.nextInt();\n long[] arr =new long[n];\n for(int i=0;i<n;i++){\n arr[i]=s.nextLong();\n }\n Arrays.sort(arr);\n int i=0;\n long c=0;\n while(i<n-1 && (arr[i]+arr[i+1])<=arr[n-1]){\n c++;\n i++;\n }\n System.out.println(c);\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": "2032C",
"submission_id": 289265882,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n \nvoid solve(){\n int n;\n cin>>n;\n int a[n];\n for(int i=0; i<n; i++)cin>>a[i];\n sort(a, a+n);\n int ans1=0;\n for(int i=n-1; i>0; i--){\n if(a[n-1]-a[i]>=a[i-1]){\n ans1=i;\n break;\n }\n }\n int ans2=0;\n for(int i=1; i<n-1; i++){\n if(a[i]+a[0]<=a[i+1]){\n ans2=n-i-1;\n break;\n }\n }\n if(ans2==0)ans2=n;\n int ans=min(ans1, ans2);\n cout<<ans<<endl;\n}\n \nint main(){ \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": "2032C",
"submission_id": 289423312,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Trinity\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $a$ of $n$ elements $a_1, a_2, \\ldots, a_n$.\n\nYou can perform the following operation any number (possibly $0$) of times:\n\n * Choose two integers $i$ and $j$, where $1 \\le i, j \\le n$, and assign $a_i := a_j$. \n\nFind the minimum number of operations required to make the array $a$ satisfy the condition:\n\n * For every pairwise distinct triplet of indices $(x, y, z)$ ($1 \\le x, y, z \\le n$, $x \\ne y$, $y \\ne z$, $x \\ne z$), there exists a non-degenerate triangle with side lengths $a_x$, $a_y$ and $a_z$, i.e. $a_x + a_y > a_z$, $a_y + a_z > a_x$ and $a_z + a_x > a_y$.\n\nInput: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($3 \\le n \\le 2 \\cdot 10^5$) — the number of elements in the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of operations required.\n\nExamples:\ninput:\n4\n7\n1 2 3 4 5 6 7\n3\n1 3 2\n3\n4 5 3\n15\n9 3 8 1 6 5 3 8 2 1 4 2 9 4 7\noutput:\n3\n1\n0\n8\n\nNote: In the first test case, one of the possible series of operations would be:\n\n * Assign $a_1 := a_4 = 4$. The array will become $[4, 2, 3, 4, 5, 6, 7]$. * Assign $a_2 := a_5 = 5$. The array will become $[4, 5, 3, 4, 5, 6, 7]$. * Assign $a_7 := a_1 = 4$. The array will become $[4, 5, 3, 4, 5, 6, 4]$. \n\nIt can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than $3$ operations.\n\nIn the second test case, we can assign $a_1 := a_2 = 3$ to make the array $a = [3, 3, 2]$.\n\nIn the third test case, since $3$, $4$ and $5$ are valid side lengths of a triangle, we don't need to perform any operation to the array.\n\nBuggy Code:\n#include <algorithm>\n#include <array>\n#include <bitset>\n#include <cassert>\n#include <chrono>\n#include <climits>\n#include <cmath>\n#include <complex>\n#include <cstring>\n#include <functional>\n#include <iomanip>\n#include <iostream>\n#include <map>\n#include <numeric>\n#include <queue>\n#include <random>\n#include <set>\n#include <vector>\nusing namespace std;\n#define int long long\nint maxi = 1e6+5;\nint mod = 1e9+7;\n \nvoid solve(void){\n int N; cin >> N;\n vector<int> a(N);\n \n for(auto &e : a) {\n cin >> e;\n }\n \n sort(a.begin(), a.end());\n int res1 = 0;\n for(int i = 0; i < N-1; i++) {\n if(a[i] + a[i+1] <= a[N-1]) {\n res1++;\n }\n else{\n break;\n }\n }\n \n int res2 = 0;\n for(int i = N-1; i > 1; i--) {\n if(a[0] + a[1] <= a[i]) {\n res2++;\n }\n else{\n break;\n }\n }\n \n \n cout << min(res1, res2) << \"\\n\";\n}\n \n \nint32_t main(){\n \n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n \n int T = 1;\n cin >> T;\n while(T--) solve();\n \n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2032C",
"submission_id": 290025793,
"task_type": "TestCase-Eval-Task2"
} |
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