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TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define endl '\\n'\n#define int long long\n#define optimize() ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)\n \nint32_t main()\n{\n optimize();\n \n int t;\n cin>>t;\n \n while(t--)\n {\n int a,b,p=0;\n cin>>a>>b;\n \n if(a>0 && b==0)\n {\n if(a%2==0) p=1;\n }\n else if(a>0 && b>0 && a%2==0)\n {\n b+=(a/2);\n \n if(b%2==0) p=1;\n }\n \n if(p) cout<<\"YES\"<<endl;\n else cout<<\"NO\"<<endl;\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279089308,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nint l;\n \nint main(){\n\tcin.tie(0)->sync_with_stdio(0);\n\t\n\tcin >> l;\n\t\n\twhile (l--){\n\t\tint a, b;\n\t\tcin >> a >> b;\n\t\tif ((a == 0 && b == 1) || (a == 1 && b == 0))\n\t\t\tcout << \"No\" << \"\\n\";\n\t\telse if (a % 2 == 0 && b % 2 == 0)\n\t\t\tcout << \"Yes\" << \"\\n\";\n\t\telse if (a == (b / 2 * 2))\n\t\t\tcout << \"Yes\" << \"\\n\";\n\t\telse \n\t\t\tcout << \"No\" << \"\\n\";\n\t}\n\t\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 287529457,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\n// Online C++ compiler to run C++ program online\n#include <iostream>\nusing namespace std;\n#include <algorithm>\n#include <map>\n#include <climits>\n#include <vector>\n#include <cmath>\n#include <stack>\n#include <deque>\n#include <queue>\n#include <set>\n#include <list>\n#include <iomanip>\n#define mod 1000000007\nstruct ing{\n long long e,s,x;\n};\n\t\nlong long gcd(long long a,long long b){\n long long temp;\n while(a > 0 && b > 0){\n temp = b;\n b = a%b;\n a = temp;\n \n }\n return max(a,b);\n}\n string alpha=\"abcdefghijklmnopqrstuvwxyz\";\n \n bool condition(pair <int,string> a,pair<int,string>b){\n if(a.first ==b.first){\n return a.second < b.second;\n }\n return a.first>b.first;\n }\n \nint main() {\n ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);\n int t = 1;\n cin >> t;\n// freopen(\"input.txt\", \"r\", stdin);\n // freopen(\"output.txt\", \"w\", stdout);\n \n \n while(t--){\nint a ,b;cin >> a >> b;\nif(a%2!=1){\n if(b==0){cout <<\"YES\\n\";continue;}\n b+=a/2;\n if(b%2==0)cout <<\"YES\\n\";\n else cout <<\"NO\\n\";\n}\nelse cout <<\"NO\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279072530,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nfor _ in range(int(input())):\n a,b = map(int, input().split())\n if (a==0 and b%2 ==0) or (a%2==0 and b==0):\n print(\"YES\")\n elif a%2==0 and abs(((a/2)-b))%2==0:\n print(\"YES\")\n else:\n print(\"NO\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 288069809,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nn = int(input())\nfor _ in range(n):\n a,b = map(int,input().split())\n if a%2==1 : # a is odd\n print(\"NO\") \n else: # a is even\n if (a//2+b)%2==0 or b==0:\n print(\"YES\")\n else:\n print(\"NO\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279463288,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\na = int(input())\nfor _ in range(a):\n b, c = map(int,input().split())\n if b % 2 != 0:\n print('NO')\n elif b == 0 and c == 1:\n print('NO')\n elif b % 2 == 0 and b // 2 <= c and (b // 2 + c) % 2 == 0:\n print('YES')\n elif b % 2 == 0 and b // 2 > c and (b - c * 2) % 2 == 0:\n print('YES')\n else:\n print('NO')\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 280898847,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nimport java.util.*;\npublic class Main {\npublic static void main(String[] args) {\n Scanner sc=new Scanner(System.in);\n int t=sc.nextInt();\n for(int i=0;i<t;i++){\n int a=sc.nextInt();\n int b=sc.nextInt();\n int target = a + 2 * b;\n if (target % 2 == 0) {\n int x = target / 2;\n if (x >= 0 && x < a + b) {\n System.out.println(\"Yes\");\n } else {\n System.out.println(\"No\");\n }\n } else {\n System.out.println(\"No\");\n }\n }\n}\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279089160,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nimport java.util.*;\npublic class main {\n public static void main(String args[]) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n while(t-->0){\n int one = sc.nextInt();\n int two = sc.nextInt();\n int sum1=1*one;\n int sum2=2*two;\n int possible=0;\n if(one>0 && one%2!=0) possible=0;\n else if(one%2==0){\n if(two==0) possible=1;\n else{\n int diff=two-one/2;\n if(Math.abs(diff)%2==0) possible=1;\n }\n }\n if(possible==1) System.out.println(\"YES\");\n else System.out.println(\"NO\");\n }\n sc.close();\n }\n}\n \n \n/*\nLOGIC ---\n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279105601,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nint main(){\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n \n int t;\n cin>>t;\n while(t--){\n int a,b;\n cin>>a>>b;\n \n if(a%2!=0) cout<<\"NO\\n\";\n \n else if(a%2==0 && b%2==0) cout<<\"YES\\n\";\n \n else if(a==0 && b%2==0) cout<<\"YES\\n\";\n \n else if (a==0 && b%2!=0) cout<<\"NO\\n\";\n \n else if(b%2!=0){\n if((a/2) %2!=0) cout<<\"YES\\n\";\n \n }\n \n else cout<<\"NO\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279219234,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ll long long int\nusing namespace std;\n \nconst int P = 239017;\nconst int MOD_X = 1e9 + 7;\nconst int MOD_Y = 1e9 + 9;\n \nstruct H {\n int x, y;\n H() = default;\n H(int _x): x(_x), y(_x) {}\n H(int _x, int _y): x(_x), y(_y) {}\n \n inline H operator+(const H& h) const {\n return H(add(x, h.x, MOD_X), add(y, h.y, MOD_Y));\n }\n inline H operator-(const H& h) const {\n return H(sub(x, h.x, MOD_X), sub(y, h.y, MOD_Y));\n }\n inline H operator*(const H& h) const {\n return H(mul(x, h.x, MOD_X), mul(y, h.y, MOD_Y));\n }\n inline bool operator==(const H& h) const {\n return x == h.x && y == h.y;\n }\n \n static inline int add(int a, int b, int mod) {\n int res = a + b;\n return res >= mod ? res - mod : res;\n }\n \n static inline int sub(int a, int b, int mod) {\n int res = a - b;\n return res < 0 ? res + mod : res;\n }\n \n static inline int mul(int a, int b, int mod) {\n return (1LL * a * b) % mod;\n }\n};\n \nconst int N = 1e5 + 5;\nH p[N], h[N];\n \ninline H get(int l, int r) {\n return h[r] - h[l] * p[r - l];\n}\n \nvoid init(const string& s) {\n int n = s.size();\n p[0] = H(1);\n for (int i = 0; i < n; i++) {\n h[i + 1] = h[i] * P + s[i];\n p[i + 1] = p[i] * P;\n }\n}\n \n//first check whether input is taken correctly or not\n//then check the function which you are calling is called with a correct name\n//int to long long\n \n \nvoid csolve() {\n// int n;\n// cin >> n;\n// vector<int>v(n);\n// for (int i = 0 ; i < n ; i++) {\n \n// }\n int a, b;\n cin >> a >> b;\n if (a == 0) {\n if (b % 2 == 0) {\n cout << \"YES\\n\";\n } else {\n cout << \"NO\\n\";\n }\n return ;\n }\n if (b == 0) {\n if (a % 2 == 0) {\n cout << \"YES\\n\";\n } else {\n cout << \"NO\\n\";\n }\n return ;\n }\n if (a & 1) {\n cout << \"NO\\n\";\n return ;\n }\n a /= 2;\n b += a;\n if (b % 2 == 0) {\n cout << \"YES\\n\";\n } else {\n cout << \"NO\\n\";\n }\n}\n \nint main() {\n int t;\n cin >> t;\n while (t--) {\n csolve();\n }\n // string s = \"examplesfdexample\";\n // init(s);\n \n // // Example: Get the hash of the substring \"exam\"\n // int l = 0, r = 6; // \"exam\" corresponds to the substring [0, 4)\n // H hash_value = get(l, r);\n // cout << \"Hash value of substring 'exam': (\" << hash_value.x << \", \" << hash_value.y << \")\" << endl;\n // H hash_value2 = get(10, 16);\n // cout << \"Hash value of substring 'exam': (\" << hash_value2.x << \", \" << hash_value2.y << \")\" << 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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279078972,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nimport sys\n \nlines = sys.stdin.readlines()\nlines.remove(lines[0])\n \nfor line in lines:\n t = [int(p) for p in line.split(\" \")]\n \n if (((t[0]/2 + t[1]) % 2 == 0) or t[1] % 2 == 0) and (t[0]%2 == 0) and (t[1] + t[0] != 1):\n print(\"YES\")\n else:\n print(\"NO\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 281460300,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nfor test in range(int(input())):\n a,b=map(int,input().split())\n if a!=0 and a%2==0 :\n print(\"YES\")\n else:\n print(\"NO\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279288833,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class GS {\n public static String func(int a,int b){\n if(a*2==b )return \"YES\";\n if((a==0 && b%2==0) || (b==0 && a%2==0) )return \"YES\";\n if( b%2==1 && a==2)return \"YES\";\n return \"NO\";\n }\n \n public static void main(String[] args) {\n Scanner sc=new Scanner(System.in);\n int n=sc.nextInt();\n for(int i=0;i<n;i++){\n int a= sc.nextInt();\n int b=sc.nextInt();\n System.out.println(func(a,b));\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279098167,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nimport java.util.*;\n \n public class mine {\n public static void main(String[] args) {\n Scanner s=new Scanner(System.in);\n int t=s.nextInt();\n for(int i=1;i<=t;i++)\n {\n int a=s.nextInt();\n int b=s.nextInt();\n if((a>0 && a%2==0) && (b%2==0 || (a+(2*b))%2==0))\n {\n System.out.println(\"Yes\");\n }\n else\n {\n System.out.println(\"No\");\n }\n \n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279430014,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nint main() {\n int N;\n cin >> N;\n for(int i=0; i<N; i++) {\n int a, b;\n cin >> a >> b;\n if (a > 2 * b) {\n a = a - 2 *b;\n b = 0;\n }else{\n b = b - a / 2;\n a = a % 2;\n }\n a = a % 2;\n b = b % 2;\n if(a==0 && b == 0) cout << \"YES\" << endl;\n else cout << \"NO\" << 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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 281882462,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n \n#define forn(iter,size) for(int iter = 0; iter < int(size); ++iter)\n \ntypedef long long ll;\ntypedef double du;\nconst int \n INF = (int)1e9 + 7,\n MOD = (int)1e9 + 7;\n \nvoid solve(){\n int a,b;cin >> a >> b;\n if(b * 2 >= a){\n b -= a / 2;\n if(b % 2 != 0){cout << \"NO\\n\";return;}\n }else if(a > b * 2){\n a -= b * 2;\n if(a % 2 != 0){cout << \"NO\\n\";return;}\n }\n cout << \"YES\\n\";\n} \n \nint main(){\n //ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279100830,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nt = int(input())\nfor i in range(t):\n a, b = map(int, input().split())\n array = [1 for i in range(a)] + [2 for j in range(b)]\n if a==0 and b%2!=0 or a%2!=0 and b==0:\n print('NO')\n elif a==0 and b%2==0 or a%2==0 and b==0:\n print('YES')\n elif a%2==0 and (b-a//2)%2==0:\n print('YES')\n else:\n print('NO')\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279099293,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nx=int(input())\nmylist=[]\n \nfor _ in range(x):\n y = input()\n (a,b) = tuple(y.split())\n a = int(a)\n b = int(b)\n \n if a%2 == 1:\n print('NO')\n else:\n if b%2 == 0:\n print('YES')\n else:\n if a%4 == 2:\n print('YES')\n else:\n print('NO')\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279215465,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nimport java.util.*;\npublic class A_Sakurako_s_Exam {\n \n public static void main(String[] args) {\n Scanner sc=new Scanner(System.in);\n int tc=sc.nextInt();\n for(int io=0;io<tc;io++){\n int n=sc.nextInt(),m=sc.nextInt();\n if(n%2!=0){\n System.out.println(\"NO\");\n }else if((m%2)*2 == n){\n System.out.println(\"YES\");\n }else if(m*2==n){\n System.out.println(\"YES\");\n }else if(m%2==0 && n%2==0) {\n System.out.println(\"YES\");\n }else{\n System.out.println(\"NO\");\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279116592,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Exam\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.\n\nIn an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.\n\nSakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.\n\nInput: The first line contains a single integer $t$ ($1\\le t\\le 100$) — the number of test cases.\n\nThe only line of each test case contains two integers $a$ and $b$ ($0\\le a,b<10$) — the number of '1's and the number of '2's in the array.\n\nOutput: For each test case, output \"Yes\" if you can make the sum of the entire array equal to $0$, and \"No\" otherwise.\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n0 1\n0 3\n2 0\n2 3\n3 1\noutput:\nNO\nNO\nYES\nYES\nNO\n\nNote: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result; 3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$); 4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);\n\nBuggy Code:\nimport java.util.*;\npublic class Main {\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 ones = in.nextInt();\n int twos = in.nextInt();\n if (twos == 0 && ones % 2 == 0) {\n System.out.println(\"Yes\");\n }\n else if (twos % 2 == 0 && ones== 0){\n System.out.println(\"YES\");\n }\n else{\n int sum = ones + (2 * twos);\n if (sum % 2 != 0) {\n System.out.println(\"NO\");\n } else {\n ones /= 2;\n int total = ones + twos;\n if (total % 2 == 0) {\n System.out.println(\"YES\");\n } else {\n System.out.println(\"NO\");\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008A",
"submission_id": 279126934,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define ll long long\n#define ull unsigned long long\n#define db double\n#define ldb long double\n#define max(A,B) (A>B? A:B)\n#define min(A,B) (A<B? A:B)\n#define yes cout<<\"Yes\\n\"\n#define no cout<<\"No\\n\"\n#define pir pair<ll,ll>\n#define mpr(A,B) make_pair(A,B)\nusing namespace std;\n \nconst int N=2e5+10;\nconst int mod=998244353;\n \nint mul(int a,int b){\n return (ll)a*b%mod;\n}\nint pls(int a,int b){\n if(a+b>=mod) return a+b-mod;\n if(a+b<0) return a+b+mod;\n return a+b;\n}\nint qpow(int a,int b){\n int ret=1;\n while(b){\n if(b&1) ret=mul(ret,a);\n a=mul(a,a);\n b>>=1;\n }\n return ret;\n}\n \nint n,a[N];\nll s[N];\n \nvoid slv(){\n cin>>n;\n for(int i=1;i<=n;i++) cin>>a[i];\n s[0]=0;\n for(int i=1;i<=n;i++) s[i]=s[i-1]+a[i];\n ll c=0;\n for(int i=0;i<=n;i++) c=max(c,abs(s[i])+s[n]-s[i]);\n // cout<<\"c=\"<<c<<'\\n';\n int ans=0,cnt=0;\n for(int i=0;i<=n;i++){\n if(s[i]>=0&&i>0) cnt++;\n if(c==s[n]-2*s[i]){\n ans=pls(ans,qpow(2,n-i+cnt));\n // cout<<\" pls: i=\"<<i<<\" p=\"<<n-i+cnt<<'\\n';\n }\n }\n cout<<ans<<'\\n';\n return;\n}\n \nint main(){\n ios::sync_with_stdio(false);\n cin.tie(0);\n cout.tie(0);\n int T;\n cin>>T;\n while(T--){\n slv();\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 269166266,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\nusing namespace std;\n \n#include <iostream>\n#include <vector>\nusing namespace std;\n \nvoid solve(){\n int n;\n cin>>n;\n vector<long long> v;\n int a;\n cin>>a;\n v.push_back(a);\n for(int i=1;i<n;i++){\n long long b;\n cin>>b;\n v.push_back(b);\n }\n long long ans=0;\n long long nAns=0;\n long long ansWays=1;\n long long nWays=1;\n for(int i=0;i<v.size();i++){\n if(v[i]==0){\n ansWays*=2;\n nWays*=2;\n ansWays%=998244353;\n nWays%=998244353;\n }else if(v[i]<0){\n if(ansWays+v[i]>=0){\n ansWays*=2;\n ansWays%=998244353;\n }\n if(nAns+v[i]>=0){\n nWays*=2;\n nWays%=998244353;\n }\n if(abs(nAns+v[i])==ans+v[i]){\n ansWays+=nWays;\n ansWays%=998244353;\n }else if(abs(nAns+v[i])>abs(ans+v[i])){\n ansWays=nWays;\n }\n nAns+=v[i];\n ans=max(ans+v[i],abs(nAns));\n }else{\n nAns+=v[i];\n if(nAns>=0){\n nWays*=2;\n nWays%=998244353;\n }\n ans=ans+v[i];\n if(ans>=0){\n ansWays*=2;\n ansWays%=998244353;\n }\n }\n }\n if(abs(nAns)>ans){\n cout<<nWays<<\"\\n\";//\"---\"<<endl;\n }else{\n cout<<ansWays<<\"\\n\";//\"---\"<<endl;\n }\n //cout<<max(abs(nAns),ans)<<\"\\n\";\n}\n \nint main(){\n int t;\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264904227,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define int long long int\n#define yes cout << \"YES\" << endl;\n#define no cout << \"NO\" << endl;\n#define neg cout << -1 << endl;\n#define pb push_back\n#define ss second\n#define ff first\n \nconst int md=998244353;\nconst int N = 2e5+5;\nconst int INF=1e18;\n \nvoid solve()\n{\n int n;\n cin>>n;\n vector<int>v(n+1, 0), suffix(n+1, 0);\n for(int i=0; i<n; i++) cin>>v[i];\n suffix[n-1]=v[n-1];\n for(int i=n-2; i>=0; i--) suffix[i]=suffix[i+1]+v[i];\n int ans=abs(suffix[0]), total=0;\n int cnt=0, it=1;\n for(int i=0; i<n; i++)\n {\n total+=v[i];\n int mx=abs(abs(total)+suffix[i+1]);\n ans=max(ans, mx);\n }\n int nig=0;\n total=0;\n for(int i=0; i<n; i++)\n {\n total+=v[i];\n int mx=abs(abs(total)+suffix[i+1]);\n //cout<<\"mx \"<<mx<<endl;\n if(mx==ans and v[i]<0)\n {\n //cout<<\"vi \"<<v[i]<<\" \";\n nig++;\n }\n else if(v[i]>=0)\n {\n it*=2;\n it%=md;\n }\n }\n \n if(nig>0)\n {\n nig=(nig*(nig+1))/2;\n nig%=md;\n it*=nig;\n it%=md;\n }\n \n cout<<it<<endl;\n}\n \nsigned main()\n{\n ios_base::sync_with_stdio(false);\n cin.tie(0);\n cout.tie(0);\n \n int test = 1;\n cin >> test;\n \n for (int tc = 1; tc <= test; tc++)\n {\n solve();\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264957382,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nimport sys\n \n \ninput = lambda: sys.stdin.readline().rstrip()\n \n \ndef get_number():\n return int(input())\n \n \ndef get_numbers():\n return list(map(int, input().split()))\n \n \ndef get_word():\n return input()\n \n \ndef solve():\n n = get_number()\n a = get_numbers()\n \n positives = [0] * n\n running_sum = [0] * n\n for i in range(n):\n running_sum[i] = a[i] + running_sum[i - 1] if i > 0 else a[i]\n positives[i] = positives[i - 1] + (a[i] >= 0) if i > 0 else a[i] >= 0\n \n max_negative = 0\n idxs = []\n for i, s in enumerate(running_sum):\n if s < max_negative:\n max_negative = s\n idxs = [i]\n elif s == max_negative:\n idxs.append(i)\n \n mod = 998_244_353\n if not idxs:\n print(pow(2, n, mod))\n return\n \n total = 0\n \n for idx in idxs:\n s = 1\n s *= pow(2, positives[idx], mod)\n s *= pow(2, n - idx - 1, mod)\n total += s\n \n print(total % mod)\n \n \ndef main():\n for _ in range(get_number()):\n solve()\n \n \nif __name__ == \"__main__\":\n main()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264908887,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nimport sys\nimport math\nimport copy\nfrom collections import deque\nfrom collections import defaultdict as ddic\n#from atcoder import fenwicktree\n \nclass UndirectedGraph:\n def __init__(self, vertices):\n self.vertices = vertices\n self.adj = {i: [] for i in range(self.vertices)}\n \n def addEdge(self, u, v):\n self.adj[u].append(v)\n self.adj[v].append(u)\n \n def DFS(self, start):\n seen = set()\n stack = [start]\n while len(stack) > 0:\n curr = stack.pop()\n seen.add(curr)\n for elem in self.adj[curr]:\n if elem not in seen:\n stack.append(elem)\n return seen\n \n def BFS(self, u):\n visited = [False for i in range(self.vertices + 1)]\n distance = [-1 for i in range(self.vertices + 1)]\n \n distance[u] = 0\n queue = deque()\n queue.append(u)\n visited[u] = True\n \n while queue:\n front = queue.popleft()\n \n for i in self.adj[front]:\n if not visited[i]:\n visited[i] = True\n distance[i] = distance[front]+1\n queue.append(i)\n maxDis = 0\n \n for i in range(self.vertices):\n if distance[i] > maxDis:\n maxDis = distance[i]\n nodeIdx = i\n return distance\n \n def LongestPathLength(self):\n node, Dis = self.BFS(0)\n node_2, LongDis = self.BFS(node)\n \n print('Longest path is from', node, 'to', node_2, 'of length', LongDis)\n \nclass Fenwick:\n def __init__(self, arr):\n self.length = len(arr)\n self.bit = [0]*(self.length+1) \n for i in range(self.length): \n self.updateBit(i, arr[i])\n \n def getSum(self, i): \n s = 0\n i = i+1\n while i > 0: \n s += self.bit[i]\n i -= i & (-i)\n return s\n \n def getRangeSum(self, left, right):\n return self.getSum(right) - self.getSum(left-1)\n \n def updateBit(self, i, v):\n i += 1\n while i <= self.length: \n self.bit[i] += v\n i += i & (-i)\n \nclass DSU:\n def __init__(self, n):\n self.parent = [None for _ in range(n)]\n self.size = [0 for _ in range(n)]\n \n def findSet(self, v):\n if v == self.parent[v]:\n return v\n p = self.findSet(self.parent[v])\n self.parent[v] = p\n return p\n \n def makeSet(self, v):\n self.parent[v] = v\n self.size[v] = 1\n \n def unionSets(self, a, b):\n a = self.findSet(a)\n b = self.findSet(b)\n if (a != b):\n if self.size[a] < self.size[b]:\n a, b = b, a\n self.parent[b] = a\n self.size[a] += self.size[b]\n \ndef factors(n):\n result = set()\n for i in range(1, math.floor(math.sqrt(n))+1):\n if n%i == 0:\n result.add(i)\n result.add(n//i)\n return result\n \n \ndef main():\n fimp = sys.stdin.readline\n MOD = 10**9+7\n iprint = lambda x: '\\n'.join(map(str, x))\n sprint = lambda x: '\\n'.join(x)\n initl = lambda x,y: [y]*x\n INF = float('inf')\n NINF = float('-inf')\n ilist = lambda: list(map(int, fimp().split()))\n slist = lambda: list(input())\n slists = lambda: input().split()\n ivars = lambda: map(int, fimp().split())\n ipt = lambda: int(fimp())\n out = []\n \n mod = 998244353\n t = ipt()\n for _ in range(t):\n n = ipt()\n a = ilist()\n c = 0\n rs = [0 for _ in range(n)]\n rs[0] = a[0]\n for i in range(1, n):\n rs[i] = rs[i-1]+a[i]\n l = 0\n count = 0\n currCount = 0\n add = False\n numNegatives = 0\n for i in range(n):\n if a[i] < 0:\n currCount+= 1\n if rs[i]-l < 0 and c+rs[i]-l < 0:\n numNegatives+= 1\n count+= currCount\n currCount = 0\n c+= rs[i]-l\n l = rs[i]\n elif i == n-1:\n if rs[i]-l > 0 and c < 0:\n c = abs(c)\n c+= rs[i]-l\n if rs[i]-l < 0:\n count+= currCount\n elif rs[i]-l == 0:\n count+= currCount\n add = True\n \n ans = pow(2, n-count, mod)\n if add:\n if numNegatives > 0:\n currCount+= 1\n addition = pow(2, currCount, mod)-1\n ans = (ans*addition)%mod\n \n print(ans)\n \n \n \n \nif __name__ == \"__main__\":\n main()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264942175,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nMOD = 998244353\ntot = int(input())\nfor _ in range(tot):\n n = int(input())\n nums = list(map(int, input().split(' ')))\n mx = 0\n mn = 0\n s = 0\n for x in nums:\n mx = abs(mx + x)\n s += x\n mn = min(mn, s)\n if mn >= 0:\n print(1 << n)\n continue\n ans = 0\n s = 0\n less_than_zero = 0\n for i, x in enumerate(nums):\n s += x\n if s == mn:\n ans = (ans + pow(2, n-1-less_than_zero)) % MOD\n if s < 0:\n less_than_zero += 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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 284041097,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nimport java.lang.*; // imports everything\nimport java.util.*;\n \npublic class C_2_Magnitude_Hard_Version\n{\n //static ArrayList<Integer> a=new ArrayList<>();\n //static int max;\n static HashMap<Long,Long> map=new HashMap<>();\n static long a[];\n static long k=998244353;\n static long max=Long.MIN_VALUE;\n static int n;\n public static void main (String[] args) throws java.lang.Exception\n {\n Scanner sc=new Scanner(System.in);\n int caseNo,i,j,N;\n N=sc.nextInt();\n for(caseNo=0;caseNo<N;caseNo++)\n {\n n=sc.nextInt();\n a=new long[n];\n for(i=0;i<n;i++)\n {\n a[i]=sc.nextLong();\n }\n calculate(0,0);\n System.out.println(map.get(max));\n map.clear();\n max=Long.MIN_VALUE;\n }\n }\n \n public static void calculate(int i,long sum)\n {\n if(i==n)\n {\n if(sum<max)\n {\n return ;\n }\n map.put(sum,(map.getOrDefault(sum,(long)0)+1)%k);\n max=Math.max(max,sum);\n return ;\n }\n calculate(i+1,sum+a[i]);\n calculate(i+1,Math.abs(sum+a[i]));\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264939454,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class C {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int tes = scanner.nextInt();\n \n for(int t = 0; t < tes; t++){\n int n = scanner.nextInt();\n long[] a = new long[n];\n long[] b = new long[n];\n \n b[0] = scanner.nextInt();\n a[0] = Math.max(b[0], -b[0]);\n \n \n long[] ac = new long[n];\n long[] bc = new long[n]; \n \n if(b[0] >= 0){\n ac[0] = 2;\n bc[0] = 2;\n }else{\n ac[0] = 1;\n bc[0] = 1;\n }\n \n for(int i = 1; i < n; i++){\n long x = scanner.nextLong();\n long cnt = 0;\n \n a[i] = Math.max(a[i - 1] + x, Math.max(Math.abs(x + a[i - 1]), Math.abs(x + b[i - 1])));\n b[i] = b[i - 1] + x;\n \n if(a[i - 1] != b[i - 1]){\n if(a[i] == a[i - 1] + x)cnt++;\n if(a[i] == Math.abs(a[i - 1] + x))cnt++;\n if(a[i] == Math.abs(b[i - 1] + x)){\n ac[i] = ac[i - 1] * cnt + bc[i - 1];\n }else{\n ac[i] = ac[i - 1] * cnt;\n }\n }else{\n if(a[i] == a[i - 1] + x)cnt++;\n if(a[i] == Math.abs(a[i - 1] + x))cnt++;\n ac[i] = ac[i - 1] * cnt;\n }\n \n if(x >= 0)bc[i] = bc[i - 1] * 2;\n else bc[i] = bc[i - 1];\n \n }\n \n System.out.println(ac[n - 1]);\n \n \n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264941595,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\ntypedef long long ll;\nconst ll N = 2e5;\nconst ll mod = 998244353;\nconst ll inf = 1e18;\nll v[N];\npair<ll,ll> nr[4];\nvoid solve(){\n ll n,mx = 0,mn = 0;\n ll m1 = 1,m2 = 1;\n cin>>n;\n for(ll i = 0;i < n;i++){\n cin>>v[i];\n int cnt;\n if(mx == mn){\n nr[0] = {mx + v[i],m1};\n nr[1] = {abs(mx + v[i]),m2};\n cnt = 2;\n }else{\n nr[0] = {mx + v[i],m1};\n nr[1] = {abs(mx + v[i]),m2};\n nr[2] = {mn + v[i],m1};\n nr[3] = {abs(mn + v[i]),m2};\n cnt = 4;\n }\n ll mx2 = -inf,mn2 = inf,m11,m22;\n for(int j = 0;j < cnt;j++){\n if(mx2 < nr[j].first){\n mx2 = nr[j].first;\n m11 = nr[j].second;\n }else if(mx2 == nr[j].first){\n m11+=nr[j].second;\n m11%=mod;\n }\n if(mn2 > nr[j].first){\n mn2 = nr[j].first;\n m22 = nr[j].second;\n }else if(mn2 == nr[j].first){\n m22+=nr[j].second;\n m22%=mod;\n }\n }\n mn = mn2;\n mx = mx2;\n m1 = m11;\n m2 = m22;\n }\n cout<<m1<<'\\n';\n}\nint main(){\n ll t;\n cin>>t;\n while(t--)solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264890699,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define int long long\nlong long mod = 998244353;\nsigned main(){\n ios_base::sync_with_stdio(false);cin.tie(0);\n int t;\n cin>>t;\n long long po[200001];\n po[0] = 1;\n for(int i = 1;i<=2e5;i++){\n po[i] = (po[i-1]*2)%mod;\n }\n while(t--){\n int n;\n cin>>n;\n int arr[n];\n long long suf[n+1];\n suf[n] = 0;\n for(int i = 0;i<n;i++){\n cin>>arr[i];\n }\n for(int i = n-1;i>=0;i--){\n suf[i] = suf[i+1]+arr[i];\n }\n int sum = 0;\n long long ma = 0 , nah = 1;\n long long all = 0;\n for(int i = 0;i<=n;i++){\n int sum2 = abs(sum)+suf[i];\n ma = max(ma,sum2);\n if(i<n){\n sum+=arr[i];\n }\n }\n sum = 0;\n bool taked = 0;\n for(int i = 0;i<=n;i++){\n int sum2 = abs(sum)+suf[i];\n if(ma==sum2){\n if(sum>=0){\n all += (nah*po[n-i])%mod;\n taked = 1;\n break;\n }else{\n all += (nah*po[n-i])%mod;\n taked = 1;\n }\n }\n if(i<n){\n sum+=arr[i];\n if(sum>=0)nah*=2;\n nah%=mod;\n }\n }\n cout<<all<<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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264916121,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nimport sys\nimport math\nimport heapq\nfrom collections import deque\n \ndef II():\n\treturn int(sys.stdin.readline())\n \ndef LI():\n\treturn list(map(int, sys.stdin.readline().split()))\n \ndef MI():\n\treturn map(int, sys.stdin.readline().split())\n \ndef SI():\n\treturn sys.stdin.readline().strip()\n \ndef ask(l,x):\n # sys.stdout.flush()\n print(\"?\",l,x)\n sys.stdout.flush()\n x = II()\n return x\n \ndef answer(m):\n # sys.stdout.flush()\n print(\"!\", m)\n sys.stdout.flush()\n \ndef lcm(a, b):\n return abs(a*b) // math.gcd(a, b)\ndef md(x,y,x1,y1):\n return(abs(x-x1) + abs(y-y1))\n \nmod = 998244353\n \nfor _ in range(II()):\n n = II()\n a = LI()\n if 0 in a:\n print(pow(2, n, mod))\n continue\n c = 0\n c2 = 0\n ans = 1\n boo = False\n cc = 0\n for i in a:\n c+=i\n c2+=i\n if i > 0:\n ans*=2\n ans%=mod\n else:\n if abs(c2) > c:\n boo = True\n cc+=1\n if c == abs(c2) and boo:\n ans*=3*pow(2, cc-1, mod)\n boo = False\n ans%=mod\n cc = 0\n c = max(abs(c), abs(c2))\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264917850,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nimport sys\ninput = sys.stdin.readline\nINF = 998244353\n \nfor _ in range(int(input())):\n n = int(input())\n a = [*map(int, input().split())]\n c = 0\n prefix_sum = [0]\n for i in a:\n prefix_sum.append(prefix_sum[-1] + i)\n \n res = prefix_sum[-1]\n idx = [-1]\n for i in range(1, n + 1):\n if prefix_sum[i] < 0:\n if abs(prefix_sum[-1] - 2 * prefix_sum[i]) > res:\n idx = [i]\n res = abs(prefix_sum[-1] - 2 * prefix_sum[i])\n elif abs(prefix_sum[-1] - 2 * prefix_sum[i]) == res:\n idx.append(i)\n plus = 0\n for i in prefix_sum[1:]:\n if i >= 0:\n plus += 1\n cnt = ((pow(2, len(idx), INF) - 1) * pow(2, plus, INF)) % INF\n \n print(cnt)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264913002,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n private static BufferedReader br;\n private static StringTokenizer toker = null;\n private static int mod = 998244353;\n private static String next() throws IOException {\n if (br == null) br = new BufferedReader(new InputStreamReader(System.in));\n while (toker == null || !toker.hasMoreTokens()) {\n toker = new StringTokenizer(br.readLine());\n \n }\n return toker.nextToken();\n }\n private static int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n private static long nextLong() throws IOException {\n return Long.parseLong(next());\n }\n public static void main(String[] args) throws IOException {\n int t = nextInt();\n StringBuilder sb = new StringBuilder();\n for (int g = 0; g < t; ++g) {\n int n = nextInt();\n int[] arr = new int[n];\n for (int i = 0; i < n; ++i) arr[i] = nextInt();\n long[][] db = new long[n][2];\n db[0][0] = Math.abs(arr[0]);\n db[0][1] = arr[0];\n // System.out.println(0 + \", \" + db[0][0] + \", \" + db[0][1]);\n for (int i = 1; i < n; ++i) {\n db[i][0] = Math.max(Math.abs(db[i - 1][0] + arr[i]), Math.abs(db[i - 1][1] + arr[i]));\n db[i][1] = db[i - 1][1] + arr[i];\n // System.out.println(i + \", \" + db[i][0] + \", \" + db[i][1]);\n }\n long[][] db2 = new long[n][2];\n db2[0][0] = arr[0] >= 0 ? 2 : 1;\n db2[0][1] = arr[0] >= 0 ? 2 : 1;\n // System.out.println(0 + \", \" + db2[0][0] + \", \" + db2[0][1]);\n for (int i = 1; i < n; ++i) {\n if (db[i - 1][0] + arr[i] == db[i][0]) db2[i][0] = (db2[i][0] + db2[i - 1][0]) % mod;\n if (Math.abs(db[i - 1][0] + arr[i]) == db[i][0]) db2[i][0] = (db2[i][0] + db2[i - 1][0]) % mod;\n if (db[i - 1][0] != db[i - 1][1] && Math.abs(db[i - 1][1] + arr[i]) == db[i][0]) db2[i][0] = (db2[i][0] + db2[i - 1][1]) % mod;\n db2[i][1] = (db2[i][1] + (db[i - 1][0] + arr[i] >= 0 ? 2 : 1) * db2[i - 1][1]) % mod;\n // System.out.println(i + \", \" + db2[i][0] + \", \" + db2[i][1]);\n }\n sb.append(db2[n - 1][0]).append(\"\\n\");\n \n \n }\n System.out.println(sb);\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 265120040,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.util.Arrays;\n \npublic class MagnitudeHard {\n final static int mod = 998244353;\n public static void main(String[] args){\n try(BufferedReader br = new BufferedReader(new InputStreamReader(System.in))){\n int t = Integer.parseInt(br.readLine());\n \n for(int i=0; i<t; i++){\n int n = Integer.parseInt(br.readLine());\n int[] inputArr = Arrays.stream(br.readLine().trim().split(\"\\\\s+\")).mapToInt(Integer::parseInt).toArray();\n System.out.println(solve(n, inputArr));\n }\n }\n catch(IOException ex){\n throw new RuntimeException(ex);\n }\n }\n \n static long solve(int n, int[] input){\n \n long[][] minmax = new long[n+1][2]; // 0: min, 1: max\n long[][] now = new long[n+1][2]; // 0: minWays, 1: maxWays\n \n for(int i=1; i<=n; i++){\n minmax[i][0] = minmax[i-1][0] + input[i-1];\n minmax[i][1] = Math.max(Math.abs(minmax[i-1][0] + input[i-1]), Math.abs(minmax[i-1][1] + input[i-1]));\n }\n \n now[0][0] = 1;\n now[0][1] = 1;\n \n for(int i=1; i<=n; i++){\n if(minmax[i-1][0] == minmax[i-1][1]){\n if(minmax[i][0] == minmax[i-1][0] + input[i-1]){\n now[i][0] += now[i-1][0];\n }\n if(minmax[i][0] == Math.abs(minmax[i-1][0] + input[i-1])){\n now[i][0] += now[i-1][0];\n }\n if(minmax[i][1] == minmax[i-1][0] + input[i-1]){\n now[i][1] += now[i-1][0];\n }\n if(minmax[i][1] == Math.abs(minmax[i-1][0] + input[i-1])){\n now[i][1] += now[i-1][0];\n }\n }\n else{\n if(minmax[i][0] == minmax[i-1][0] + input[i-1]){\n now[i][0] += now[i-1][0];\n }\n if(minmax[i][0] == Math.abs(minmax[i-1][0] + input[i-1])){\n now[i][0] += now[i-1][0];\n }\n if(minmax[i][1] == minmax[i-1][0] + input[i-1]){\n now[i][1] += now[i-1][0];\n }\n if(minmax[i][1] == Math.abs(minmax[i-1][0] + input[i-1])){\n now[i][1] += now[i-1][0];\n }\n \n if(minmax[i][0] == minmax[i-1][1] + input[i-1]){\n now[i][0] += now[i-1][0];\n }\n if(minmax[i][0] == Math.abs(minmax[i-1][1] + input[i-1])){\n now[i][0] += now[i-1][0];\n }\n if(minmax[i][1] == minmax[i-1][1] + input[i-1]){\n now[i][1] += now[i-1][0];\n }\n if(minmax[i][1] == Math.abs(minmax[i-1][1] + input[i-1])){\n now[i][1] += now[i-1][0];\n }\n }\n \n now[i][0] %= mod;\n now[i][1] %= mod;\n }\n \n return now[n][1] % mod;\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 265201659,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\nconst int MOD = 998244353;\nconst int MAX = 2 * 1e5;\nint N;\nlong long A[MAX + 3], pfx[MAX + 3];\nint nk, pk;\nint pw[MAX + 3];\n \n \nvoid test_case() {\n nk = 0, pk = 0;\n long long minn = 1e18;\n cin >> N;\n for(int i = 1; i <= N; i++) {\n cin >> A[i];\n pfx[i] = pfx[i - 1] + A[i];\n minn = min(minn, pfx[i]);\n }\n \n for(int i = 1; i <= N; i++) {\n if(pfx[i] >= 0) pk++;\n else {\n if(minn == pfx[i]) nk++;\n }\n }\n \n if(minn >= 0) nk = 1;\n cout << (1LL * (1LL * pw[nk] - 1) * pw[pk]) % MOD << '\\n';\n}\n \nint main()\n{\n //freopen(\"file.in\", \"r\", stdin);\n //freopen(\"file.out\", \"w\", stdout);\n \n int T;\n cin >> T;\n pw[0] = 1;\n for(int i = 1; i <= 2 * 1e5; i++) {\n pw[i] = (1LL * pw[i - 1] * 2) % MOD;\n }\n while(T--) {\n test_case();\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 265716902,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define rep(i, a, b) for (long long i = a; i < b; i++)\n \nint main()\n{\n ios::sync_with_stdio(0);\n cin.tie(0);\n long long T;\n cin >> T;\n vector<long long> sol(T);\n rep(I, 0, T)\n {\n long long kl=998244353;\n long long n;\n cin >> n;\n vector<long long> a(n);\n rep(i, 0, n)\n {\n cin >> a[i];\n }\n vector <long long> pref(n);\n pref[0]=a[0];\n long long min=a[0];\n long long id=0;\n rep(i,1,n){\n pref[i]=a[i]+pref[i-1];\n if(min > pref[i]){\n min=pref[i];\n id=i;\n }\n }\n if(min>=0){\n long long pp=1;\n rep(i,0,n){\n pp=(pp*2)%kl;\n }\n sol[I]=pp;\n continue;\n }\n vector <long long> pr;\n int neg_count=0;\n rep(i,0,n){\n if(pref[i]<0){\n neg_count++;\n }\n if(min==pref[i]){\n pr.push_back(neg_count);\n }\n }\n long long inter_sum=1;\n rep(j,0,n-neg_count){\n inter_sum=(inter_sum*2)%kl;\n }\n long long sum=inter_sum;\n for(int i=pr.size()-2;i>-1;i--){\n long long diff=pr[i+1]-pr[i];\n while(diff>0){\n inter_sum=(inter_sum*2)%kl;\n diff--;\n }\n sum=(inter_sum+sum)%kl;\n }\n sol[I]=sum;\n }\n rep(i, 0, T)\n {\n cout << sol[i] << \"\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 265007914,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nfrom collections import deque\nt=int(input())\nfor _ in range(t):\n n=int(input())\n a=list(map(int,input().split()))\n c=0\n \n \n \n # for i in range(len(a)):\n # if a[i]*c>=0:\n # c+=a[i]\n # else:\n # c=abs(c)+a[i]\n \n \n \n \n # print(abs(c))\n sum=0\n l=deque()\n for i in range(n):\n sum+=a[i]\n l.append(sum)\n maxi=l[-1]\n for i in range(len(l)):\n if l[i]<0:\n \n maxi=max(maxi,l[-1]+2*abs(l[i]))\n cnt=0\n ptr=-1\n for i in range(len(l)):\n if l[-1]+2*abs(l[i]) == maxi and l[i]<0:\n cnt+=1\n ptr=i\n ans=1\n for i in range(len(l)):\n \n if l[i]>=0:\n ans*=2\n \n \n \n if cnt==0:\n print(ans%998244353)\n else:\n ans2=(2**cnt)-1\n for i in range(ptr+1,len(l)):\n if l[i]<0:\n ans2*=2\n print((ans*ans2)%998244353)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 266371657,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nfrom math import ceil\ndef li():return [int(i)for i in input().split()]\nmod=998244353\nfor _ in range(int(input())):\n\tn,=li()\n\ta=li()\n\tp,s=[],0\n\tfor i in a:s+=i;p+=[s]\n\tmi,q=min(p),[]\n\tif mi>=0:print(\" \",pow(2,n,mod));continue\n\tfor i in range(n):\n\t\tif p[i]==mi:q+=[i]\n\tq=q[::-1]\n\ts=0\n\tans,f=0,1\n\tfor i in range(n):\n\t\tif not q:break\n\t\tif i==q[-1]:\n\t\t\tans+=f*pow(2,n-1-i,mod)\n\t\t\tq.pop()\n\t\telse:\n\t\t\tif p[i]>=0:f=(f*2)%mod\n\tprint(\" \",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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 284490701,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n \n static long MOD = 998244353;\n \n public static void main(String[] args) throws IOException {\n BufferedReader f = new BufferedReader(new InputStreamReader(System.in));\n PrintWriter out = new PrintWriter(System.out);\n StringTokenizer st;\n int t = Integer.parseInt(f.readLine());\n while (t-- > 0) {\n int n = Integer.parseInt(f.readLine());\n int[] arr = new int[n];\n st = new StringTokenizer(f.readLine());\n for (int i = 0; i < n; i++) {\n arr[i] = Integer.parseInt(st.nextToken());\n }\n long[][] ans = new long[n][2];\n long[][] dp = new long[n][2];\n ans[0][0] = ans[0][1] = arr[0];\n ans[0][1] = Math.abs(arr[0]);\n dp[0][0] = dp[0][1] = 2;\n if (arr[0] < 0) dp[0][0] = dp[0][1] = 1;\n for (int i = 1; i < n; i++) {\n if (arr[i] == 0) {\n ans[i][0] = ans[i - 1][0];\n ans[i][1] = ans[i - 1][1];\n dp[i][0] = dp[i - 1][0] % MOD;\n if (dp[i][0] >= 0) {\n dp[i][0] *= 2;\n dp[i][0] %= MOD;\n }\n dp[i][1] = dp[i - 1][1] * 2 % MOD;\n }\n else {\n ans[i][0] = ans[i - 1][0] + arr[i];\n dp[i][0] = dp[i - 1][0];\n if (Math.abs(ans[i - 1][0] + arr[i]) == ans[i - 1][0] + arr[i]) {\n dp[i][0] *= 2;\n dp[i][0] %= MOD;\n }\n ans[i][1] = Math.max(Math.abs(ans[i - 1][0] + arr[i]), Math.abs(ans[i - 1][1] + arr[i]));\n int count = 0;\n int count2 = 0;\n if (ans[i - 1][0] + arr[i] == ans[i][1]) {\n count2++;\n }\n if (ans[i - 1][1] + arr[i] == ans[i][1]) {\n count++;\n }\n if (Math.abs(ans[i - 1][0] + arr[i]) == ans[i][1]) {\n count2++;\n }\n if (Math.abs(ans[i - 1][1] + arr[i]) == ans[i][1]) {\n count++;\n }\n if (ans[i - 1][0] == ans[i - 1][1]) {\n dp[i][1] = dp[i - 1][1] * count;\n dp[i][1] %= MOD;\n }\n else {\n dp[i][1] = dp[i - 1][1] * count + dp[i - 1][0] * count2;\n dp[i][1] %= MOD;\n }\n }\n }\n out.println(dp[n - 1][1]);\n }\n f.close();\n out.close();\n }\n \n public static void sort(int[] arr) {\n Random r = new Random();\n for (int i = 0; i < arr.length; i++) {\n int x = r.nextInt(arr.length);\n int temp = arr[x];\n arr[x] = arr[i];\n arr[i] = temp;\n }\n Arrays.sort(arr);\n }\n \n public static void sort(int[][] arr) {\n Random r = new Random();\n for (int i = 0; i < arr.length; i++) {\n int x = r.nextInt(arr.length);\n int[] temp = arr[x];\n arr[x] = arr[i];\n arr[i] = temp;\n }\n Arrays.sort(arr, Comparator.comparingInt(a -> a[0]));\n }\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264894077,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Magnitude (Hard Version)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.\n\nYou are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:\n\n * Option $1$: set $c$ to $c + a_i$. * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$. \n\nLet the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.\n\nSince the answer may be large, output it modulo $998\\,244\\,353$.\n\nInput: The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe first line of each test case contains a single integer $n$ ($2 \\leq n \\leq 2 \\cdot 10^5$).\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($-10^9 \\leq a_i \\leq 10^9$).\n\nThe sum of $n$ over all test cases does not exceed $3 \\cdot 10^5$.\n\nOutput: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\\,244\\,353$.\n\nExamples:\ninput:\n5\n4\n2 -5 3 -3\n8\n1 4 3 4 1 4 3 4\n3\n-1 -2 -3\n4\n-1000000000 1000000000 1000000000 1000000000\n4\n1 9 8 4\noutput:\n12\n256\n1\n8\n16\n\nNote: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \\cdot 2 = 4$ ways for them. In total, we have $3 \\cdot 4 = 12$ ways.\n\nIn the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\nimport java.util.function.*;\nimport java.util.stream.*;\n \npublic class Main {\n \n static final long M = 998244353;\n \n public static void main(String[] args) {\n FastReader sc = new FastReader();\n long start1 = System.currentTimeMillis();\n \n ArrayList<Long> p2 = new ArrayList<>();\n long d = 1;\n for (int i = 0; i <= 2 * 1e5; i++) {\n p2.add(d);\n d = (d * 2) % M;\n }\n \n int t = sc.nextInt();\n while (t-- > 0) {\n int n = sc.nextInt();\n ArrayList<Long> v = new ArrayList<>();\n scn(v, n, sc);\n long mn = Long.MAX_VALUE;\n \n long a = 0;\n int c = 0;\n for (int i = 0; i < n; i++) {\n a += v.get(i);\n mn = Math.min(mn, a);\n if (a < 0) {\n c++;\n }\n }\n \n if (mn >= 0) {\n System.out.println(p2.get(n));\n } else if (c == n) {\n System.out.println(1);\n } else {\n System.out.println(((p2.get(c) - 1) * (p2.get(n - c))) % M);\n }\n }\n \n long stop1 = System.currentTimeMillis();\n System.err.println(\"Time: \" + (stop1 - start1) + \" ms\");\n }\n \n static void scn(ArrayList<Long> v, int n, FastReader sc) {\n for (int i = 0; i < n; i++) {\n v.add(sc.nextLong());\n }\n }\n \n static long modexp(long x, long n) {\n if (n == 0) {\n return 1;\n } else if (n % 2 == 0) {\n return modexp((x % M * x % M) % M, (n / 2) % M) % M;\n } else {\n return (((x % M) * (modexp((x % M * x % M) % M, ((n - 1) / 2) % M) % M)) % M);\n }\n }\n \n static long get(long a, long b) {\n long c = gcd(a, b) % M;\n a = a / c;\n a %= M;\n b = b / c;\n b %= M;\n long d = modexp(b, M - 2) % M;\n long ans = ((a % M) * (d % M)) % M;\n return ans % M;\n }\n \n static long gcd(long a, long b) {\n if (b == 0) return a;\n return gcd(b, a % b);\n }\n \n static long power(long a, long b) {\n long result = 1;\n while (b > 0) {\n if ((b & 1) == 1) {\n result = (result * a) % M;\n }\n a = (a * a) % M;\n b >>= 1;\n }\n return result;\n }\n \n static long nCr(long n, long r) {\n if (n - r == 1) {\n return n;\n }\n if (r > n) return 0;\n if (r == 0 || n == r) return 1;\n double res = 0;\n for (int i = 0; i < Math.min(r, n - r); i++) {\n res += Math.log(n - i) - Math.log(i + 1);\n }\n return Math.round(Math.exp(res)) % M;\n }\n \n static long pr(long a, long b) {\n return ((a % M) * (b % M)) % M;\n }\n \n static boolean ch(long n) {\n while (n != 0) {\n if (n % 10 < 5) {\n return false;\n }\n n /= 10;\n }\n return true;\n }\n \n static boolean regu(String s) {\n int i = -1;\n for (char ch : s.toCharArray()) {\n if (ch == '(' || ch == '{' || ch == '[') {\n i++;\n s = s.substring(0, i) + ch + s.substring(i + 1);\n } else {\n if (i >= 0 && ((s.charAt(i) == '(' && ch == ')') || (s.charAt(i) == '{' && ch == '}') || (s.charAt(i) == '[' && ch == ']'))) {\n i--;\n } else {\n return false;\n }\n }\n }\n return i == -1;\n }\n \n static int MSBPosition(long N) {\n int msb_p = -1;\n while (N > 0) {\n N >>= 1;\n msb_p++;\n }\n return msb_p;\n }\n \n static long findBitwiseOR(long L, long R) {\n long res = 0;\n int msb_p1 = MSBPosition(L);\n int msb_p2 = MSBPosition(R);\n while (msb_p1 == msb_p2) {\n long res_val = 1 << msb_p1;\n res += res_val;\n L -= res_val;\n R -= res_val;\n msb_p1 = MSBPosition(L);\n msb_p2 = MSBPosition(R);\n }\n msb_p1 = Math.max(msb_p1, msb_p2);\n for (int i = msb_p1; i >= 0; i--) {\n long res_val = 1 << i;\n res += res_val;\n }\n return res;\n }\n \n static boolean checkPalindrome(int n) {\n int reverse = 0;\n int temp = n;\n while (temp != 0) {\n reverse = (reverse * 10) + (temp % 10);\n temp /= 10;\n }\n return reverse == n;\n }\n \n static long uro(long n, long k) {\n if (n % k == 0) {\n return n / k;\n } else {\n return n / k + 1;\n }\n }\n \n static long lis(ArrayList<Long> arr, int n) {\n ArrayList<Long> v = new ArrayList<>(arr);\n ArrayList<Long> ans = new ArrayList<>();\n ans.add(arr.get(0));\n for (int i = 1; i < n; i++) {\n int pos = upperBound(ans, v.get(i));\n if (pos < ans.size()) {\n ans.set(pos, v.get(i));\n } else {\n ans.add(v.get(i));\n }\n }\n return ans.size();\n }\n \n static int upperBound(ArrayList<Long> list, long value) {\n int low = 0;\n int high = list.size();\n while (low < high) {\n int mid = (low + high) / 2;\n if (list.get(mid) <= value) {\n low = mid + 1;\n } else {\n high = mid;\n }\n }\n return low;\n }\n \n static void pridiv(long n, ArrayList<Long> v) {\n for (int i = 1; i <= Math.sqrt(n); i++) {\n if (n % i == 0) {\n if (n / i == i) {\n v.add((long) i);\n } else {\n v.add((long) i);\n v.add(n / i);\n }\n }\n }\n }\n \n static void scv(ArrayList<Long> v, int n) {\n for (int i = 0; i < n; i++) {\n System.out.print(v.get(i) + \" \");\n }\n System.out.println();\n }\n \n static class FastReader {\n BufferedReader br;\n StringTokenizer st;\n \n public FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() {\n while (st == null || !st.hasMoreElements()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine() {\n String str = \"\";\n try {\n str = br.readLine();\n } catch (IOException e) {\n e.printStackTrace();\n }\n return str;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1984C2",
"submission_id": 264933390,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\n#include <cstdio>\nusing namespace std;\n#define __IN freopen(\"a.in\",\"r\",stdin)\n#define __OUT freopen(\"k.out\",\"w\",stdout)\n#define ll long long\nconst int N = 2e5 + 10;\n \n ll a[N], b[N], t, n;\n \nint solve(ll v, ll pos)\n{\n // printf(\"-%lld-\\n\", v);\n for(int i=1; i<=n; i++) b[i] = a[i];\n // for (int i=1;i<=n;i++) printf(\"%lld \", b[i]);\n // printf(\"\\n\");\n for(int i=0; i<n; i++)\n {\n ll p = (pos + i - 1) % n + 1;\n ll pi = (pos + i) % n + 1;\n if (b[p] < v) return false;\n ll k = (b[p] - v) / 2;\n b[p] -= 2 * k;\n b[pi] += k;\n }\n \n for (int i=1;i<=n;i++)\n {\n if (b[i] != b[1]) return false;\n }\n return true;\n}\nint main()\n{\n // __IN;__OUT;\n scanf(\"%lld\", &t);\n while(t--)\n {\n scanf(\"%lld\", &n);\n ll pos = 0, mx = 0, mi = 1e9+10, sum = 0;\n for (int i=1;i<=n;i++) \n {\n scanf(\"%lld\", a+i);\n if (a[i] > mx)\n {\n pos = i;\n mx = a[i];\n }\n if (a[i] < mi)\n {\n mi = a[i];\n }\n sum += a[i];\n }\n ll l = mi, r = mx, res = -1;\n while(l <= r)\n {\n ll mid = (l+r) / 2 ;\n if (solve(mid, pos)) \n {\n res = mid;\n l = mid + 1;\n }\n else \n {\n r = mid - 1;\n }\n }\n printf(\"%lld\\n\", res == -1 ? -1 : sum-(res*n));\n // printf(\"%lld \", res);\n // for (int i=1;i<=n;i++) printf(\"%lld \", a[i]);\n // printf(\"\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 294578885,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\n#include <cstdio>\n#include <algorithm>\nusing namespace std;\n \nint t;\nint n;\nint a[300005];\nint b[300006];\nint chk(int k){\n\tint i,res=0,c;\n\tfor(i=1;i<=n;i++) a[i]=b[i];\n\ta[1]-=k*2;\n\ta[2]+=k;\n\tres+=k;\n\tfor(i=2;i<n;i++){\n\t\tif((a[i]-a[1])%2==1||a[i]<a[1]) return -1;\n\t\telse{\n\t\t\tc=(a[i]-a[1])/2;\n\t\t\ta[i]=a[1];\n\t\t\ta[i+1]+=c;\n\t\t\tres+=c;\n\t\t}\n\t}\n\tif(a[n]==a[1]) return res;\n\telse return -1;\n}\nint slv(){\n\tint l,r,mid,res,cnt;\n\tres=cnt=-1;\n\tl=0;\n\tr=(b[n]-b[1])/2;\n\twhile(l<=r){\n\t\tmid=l+r>>1;\n\t\tcnt=chk(mid);\n\t\tif(cnt!=-1){\n\t\t\tres=cnt;\n\t\t\tr=mid-1;\n\t\t}else l=mid+1;\n\t}\n\treturn res;\n}\n \nint main(){\n\tint i,j,r1,r2;\n\tscanf(\"%d\",&t);\n\twhile(t--){\n\t\tscanf(\"%d\",&n);\n\t\tj=1;\n\t\tfor(i=1;i<=n;i++){\n\t\t\tscanf(\"%d\",&a[i]);\n\t\t\tif(i!=1&&a[i]<=a[j]) j=i;\n\t\t}\n\t\tj=j%n+1;\n\t\tfor(i=1,j;i<=n;i++,j=j%n+1) b[i]=a[j];\n\t\tr1=slv();\n\t\tb[1]++;\n\t\tb[n]-=2;\n\t\tr2=slv();\n\t\tif(r2!=-1) r2++;\n\t\tif(r1==-1&&r2==-1) puts(\"-1\");\n\t\telse if(r1==-1) printf(\"%d\\n\",r2);\n\t\telse if(r2==-1) printf(\"%d\\n\",r1);\n\t\telse printf(\"%d\\n\",min(r1,r2));\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292191021,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\n#include <iostream>\nusing namespace std;\n \nlong long test, n, cnt, a[200010], tmp;\npair<long long,long long> maxx;\n \nint main() {\n\tios::sync_with_stdio(0);\n\tcin.tie(0); cout.tie(0);\n\tcin >> test;\n\twhile (test--) {\n\t\tcin >> n;\n\t\tcnt = 0;\n\t\tfor (int i = 0; i < n; i++) cin >> a[i];\n\t\tmaxx = {0,0};\n\t\tfor (int i = 0; i < n; i++) maxx = max(maxx,{a[i],i});\n\t\tfor (int i = 1; i <= 40; i++) {\n\t\t\tfor (int j = 0; j < n; j++) {\n\t\t\t\tint ii = (maxx.second+j)%n;\n\t\t\t\tif (a[ii] < 2) continue;\n\t\t\t\ttmp = max(0LL,max((a[ii]-a[(ii+1)%n]+2)/3,(a[ii]-a[(ii-1+n)%n]+1)/2));\n\t\t\t\tcnt += tmp;\n\t\t\t\ta[ii] -= tmp*2;\n\t\t\t\ta[(ii+1)%n] += tmp;\n\t\t\t}\n\t\t}\n\t\tfor (int i = 1; i < n; i++) {\n\t\t\tif (a[i] != a[i-1]) {\n\t\t\t\tcout << -1 << endl;\n\t\t\t\tgoto skip;\n\t\t\t}\n\t\t}\n\t\tcout << cnt << endl;\n\t\tskip:;\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292176268,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\ndef check(L,num,id,n):\n M = [x for x in L]\n for i in range(id,id+2*n):\n id1 = (i%n)\n id2 = (i+1)%n\n x1 = M[i%n]\n x2 = M[(i+1)%n]\n a = x1-num\n if(a%2 == 0):\n M[id1] = num\n M[id2] += a//2\n else:\n if(num > 0):\n M[id1] = num-1\n M[id2] += a//2 + 1\n else:\n return False\n # print(M,id,num,id1)\n if(len(set(M)) == 1):\n return True\n return False \n \nfor _ in range(int(input())):\n n = int(input())\n L = list(map(int,input().split()))\n s = sum(L)\n m = max(L)\n id =L.index(m)\n l = 0\n r = s//n\n if(len(set(L)) == 1):\n print(0)\n continue\n f = 0\n ans1 = -1\n while(l < r):\n # print(l,r)\n mid = (l+r)//2\n if(mid == l):break\n if(check(L,mid,id,n)):\n f = 1\n ans1 = mid\n l = mid+1\n else:\n r = mid\n if(check(L,l,id,n)):\n f = 1\n ans1 = l\n if(check(L,r,id,n)):\n f = 1\n ans1 = r\n if(f):\n ans = sum(L)-ans1*n\n print(ans)\n else:\n print(-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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292175571,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nimport math\n \n \ndef check(a, eq, idx_max):\n a = list(a)\n cnt = q = 0\n i = idx_max\n \n while q < len(a):\n if i == len(a): i = 0\n if a[i] <= eq:\n q += 1; continue\n op = (a[i] - eq) // 2\n if (a[i] - eq) % 2 > 0: op += 1\n cnt += op\n if i + 1 < len(a):\n a[i] -= 2 * op; a[i + 1] += op\n else:\n a[i] -= 2 * op; a[0] += op\n i += 1\n \n if len(set(a)) == 1:\n return len(a), cnt\n \n cnt_eq = 0; flag = 0\n for i in range(len(a)):\n cnt_eq += 1 if a[i] == eq else 0\n if a[i] < 0: flag = 1\n \n if flag: cnt_eq = 0\n return cnt_eq, cnt\n \n \ndef solve():\n n = int(input())\n a = list(map(int, input().split()))\n \n max_ = max(a); idx_max = a.index(max_)\n if len(set(a)) == 1:\n print(0)\n return\n \n l = 0; r = max_\n ans = math.inf\n while l <= r:\n mid = (l + r) // 2\n cnt_eq, cnt_op = check(a, mid, idx_max)\n if cnt_eq < len(a):\n r = mid - 1\n else:\n l = mid + 1\n ans = min(cnt_op, ans)\n \n if ans == math.inf:\n print(-1)\n return\n \n print(ans)\n \n \nfor _ in range(int(input())):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292311642,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nimport math\n \n \ndef check(a, eq, idx_max):\n a = list(a); sum_ = sum(a)\n cnt = 0; i = idx_max\n \n _5n = len(a) * 3; cnt_step = 0\n while sum_ > eq * len(a):\n cnt_step += 1\n if cnt_step >= _5n: break\n \n if i == len(a): i = 0\n if a[i] <= eq: continue\n \n op = (a[i] - eq) // 2\n if (a[i] - eq) % 2 > 0: op += 1\n cnt += op\n \n a[i] -= 2 * op; sum_ -= 2 * op; sum_ += op\n if i + 1 < len(a): a[i + 1] += op\n else: a[0] += op\n \n i += 1\n \n cnt_eq = 0\n for i in range(len(a)):\n cnt_eq += 1 if a[i] == eq else 0\n \n return cnt_eq, cnt\n \n \ndef solve():\n n = int(input())\n a = list(map(int, input().split()))\n \n max_ = max(a); idx_max = a.index(max_)\n if len(set(a)) == 1:\n print(0)\n return\n \n l = 0; r = max_\n ans = math.inf\n while l <= r:\n mid = (l + r) // 2\n cnt_eq, cnt_op = check(a, mid, idx_max)\n if cnt_eq < len(a):\n r = mid - 1\n else:\n l = mid + 1\n ans = min(cnt_op, ans)\n \n if ans == math.inf:\n print(-1)\n return\n \n print(ans)\n \n \nfor _ in range(int(input())):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292313616,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\nimport java.lang.*;\nimport java.util.function.LongConsumer;\n \npublic class CP\n{\n static long mod = (long)(1e9+7);\n static int minInt = Integer.MIN_VALUE;\n static int maxInt = Integer.MAX_VALUE;\n static long minLong = Long.MIN_VALUE;\n static long maxLong = Long.MAX_VALUE;\n \n /*-----------------------------------------------------------------------MAIN CODE HERE----------------------------------------------------------------------------*/\n static void time_is_running(PrintWriter out,FastReader sc) throws IOException\n {\n int n = sc.ni(),arr[] = new int[n];\n for(int j=0;j<n;j++) arr[j] = sc.ni();\n \n int s=0,e=(int)1e9,ans=maxInt;\n while(s<=e){\n int mid = s+((e-s)/2);\n \n int tmp = check(arr,mid);\n if(tmp==-1){\n e=mid-1;\n }\n else{\n ans=Math.min(ans,tmp);\n s=mid+1;\n }\n }\n ans=ans==maxInt?-1:ans;\n \n out.println(ans);\n }\n \n static int check(int a[],int k){\n int maxInd=0;\n int arr[] = Arrays.copyOf(a,a.length);\n for(int j=0;j<arr.length;j++){\n if(arr[maxInd]<arr[j]) maxInd=j;\n }\n \n if(arr[maxInd]<k) return -1;\n int totalOps=0;\n for(int j=0;j<arr.length;j++){\n int ind = (j+maxInd)%arr.length;\n if(arr[ind]<=k){\n totalOps+=(k-arr[ind]);\n arr[(ind-1+arr.length)%arr.length]-=(2*(k-arr[ind]));\n arr[ind]=k;\n }\n else{\n int ops = (arr[ind]-k+1)/2;\n totalOps+=ops;\n arr[ind]-=(2*ops);\n arr[(ind+1)%arr.length]+=ops;\n }\n }\n \n for(int e:arr){\n if(e!=k) return -1;\n }\n \n return totalOps;\n }\n /*----------------------------------------------------------------------------END---------------------------------------------------------------------------------*/\n \n public static void main(String[] args) throws IOException\n {\n OutputStream outputStream = System.out;\n PrintWriter out = new PrintWriter(outputStream);\n FastReader sc = new FastReader();\n int n = sc.ni();\n for(int i=1;i<=n;i++)\n time_is_running(out,sc);\n out.close();\n }\n \n \n /*-----------------------------------------------------------------------SKIP THIS PART----------------------------------------------------------------------------*/\n static class FastReader {\n BufferedReader br;\n StringTokenizer st;\n// public FastReader() throws FileNotFoundException {\n// br = new BufferedReader(new FileReader(\"input.txt\"));\n// }\n public FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n String next() {\n while (st == null || !st.hasMoreElements()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n int ni() {return Integer.parseInt(next());}\n long nl() {return Long.parseLong(next());}\n double nd() {return Double.parseDouble(next());}\n char nextChar(){return next().charAt(0);}\n String nextLine() {\n String str = \"\";\n try {\n if (st.hasMoreTokens()) {\n str = st.nextToken(\"\\n\");\n } else {\n str = br.readLine();\n }\n } catch (IOException e) {\n e.printStackTrace();\n }\n return str;\n }\n }\n /*-------------------------------------------------------------------------END------------------------------------------------------------------------------*/\n}\nclass Pair\n{\n int a,b,c;\n Pair(int a1,int b1,int c1)\n {\n this.a=a1;\n this.b=b1;\n this.c=c1;\n }\n}\nclass Pair2\n{\n long a,b;\n Pair2(long a1,long b1)\n {\n this.a=a1;\n this.b=b1;\n }\n}\nclass comparingLong implements Comparator<Pair2>{\n public int compare(Pair2 s1, Pair2 s2){\n // We are returning the object in asending order\n if (s1.a < s2.a)\n return -1;\n else if (s1.a > s2.a)\n return 1;\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 295861618,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class Solution {\n \n public static void main(String[] args) {\n int t, n, carry, k, tmp;\n long sum;\n boolean canBe;\n Scanner sc = new Scanner(System.in);\n int[] arr = new int[2*100000];\n t = sc.nextInt();\n while (t-- != 0) {\n n = sc.nextInt();\n sum = 0;\n for (int i = 0; i < n; i++) {\n arr[i] = sc.nextInt();\n sum += arr[i];\n }\n carry = 0;\n for (int i = 0; i < n; i++) {\n tmp = (carry+arr[i])%2;\n carry = (carry + arr[i])/2;\n arr[i] = tmp;\n }\n k = 0;\n while (carry != 0) {\n tmp = (carry + arr[k])%2;\n carry = (carry + arr[k])/2;\n arr[k] = tmp;\n k = (k+1)%n;\n }\n canBe = true;\n for (int i = 1; i < n; i++) {\n if (arr[i] != arr[i-1]) {\n System.out.println(-1);\n canBe = false;\n break;\n }\n }\n if (canBe) {\n System.out.println(sum%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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 295382725,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n//#define map unordered_map\n#define re register\n#define ll long long\n#define cll const ll\n#define forl(i,a,b) for(re ll (i)=(a);i<=(b);(i)++)\n#define forr(i,a,b) for(re ll (i)=(a);i>=(b);(i)--)\n#define forll(i,a,b,c) for(re ll (i)=(a);i<=(b);(i)+=(c))\n#define forrr(i,a,b,c) for(re ll (i)=(a);i>=(b);(i)-=(c))\n#define forL(i,a,b,c) for(re ll (i)=(a);((i)<=(b)) && (c);(i)++)\n#define forR(i,a,b,c) for(re ll (i)=(a);((i)>=(b)) && (c);(i)--)\n#define forLL(i,a,b,c,d) for(re ll (i)=(a);((i)<=(b)) && (d);(i)+=(c))\n#define forRR(i,a,b,c,d) for(re ll (i)=(a);((i)>=(b)) && (d);(i)-=(c))\n#define pii pair<ll,ll>\n#define mid ((l+r)>>1)\n#define lowbit(x) (x&-x)\n#define pb push_back\n#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);\n#define endl '\\n'\n#define QwQ return 0;\nll _t_;\nvoid _clear(){}\nll n;\nll a[1000010];\nll L,R;\nll b[1000010];\nll nxt(ll x)\n{\n if(x==n)\n return 1;\n return x+1;\n}\nll check(ll Mid)\n{\n ll sum=0;\n forl(i,1,n)\n b[i]=a[i];\n forl(_,1,(n<=30)?30:1)\n forl(i,1,n)\n if(b[i]>Mid)\n {\n ll need=(b[i]-Mid+1)/2;\n sum+=need;\n b[i]-=need*2;\n b[nxt(i)]+=need;\n }\n forl(i,1,n)\n if(b[1]!=b[i] || b[i]<0)\n return 1e18;\n return sum;\n}\nvoid solve()\n{\n _clear();\n cin>>n;\n forl(i,1,n)\n cin>>a[i];\n L=0,R=2e9;\n while(L<R)\n {\n ll Mid=(L+R+1)/2;\n if(check(Mid)!=1e18)\n L=Mid;\n else\n R=Mid-1;\n }\n if(check(L)!=1e18)\n cout<<check(L)<<endl;\n else\n cout<<-1<<endl;\n}\nint main()\n{\n// freopen(\"tst.txt\",\"r\",stdin);\n// freopen(\"sans.txt\",\"w\",stdout);\n IOS;\n _t_=1;\n cin>>_t_;\n while(_t_--)\n solve();\n QwQ;\n}\n/*\n1\n2\n99999999 100000002\n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 300266298,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nconst int N = 2e5+10;\n \nint n;\nint a[N];\nint b[N];\nint check(int x) {\n\tfor(int i = 1;i <= n;i++) b[i] = a[i];\n\tint sum = 0;\n\tfor(int i = 1;i <= n;i++) {\n\t\tint rd = b[i]-x;\n\t\tif(rd <= 0) continue;\n\t\telse {\n\t\t\tif(rd%2 == 1) rd++;\n\t\t\trd /= 2;\n\t\t\tb[i] -= 2*rd;\n\t\t\tb[i%n+1] += rd;\n\t\t\tsum += rd;\n\t\t}\n\t}\n\tfor(int i = 1;i <= n;i++) {\n\t\tint rd = b[i]-x;\n\t\tif(rd <= 0) continue;\n\t\telse {\n\t\t\tif(rd%2 == 1) rd++;\n\t\t\trd /= 2;\n\t\t\tb[i] -= 2*rd;\n\t\t\tb[i%n+1] += rd;\n\t\t\tsum += rd;\n\t\t}\n\t}\n\tif(b[1] > x) return -1;\n\tint tmp = x-b[1];\n\tsum += n*tmp;\n\tfor(int i = 1;i <= n;i++) if(b[i]-tmp != x) return -1;\n\treturn sum;\n}\nsigned main() {\n\tios::sync_with_stdio(false);\n\tcin.tie(0);\n\tcout.tie(0);\n\tint T;\n\tcin >> T;\n\twhile(T--) {\n\t\tcin >> n;\n\t\tint maxn = 0;\n\t\tfor(int i = 1;i <= n;i++) cin >> a[i], maxn = max(maxn, a[i]);\n\t\tint l = 0, r = maxn+1, ans = -1;\n\t\twhile(l <= r) {\n\t\t\tint mid = (l+r)/2;\n\t\t\tint x = check(mid);\n\t\t\tif(x < 0) r = mid-1;\n\t\t\telse ans = x, l = mid+1;\n\t\t}\n\t\tcout << ans << '\\n';\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292297704,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\n\"\"\"\nLet m be the final equal value.\nLet x_i be the number of times we subtract 2 from a_i. \nWe have the following equation:\na_0 - 2x_0 + x_{n-1} = m\na_1 - 2x_1 + x_0 = m\na_2 - 2x_2 + x_1 = m\n...\na_{n-1} - 2x_{n-1} + x_{n-2} = m \n \nDef: We say m is feasible if the above system has a solution x where each x_i is non-negative integer.\nFurthermore, the total number of operations is sum(x). From the above, sum(x) = sum(a) - n*m.\nThus, we minimize sum(x) by maximizing m such that m is feasible.\nNote that if m is feasible then m+1 is feasible by adding to each x_i. \nHence, we can do binary search on m if we can tell whether m is feasible or not.\n \nOnce we fix m, since we have n linear equations and n variables, \nWe can solve for each x_i which has the following form (exercise):\nLet y_i = a_i - m, then \nx_i = (2^{n-1} y_i + 2^{n-2} y_{i-1 % n} + a^{n-3} y_{i-2 % n} + ... + 2^0 y_{i-(n-1) % n} - m) / (2^n - 1)\nIf all x_i are non-negative integers, then m is feasible.\n\"\"\"\n \ndef is_feasible(a, m):\n n = len(a)\n x = [0] * n\n y = [a[i] - m for i in range(n)]\n for i in range(n):\n for j in range(n):\n x[i] += (2 ** (n - 1 - j)) * y[(i - j) % n] \n x[i] = x[i] / (2 ** n - 1)\n if not(x[i].is_integer()) or (x[i] < 0):\n return False, x\n \n return True, x\n \nif __name__ == '__main__':\n num_tests = int(input())\n for i in range(num_tests):\n n = int(input())\n a = list(map(int, input().split()))\n \n l, r = 0, max(a)\n ans = -1 \n while l <= r:\n m = (l + r) // 2\n feasible, x = is_feasible(a, m)\n if feasible:\n ans = sum(x)\n l = m + 1\n else:\n r = m - 1\n print(int(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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 293476202,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nfor _ in range(int(input())):\n n = int(input())\n a = list(map(int, input().split()))\n mn = min(a)\n for i in range(0, n):\n if a[i] == mn:\n if i != n - 1:\n a = a[i + 1: ] + a[: i + 1]\n break\n t = 0\n for i in range(0, n):\n t += (a[i] << i)\n if t % ((1 << n) - 1):\n print(\"-1\")\n continue\n S = t // ((1 << n) - 1)\n print(sum(a) - S * n)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292211607,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;\n public static void main(String args[]){\n try {\n PrintWriter o = new PrintWriter(System.out);\n boolean multiTest = true;\n // init\n if(multiTest) {\n int t = nextInt(), loop = 0;\n while (loop < t) {loop++;solve(o);}\n } else solve(o);\n o.close();\n } catch (Exception e) {e.printStackTrace();}\n }\n static int n;\n static long sum, record;\n static int[] a;\n static void solve(PrintWriter o) {\n try {\n n = nextInt();\n a = new int[n];\n sum = 0l;\n record = 0l;\n for(int i=0;i<n;i++) {\n a[i] = nextInt(); // 1e9\n sum += a[i];\n }\n int l = 0, r = (int)1e9+1;\n while(l < r) {\n int m = (l+r)>>1;\n if(check(m)) l = m+1;\n else r = m;\n }\n if(l == 0) {\n o.println(-1);\n return;\n }\n l--;\n check(l);\n o.println(record);\n } catch (Exception e) {\n e.printStackTrace();\n }\n }\n static boolean check(int m) {\n int[] b = a.clone();\n long remain = sum-1l*m*n, use = 0l;\n if(remain < 0) return false;\n for(int i=0;i<3*n;i++) {\n long d = b[i%n]-m;\n if(d > 0) {\n long t = d/2;\n int md = (int)(d%2);\n use += t;\n b[i%n] = m+md;\n b[(i+1)%n] += t;\n }\n if(use > remain) {\n return false;\n }\n }\n for(int i=0;i<n-1;i++) if(b[i] != b[i+1]) return false;\n use += 1l*(b[0]-m)*n;\n record = use;\n return use <= remain;\n }\n public static int upper_bound(long[] preSum, long val){\n int l = 0, r = preSum.length;\n while(l < r){\n int mid = l + (r - l) / 2;\n if(preSum[mid] <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int upper_bound(List<long[]> li, long val){\n int l = 0, r = li.size();\n while(l < r){\n int mid = l + (r - l) / 2;\n if(li.get(mid)[0] <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int upper_bound(List<Integer> li, int val){\n int l = 0, r = li.size();\n while(l < r){\n int mid = l + (r - l) / 2;\n if(li.get(mid) <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int upper_bound(int[] a, int val){\n int l = 0, r = a.length;\n while(l < r){\n int mid = l + (r - l) / 2;\n if(a[mid] <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int lower_bound(List<Integer> li, int val){\n int l = 0, r = li.size();\n while(l < r){\n int mid = l + (r - l) / 2;\n if(li.get(mid) < val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int lower_bound(int[] a, int val){\n int l = 0, r = a.length;\n while(l < r){\n int mid = l + (r - l) / 2;\n if(a[mid] < val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static long gcd(long a, long b) {\n return b == 0 ? a : gcd(b, a%b);\n }\n public static long[] extgcd(long a, long b) {\n if(b == 0) return new long[]{1, 0};\n long[] it = extgcd(b, a%b);\n long x = it[1], y = it[0];\n y -= a/b*x;\n return new long[]{x, y};\n }\n public static long lcm(long a, long b){\n return a / gcd(a,b)*b;\n }\n public static long qpow(long a, long n, int md){\n a %= md;\n long ret = 1l;\n while(n > 0){\n if((n & 1) == 1) ret = ret * a % md;\n n >>= 1;\n a = a * a % md;\n }\n return ret;\n }\n public static class FenWick {\n int n;\n long[] a;\n long[] tree;\n public FenWick(int n){\n this.n = n;\n a = new long[n+1];\n tree = new long[n+1];\n }\n private void add(int x, long val){\n while(x <= n){\n tree[x] += val;\n x += x&-x;\n }\n }\n private void addMx(int x, long val) {\n a[x] += val;\n tree[x] = a[x];\n while(x <= n) {\n for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);\n x += x&-x;\n }\n }\n private long query(int x){\n long ret = 0l;\n while(x > 0){\n ret += tree[x];\n x -= x&-x;\n }\n return ret;\n }\n private long queryMx(int l, int r) {\n long res = 0l;\n while(l <= r) {\n if(r-(r&-r) >= l) {\n res = Math.max(res, tree[r]);\n r -= r&-r;\n }\n else {\n res = Math.max(res, a[r]);\n r--;\n }\n }\n return res;\n }\n }\n static class Pair {\n Integer p1, p2;\n public Pair(Integer p1, Integer p2) {\n this.p1 = p1;\n this.p2 = p2;\n }\n @Override\n public int hashCode() {\n int prime = 31, ret = 0;\n ret = ret*prime+p1.hashCode();\n ret = ret*prime+p2.hashCode();\n return ret;\n }\n @Override\n public boolean equals(Object o) {\n if(o instanceof Pair pair) {\n return p1.equals(pair.p1) && p2.equals(pair.p2);\n }\n return false;\n }\n }\n static class DSU {\n int[] parent;\n int[] size;\n public DSU(int n) {\n parent = new int[n];\n size = new int[n];\n for(int i=0;i<n;i++) {\n parent[i] = i;\n size[i] = 1;\n }\n }\n public int find(int x) {\n if(parent[x] != x) parent[x] = find(parent[x]);\n return parent[x];\n }\n public void union(int x, int y) {\n int root_x = find(x);\n int root_y = find(y);\n if(root_x == root_y) return;\n parent[root_x] = root_y;\n size[root_y] += size[root_x];\n size[root_x] = 0;\n }\n }\n private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));\n private static StringTokenizer tokenizer = new StringTokenizer(\"\");\n private static String next() throws IOException{\n while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}\n return tokenizer.nextToken();\n }\n public static int nextInt() throws IOException {return Integer.parseInt(next());}\n public static Long nextLong() throws IOException {return Long.parseLong(next());}\n public static double nextDouble() throws IOException {return Double.parseDouble(next());}\n public static char nextChar() throws IOException {return next().toCharArray()[0];}\n public static String nextString() throws IOException {return next();}\n public static String nextLine() throws IOException {return reader.readLine();}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292717882,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;\n public static void main(String args[]){\n try {\n PrintWriter o = new PrintWriter(System.out);\n boolean multiTest = true;\n // init\n if(multiTest) {\n int t = nextInt(), loop = 0;\n while (loop < t) {loop++;solve(o);}\n } else solve(o);\n o.close();\n } catch (Exception e) {e.printStackTrace();}\n }\n static int n;\n static long sum;\n static int[] a;\n static void solve(PrintWriter o) {\n try {\n n = nextInt();\n a = new int[n];\n sum = 0l;\n for(int i=0;i<n;i++) {\n a[i] = nextInt(); // 1e9\n sum += a[i];\n }\n int l = 0, r = (int)1e9+1;\n while(l < r) {\n int m = (l+r)>>1;\n if(check(m)) l = m+1;\n else r = m;\n }\n if(l == 0) {\n o.println(-1);\n return;\n }\n l--;\n o.println(sum-1l*l*n);\n } catch (Exception e) {\n e.printStackTrace();\n }\n }\n static boolean check(int m) {\n int[] b = a.clone();\n if(sum < 1l*n*m) return false;\n long tot = sum;\n for(int i=0;;i++) {\n long d = b[i%n]-m;\n if(d > 0) {\n long t = (d+1)/2;\n int md = (int)d%2;\n if(md > 0) b[i%n] = m-1;\n else b[i%n] = m;\n b[(i+1)%n] += t;\n tot -= t;\n }\n if(tot <= 1l*n*m) break;\n }\n for(int i=0;i<n-1;i++) if(b[i] != b[i+1]) return false;\n return true;\n }\n public static int upper_bound(long[] preSum, long val){\n int l = 0, r = preSum.length;\n while(l < r){\n int mid = l + (r - l) / 2;\n if(preSum[mid] <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int upper_bound(List<long[]> li, long val){\n int l = 0, r = li.size();\n while(l < r){\n int mid = l + (r - l) / 2;\n if(li.get(mid)[0] <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int upper_bound(List<Integer> li, int val){\n int l = 0, r = li.size();\n while(l < r){\n int mid = l + (r - l) / 2;\n if(li.get(mid) <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int upper_bound(int[] a, int val){\n int l = 0, r = a.length;\n while(l < r){\n int mid = l + (r - l) / 2;\n if(a[mid] <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int lower_bound(List<Integer> li, int val){\n int l = 0, r = li.size();\n while(l < r){\n int mid = l + (r - l) / 2;\n if(li.get(mid) < val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int lower_bound(int[] a, int val){\n int l = 0, r = a.length;\n while(l < r){\n int mid = l + (r - l) / 2;\n if(a[mid] < val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static long gcd(long a, long b) {\n return b == 0 ? a : gcd(b, a%b);\n }\n public static long[] extgcd(long a, long b) {\n if(b == 0) return new long[]{1, 0};\n long[] it = extgcd(b, a%b);\n long x = it[1], y = it[0];\n y -= a/b*x;\n return new long[]{x, y};\n }\n public static long lcm(long a, long b){\n return a / gcd(a,b)*b;\n }\n public static long qpow(long a, long n, int md){\n a %= md;\n long ret = 1l;\n while(n > 0){\n if((n & 1) == 1) ret = ret * a % md;\n n >>= 1;\n a = a * a % md;\n }\n return ret;\n }\n public static class FenWick {\n int n;\n long[] a;\n long[] tree;\n public FenWick(int n){\n this.n = n;\n a = new long[n+1];\n tree = new long[n+1];\n }\n private void add(int x, long val){\n while(x <= n){\n tree[x] += val;\n x += x&-x;\n }\n }\n private void addMx(int x, long val) {\n a[x] += val;\n tree[x] = a[x];\n while(x <= n) {\n for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);\n x += x&-x;\n }\n }\n private long query(int x){\n long ret = 0l;\n while(x > 0){\n ret += tree[x];\n x -= x&-x;\n }\n return ret;\n }\n private long queryMx(int l, int r) {\n long res = 0l;\n while(l <= r) {\n if(r-(r&-r) >= l) {\n res = Math.max(res, tree[r]);\n r -= r&-r;\n }\n else {\n res = Math.max(res, a[r]);\n r--;\n }\n }\n return res;\n }\n }\n static class Pair {\n Integer p1, p2;\n public Pair(Integer p1, Integer p2) {\n this.p1 = p1;\n this.p2 = p2;\n }\n @Override\n public int hashCode() {\n int prime = 31, ret = 0;\n ret = ret*prime+p1.hashCode();\n ret = ret*prime+p2.hashCode();\n return ret;\n }\n @Override\n public boolean equals(Object o) {\n if(o instanceof Pair pair) {\n return p1.equals(pair.p1) && p2.equals(pair.p2);\n }\n return false;\n }\n }\n static class DSU {\n int[] parent;\n int[] size;\n public DSU(int n) {\n parent = new int[n];\n size = new int[n];\n for(int i=0;i<n;i++) {\n parent[i] = i;\n size[i] = 1;\n }\n }\n public int find(int x) {\n if(parent[x] != x) parent[x] = find(parent[x]);\n return parent[x];\n }\n public void union(int x, int y) {\n int root_x = find(x);\n int root_y = find(y);\n if(root_x == root_y) return;\n parent[root_x] = root_y;\n size[root_y] += size[root_x];\n size[root_x] = 0;\n }\n }\n private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));\n private static StringTokenizer tokenizer = new StringTokenizer(\"\");\n private static String next() throws IOException{\n while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}\n return tokenizer.nextToken();\n }\n public static int nextInt() throws IOException {return Integer.parseInt(next());}\n public static Long nextLong() throws IOException {return Long.parseLong(next());}\n public static double nextDouble() throws IOException {return Double.parseDouble(next());}\n public static char nextChar() throws IOException {return next().toCharArray()[0];}\n public static String nextString() throws IOException {return next();}\n public static String nextLine() throws IOException {return reader.readLine();}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292724574,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define int long long\n#define endl '\\n'\nusing namespace std;\nusing PII = pair<int, int>;\nconst int N = 2e5 + 10;\nint a[N], b[N];\nint n;\nbool check(int x)\n{\n for (int i = 1; i <= n; i++)\n {\n b[i] = a[i] - x;\n cerr << b[i] << ' ';\n }\n bool flag = true;\n while (flag)\n {\n flag = false;\n for (int i = 1; i <= n; i++)\n {\n if (b[i] > 0)\n {\n flag = true;\n int nxt = (i % n) + 1;\n b[nxt] += (b[i] + 1) / 2;\n b[i] -= (b[i] + 1) / 2 * 2;\n }\n }\n }\n for (int i = 1; i <= n; i++)\n {\n if (b[i])\n return false;\n }\n return true;\n}\nvoid solve()\n{\n int sums = 0;\n cin >> n;\n for (int i = 1; i <= n; i++)\n {\n cin >> a[i];\n sums += a[i];\n }\n int l = 1, r = *max_element(a + 1, a + 1 + n), ans = -1, cnt = 0;\n while (l <= r)\n {\n cnt++;\n int mid = l + r >> 1;\n cerr << mid << endl;\n if (check(mid))\n {\n ans = mid;\n l = mid + 1;\n }\n else\n r = mid - 1;\n if (cnt > 3)\n break;\n }\n if (ans == -1)\n cout << ans << endl;\n else\n cout << sums - ans * n << endl;\n}\n \nsigned main()\n{\n ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);\n int T = 1;\n cin >> T;\n while (T--)\n solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292598286,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define int long long\n \nvoid solve() {\n int n; cin >> n;\n vector<int> a(n);\n for (int i = 0; i < n; i++) cin >> a[i];\n \n int mx = -1e9, ix = 0;\n for (int i = 0; i < n; i++)\n if (a[i] > mx) mx = a[i], ix = i;\n \n if (ix != 0) rotate(a.begin(), a.begin() + ix, a.end());\n \n int ans = 0;\n int times = 300*n, i = 0;\n while (times--) {\n int c = (a[i] - a[(i+1)%n] + 2)/3;\n c = max(c, 0LL);\n a[i] -= 2*c; a[(i+1)%n] += c;\n if (a[i] < 0) break;\n ans += c;\n i = rand()%n;\n }\n \n for (int i = 0; i < n; i++)\n if (a[i] < 0 or a[i] != a[0]) {\n cout << -1 << \"\\n\";\n return;\n }\n \n cout << ans << \"\\n\";\n}\n \nsigned main() {\n ios::sync_with_stdio(false);\n cin.tie(0);\n \n srand(time(NULL));\n \n #ifndef ONLINE_JUDGE\n freopen(\"in.txt\", \"r\", stdin);\n freopen(\"out.txt\", \"w\", stdout);\n #endif\n \n int tt; cin >> tt;\n while (tt--)\n solve();\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292226139,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nimport sys\n \ndef input(): \n return sys.stdin.readline().strip()\n \ndef getints(): \n return map(int, sys.stdin.readline().strip().split())\n \ndef con(num, l):\n l = list(l)\n ans = 0\n for i in range(len(l)):\n if l[i] > num:\n t = l[i] - num\n if t % 2 != 0: \n t += 1\n l[i] -= t\n l[(i + 1) % len(l)] += t // 2\n ans += t // 2\n for i in range(len(l)):\n if l[i] > num:\n t = l[i] - num\n if t % 2 != 0: \n t += 1\n l[i] -= t\n l[(i + 1) % len(l)] += t // 2\n ans += t // 2\n for i in range(len(l) - 1, -1, -1):\n if l[i] <= num:\n l[i - 1] -= 2 * (num - l[i])\n ans += (num - l[i])\n l[i] = num\n if l[-1] == num: \n return True, ans\n else: \n return False, -1\n \nfor _ in range(int(input())):\n n = int(input())\n l = list(getints())\n \n lo = 0\n hi = max(l)\n \n while lo != hi:\n mid = (lo + hi + 1) // 2\n if con(mid, l)[0]: \n lo = mid\n else: \n hi = mid - 1\n print(con(lo, l)[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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 296011937,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nfrom __future__ import annotations\n \nimport bisect # noqa\nimport heapq # noqa\nimport math # noqa\nimport random # noqa\nimport statistics # noqa\nimport sys\nfrom bisect import bisect_left as lower_bound\nfrom bisect import bisect_right as upper_bound\nfrom collections import Counter, defaultdict, deque # noqa\nfrom copy import deepcopy # noqa\nfrom datetime import datetime # noqa\nfrom fractions import Fraction # noqa\nfrom functools import lru_cache # noqa\nfrom itertools import ( # noqa\n permutations, combinations # difference is that order matters for permutations but not for combinations\n)\nfrom typing import Iterator # noqa\n \n# rewrite input\n# sys.stdin = open(\"cutting.in\", \"r\")\n# sys.stdout = open(\"cutting.out\", \"w\")\nuser_input = sys.stdin.readline\n \n# allow big recursion\nsys.setrecursionlimit(10**5)\n \n# Magic no recursion\nfrom types import GeneratorType # noqa\ndef bootstrap(f, stack=[]): # noqa\n def wrappedfunc(*args, **kwargs):\n if stack:\n return f(*args, **kwargs)\n to = f(*args, **kwargs)\n while True:\n if type(to) is GeneratorType:\n stack.append(to)\n to = next(to)\n else:\n stack.pop()\n if not stack:\n break\n to = stack[-1].send(to)\n return to\n \n return wrappedfunc\n \ndef get_int() -> int:\n return int(user_input())\n \ndef line_integers() -> list[int]:\n return list(map(int, user_input().split()))\n \ndef read_matrix(n: int) -> list[list[int]]:\n return [line_integers() for _ in range(n)]\n \ndef read_matrix_str(n: int) -> list[list[str]]:\n return [get_string(as_list=True) for _ in range(n)]\n \ndef read_matrix_str_blank_separated(n: int) -> list[list[str]]:\n return [get_string(as_list=False).split() for _ in range(n)]\n \ndef get_string(as_list: bool = False) -> str | list[str]:\n input_string = user_input()\n if as_list:\n return list(input_string[: len(input_string) - 1])\n else:\n return input_string[: len(input_string) - 1]\n \ndef invr() -> Iterator[int]:\n return map(int, user_input().split())\n \ndef partitions(partitions_n: int):\n partitions_a = [0 for _ in range(partitions_n + 1)]\n k_ = 1\n y_ = partitions_n - 1\n while k_ != 0:\n x_ = partitions_a[k_ - 1] + 1\n k_ -= 1\n while 2 * x_ <= y_:\n partitions_a[k_] = x_\n y_ -= x_\n k_ += 1\n l_ = k_ + 1\n while x_ <= y_:\n partitions_a[k_] = x_\n partitions_a[l_] = y_\n yield partitions_a[: k_ + 2]\n x_ += 1\n y_ -= 1\n partitions_a[k_] = x_ + y_\n y_ = x_ + y_ - 1\n yield partitions_a[: k_ + 1]\n \ndef gen_suffix_array(s_: str) -> list[int]:\n def radix_sort(p_: list[int], c_: list[int]) -> list[int]:\n count: list[int] = [0 for _ in range(len(p_))]\n for x in p_:\n count[c_[x]] += 1\n new_p: list[int] = p_.copy()\n pos: list[int] = [0 for _ in range(len(p_))]\n for i in range(1, len(p_)):\n pos[i] = pos[i-1] + count[i-1]\n for x in p_:\n new_p[pos[c_[x]]] = x\n pos[c_[x]] += 1\n \n return new_p\n s_ += \"$\"\n n: int = len(s_)\n a: list[tuple[str, int]] = [(s_[i], i) for i in range(n)]\n a.sort(key=lambda x: x[0])\n p: list[int] = [a[i][1] for i in range(n)]\n c: list[int] = [0] * n\n c[p[0]] = 0\n for i in range(1, n):\n if a[i][0] == a[i-1][0]:\n c[p[i]] = c[p[i-1]]\n else:\n c[p[i]] = c[p[i-1]] + 1\n k: int = 1\n while (1 << k) <= 2*n:\n p = [(p[i] - (1 << (k-1))) % n for i in range(n)]\n p = radix_sort(p, c)\n \n c_new: list[int] = [0] * n\n prev: tuple[int, int] = (c[p[0]], c[(p[0] + (1 << (k-1))) % n])\n for i in range(1, n):\n curr: tuple[int, int] = (c[p[i]], c[(p[i] + (1 << (k-1))) % n])\n if prev == curr:\n c_new[p[i]] = c_new[p[i-1]]\n else:\n c_new[p[i]] = c_new[p[i-1]] + 1\n prev = curr\n c = c_new\n k += 1\n if c_new[p[n-1]] == n - 1:\n break\n return p\n \ndef gen_suffix_array_and_lcp(s_: str) -> tuple[list[int], list[int]]:\n def radix_sort(p_: list[int], c_: list[int]) -> list[int]:\n count: list[int] = [0 for _ in range(len(p_))]\n for x in p_:\n count[c_[x]] += 1\n new_p: list[int] = p_.copy()\n pos: list[int] = [0 for _ in range(len(p_))]\n for i in range(1, len(p_)):\n pos[i] = pos[i-1] + count[i-1]\n for x in p_:\n new_p[pos[c_[x]]] = x\n pos[c_[x]] += 1\n \n return new_p\n s_ += \"$\"\n n: int = len(s_)\n a: list[tuple[str, int]] = [(s_[i], i) for i in range(n)]\n a.sort(key=lambda x: x[0])\n p: list[int] = [a[i][1] for i in range(n)]\n c: list[int] = [0] * n\n c[p[0]] = 0\n for i in range(1, n):\n if a[i][0] == a[i-1][0]:\n c[p[i]] = c[p[i-1]]\n else:\n c[p[i]] = c[p[i-1]] + 1\n k: int = 1\n while (1 << k) <= 2*n:\n p = [(p[i] - (1 << (k-1))) % n for i in range(n)]\n p = radix_sort(p, c)\n \n c_new: list[int] = [0] * n\n prev: tuple[int, int] = (c[p[0]], c[(p[0] + (1 << (k-1))) % n])\n for i in range(1, n):\n curr: tuple[int, int] = (c[p[i]], c[(p[i] + (1 << (k-1))) % n])\n if prev == curr:\n c_new[p[i]] = c_new[p[i-1]]\n else:\n c_new[p[i]] = c_new[p[i-1]] + 1\n prev = curr\n c = c_new\n k += 1\n if c_new[p[n-1]] == n - 1:\n break\n \n lcp: list[int] = [0] * (n-1)\n k: int = 0\n for i in range(n-1):\n # i => where do we start (in the string) this suffix. we never start the suffix at n-1.\n # because the suffix at n-1 is $ and it is the first of the suffix array so we don't have a previous suffix to \n # compare it to.\n pi: int = c[i] # pi => position of the suffix i in the suffix array.\n # pi is never 0 because the suffix at 0 is $\n j: int = p[pi - 1] # j => where does the previous (in the suffix array) suffix start (in the string).\n while s_[i+k] == s_[j+k]:\n k += 1\n lcp[pi-1] = k\n k = max(k-1, 0)\n \n return p, lcp\n \ndef get_lps(string: str) -> list[int]:\n lps: list[int] = [0] * len(string)\n length: int = 0\n i: int = 1\n while i < len(string):\n if string[i] == string[length]:\n length += 1\n lps[i] = length\n i += 1\n else:\n if length != 0:\n length = lps[length - 1]\n else:\n lps[i] = 0\n i += 1\n return lps\n \ndef create_string_hash(string: str, n: int, p: int, mod: int) -> tuple[list[int], list[int]]:\n hash_prefixes: list[int] = [0] * (n + 1)\n powers: list[int] = [1] * (n + 1)\n for i in range(1, n + 1):\n powers[i] = (powers[i - 1] * p) % mod\n for i in range(n):\n hash_prefixes[i + 1] = (hash_prefixes[i] * p + (ord(string[i]) - ord(\"a\") + 1)) % mod\n return hash_prefixes, powers\n \ndef get_hash(hash_prefixes: list[int], powers: list[int], mod: int, i: int, j: int) -> int:\n \"\"\"\n Returns the hash of the substring s[i:j] (j excluded) \n \"\"\"\n return (hash_prefixes[j] - hash_prefixes[i] * powers[j-i]) % mod\n \ndef make_nCr_mod(max_n: int, mod: int = 10**9+7):\n max_n = min(max_n, mod - 1)\n \n fact, inv_fact = [0] * (max_n + 1), [0] * (max_n + 1)\n fact[0] = 1\n for i in range(max_n):\n fact[i + 1] = fact[i] * (i + 1) % mod\n \n inv_fact[-1] = pow(fact[-1], mod - 2, mod)\n for i in reversed(range(max_n)):\n inv_fact[i] = inv_fact[i + 1] * (i + 1) % mod\n \n def nCr_mod(n, r):\n res = 1\n while n or r:\n a, b = n % mod, r % mod\n if a < b:\n return 0\n res = res * fact[a] % mod * inv_fact[b] % mod * inv_fact[a - b] % mod\n n //= mod\n r //= mod\n return res\n \n return nCr_mod\n \ndef sum_from_k_to_n(k_: int, n_: int) -> int:\n # k inclusive, n inclusive\n return (n_ * (n_ + 1) - k_ * (k_ - 1)) // 2\n \nclass FenwickTree:\n def __init__(self, x):\n bit = self.bit = list(x)\n size = self.size = len(bit)\n for i in range(size):\n j = i | (i + 1)\n if j < size:\n bit[j] += bit[i]\n \n def update(self, idx, x):\n \"\"\"updates bit[idx] += x\"\"\"\n while idx < self.size:\n self.bit[idx] += x\n idx |= idx + 1\n \n def __call__(self, end):\n \"\"\"calc sum(bit[:end])\"\"\"\n x = 0\n while end:\n x += self.bit[end - 1]\n end &= end - 1\n return x\n \n def find_kth(self, k):\n \"\"\"Find largest idx such that sum(bit[:idx]) <= k\"\"\"\n idx = -1\n for d in reversed(range(self.size.bit_length())):\n right_idx = idx + (1 << d)\n if right_idx < self.size and self.bit[right_idx] <= k:\n idx = right_idx\n k -= self.bit[idx]\n return idx + 1, k\n \nclass SortedList:\n block_size = 700\n \n def __init__(self, iterable=()):\n self.macro = []\n self.micros = [[]]\n self.micro_size = [0]\n self.fenwick = FenwickTree([0])\n self.size = 0\n for item in iterable:\n self.insert(item)\n \n def insert(self, x):\n i = lower_bound(self.macro, x)\n j = upper_bound(self.micros[i], x)\n self.micros[i].insert(j, x)\n self.size += 1\n self.micro_size[i] += 1\n self.fenwick.update(i, 1)\n if len(self.micros[i]) >= self.block_size:\n self.micros[i:i + 1] = self.micros[i][:self.block_size >> 1], self.micros[i][self.block_size >> 1:]\n self.micro_size[i:i + 1] = self.block_size >> 1, self.block_size >> 1\n self.fenwick = FenwickTree(self.micro_size)\n self.macro.insert(i, self.micros[i + 1][0])\n \n def pop(self, k=-1):\n i, j = self._find_kth(k)\n self.size -= 1\n self.micro_size[i] -= 1\n self.fenwick.update(i, -1)\n return self.micros[i].pop(j)\n \n def __getitem__(self, k):\n i, j = self._find_kth(k)\n return self.micros[i][j]\n \n def count(self, x):\n return self.upper_bound(x) - self.lower_bound(x)\n \n def __contains__(self, x):\n return self.count(x) > 0\n \n def lower_bound(self, x):\n i = lower_bound(self.macro, x)\n return self.fenwick(i) + lower_bound(self.micros[i], x)\n \n def upper_bound(self, x):\n i = upper_bound(self.macro, x)\n return self.fenwick(i) + upper_bound(self.micros[i], x)\n \n def _find_kth(self, k):\n return self.fenwick.find_kth(k + self.size if k < 0 else k)\n \n def __len__(self):\n return self.size\n \n def __iter__(self):\n return (x for micro in self.micros for x in micro)\n \n def __repr__(self):\n return str(list(self))\n \nclass RangeQuery:\n def __init__(self, data, func):\n self.func = func\n self._data = _data = [data]\n i = 1\n n = len(_data[0])\n while 2 * i <= n:\n prev = _data[-1]\n _data.append(\n [\n func(prev[j], prev[j + i])\n for j in range(n - 2 * i + 1)\n ]\n )\n i <<= 1\n \n def query(self, start, stop):\n \"\"\"func of data[start, stop)\"\"\"\n if start >= stop:\n return 0 \n depth = (stop - start).bit_length() - 1\n return self.func(self._data[depth][start], self._data[depth][stop - (1 << depth)])\n \n def __getitem__(self, idx):\n return self._data[0][idx]\n \nclass Treap:\n def __init__(self, key: int):\n self.key = key # BST in regards to key (IF INORDER TRAVERSAL = SORTED ORDER)\n self.priority = random.randint(0, 10**6) # heap in regards to priority\n self.size = 1\n self.left = self.right = None\n \ndef split_treap(my_treap: Treap | None, n_in_left: int) -> tuple[Treap | None, Treap | None]:\n # returns left_treap, right_treap roots\n # n_in_left is the number of nodes in the left treap\n if my_treap is None:\n return None, None\n if n_in_left <= 0:\n return None, my_treap\n if my_treap.left is not None and my_treap.left.size >= n_in_left:\n # it means the my_treap node (root of my current treap) won't be in the left_treap\n left_treap, my_treap.left = split_treap(my_treap.left, n_in_left)\n my_treap.size -= n_in_left\n return left_treap, my_treap\n else:\n # it means my_treap.left.size > n_in_left, so the my_treap node (root of my current treap) will be in the\n # left_treap\n additional_to_split = n_in_left - ((my_treap.left.size if my_treap.left is not None else 0) + 1)\n my_treap.right, right_treap = split_treap(my_treap.right, additional_to_split)\n my_treap.size = n_in_left\n return my_treap, right_treap\n \ndef merge_treap(treap_left: Treap | None, treap_right: Treap | None) -> Treap | None:\n # IF YOU WANT INORDER TRAVERSAL = SORTED ORDER REGARDING KEY for your resulting Treap:\n # Then we assume treap_left and treap_right are ordered (all keys in treap_left are smaller than the keys in\n # treap_right)\n # IF YOU DON'T CARE ABOUT INORDER TRAVERSAL = SORTED ORDER REGARDING KEY for your resulting Treap, then all good\n if treap_left is None:\n return treap_right\n if treap_right is None:\n return treap_left\n if treap_left.priority > treap_right.priority:\n treap_left.size += treap_right.size\n treap_left.right = merge_treap(treap_left.right, treap_right)\n return treap_left\n else:\n treap_right.size += treap_left.size\n treap_right.left = merge_treap(treap_left, treap_right.left)\n return treap_right\n \ndef inorder_traversal(treap: Treap | None) -> list[int]:\n if treap is None:\n return []\n return inorder_traversal(treap.left) + [treap.key] + inorder_traversal(treap.right)\n \n \ndef solve() -> None:\n \"\"\"\n Additional test case\n 1\n 3\n 15 1 1\n \n 8\n \"\"\"\n n: int = get_int()\n a: list[int] = line_integers()\n a_copy = a.copy()\n \n def is_feasible(target: int) -> bool:\n a = a_copy.copy()\n for _ in range(5):\n for i in range(n):\n a_i: int = a[i]\n if a_i > target:\n a[i] -= 2 * (((a_i - target) // 2) + (a_i - target) % 2)\n \n nxt_idx: int = (i + 1) % n\n a[nxt_idx] += 1 * (((a_i - target) // 2) + (a_i - target) % 2)\n \n return all([x == target for x in a])\n \n \n targets: list[int] = [i * n for i in range((sum(a) // n) + 1)]\n l_ptr: int = -1\n r_ptr: int = len(targets) - 1\n while l_ptr < r_ptr:\n mid: int = (l_ptr + r_ptr + 1) // 2\n if is_feasible(targets[mid] // n):\n l_ptr = mid\n else:\n r_ptr = mid - 1\n \n if r_ptr == -1:\n print(-1)\n return\n print(sum(a) - targets[l_ptr])\n \n \nif __name__ == \"__main__\":\n n_tests: int = get_int()\n test_nb: int\n for test_nb in range(n_tests):\n solve()\n \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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292211013,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;\n public static void main(String args[]){\n try {\n PrintWriter o = new PrintWriter(System.out);\n boolean multiTest = true;\n // init\n if(multiTest) {\n int t = nextInt(), loop = 0;\n while (loop < t) {loop++;solve(o);}\n } else solve(o);\n o.close();\n } catch (Exception e) {e.printStackTrace();}\n }\n static int n;\n static long sum, record;\n static int[] a;\n static void solve(PrintWriter o) {\n try {\n n = nextInt();\n a = new int[n];\n sum = 0l;\n record = 0l;\n for(int i=0;i<n;i++) {\n a[i] = nextInt(); // 1e9\n sum += a[i];\n }\n int l = 0, r = (int)1e9+1;\n while(l < r) {\n int m = (l+r)>>1;\n if(check(m)) l = m+1;\n else r = m;\n }\n if(l == 0) {\n o.println(-1);\n return;\n }\n l--;\n check(l);\n o.println(record);\n } catch (Exception e) {\n e.printStackTrace();\n }\n }\n static boolean check(int m) {\n int[] b = a.clone();\n long remain = sum-1l*m*n, use = 0l;\n if(remain < 0) return false;\n for(int i=0;i<4*n;i++) {\n long d = b[i%n]-m;\n if(d > 0) {\n long t = d/2;\n int md = (int)(d%2);\n use += t;\n b[i%n] = m+md;\n b[(i+1)%n] += t;\n }\n if(use > remain) {\n return false;\n }\n }\n for(int i=0;i<n-1;i++) if(b[i] != b[i+1]) return false;\n use += 1l*(b[0]-m)*n;\n record = use;\n return use <= remain;\n }\n public static int upper_bound(long[] preSum, long val){\n int l = 0, r = preSum.length;\n while(l < r){\n int mid = l + (r - l) / 2;\n if(preSum[mid] <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int upper_bound(List<long[]> li, long val){\n int l = 0, r = li.size();\n while(l < r){\n int mid = l + (r - l) / 2;\n if(li.get(mid)[0] <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int upper_bound(List<Integer> li, int val){\n int l = 0, r = li.size();\n while(l < r){\n int mid = l + (r - l) / 2;\n if(li.get(mid) <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int upper_bound(int[] a, int val){\n int l = 0, r = a.length;\n while(l < r){\n int mid = l + (r - l) / 2;\n if(a[mid] <= val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int lower_bound(List<Integer> li, int val){\n int l = 0, r = li.size();\n while(l < r){\n int mid = l + (r - l) / 2;\n if(li.get(mid) < val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static int lower_bound(int[] a, int val){\n int l = 0, r = a.length;\n while(l < r){\n int mid = l + (r - l) / 2;\n if(a[mid] < val) l = mid + 1;\n else r = mid;\n }\n return l;\n }\n public static long gcd(long a, long b) {\n return b == 0 ? a : gcd(b, a%b);\n }\n public static long[] extgcd(long a, long b) {\n if(b == 0) return new long[]{1, 0};\n long[] it = extgcd(b, a%b);\n long x = it[1], y = it[0];\n y -= a/b*x;\n return new long[]{x, y};\n }\n public static long lcm(long a, long b){\n return a / gcd(a,b)*b;\n }\n public static long qpow(long a, long n, int md){\n a %= md;\n long ret = 1l;\n while(n > 0){\n if((n & 1) == 1) ret = ret * a % md;\n n >>= 1;\n a = a * a % md;\n }\n return ret;\n }\n public static class FenWick {\n int n;\n long[] a;\n long[] tree;\n public FenWick(int n){\n this.n = n;\n a = new long[n+1];\n tree = new long[n+1];\n }\n private void add(int x, long val){\n while(x <= n){\n tree[x] += val;\n x += x&-x;\n }\n }\n private void addMx(int x, long val) {\n a[x] += val;\n tree[x] = a[x];\n while(x <= n) {\n for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);\n x += x&-x;\n }\n }\n private long query(int x){\n long ret = 0l;\n while(x > 0){\n ret += tree[x];\n x -= x&-x;\n }\n return ret;\n }\n private long queryMx(int l, int r) {\n long res = 0l;\n while(l <= r) {\n if(r-(r&-r) >= l) {\n res = Math.max(res, tree[r]);\n r -= r&-r;\n }\n else {\n res = Math.max(res, a[r]);\n r--;\n }\n }\n return res;\n }\n }\n static class Pair {\n Integer p1, p2;\n public Pair(Integer p1, Integer p2) {\n this.p1 = p1;\n this.p2 = p2;\n }\n @Override\n public int hashCode() {\n int prime = 31, ret = 0;\n ret = ret*prime+p1.hashCode();\n ret = ret*prime+p2.hashCode();\n return ret;\n }\n @Override\n public boolean equals(Object o) {\n if(o instanceof Pair pair) {\n return p1.equals(pair.p1) && p2.equals(pair.p2);\n }\n return false;\n }\n }\n static class DSU {\n int[] parent;\n int[] size;\n public DSU(int n) {\n parent = new int[n];\n size = new int[n];\n for(int i=0;i<n;i++) {\n parent[i] = i;\n size[i] = 1;\n }\n }\n public int find(int x) {\n if(parent[x] != x) parent[x] = find(parent[x]);\n return parent[x];\n }\n public void union(int x, int y) {\n int root_x = find(x);\n int root_y = find(y);\n if(root_x == root_y) return;\n parent[root_x] = root_y;\n size[root_y] += size[root_x];\n size[root_x] = 0;\n }\n }\n private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));\n private static StringTokenizer tokenizer = new StringTokenizer(\"\");\n private static String next() throws IOException{\n while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}\n return tokenizer.nextToken();\n }\n public static int nextInt() throws IOException {return Integer.parseInt(next());}\n public static Long nextLong() throws IOException {return Long.parseLong(next());}\n public static double nextDouble() throws IOException {return Double.parseDouble(next());}\n public static char nextChar() throws IOException {return next().toCharArray()[0];}\n public static String nextString() throws IOException {return next();}\n public static String nextLine() throws IOException {return reader.readLine();}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 292720757,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Make It Equal\ntime_limit_ms: 2000\nmemory_limit_mb: 512\nDescription: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.\n\nYou can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \\bmod n) + 1}$$$ by $$$1$$$.\n\nAfter you perform the operations, all elements of the array should be non-negative equal integers.\n\nYour task is to calculate the minimum number of operations you have to perform.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\le t \\le 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\dots, a_n$$$ ($$$1 \\le a_i \\le 10^9$$$).\n\nAdditional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \\cdot 10^5$$$.\n\nOutput: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.\n\nExamples:\ninput:\n3\n2\n1 1\n3\n1 3 2\n4\n2 1 2 6\noutput:\n0\n-1\n3\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\nimport java.lang.*;\nimport java.util.function.LongConsumer;\n \npublic class CP\n{\n static long mod = (long)(1e9+7);\n static int minInt = Integer.MIN_VALUE;\n static int maxInt = Integer.MAX_VALUE;\n static long minLong = Long.MIN_VALUE;\n static long maxLong = Long.MAX_VALUE;\n \n /*-----------------------------------------------------------------------MAIN CODE HERE----------------------------------------------------------------------------*/\n static void time_is_running(PrintWriter out,FastReader sc) throws IOException\n {\n int n = sc.ni(),arr[] = new int[n];\n for(int j=0;j<n;j++) arr[j] = sc.ni();\n \n int s=0,e=(int)1e9,ans=maxInt;\n while(s<=e){\n int mid = s+((e-s)/2);\n \n int tmp = check(arr,mid);\n// out.println(mid+\" \"+tmp);\n if(tmp==-1){\n e=mid-1;\n }\n else{\n ans=Math.min(ans,tmp);\n s=mid+1;\n }\n }\n ans=ans==maxInt?-1:ans;\n \n out.println(ans);\n }\n \n static int check(int a[],int k){\n int maxInd=0;\n long sum=0;\n int arr[] = Arrays.copyOf(a,a.length);\n for(int j=0;j<arr.length;j++){\n if(arr[maxInd]<arr[j]) maxInd=j;\n sum+=arr[j];\n }\n \n if(arr[maxInd]<k) return -1;\n int totalOps=0,ind=maxInd;\n while(sum>(k*1l*a.length)){\n if(arr[ind]<=k){\n totalOps+=(k-arr[ind]);\n sum-=(k-arr[ind]);\n arr[(ind-1+arr.length)%arr.length]-=(2*(k-arr[ind]));\n arr[ind]=k;\n }\n else{\n int ops = (arr[ind]-k+1)/2;\n sum-=ops;\n totalOps+=ops;\n arr[ind]-=(2*ops);\n arr[(ind+1)%arr.length]+=ops;\n }\n ind=(ind+1)%arr.length;\n }\n \n for(int e:arr){\n if(e!=k) return -1;\n }\n \n return totalOps;\n }\n /*----------------------------------------------------------------------------END---------------------------------------------------------------------------------*/\n \n public static void main(String[] args) throws IOException\n {\n OutputStream outputStream = System.out;\n PrintWriter out = new PrintWriter(outputStream);\n FastReader sc = new FastReader();\n int n = sc.ni();\n for(int i=1;i<=n;i++)\n time_is_running(out,sc);\n out.close();\n }\n \n \n /*-----------------------------------------------------------------------SKIP THIS PART----------------------------------------------------------------------------*/\n static class FastReader {\n BufferedReader br;\n StringTokenizer st;\n// public FastReader() throws FileNotFoundException {\n// br = new BufferedReader(new FileReader(\"input.txt\"));\n// }\n public FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n String next() {\n while (st == null || !st.hasMoreElements()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n int ni() {return Integer.parseInt(next());}\n long nl() {return Long.parseLong(next());}\n double nd() {return Double.parseDouble(next());}\n char nextChar(){return next().charAt(0);}\n String nextLine() {\n String str = \"\";\n try {\n if (st.hasMoreTokens()) {\n str = st.nextToken(\"\\n\");\n } else {\n str = br.readLine();\n }\n } catch (IOException e) {\n e.printStackTrace();\n }\n return str;\n }\n }\n /*-------------------------------------------------------------------------END------------------------------------------------------------------------------*/\n}\nclass Pair\n{\n int a,b,c;\n Pair(int a1,int b1,int c1)\n {\n this.a=a1;\n this.b=b1;\n this.c=c1;\n }\n}\nclass Pair2\n{\n long a,b;\n Pair2(long a1,long b1)\n {\n this.a=a1;\n this.b=b1;\n }\n}\nclass comparingLong implements Comparator<Pair2>{\n public int compare(Pair2 s1, Pair2 s2){\n // We are returning the object in asending order\n if (s1.a < s2.a)\n return -1;\n else if (s1.a > s2.a)\n return 1;\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038B",
"submission_id": 295862537,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\n#include<cstdio>\nusing namespace std;\nint a[500005],T,n,ans[500005],st[500005],lst;\ninline int Mx(int x,int y){return x>y?x:y;}\ninline int Mn(int x,int y){return x<y?x:y;}\nint main(){\n scanf(\"%d\",&T);\n while(T--){\n scanf(\"%d\",&n);\n for(int i=1;i<=n;i++){\n scanf(\"%d\",&a[i]);\n ans[i]=a[i]<ans[i-1]?ans[i-1]:a[i];\n }\n // int cnt=0;\n lst=0;\n for(int i=n;i;i--){\n st[i]=a[i];\n while(ans[i]==ans[i-1]){\n ans[i-1]=ans[i];\n st[i-1]=Mn(st[i],a[i-1]);\n i--;\n }\n if(a[i]>st[lst])for(int j=i;j<lst;j++)ans[j]=ans[lst];\n lst=i;\n }\n for(int i=1;i<=n;i++)printf(\"%d \",ans[i]);printf(\"\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 291630527,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\n#include <iostream>\n#include <cstring>\n#include <algorithm>\n#include <cmath>\n \nusing namespace std;\n \nconst int N=500010;\nint n;\nint a[N];\nint rpos[N];\nint mx[N],mxpos[N];\nint ans[N];\nint tr[1000010];\nint f[N];\n \ninline int lowbit(int x)\n{\n\treturn x&(-x);\n}\n \nint query(int x)\n{\n\tint mx=0;\n\twhile (x>0)\n\t{\n\t\tmx=max(mx,tr[x]);\n\t\tx-=lowbit(x);\n\t}\n\treturn mx;\n}\n \nvoid modify(int x,int v)\n{\n\twhile (x<=n)\n\t{\n\t\ttr[x]=max(tr[x],v);\n\t\tx+=lowbit(x);\n\t}\n}\n \nint main()\n{\n\tint T;\n\tcin >> T;\n\twhile (T--)\n\t{\n\t\tcin >> n;\n\t\tfor (int i=1;i<=n;i++) \n\t\t{\n\t\t\tcin >> a[i];\n\t\t}\n\t\t\n\t\tfor (int i=1;i<=n;i++)\n\t\t{\n\t\t\tif (a[i]>mx[i-1])\n\t\t\t{\n\t\t\t\tmx[i]=a[i];\n\t\t\t\tmxpos[i]=i;\n\t\t\t}\n\t\t\telse \n\t\t\t{\n\t\t\t\tmx[i]=mx[i-1];\n\t\t\t\tmxpos[i]=mxpos[i-1];\n\t\t\t}\n\t\t}\n\t\t\n\t\tmemset(tr,0,sizeof(tr));\n\t\tfor (int i=n;i>=1;i--)\n\t\t{\n\t\t\trpos[i]=max(i,query(a[i]-1));\n\t\t\tmodify(a[i],i);\n\t\t}\n\t\t\n\t\t//memset(f,-1,sizeof(f));\n\t\tfor (int i=1;i<=n;i++)\n\t\t{\n\t\t\tint j=i;\n\t\t\tint mx=a[i];\n\t\t\twhile (mxpos[rpos[j]]!=j)\n\t\t\t{\n\t\t\t\tj=mxpos[rpos[j]];\n\t\t\t\tmx=max(mx,a[j]);\n\t\t\t}\n\t\t\tcout << mx << \" \";\n\t\t}\n\t\tcout << \"\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 291721609,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\n#include<iostream>\nusing namespace std;\nint main()\n{\n\tint n;\n\tcin>>n;\n\tfor (int p=0;p<n;p++)\n\t{\n\t\t\n\t\tif(p==0)cout<<3<<\" \"<<3<<\" \"<<3<<\" \"<<4<<endl;\n\t\tif(p==1)cout<<5<<\" \"<<5<<\" \"<<5<<\" \"<<5<<\" \"<<5<<endl;\n\t\tif(p==2)cout<<2<<\" \"<<2<<\" \"<<2<<\" \"<<3<<endl;\n\t\tif(p==3)cout<<1<<\" \"<<1<<\" \"<<3<<\" \"<<3<<endl;\n\t\tif(p==4)cout<<8<<\" \"<<8<<\" \"<<8<<\" \"<<8<<\" \"<<8<<\" \"<<8<<\" \"<<8<<\" \"<<8<<endl;\n\t\t\n\t}\t\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 295374006,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nimport random\n \n \ndef adjust(list_n,xiabiao,flag):\n global list_max,index_list\n max_tmp=list_n[xiabiao]\n #print(list_max)\n list_max.append(max_tmp)\n for j in range(len(list_n)):\n if xiabiao>j and list_n[xiabiao]<list_n[j] and flag==1:\n list_max.append(list_n[j])\n index_list.append(j)\n adjust(list_n, j, 1)\n if xiabiao<j and list_n[xiabiao]>list_n[j] and flag ==0:\n index_list.append(j)\n adjust(list_n, j, 0)\n return\n \nif __name__ == '__main__':\n num=int(input())\n for i in range(num):\n # 获取总个数\n n=int(input())\n # 获取数组\n user_input = input()\n number_strings = user_input.split()\n list_n = [int(n0) for n0 in number_strings]\n #print(list_n)\n for w in range(len(list_n)) :\n list_max=[list_n[w]]\n index_list=[w]\n for k in range(len(list_n)):\n adjust(list_n, w, 0)\n for k in range(len(list_n)):\n adjust(list_n, max(index_list), 1)\n \n for ww in range(len(list_n)*5):\n random1= random.randint(min(index_list), max(index_list))\n random2 = random.randint(min(index_list), max(index_list))\n for k in range(len(list_n)) :\n adjust(list_n,random1,0)\n for k in range(len(list_n)) :\n adjust(list_n,random2, 1)\n print(max(list_max),end=' ')\n print()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 295369937,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\ndef solve_rabbit_jumps():\n t = int(input())\n \n for _ in range(t):\n n = int(input())\n a = list(map(int, input().split()))\n \n def dfs(start, visited):\n max_height = a[start]\n visited[start] = True\n \n # 向左跳（跳到更高的树）\n for j in range(start-1, -1, -1):\n if not visited[j] and a[j] > a[start]:\n max_height = max(max_height, dfs(j, visited))\n \n # 向右跳（跳到更矮的树）\n for j in range(start+1, n):\n if not visited[j] and a[j] < a[start]:\n max_height = max(max_height, dfs(j, visited))\n \n return max_height\n \n # 对每个起始位置计算最大高度\n result = []\n for i in range(n):\n visited = [False] * n\n max_height = dfs(i, visited)\n result.append(max_height)\n \n print(*result)\n \n# 运行程序\nsolve_rabbit_jumps()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 295361096,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nimport random\n \n \ndef adjust(list_n,xiabiao,flag):\n global list_max,index_list\n max_tmp=list_n[xiabiao]\n #print(list_max)\n list_max.append(max_tmp)\n for j in range(len(list_n)):\n if xiabiao>j and list_n[xiabiao]<list_n[j] and flag==1:\n list_max.append(list_n[j])\n index_list.append(j)\n adjust(list_n, j, 1)\n if xiabiao<j and list_n[xiabiao]>list_n[j] and flag ==0:\n index_list.append(j)\n adjust(list_n, j, 0)\n return\n \nif __name__ == '__main__':\n num=int(input())\n for i in range(num):\n # 获取总个数\n n=int(input())\n # 获取数组\n user_input = input()\n number_strings = user_input.split()\n list_n = [int(n0) for n0 in number_strings]\n #print(list_n)\n for w in range(len(list_n)) :\n list_max=[list_n[w]]\n index_list=[w]\n for k in range(len(list_n)):\n adjust(list_n, w, 0)\n for k in range(len(list_n)):\n adjust(list_n, max(index_list), 1)\n \n for ww in range(len(list_n)*10):\n random1= random.randint(min(index_list), max(index_list))\n random2 = random.randint(min(index_list), max(index_list))\n for k in range(len(list_n)) :\n adjust(list_n,random1,0)\n for k in range(len(list_n)) :\n adjust(list_n,random2, 1)\n print(max(list_max),end=' ')\n print()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 295369607,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class codeforces {\n \n static int MOD = 998244353 ;\n static long modInverse(long x, int MOD) {\n return power(x, MOD - 2, MOD);\n }\n \n static long power(long base, long exp, int MOD) {\n long result = 1;\n while (exp > 0) {\n if ((exp & 1) == 1) { // If exp is odd\n result = (result * base) % MOD;\n }\n base = (base * base) % MOD;\n exp >>= 1; // Divide exp by 2\n }\n return result;\n }\n \n static int max = 0 ;\n static void dfs(int i, int p,int[] a ,int[] visited, int[] dp, int n) {\n visited[i] = 1 ;\n \n max = Math.max(max, a[i]) ;\n \n for (int j = 0; j < i; j++) {\n if(a[j] > a[i] && visited[j] == 0) dfs(j,i,a,visited,dp,n) ;\n }\n for (int j = i+1; j < n; j++) {\n if(a[j] < a[i] && visited[j] == 0) dfs(j,i,a,visited,dp,n) ;\n }\n \n dp[i] = max ;\n }\n \n public static void main(String[] args) throws IOException {\n Scanner sc = new Scanner(System.in);\n \n int t = sc.nextInt(); // Number of test cases\n while (t-- > 0) {\n int n = sc.nextInt() ;\n max = 0 ;\n int[] a = new int[n] ;\n for (int i = 0; i < n; i++) {\n a[i] = sc.nextInt() ;\n }\n \n int[] ans = new int[n] , v = new int[n] ;\n for (int i = 0; i < n; i++) {\n if(v[i] == 0) dfs(i,-1,a,v,ans,n) ;\n }\n for(int x : ans) System.out.print(x+\" \");\n System.out.println();\n }\n sc.close();\n }\n \n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 302715491,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class CodeForcesSolution {\n static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n static StringTokenizer st;\n static int MOD = (int) 998244353, n;\n static int[] a;\n static List<Integer> lst;\n static int [][] minTable;\n public static void main(String[] args) throws IOException {\n int t = readInt();\n while (t-- > 0) {\n solve();\n }\n }\n static void solve() throws IOException {\n n = readInt();\n a = new int[n]; lst = new ArrayList<>();\n for (int i = 0; i < n; i++) {\n a[i] = readInt();\n lst.add(i);\n }\n minTable = new int[n+1][(int) (Math.log(n)/Math.log(2))+1];\n for (int i = 1; i <= n; i++) {\n minTable[i][0] = a[i - 1];\n }\n for (int i = 1; (1<<i) <= n; i++) {\n for (int j = 1; j+(1<<i)-1 <= n; j++) {\n minTable[j][i] = Math.min(minTable[j][i-1], minTable[j+(1<<(i-1))][i-1]);\n }\n }\n lst.sort((f, s) -> {\n if (a[f] == a[s]) return f - s;\n else return a[s] - a[f];\n });\n List<int[]> cutoff = new ArrayList<>();\n int j = lst.get(0), j2 = n - 1, v = a[lst.get(0)];\n for (int i = 0; i < n; i++) {\n if (j <= lst.get(i) || minQuery(j + 1, j2) < a[lst.get(i)]) {\n j = Math.min(j, lst.get(i));\n } else {\n cutoff.add(new int[] {j, v});\n j2 = j;\n j = lst.get(i);\n v = a[lst.get(i)];\n }\n }\n cutoff.add(new int[] {j, v});\n int i = n - 1;\n int[] ans = new int[n];\n for (int[] c : cutoff) {\n while (i >= c[0]) {\n ans[i--] = c[1];\n }\n }\n printArray(ans, n);\n }\n static int minQuery(int l, int r) {\n if (l > r) return 1<<30;\n int j = (int) (Math.log(r-l+1)/Math.log(2));\n return Math.min(minTable[l][j], minTable[r-(1<<j)+1][j]);\n }\n static void printArray(int[] arr, int size) {\n for(int i = 0; i < size; i++) {\n System.out.print(arr[i]);\n if (i < size - 1) System.out.print(\" \");\n }\n System.out.println();\n }\n static String next() throws IOException {\n while (st == null || !st.hasMoreTokens())\n st = new StringTokenizer(br.readLine().trim());\n return st.nextToken();\n }\n static long readLong() throws IOException {\n return Long.parseLong(next());\n }\n static int readInt() throws IOException {\n return Integer.parseInt(next());\n }\n static double readDouble() throws IOException {\n return Double.parseDouble(next());\n }\n static char readCharacter() throws IOException {\n return next().charAt(0);\n }\n static String readLine() throws IOException {\n return br.readLine().trim();\n }\n static int readALotInt() throws IOException{\n int x = 0, c;\n while((c = br.read()) != ' ' && c != '\\n')\n x = x * 10 + (c - '0');\n return x;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 303170462,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\ntypedef long long ll;\nusing namespace std;\n#define endl '\\n';\n#define int ll\n#define double long double\n \nvector<int> ans;\nvector<vector<int>> gr;\nvector<int> used;\n \nvoid dfs(int i)\n{\n used[i] = 1;\n for (auto c : gr[i])\n {\n if (!used[c])\n {\n dfs(c);\n }\n ans[i] = max(ans[i], ans[c]);\n }\n}\n \nvoid solve()\n{\n int n;\n cin >> n;\n vector<int> a(n+1);\n \n for (int i = 1; i <= n; ++i)\n {\n cin >> a[i];\n // a[i]--;\n }\n \n ans.clear();\n ans.resize(n+1);\n gr.clear();\n gr.resize(n + 2);\n used.clear();\n used.resize(n + 2);\n for (int i = 1; i <= n; ++i)\n {\n ans[i] = a[i];\n }\n vector<pair<int,int>> pref(n + 1);\n vector<int> suf(n + 2, a[n]);\n for (int i = 1; i <= n; ++i)\n {\n if (pref[i-1].first<a[i]){\n pref[i]={a[i],i};\n }\n else{\n pref[i]=pref[i-1];\n }\n // pref[i] = max(pref[i - 1], a[i - 1]);\n }\n for (int i = n; i >= 1; --i)\n {\n suf[i] = min(suf[i + 1], a[i]);\n }\n suf[0] = suf[1];\n //for (auto c : suf)\n //{\n // cout << c << \" \";\n //}\n \n //cout<<endl;\n \n \n for (int i = 1; i <=n; ++i)\n {\n // cout<<a[i]+1<<\" \"<<suf[i+2]+1<<\" \"<<pref[i]+1<<endl;\n if (i + 1 < suf.size() && suf[i+1] < a[i])\n {\n // int l=i+1;\n int l = i + 1;\n int r = suf.size();\n while (r - l > 1)\n {\n int mid = (r + l) / 2;\n if (suf[mid] <= a[i])\n {\n l = mid;\n }\n else\n {\n r = mid;\n }\n }\n // cout << l << \" \" << a[i] + 1 << \" \" << suf[l] + 1 << endl;\n //cout<<l<<\" \"<<suf[l]<<endl;\n if (suf[l] < a[i] && l<=n)\n {\n \n \n //cout<<\"suf\"<<i+1<<\" \"<<l<<endl;\n gr[i].push_back(l);\n \n // gr[suf[l]].push_back(a[i]);\n }\n }\n if (i >= 1 && pref[i].first > a[i])\n {\n // cout<<a[i]+1<<\" \"<<pref[i]+1<<endl;\n //cout<<\"pr\"<<i+1<<\" \"<<pref[i].second+1<<endl;\n gr[i].push_back(pref[i].second);\n // gr[pref[i]].push_back(a[i]);\n }\n \n dfs(i);\n }\n for (int k=0;k<n;++k){\n for (int i=1;i<=n;++i){\n dfs(i);\n }\n }\n \n \n // for (auto c:gr[5]){\n // cout<<c+1<<\"w\";\n // }\n // cout<<endl;\n // for (int i = 0; i < n; ++i)\n // {\n // cout << i + 1 << \" \" << ans[i] + 1 << endl;\n // }\n // for ()\n \n for (int i = 1; i <= n; ++i)\n {\n cout << ans[i] << \" \";\n }\n cout << endl;\n}\n \nsigned main()\n{\n int t;\n cin >> t;\n while (t--)\n {\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 293554698,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\n#pragma GCC optimize(\"O3,unroll-loops\")\n#include<bits/stdc++.h>\n// #include <atcoder/all>\n// #include<ext/pb_ds/assoc_container.hpp>\n// #include<ext/pb_ds/tree_policy.hpp>\nusing namespace std;\n \n// using namespace __gnu_pbds;\nstruct custom_hash {\n static uint64_t splitmix64(uint64_t x) {\n x += 0x9e3779b97f4a7c15;\n x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;\n x = (x ^ (x >> 27)) * 0x94d049bb133111eb;\n return x ^ (x >> 31);\n }\n \n size_t operator()(uint64_t x) const {\n static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();\n return splitmix64(x + FIXED_RANDOM);\n }\n};\n \n \n \n// https://docs.google.com/document/d/1VuLkJfkaBBFdcV-BPtT2JdTDrD2lC3POFj6NPTl_Pzo/edit?addon_store&tab=t.0\n// typedef tree<int, null_type, greater<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; // find_by_order, order_of_key\n//find by order-> element at ith idx(starting from 0)//gives iterator to that index//if not present *it gives 0\n//order of key-> number of element less than a number\n \n// erase function for ordered multiset\n// void myerase(pbds &t, int v){\n// int rank = t.order_of_key(v);//Number of elements that are less than v in t\n// pbds::iterator it = t.find_by_order(rank); //Iterator that points to the (rank+1)th element in t\n// t.erase(it);\n// }\n \n \n \n#define int long long int \n// #define ll long long int\n#define loop(var, s, e) for(int var = s; var < e; var++)\n#define rloop(var, e, s) for(int var = e; var>= s; var--)\n#define v vector<int>\n// #define vll vector<long long int>\n#define printyes cout<<\"YES\\n\"\n#define printno cout<<\"NO\\n\"\n//COMMENT THIS FOR INTERACTIVE PROBLEMS...............\n// #define endl '\\n'\n#define mod (int)(998244353)\n#define inf 1e18\n#define dbg(x) cerr<<\"value of \"<<#x<<\" is => \"<<x<<endl;\n#define all(x) x.begin(), x.end()\n#define vpii vector<pair<int, int>>\n#define pii pair<int,int>\n#define pll pair<ll, ll>\n#define ff first\n#define ss second\n#define pb push_back\n#define vpll vector<pll>\n#define here(i) cerr<<\"here \"<<i<<\" \"<<endl;\n#define rt return;\n#define mx max_element\n#define mn min_element\n#define asum accumulate\nmt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());\ntemplate <class A, class B> ostream& operator << (ostream& out, const pair<A, B> &a) {\n\treturn out << \"(\" << a.ff << \", \" << a.ss << \")\";\n}\n \ntemplate <class A> ostream& operator << (ostream& out, const vector<A> &ve) {\n\tout << \"[\";\n\tloop(i, 0, ve.size()) {\n\t\tif(i) out << \", \";\n\t\tout << ve[i];\n\t}\n\treturn out << \"]\";\n}\n \n //U R D L\nint delrow[4] = {-1,0,1,0};\nint delcol[4] = {0,1,0,-1};\n// unordered_map<char, int> mpp = {{'U', 0}, {'R', 1}, {'D', 2}, {'L', 3}};\n \nnamespace bit {\n int setBit(int n, int bitNo) {\n n = (n|(1LL<<bitNo));\n return n;\n }\n \n int currBit(int n, int bitNo) {\n return ((n&(1ll<<bitNo)) == 0)?0:1;\n }\n \n int unSetBit(int n, int bitNo) {\n n = (n&(~(1ll<<bitNo)));\n return n;\n }\n /*\n __lg() returns the highest set bit index in a number i.e for 100 it will be 2 and for 000 will be -1\n \n n&(n - 1) clears the first set bit\n \n n&(-n) give the first set bit => 101100 -> 000100\n \n n&(n + 1) clear all trailing ones\n \n n|(n + 1) set the first unset bit\n */\n}\n//*************************************************CHECK THE CONSTRAINTS**********************************************************//\nint extendedEuclidean(int a, int b, int&x, int& y) {\n if(b == 0) {\n x = 1;\n y = 0;\n return a;\n }\n int x1, y1;\n int d = extendedEuclidean(b, a%b, x1, y1);\n x = y1;\n y = x1 - y1*(a/b);\n return d;\n}\n \n \nint f(int a, int b) {\n if(b == 0)return 1;\n int res = f(a, b/2);\n if(b%2 != 0) {\n return (((res)%mod* (res)%mod)%mod * a)%mod;\n }\n else {\n return ((res)%mod * (res)%mod)%mod;\n }\n}\n \n \nint _f(int a, int b) {\n if(b == 0)return 1;\n int res = f(a, b/2);\n if(b%2 != 0) {\n return (((res)* (res))* a);\n }\n else {\n return ((res) * (res));\n }\n}\n \n \nint inv(int a, int b){\n return 1<a ? b - inv(b%a,a)*b/a : 1;\n}\n \n \nvector<bool> sieve(int n) {\n vector<bool> is_prime(n+1, 1);\n // v prime;\n // v dp(n + 1, 0);\n is_prime[0] = 0;\n is_prime[1] = 0;\n for(int i = 2; i <= n; i++) {\n if(is_prime[i] == 1) {\n // prime.pb(i);\n for(int j = i*i; j <= n; j+=i) {\n is_prime[j] = 0;\n }\n }\n }\n return is_prime;\n}\n// void *memcpy(void *dest, const void *src, size_t n);\nint ceilDiv(int a, int b) {\n double _a = a;\n int ans = ceil(_a/b);\n return ans;\n}\n \n \n// first write complete pseudo code on paper then code dont care about the time!!!!!!\n \nvoid pikachu(int tt, int tc) {\n int n;\n cin>>n;\n v arr(n);\n loop(i, 0, n)cin>>arr[i];\n int min_so_far = inf;\n v lastMinIndex(n + 1, inf);\n rloop(i, n - 1, 0) {\n if(arr[i] < min_so_far) {\n min_so_far = arr[i];\n loop(j, arr[i] + 1, n + 1) {\n if(lastMinIndex[j] != inf)break;\n lastMinIndex[j] = i;\n }\n }\n }\n \n v ans(n, inf);\n v prevMaxIndex(n);\n prevMaxIndex[0] = 0;\n loop(i, 1, n) {\n if(arr[i] <= arr[prevMaxIndex[i - 1]])prevMaxIndex[i] = prevMaxIndex[i - 1];\n else prevMaxIndex[i] = i;\n }\n auto dfs = [&](int src, auto& self)->int {\n if(ans[src] != inf)return ans[src];\n ans[src] = arr[src];\n if(lastMinIndex[arr[src]] != inf)ans[src] = max(ans[src], self(lastMinIndex[arr[src]], self));\n if(prevMaxIndex[src] != src)ans[src] = max(ans[src], self(prevMaxIndex[src], self));\n return ans[src];\n };\n loop(i, 0, n)dfs(i, dfs);\n for(auto it: ans)cout<<it<<\" \";\n cout<<endl;\n} \n \n \n \n \nint32_t main() {\n auto begin = std::chrono::high_resolution_clock::now();\n ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n #ifndef ONLINE_JUDGE\n freopen(\"input.txt\", \"r\", stdin);\n freopen(\"output.txt\", \"w\", stdout);\n #endif\n int gsBall = 1;\n cin>>gsBall;\n int tt = gsBall;\n int tc = 1; \n cout<<setprecision(20);\n while(gsBall--) {\n pikachu(tt, tc);\n tc++;\n }\n auto end = std::chrono::high_resolution_clock::now();\n auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);\n cerr << \"Time measured: \" << elapsed.count() * 1e-9 << \" seconds.\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 300287663,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nimport math\nfrom sys import stdout\n \n \ndef ii():\n return int(input())\n \n \ndef mi():\n return map(int, input().split())\n \n \ndef li():\n return list(mi())\n \n \nclass UF:\n \n def __init__(self, n):\n self.p = [i for i in range(n)]\n self.ranks = [0] * n\n \n def find(self, x):\n if x == self.p[x]: return x\n self.p[x] = self.find(self.p[x])\n return self.p[x]\n \n def union(self, x, y):\n x, y = self.find(x), self.find(y)\n if x == y: return\n if self.ranks[x] > self.ranks[y]: self.p[y] = x\n elif self.ranks[x] < self.ranks[y]: self.p[x] = y\n else: self.p[y] = x; self.ranks[x] += 1\n \n \ndef bin_right(arr, curr, uf):\n l = 0; r = len(arr) - 1\n while l <= r:\n mid = (l + r) // 2\n if arr[mid][0] < curr[0]:\n if arr[mid][1] > curr[1]:\n uf.union(curr[1], arr[mid][1])\n r = mid - 1\n \n return uf\n \n \ndef bin_left(arr, curr, uf):\n l = 0; r = len(arr) - 1\n while l <= r:\n mid = (l + r) // 2\n if arr[mid][0] > curr[0]:\n if arr[mid][1] < curr[1]: uf.union(curr[1], arr[mid][1])\n l = mid + 1\n return uf\n \n \ndef solve():\n n = ii(); a = li(); uf = UF(n)\n array = [(a[i], i) for i in range(n)]; array.sort()\n \n for i in range(n):\n uf = bin_left(array, (a[i], i), uf)\n uf = bin_right(array, (a[i], i), uf)\n \n segments = []; i = 0; j = 0\n while j < n:\n while j < n and uf.p[i] == uf.p[j]: j += 1\n segments.append((i, j - 1)); i = j\n \n ans = list()\n for s in segments:\n curr = max(a[s[0]: s[1] + 1])\n ans += [f'{curr}'] * (s[1] - s[0] + 1)\n print(' '.join(ans))\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 293125585,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nimport sys\ninput = sys.stdin.readline\nfrom collections import deque\nfrom bisect import bisect_left\n \ndef dfs(i):\n if dp[i] != -1: return dp[i]\n \n dp[i] = li[i]\n dp[i] = max(dp[i], dfs(b[i]))\n dp[i] = max(dp[i], dfs(f[i]))\n \n return dp[i]\n \ndef solve():\n global n, b, f, dp, li\n \n n = int(input())\n li = list(map(int, input().split()))\n a = [(li[i], i) for i in range(n)]\n dp = [-1]*n\n ans = [0]*n\n f = [0]*n\n b = [0]*n\n \n st = deque()\n st.appendleft((int(1e9), -1))\n for i in range(n - 1, -1, -1):\n if li[i] < st[0][0]:\n st.appendleft((li[i], i))\n \n j = bisect_left(st, (li[i], 1e9)) - 1\n if li[i] == st[j][0]:\n f[i] = i\n else:\n f[i] = st[j][1]\n \n # print(f)\n \n a.sort(reverse=True)\n r = n\n for i in range(n):\n for j in range(a[i][1], r):\n b[j] = a[i][1]\n r = min(r, a[i][1])\n \n # print(*f)\n # print(*b)\n \n for i in range(n):\n ans[i] = dfs(i)\n \n print(*ans)\n \nfor _ in range(int(input())):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 291851954,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Main {\n static int maxHeight(int index, int n, int[] heights, int[] memo, boolean[] visited) {\n if (visited[index]) {\n return memo[index];\n }\n visited[index] =true;\n int max = heights[index];\n max = Math.max(max, backJumps(index, n, heights, memo, visited));\n max = Math.max(max, forwardJumps(index, n, heights, memo, visited));\n \n memo[index] = max;\n visited[index] = false;\n return max;\n }\n \n static int backJumps(int index, int n, int[] heights, int[] memo, boolean[] visited) {\n int max = heights[index];\n for (int i = index-1; i > -1; i--) {\n if (heights[i] > heights[index]) {\n int height = maxHeight(i, n, heights, memo, visited);\n max = Math.max(max, height);\n }\n }\n return max;\n }\n \n static int forwardJumps(int index, int n, int[] heights, int[] memo, boolean[] visited) {\n int max = heights[index];\n for (int i = index + 1; i < n; i++) {\n if (heights[i] < heights[index]) {\n int height = maxHeight(i, n, heights, memo, visited);\n max = Math.max(max, height);\n }\n }\n return max;\n }\n \n public static void main(String[] args) {\n Scanner scan = new Scanner(System.in);\n \n int t = scan.nextInt();\n \n for (int test = 0; test < t; test++) {\n int n = scan.nextInt();\n int[] heights = new int[n];\n int[] memo = new int[n];\n boolean[] visited = new boolean[n];\n \n for (int i = 0; i < n; i++) {\n heights[i] = scan.nextInt();\n memo[i] = -1;\n visited[i] = false; \n }\n \n for (int i = 0; i<n; i++) {\n System.out.print(maxHeight(i, n, heights, memo, visited) + \" \");\n }\n System.out.println();\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 291654015,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Forked {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt(); // Number of test cases\n while (t-- > 0) {\n int n = sc.nextInt(); // Length of permutation\n int[] p = new int[n];\n for (int i = 0; i < n; i++) {\n p[i] = sc.nextInt();\n }\n \n for(int i = 0; i < n ; i++){\n int currval = p[i];\n int max = currval;\n int minidx = -1;\n \n for(int j = i + 1 ; j < n ; j++){\n max = Math.max(max, p[j]);\n if(p[j] < currval){\n minidx = j;\n currval = max;\n }\n }\n if(minidx!=-1){\n for(int k = i ; k<=minidx; k++){\n p[k] = currval;\n }\n i = minidx;\n }\n else{\n break;\n }\n }\n int maxidx = n-1;\n int max = p[n-1];\n for(int i = n-2 ; i >= 0 ; i--){\n if(p[i] > max){\n max = p[i];\n maxidx = i;\n }\n }\n for(int i = n-1 ; i>= maxidx ; i--){\n p[i] = max;\n }\n for(int i = 0; i < n ; i++){\n System.out.print(p[i] + \" \");\n }\n System.out.println();\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 291766822,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#include <cmath>\n#include <complex>\n#include<string>\n#include <ext/pb_ds/assoc_container.hpp>\n//#include \"testlib.h\"\n//freopen(\"orxor.in\", \"r\", stdin);\nusing namespace std;\n//using namespace __gnu_pbds;\n//template <class T>\n//using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\n#define ll long long\n#define pb push_back\n#define F first\n#define S second\n#define fast ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)\n#define dcm(a) setprecision(a)<<fixed4\n#define double long double\nll const md= 1e9+7;\ndouble const EPS=1e-9;\nint const LOG=22;\nusing uint = unsigned int;\nusing ull = unsigned long long;\n//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());\n#define clr(memo, a) memset(memo,a,sizeof(memo))\n#define PI acos(-1)\n#define endl '\\n'\nstruct segtree\n{\n vector<ll> value;\n vector<ll> op;\n ll NEUTRAL=0;\n ll NO_OP=0;\n int sz=1;\n void init(int n)\n {\n sz=1;\n while(sz<n)\n sz*=2;\n value.assign(2*sz,0);\n op.assign(2*sz,0);\n }\n ll modify_op(ll a,ll b,ll len)\n {\n if(b==NO_OP)\n return a;\n return (b*len);\n }\n ll calcval(ll a,ll b)\n {\n return max(a,b);\n }\n void apply_mod(ll &a,ll b,ll len)\n {\n a=modify_op(a,b,len);\n }\n void propagte(int x,int lx,int rx)\n {\n if(rx-lx==1)\n return;\n int m=(rx+lx)/2;\n apply_mod(value[2*x+1],op[x],m-lx);\n apply_mod(value[2*x+2],op[x],rx-m);\n apply_mod(op[2*x+1],op[x],1);\n apply_mod(op[2*x+2],op[x],1);\n op[x]=NO_OP;\n }\n void modify(int l,int r,ll v,int x,int lx,int rx)\n {\n propagte(x,lx,rx);\n if(lx>=r||l>=rx)return;\n if(lx>=l&&rx<=r)\n {\n apply_mod(op[x],v,1);\n apply_mod(value[x],v,rx-lx);\n return;\n }\n int m=(rx+lx)/2;\n modify(l,r,v,2*x+1,lx,m);\n modify(l,r,v,2*x+2,m,rx);\n value[x]=calcval(value[2*x+1],value[2*x+2]);\n \n }\n void modify(int l,int r,ll v)\n {\n modify(l,r,v,0,0,sz);\n }\n ll calc(int l,int r,int x,int lx,int rx)\n {\n propagte(x,lx,rx);\n if(lx>=r||l>=rx)\n return NEUTRAL;\n if(lx>=l&&rx<=r)\n {\n return value[x];\n }\n int m=(rx+lx)/2;\n ll c1=calc(l,r,2*x+1,lx,m);\n ll c2=calc(l,r,2*x+2,m,rx);\n return calcval(c1,c2);\n }\n \n ll calc(int l,int r)\n {\n return calc(l,r,0,0,sz);\n }\n void build(vector<ll> &a,int x,int lx,int rx)\n {\n if(rx-lx==1)\n {\n \n if(lx<a.size())\n value[x]=a[lx];\n return;\n }\n int md=(lx+rx)/2;\n build(a,2*x+1,lx,md);\n build(a,2*x+2,md,rx);\n value[x]=calcval(value[2*x+1],value[2*x+2]);\n }\n void build(vector<ll> &a)\n {\n build(a,0,0,sz);\n }\n \n};\nvoid solve(int testCase)\n{\n int n;\n cin>>n;\n segtree seg;\n seg.init(n);\n vector<int> ans(n+1);\n vector<vector<int> > adj(n+5);\n int mx=0;\n for(int i=0; i<n; i++)\n {\n int x;\n cin>>x;\n mx=max(mx,x);\n ans[i]=mx;\n adj[x].push_back(i);\n }\n for(int i=1; i<=n; i++)\n {\n for(int j:adj[i])\n {\n int s=seg.calc(j+1,n);\n ans[j]=max(ans[j],s);\n }\n for(int j:adj[i])\n {\n seg.modify(j,j+1,ans[j]);\n if(ans[j]>i){\n adj[ans[j]].push_back(j);\n }\n }\n }\n for(int i=0; i<n; i++)\n cout<<ans[i]<<\" \";\n cout<<endl;\n}\n \n \nvoid precalc()\n{\n \n}\n \nint main()\n{\n \n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n cerr.tie(0);\n int testCase=1;\n precalc();\n //freopen(\"permutations.in\", \"r\", stdin);\n cin>>testCase;\n //freopen(\"test.txt\", \"w\", stdout);\n for(int i=1; i<=testCase; i++)\n solve(i);\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 291625458,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ff cout.flush()\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\n#define FAST ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0);\n#define ll long long\n#define str string\nusing ld = long double;\n#define pb push_back\n#define pf push_front\n#define pob pop_back()\n#define pof pop_front()\n#define in insert\n#define fi first\n#define se second\n#define begin begin()\n#define end end()\n#define size size()\n#define back back()\n#define rend rend()\n#define rbegin rbegin()\n#define ret return\n#define ull unsigned long long\n#define all(a) a.begin , a.end\n#define gcd(a , b) __gcd(a , b)\n#define lcm(a , b) (a * b) / gcd(a , b)\n#define friopen freopen (\"input.txt\", \"r\", stdin); freopen(\"output.txt\", \"w\", stdout);\nusing namespace std;\nusing namespace __gnu_pbds;\ntypedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> pbds;\n \n/// ERASE\n/*\n \nstr -- a.erase(index nachala , kolichestvo simvolov);\n \n \ndeque , vector -- a.erase(a.begin + index nachala , a.begin + index konca + 1); \n \n \n*/\n \n \n \n///------------\n \nconst ll mod = 1e9 + 7 , mod2 = 998244353;\nconst ll inf = 1e18;\nconst ld esp = 0.0000001 , Pi = 3.1415926535897932384626433832795;\n \nll binpow(ll x , ll y)\n{\n if (y == 0) ret 1;\n else if (y == 1) ret x;\n else if (y % 2 == 1) ret binpow(x , y - 1) * x;\n else ret binpow(x , y / 2) * binpow(x , y / 2);\n}\n \nll qpow(ll a , ll b , ll md){\n\ta = (a + md)%md;\n\tll ans = 1;\n\twhile(b > 0){\n\t\tif(b&1){\n\t\t\tans = (ans*a)%md;\n\t\t}\n\t\ta = (a * a)%md;\n\t\tb>>=1;\n\t}\n\treturn ans;\n}\n \nll div(ll a, ll b, ll mod) {\n ret (a % mod) * (qpow(b, mod-2 , mod)) % mod;\n}\n \nll sum_ (ll x)\n{\n ret x * (x + 1) / 2;\n}\n \nll ask(ll x , ll y , ll a , ll b)\n{\n\tret (x == a) + (y == b);\n}\n \nld dist(ld x , ld y , ld a , ld b)\n{\n\tret sqrt(abs(x - a) * abs(x - a) + abs(y - b) * abs(y - b));\n}\n \nvoid kol_a()\n{\n\tll n , i , j , mn = 1e9 , mx = 0;\n\tcin >> n;\n\tll a[n + 12] = {} , s[n + 12] = {} , d[n + 12] = {};\n\tfor (i = 1 ; i <= n ; i ++)\n\t{\n\t\tcin >> a[i];\n\t\td[i] = d[i - 1];\n\t\tif (a[i] > mx)\n\t\t{\n\t\t\tmx = a[i];\n\t\t\td[i] = i;\n\t\t}\n\t}\n\tfor (i = n ; i >= 1 ; i --)\n\t{\n\t\ts[i] = s[i + 1];\n\t\tif (a[i] < mn)\n\t\t{\n\t\t\tmn = a[i];\n\t\t\ts[i] = i;\n\t\t}\n\t}\n\tll ans[n + 12] = {};\n\tfor (i = 1 ; i <= n ; i ++)\n\t{\n\t\tif (ans[i])\n\t\t{\n\t\t\tcout << ans[i] << \" \";\n\t\t\tcontinue;\n\t\t}\n\t\tmap<ll , ll> us;\n\t\tll ind = i;\n\t\tset<ll> v;\n\t\twhile(us[ind] == 0 && a[ind] != mx)\n\t\t{\n\t\t\tv.in(ind);\n\t\t\tus[ind] ++;\n\t\t\tll l = ind , r = n + 1 , m;\n\t\t\twhile(l + 1 < r)\n\t\t\t{\n\t\t\t\tm = (l + r) / 2;\n\t\t\t\tif (a[ind] > a[s[m]]) l = m;\n\t\t\t\telse r = m;\n\t\t\t}\n\t\t\tind = d[l];\n\t\t}\n\t\tcout << a[ind] << \" \";\n\t\tfor (auto x : v) ans[x] = a[ind];\n\t}\n\tcout << '\\n';\n}\n \nmain()\n{\n\t//cout.flush()\n\tFAST\n\t// friopen\n\tll _ = 1;\n\tcin >> _;\n while(_ --) kol_a();\n}\n \n \n/*\n \n \n \n \n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 291655848,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nvisitados = [0] \nsaida = [0] \n \ndef bfs(grafo, lista, i):\n if visitados[i] == 0:\n visitados[i] = 1\n else:\n return 0\n \n fila = [i]\n aux = [i]\n maior = lista[i]\n \n while fila:\n v_atual = fila[0]\n del fila[0]\n for vz in grafo[v_atual]:\n if visitados[vz] == 0:\n if lista[vz] > maior:\n maior = lista[vz]\n fila.append(vz)\n aux.append(vz)\n visitados[vz] = 1\n \n for valor in aux:\n saida[valor] = maior \n \nt = int(input())\nfor _ in range(t):\n n = int(input())\n visitados = [0] * n\n saida = [0] * n\n lista = list(map(int, input().split()))\n \n grafo = [[] for _ in range(n)]\n for i in range(n):\n for j in range(i, n):\n if lista[i] > lista[j]:\n grafo[i].append(j)\n grafo[j].append(i)\n \n for i in range(n):\n bfs(grafo, lista, i)\n print(saida[i], end=' ')\n print()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 292528660,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nI = lambda : list(map(int, input().split(' ')))\nR = lambda : (int(input()))\n \nimport sys\nimport math\nfrom collections import deque, defaultdict\nfrom bisect import bisect_left, bisect_right\nfrom heapq import heappop, heapify, heappush\n# from typing import List\n \ndef join_parents(tp1 : int, tp2 : int, grp_map, parents_map ):\n \n # print(tp1, tp2)\n tp1 = parents_map[tp1]\n tp2 = parents_map[tp2]\n \n if len(grp_map[tp1]) < len(grp_map[tp2]):\n tp1, tp2 = tp2, tp1\n \n if parents_map[tp1] == parents_map[tp2] : return\n \n for e in grp_map[tp2]:\n grp_map[tp1].add(e)\n parents_map[e] = parents_map[tp1]\n \n grp_map[tp2].clear()\n grp_map[tp2] = []\n \n \nfor kp in range(R()):\n \n n = R()\n a = I()\n \n max_height = []\n parents = []\n new_a = []\n \n for i in range(n):\n parents.append(i) ;\n max_height.append(set([i])) ;\n new_a.append([(a[i],i)])\n \n # if n == 500000 : print(\"kk\") ; continue\n \n while len(new_a) > 1:\n \n temp_a = []\n \n for i in range(1,len(new_a),2):\n arr1 = new_a[i-1]\n arr2 = new_a[i]\n \n arr3 = []\n \n ind1 = -1\n ind2 = 0\n \n curr_val = arr2[ind2]\n \n for j in range(len(arr1)):\n if arr2[ind2][0] < arr1[j][0]: break\n else :\n arr3.append(arr1[j])\n if j == len(arr1) - 1: j += 1\n \n # updating the max_height of arr1\n for k in range(j,len(arr1)):\n # ind = arr1[k][1]\n # max_height[ind] = max(max_height[ind], arr1[-1][0])\n join_parents(arr1[k][1], arr1[-1][1], max_height, parents)\n \n ind1 = j\n \n while ind1 < len(arr1) and ind2 < len(arr2):\n \n if arr1[ind1][0] > arr2[ind2][0]:\n arr3.append(arr2[ind2])\n \n # updating the max_height\n # curr_max_height = max( max_height[arr2[ind2][1]], max_height[arr1[ind1][1]] )\n # max_height[arr2[ind2][1]] = curr_max_height\n # max_height[arr1[ind1][1]] = curr_max_height\n join_parents(arr2[ind2][1], arr1[ind1][1], max_height, parents)\n # print(ind1, ind2, arr2[ind2], arr1[ind1])\n ind2 += 1\n \n else:\n arr3.append(arr1[ind1])\n ind1 += 1\n \n arr3 = arr3 + arr1[ind1:] + arr2[ind2:]\n temp_a.append(arr3)\n \n # print(arr3, max_height, parents)\n \n # print(\"new_a : \", new_a,'\\n', \"temp_a : \", temp_a, '\\n')\n new_a = temp_a + new_a[i+1:]\n # del temp_a\n # del arr3\n # del arr1\n \n max_val = []\n for i in range(n):\n val = -1\n for j in max_height[i]: val = max(val, a[j])\n max_val.append(val)\n \n res = []\n for i in range(n):\n res.append(max_val[parents[i]])\n \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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 306215034,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class RabbitJump {\n public static void solve() {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n \n while (t-- > 0) {\n int n = sc.nextInt();\n int[] heights = new int[n];\n \n for (int i = 0; i < n; i++) {\n heights[i] = sc.nextInt();\n }\n \n // For each starting position\n for (int start = 0; start < n; start++) {\n boolean[] visited = new boolean[n];\n int maxHeight = dfs(start, heights, visited);\n System.out.print(maxHeight + \" \");\n }\n System.out.println();\n }\n }\n \n private static int dfs(int current, int[] heights, boolean[] visited) {\n visited[current] = true;\n int maxHeight = heights[current];\n \n // Try all possible jumps\n for (int next = 0; next < heights.length; next++) {\n if (!visited[next]) {\n // Check jumping conditions\n if ((next < current && heights[next] > heights[current]) || \n (next > current && heights[next] < heights[current])) {\n maxHeight = Math.max(maxHeight, dfs(next, heights, visited));\n }\n }\n }\n \n return maxHeight;\n }\n \n public static void main(String[] args) {\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 291814472,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Penchick and Desert Rabbit\ntime_limit_ms: 3000\nmemory_limit_mb: 256\nDescription: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!\n\nWhile trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.\n\nThe rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:\n\n- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.\n- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.\n\nFor each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nInput: The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 5 \\cdot 10^5$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 5 \\cdot 10^5$$$) — the number of trees.\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\le a_i \\le n$$$) — the height of the trees.\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \\cdot 10^5$$$.\n\nOutput: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.\n\nExamples:\ninput:\n5\n4\n2 3 1 4\n5\n5 4 3 2 1\n4\n2 1 1 3\n4\n1 1 3 1\n8\n2 4 1 6 3 8 5 7\noutput:\n3 3 3 4 \n5 5 5 5 5 \n2 2 2 3 \n1 1 3 3 \n8 8 8 8 8 8 8 8\n\nNote: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.\n\n- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.\n- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.\n\nIn the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.\n\nIn the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.\n\nBuggy Code:\nimport java.util.*;\npublic class program{\n\tpublic static void main(String [] args) {\n\t\tScanner sc=new Scanner(System.in);\n\t\tint t= sc.nextInt();\n\t\twhile(t-->0) {\n\t\t\tint n=sc.nextInt();\n\t\t\tArrayList<Integer> x=new ArrayList<>();\n\t\t\tint [] prevMax=new int[n];\n\t\t\tint max=0;\n\t\t\tfor(int i=0;i<n;i++) {\n\t\t\t\tint a=sc.nextInt();\n\t\t\t\tif(a>max) {\n\t\t\t\t\tmax=a;\n\t\t\t\t}\n\t\t\t\tx.add(a);\n\t\t\t\tprevMax[i]=max;\n\t\t\t}\n\t\t\tMap<Integer,Integer> map=new HashMap<>();\n\t\t\tint [] ans=new int[n];\n\t\t\tint k=n-1;\n\t\t\tfor(int i=n-1;i>=0;i--) {\n\t\t\t\tint mxel=0;\n\t\t\t\tint a=prevMax[i];\n\t\t\t\tfor(Map.Entry<Integer,Integer> y: map.entrySet()) {\n\t\t\t\t\tif(y.getKey()<a) {\n\t\t\t\t\t\tmxel=Math.max(mxel,y.getValue());\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tint z=Math.max(mxel,a);\n\t\t\t\tmap.put(x.get(i),z);\n\t\t\t\tans[k--]=z;\n\t\t\t}\n\t\t\tfor(int i=0;i<n;i++) {\n\t\t\t\tSystem.out.print(ans[i]+\" \");\n\t\t\t}\n\t\t\tSystem.out.println();\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031D",
"submission_id": 291739154,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n/*\n Code by: linvg\n Created: 25.11.2024 14:07:16\n*/\n \n#include <bits/stdc++.h>\nusing namespace std;\n \n#define int long long\n#define all(x) (x).begin(), (x).end()\n#define sz(x) ((int)(x).size())\nstring to_upper(string a) { for (int i = 0;i < (int)a.size();++i) if (a[i] >= 'a' && a[i] <= 'z') a[i] -= 'a' - 'A'; return a; }\nstring to_lower(string a) { for (int i = 0;i < (int)a.size();++i) if (a[i] >= 'A' && a[i] <= 'Z') a[i] += 'a' - 'A'; return a; }\n \ntemplate<class T> T gcd(T a, T b) { T r; while (b != 0) { r = a % b; a = b; b = r; } return a; }\ntemplate<class T> T gcd(initializer_list<T> __l) { int a = 0; for (auto x : __l) { a = gcd(a, x); } return a; }\ntemplate<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }\ntemplate<class T> T sqr(T x) { return x * x; }\ntemplate<class T> T cube(T x) { return x * x * x; }\n \n// #define DBG\n \n#ifdef ONLINE_JUDGE\n#define dbg(...)\n#else\n#include \"debug.h\"\n#endif\n \nconst int INF = 0x3f3f3f3f3f;\nconst int MOD = 1e9 + 7;\n \nunordered_map<char, int> mapped;\n \n#define MAXLEN 1000010\nusing namespace std;\nconstexpr uint64_t mod = (1ULL << 61) - 1;\nconst uint64_t seed = chrono::system_clock::now().time_since_epoch().count();\nconst uint64_t base = mt19937_64(seed)() % (mod / 3) + (mod / 3);\nuint64_t base_pow[MAXLEN];\nint64_t modmul(uint64_t a, uint64_t b) {\n uint64_t l1 = (uint32_t)a, h1 = a >> 32, l2 = (uint32_t)b, h2 = b >> 32;\n uint64_t l = l1 * l2, m = l1 * h2 + l2 * h1, h = h1 * h2;\n uint64_t ret = (l & mod) + (l >> 61) + (h << 3) + (m >> 29) + (m << 35 >> 3) + 1;\n ret = (ret & mod) + (ret >> 61);\n ret = (ret & mod) + (ret >> 61);\n return ret - 1;\n}\nvoid init() {\n base_pow[0] = 1;\n for (int i = 1; i < MAXLEN; i++) {\n base_pow[i] = modmul(base_pow[i - 1], base);\n }\n}\nstruct PolyHash {\n /// Remove suff vector and usage if reverse hash is not required for more speed\n vector<int64_t> pref, suff;\n PolyHash() {}\n template <typename T>\n PolyHash(const vector<T>& ar) {\n if (!base_pow[0]) init();\n int n = ar.size();\n assert(n < MAXLEN);\n pref.resize(n + 3, 0), suff.resize(n + 3, 0);\n for (int i = 1; i <= n; i++) {\n pref[i] = modmul(pref[i - 1], base) + ar[i - 1] + 997;\n if (pref[i] >= mod) pref[i] -= mod;\n }\n for (int i = n; i >= 1; i--) {\n suff[i] = modmul(suff[i + 1], base) + ar[i - 1] + 997;\n if (suff[i] >= mod) suff[i] -= mod;\n }\n }\n PolyHash(const char* str)\n : PolyHash(vector<char>(str, str + strlen(str))) {\n }\n uint64_t get_hash(int l, int r) {\n int64_t h = pref[r + 1] - modmul(base_pow[r - l + 1], pref[l]);\n return h < 0 ? h + mod : h;\n }\n uint64_t rev_hash(int l, int r) {\n int64_t h = suff[l + 1] - modmul(base_pow[r - l + 1], suff[r + 2]);\n return h < 0 ? h + mod : h;\n }\n bool is_palindrome(int l, int r) {\n return get_hash(l, r) == rev_hash(l, r);\n }\n};\n \n// compare two strings (1: s < t, 0: s == t, -1: s > t)\nint compare(PolyHash& a, PolyHash&b) {\n int len = min(a.pref.size(), b.pref.size()) - 1;\n int l = 0, r = len;\n while (l < r) {\n int mid = (l + r + 1) / 2;\n if (a.get_hash(0, mid - 1) == b.get_hash(0, mid - 1)) l = mid;\n else r = mid - 1;\n }\n if (l == len) {\n if (a.pref.size() == b.pref.size()) return 0;\n return a.pref.size() < b.pref.size() ? 1 : -1;\n }\n return a.pref[l] < b.pref[l] ? 1 : -1;\n}\n \n \nvoid concat(string& s, string& t, int& start) {\n if (start == 0) {\n s = t;\n start = sz(s);\n }\n else if (start == sz(s)) {\n s += t;\n start = sz(s);\n }\n else {\n int n = sz(s);\n for (int i = 0; i < t.size(); i++, start++) {\n if (start < sz(s)) s[start] = t[i];\n else s += t[i];\n }\n }\n}\n \nconst int N = 1e6 + 7;\n \nstruct Hash {\n size_t operator()(const pair<int, int>& p) const {\n return p.first ^ p.second;\n }\n};\n \nunordered_map<int, PolyHash> mp;\n \nstruct Trie\n{\n struct Node\n {\n unordered_map<int, Node*> next;\n unordered_set<pair<int, int>, Hash> comb;\n };\n \n Node* root;\n Trie() {\n root = new Node();\n }\n \n void add(string& s, int idx) {\n Node* cur = root;\n for (int i = 0; i < s.size(); i++) {\n int c = mapped[s[i]];\n if (cur->next.count(c) == 0) {\n cur->next[c] = new Node();\n }\n cur->comb.insert({ sz(s), N - idx });\n cur = cur->next[c];\n }\n cur->comb.insert({ sz(s), N - idx });\n }\n \n int find(string& t, int start) {\n Node* cur = root;\n int idx = -1, loop = 0, mxLen = 0;\n for (; loop < start; ++loop) {\n if (cur->next.count(mapped[t[loop]]) == 0 || cur->next[mapped[t[loop]]]->comb.size() == 0) return -1;\n cur = cur->next[mapped[t[loop]]];\n }\n if (cur->comb.size() > 0) {\n int c = -1;\n for (auto& p : cur->comb) {\n if (p.first > mxLen) {\n mxLen = p.first;\n idx = N - p.second;\n c = N - p.second;\n }\n else if (p.first == mxLen) {\n if (compare(mp[c], mp[N - p.second]) == -1) {\n c = N - p.second;\n idx = N - p.second;\n }\n else if (compare(mp[c], mp[N - p.second]) == 0) {\n idx = min(idx, N - p.second);\n }\n }\n }\n }\n return idx;\n }\n \n void del(string& s, int idx) {\n Node* cur = root;\n // dbg(\"del\", s, idx);\n for (int i = 0; i < s.size(); i++) {\n int c = mapped[s[i]];\n if (cur->next.count(c) == 0) return;\n cur->comb.erase({ sz(s), N - idx });\n cur = cur->next[c];\n }\n // dbg(\"3\", cur->comb);\n cur->comb.erase({ sz(s), N - idx });\n }\n};\n \n \nvoid solve()\n{\n for (int i = 33; i <= 126; i++) {\n mapped[(char)i] = i - 33;\n }\n \n int n;\n cin >> n;\n Trie trie;\n string t;\n int start = 0;\n unordered_map<int, string> cache;\n while (n--) {\n string op;\n cin >> op;\n // dbg(\"op\", op);\n if (op == \"add\") {\n string pattern;\n int idx;\n cin >> idx >> pattern;\n vector<char> v(pattern.begin(), pattern.end());\n mp[idx] = PolyHash(v);\n cache[idx] = pattern;\n trie.add(pattern, idx);\n }\n else if (op == \"append\") {\n string pattern;\n cin >> pattern;\n concat(t, pattern, start);\n }\n else if (op == \"delete\") {\n int idx;\n cin >> idx;\n trie.del(cache[idx], idx);\n }\n else {\n int del;\n cin >> del;\n start -= min(start, del);\n }\n // dbg(\"t\", t, start);\n cout << trie.find(t, start) << '\\n';\n }\n}\n \n \nint32_t main() {\n ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);\n // int t;\n // cin >> t;\n // while (t--)\n solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 293214684,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\nclass TrieNode:\n def __init__(self):\n self.children = {}\n self.pattern_ids = []\n \nclass Trie:\n def __init__(self):\n self.root = TrieNode()\n \n def insert(self, pattern, pattern_id):\n node = self.root\n for char in pattern:\n if char not in node.children:\n node.children[char] = TrieNode()\n node = node.children[char]\n node.pattern_ids.append(pattern_id)\n node.pattern_ids.sort()\n \n def delete(self, pattern, pattern_id):\n node = self.root\n nodes = []\n chars = []\n for char in pattern:\n nodes.append(node)\n chars.append(char)\n node = node.children[char]\n node.pattern_ids.remove(pattern_id)\n if not node.children and not node.pattern_ids:\n for i in range(len(pattern)-1, -1, -1):\n char = chars[i]\n node = nodes[i]\n del node.children[char]\n if node.children or node.pattern_ids:\n break\n \n def find_matching_patterns(self, prefix):\n node = self.root\n for char in prefix:\n if char not in node.children:\n return []\n node = node.children[char]\n return self._collect_all_patterns(node)\n \n def _collect_all_patterns(self, node):\n patterns = []\n stack = [(node, \"\")]\n while stack:\n current, prefix = stack.pop()\n if current.pattern_ids:\n for pattern_id in current.pattern_ids:\n patterns.append((prefix, pattern_id))\n for char, child_node in current.children.items():\n stack.append((child_node, prefix + char))\n return patterns\n \ndef find_best_match(current_text, patterns):\n matching_patterns = trie.find_matching_patterns(current_text)\n if not matching_patterns:\n print(-1)\n return\n matching_patterns.sort(key=lambda x: (-len(x[0]), x[0], x[1]))\n best_match = matching_patterns[0][1]\n print(best_match)\n \nimport sys\ninput = sys.stdin.read\ndata = input().split()\n \nn = int(data[0])\nindex = 1\ntrie = Trie()\n \ncurrent_text = \"\"\n \nfor _ in range(n):\n action = data[index]\n index += 1\n if action == \"add\":\n pattern_id = int(data[index])\n pattern = data[index + 1]\n index += 2\n trie.insert(pattern, pattern_id)\n find_best_match(current_text, trie)\n elif action == \"delete\":\n pattern_id = int(data[index])\n index += 1\n pattern = data[index]\n index += 1\n trie.delete(pattern, pattern_id)\n find_best_match(current_text, trie)\n elif action == \"append\":\n s = data[index]\n index += 1\n current_text += s\n find_best_match(current_text, trie)\n elif action == \"backspace\":\n c = int(data[index])\n index += 1\n current_text = current_text[:-c]\n find_best_match(current_text, trie)\n \n \n# xufhzrDiUAqMIHvGLWC\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 304186383,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main{\n \n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n int n = Integer.parseInt(br.readLine());\n \n Map<String, TreeMap<Integer, String>> patterns = new HashMap<>();\n StringBuilder t = new StringBuilder();\n StringBuilder result = new StringBuilder();\n \n for (int i = 0; i < n; i++) {\n String[] parts = br.readLine().split(\" \");\n if (parts[0].equals(\"add\")) {\n int id = Integer.parseInt(parts[1]);\n String pattern = parts[2];\n patterns\n .computeIfAbsent(pattern, k -> new TreeMap<>())\n .put(id, pattern);\n } else if (parts[0].equals(\"delete\")) {\n int id = Integer.parseInt(parts[1]);\n patterns.forEach((key, value) -> value.remove(id));\n } else if (parts[0].equals(\"append\")) {\n t.append(parts[1]);\n } else if (parts[0].equals(\"backspace\")) {\n int c = Integer.parseInt(parts[1]);\n int len = t.length();\n int toDelete = Math.min(c, len);\n t.delete(len - toDelete, len);\n }\n \n // Find the best match\n int bestId = findBestMatch(patterns, t.toString());\n result.append(bestId).append('\\n');\n }\n \n System.out.print(result.toString());\n }\n \n private static int findBestMatch(Map<String, TreeMap<Integer, String>> patterns, String currentText) {\n int bestId = -1;\n String bestPattern = null;\n \n for (Map.Entry<String, TreeMap<Integer, String>> entry : patterns.entrySet()) {\n String pattern = entry.getKey();\n if (pattern.startsWith(currentText) &&\n (bestPattern == null ||\n pattern.length() > bestPattern.length() ||\n (pattern.length() == bestPattern.length() && pattern.compareTo(bestPattern) < 0))) {\n bestPattern = pattern;\n }\n }\n \n if (bestPattern != null) {\n bestId = patterns.get(bestPattern).firstKey();\n }\n \n return bestId;\n }\n}\n \n \n// 7oldVP1puzOiWmcQeXF\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 304293551,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main{\n static class TrieNode {\n Map<Character, TrieNode> children;\n int id;\n boolean isEndOfPattern;\n \n public TrieNode() {\n children = new HashMap<>();\n id = -1;\n isEndOfPattern = false;\n }\n }\n \n static class Pattern implements Comparable<Pattern> {\n int id;\n String pattern;\n \n public Pattern(int id, String pattern) {\n this.id = id;\n this.pattern = pattern;\n }\n \n @Override\n public int compareTo(Pattern other) {\n if (pattern.length() != other.pattern.length()) {\n return Integer.compare(other.pattern.length(), pattern.length());\n }\n if (!pattern.equals(other.pattern)) {\n return pattern.compareTo(other.pattern);\n }\n return Integer.compare(id, other.id);\n }\n }\n \n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n int n = Integer.parseInt(br.readLine());\n TrieNode root = new TrieNode();\n Set<Pattern> patterns = new TreeSet<>();\n StringBuilder currentText = new StringBuilder();\n \n for (int i = 0; i < n; i++) {\n String[] parts = br.readLine().split(\" \");\n if (parts[0].equals(\"add\")) {\n int id = Integer.parseInt(parts[1]);\n String pattern = parts[2];\n addPattern(root, id, pattern, patterns);\n } else if (parts[0].equals(\"delete\")) {\n int id = Integer.parseInt(parts[1]);\n deletePattern(root, id, patterns);\n } else if (parts[0].equals(\"append\")) {\n String s = parts[1];\n append(root, currentText, s, patterns);\n } else {\n int c = Integer.parseInt(parts[1]);\n backspace(root, currentText, c, patterns);\n }\n printSuggestedPattern(patterns);\n }\n }\n \n private static void addPattern(TrieNode root, int id, String pattern, Set<Pattern> patterns) {\n TrieNode node = root;\n for (char c : pattern.toCharArray()) {\n if (!node.children.containsKey(c)) {\n node.children.put(c, new TrieNode());\n }\n node = node.children.get(c);\n }\n node.id = id;\n node.isEndOfPattern = true;\n patterns.add(new Pattern(id, pattern));\n }\n \n private static void deletePattern(TrieNode root, int id, Set<Pattern> patterns) {\n TrieNode node = root;\n for (char c : getPattern(root, id).toCharArray()) {\n node = node.children.get(c);\n }\n node.id = -1;\n node.isEndOfPattern = false;\n patterns.remove(new Pattern(id, getPattern(root, id)));\n }\n \n private static String getPattern(TrieNode root, int id) {\n StringBuilder sb = new StringBuilder();\n Queue<TrieNode> queue = new LinkedList<>();\n queue.add(root);\n while (!queue.isEmpty()) {\n TrieNode node = queue.poll();\n if (node.id == id) {\n return sb.toString();\n }\n for (Map.Entry<Character, TrieNode> entry : node.children.entrySet()) {\n sb.append(entry.getKey());\n queue.add(entry.getValue());\n sb.deleteCharAt(sb.length() - 1);\n }\n }\n return \"\";\n }\n \n private static void append(TrieNode root, StringBuilder currentText, String s, Set<Pattern> patterns) {\n currentText.append(s);\n TrieNode node = root;\n for (char c : currentText.toString().toCharArray()) {\n if (!node.children.containsKey(c)) {\n node.children.put(c, new TrieNode());\n }\n node = node.children.get(c);\n }\n patterns.clear();\n dfs(node, currentText.toString(), patterns);\n }\n \n private static void backspace(TrieNode root, StringBuilder currentText, int c, Set<Pattern> patterns) {\n currentText.delete(currentText.length() - c, currentText.length());\n TrieNode node = root;\n for (char c1 : currentText.toString().toCharArray()) {\n if (!node.children.containsKey(c1)) {\n node.children.put(c1, new TrieNode());\n }\n node = node.children.get(c1);\n }\n patterns.clear();\n dfs(node, currentText.toString(), patterns);\n }\n \n private static void dfs(TrieNode node, String prefix, Set<Pattern> patterns) {\n if (node.isEndOfPattern) {\n patterns.add(new Pattern(node.id, prefix));\n }\n for (Map.Entry<Character, TrieNode> entry : node.children.entrySet()) {\n dfs(entry.getValue(), prefix + entry.getKey(), patterns);\n }\n }\n \n private static void printSuggestedPattern(Set<Pattern> patterns) {\n if (patterns.isEmpty()) {\n System.out.println(-1);\n } else {\n System.out.println(patterns.iterator().next().id);\n }\n }\n}\n \n \n// oG6DPBNkctuX49HFQEI\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 304524457,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n#include <cstdio>\n#include <algorithm>\n#include <utility>\n#include <cstring>\n#include <set>\n \nusing namespace std;\n \nint T;\nchar op[10],s[2000005];\n \nstruct Node\n{\n\tNode* np[127];\n\tNode* pre;\n\tpair<int,int> lgst;//Length,-Id.\n\tset<pair<int,int> >* ids;\n\t\n\tNode()\n\t{\n\t\tfor(int i=0;i<127;i++)\n\t\t\tnp[i] = NULL;\n\t\tpre = NULL;\n\t\tids = NULL;\n\t\tlgst = make_pair(0,0);\n\t}\n\tNode(Node* pre):pre(pre)\n\t{\n\t\tfor(int i=0;i<127;i++)\n\t\t\tnp[i] = NULL;\n\t\tids = NULL;\n\t\tlgst = make_pair(0,0);\n\t}\n};\n \ntypedef Node* LPNode;\n \nLPNode pl[1000005],nwat,head,header;\nint len[1000005];\n \nint main()\n{\n\tscanf(\"%d\",&T);\n\t\n\theader = new Node();\n\thead = new Node();\n\theader->pre = header;\n\thead->pre = header;\n\tnwat = head;\n\t\n\twhile(T--)\n\t{\n\t\tscanf(\"%s\",op);\n\t\tif(op[1]=='d')//add\n\t\t{\n\t\t\tint id,n;\n\t\t\tscanf(\"%d%s\",&id,s);\n\t\t\t\n\t\t\tLPNode t = head;\n\t\t\tn = strlen(s);\n\t\t\t\n\t\t\tfor(int i=0;i<n;i++)\n\t\t\t{\n\t\t\t\tif(t->np[s[i]]==NULL)\n\t\t\t\t{\n\t\t\t\t\tt->np[s[i]] = new Node(t);\n\t\t\t\t}\n\t\t\t\tt = t->np[s[i]];\n\t\t\t}\n\t\t\tpl[id] = t;\n\t\t\tlen[id] = n;\n\t\t\t\n\t\t\tif(t->ids==NULL)\n\t\t\t{\n\t\t\t\tt->ids = new set<pair<int,int> >();\n\t\t\t}\n\t\t\tt->ids->insert(make_pair(n,-id));\n\t\t\twhile(t->pre!=t)\n\t\t\t{\n\t\t\t\tt->lgst = make_pair(0,0);\n\t\t\t\tif(t->ids!=NULL)\n\t\t\t\t{\n\t\t\t\t\tfor(pair<int,int> v:*(t->ids))\n\t\t\t\t\t{\n\t\t\t\t\t\tt->lgst = max(t->lgst,v);\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tfor(int i=0;i<127;i++)\n\t\t\t\t{\n\t\t\t\t\tif(t->np[i]==NULL)\n\t\t\t\t\t\tcontinue;\n\t\t\t\t\tt->lgst = max(t->lgst,t->np[i]->lgst);\n\t\t\t\t}\n\t\t\t\tt = t->pre;\t\n\t\t\t}\n\t\t}\n\t\telse if(op[1]=='e')//delete\n\t\t{\n\t\t\tint id;\n\t\t\tscanf(\"%d\",&id);\n\t\t\t\n\t\t\tLPNode t = pl[id];\n\t\t\t\n\t\t\tt->ids->erase(make_pair(len[id],-id));\n\t\t\t\n\t\t\twhile(t->pre!=t)\n\t\t\t{\n\t\t\t\tt->lgst = make_pair(0,0);\n\t\t\t\tif(t->ids!=NULL)\n\t\t\t\t{\n\t\t\t\t\tfor(pair<int,int> v:*(t->ids))\n\t\t\t\t\t{\n\t\t\t\t\t\tt->lgst = max(t->lgst,v);\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tfor(int i=0;i<127;i++)\n\t\t\t\t{\n\t\t\t\t\tif(t->np[i]==NULL)\n\t\t\t\t\t\tcontinue;\n\t\t\t\t\tt->lgst = max(t->lgst,t->np[i]->lgst);\n\t\t\t\t}\n\t\t\t\tt = t->pre;\n\t\t\t}\n\t\t}\n\t\telse if(op[1]=='p')//append\n\t\t{\n\t\t\tscanf(\"%s\",s);\n\t\t\t\n\t\t\tint n = strlen(s);\n\t\t\t\n\t\t\tfor(int i=0;i<n;i++)\n\t\t\t{\n\t\t\t\tif(nwat->np[s[i]]==NULL)\n\t\t\t\t{\n\t\t\t\t\tnwat->np[s[i]] = new Node(nwat);\n\t\t\t\t}\n\t\t\t\tnwat = nwat->np[s[i]];\n\t\t\t}\n\t\t\t\n\t\t\tLPNode t = nwat;\n\t\t\t\n\t\t\twhile(t->pre!=t)\n\t\t\t{\n\t\t\t\tt->lgst = make_pair(0,0);\n\t\t\t\tif(t->ids!=NULL)\n\t\t\t\t{\n\t\t\t\t\tfor(pair<int,int> v:*(t->ids))\n\t\t\t\t\t{\n\t\t\t\t\t\tt->lgst = max(t->lgst,v);\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tfor(int i=0;i<127;i++)\n\t\t\t\t{\n\t\t\t\t\tif(t->np[i]==NULL)\n\t\t\t\t\t\tcontinue;\n\t\t\t\t\tt->lgst = max(t->lgst,t->np[i]->lgst);\n\t\t\t\t}\n\t\t\t\tt = t->pre;\n\t\t\t}\n\t\t}\n\t\telse //op[1]=='a' backspace\n\t\t{\n\t\t\tint n;\n\t\t\tscanf(\"%d\",&n);\n\t\t\t\n\t\t\twhile(n--)\n\t\t\t\tnwat = nwat->pre;\n\t\t\tif(nwat==header)\n\t\t\t\tnwat = head;\n\t\t}\n\t\t\n\t\tif(nwat->lgst.first)//怎么有id为0的。。。注意。。。 \n\t\t{\n\t\t\tprintf(\"%d\\n\",-nwat->lgst.second);\n\t\t}\n\t\telse\n\t\t{\n\t\t\tputs(\"-1\");\n\t\t}\n\t}\n\t\n\treturn 0;\n}\n \n/*\n \nWhy????????\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 293350728,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n#include <cstdio>\n#include <algorithm>\n#include <utility>\n#include <cstring>\n#include <set>\n \nusing namespace std;\n \nint T;\nchar op[10],s[2000005];\n \nstruct Node\n{\n\tNode* np[127];\n\tNode* pre;\n\tpair<int,int> lgst;//Length,-Id.\n\tset<pair<int,int> >* ids;\n\t\n\tNode()\n\t{\n\t\tfor(int i=0;i<127;i++)\n\t\t\tnp[i] = NULL;\n\t\tpre = NULL;\n\t\tids = NULL;\n\t\tlgst = make_pair(0,0);\n\t}\n\tNode(Node* pre):pre(pre)\n\t{\n\t\tfor(int i=0;i<127;i++)\n\t\t\tnp[i] = NULL;\n\t\tids = NULL;\n\t\tlgst = make_pair(0,0);\n\t}\n};\n \ntypedef Node* LPNode;\n \nLPNode pl[1000005],nwat,head,header;\nint len[1000005];\n \nint main()\n{\n\tscanf(\"%d\",&T);\n\t\n\theader = new Node();\n\thead = new Node();\n\theader->pre = header;\n\thead->pre = header;\n\tnwat = head;\n\t\n\twhile(T--)\n\t{\n\t\tscanf(\"%s\",op);\n\t\tif(op[1]=='d')//add\n\t\t{\n\t\t\tint id,n;\n\t\t\tscanf(\"%d%s\",&id,s);\n\t\t\t\n\t\t\tLPNode t = head;\n\t\t\tn = strlen(s);\n\t\t\t\n\t\t\tfor(int i=0;i<n;i++)\n\t\t\t{\n\t\t\t\tif(t->np[s[i]]==NULL)\n\t\t\t\t{\n\t\t\t\t\tt->np[s[i]] = new Node(t);\n\t\t\t\t}\n\t\t\t\tt = t->np[s[i]];\n\t\t\t}\n\t\t\tpl[id] = t;\n\t\t\tlen[id] = n;\n\t\t\t\n\t\t\tif(t->ids==NULL)\n\t\t\t{\n\t\t\t\tt->ids = new set<pair<int,int> >();\n\t\t\t}\n\t\t\tt->ids->insert(make_pair(n,id));\n\t\t\twhile(t->pre!=t)\n\t\t\t{\n\t\t\t\tt->lgst = make_pair(0,0);\n\t\t\t\tif(t->ids!=NULL)\n\t\t\t\t{\n\t\t\t\t\tfor(pair<int,int> v:*(t->ids))\n\t\t\t\t\t{\n\t\t\t\t\t\tif(t->lgst.first<v.first)\n\t\t\t\t\t\t{\n\t\t\t\t\t\t\tt->lgst = v;\n\t\t\t\t\t\t}\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tfor(int i=0;i<127;i++)\n\t\t\t\t{\n\t\t\t\t\tif(t->np[i]==NULL)\n\t\t\t\t\t\tcontinue;\n\t\t\t\t\tif(t->lgst.first<t->np[i]->lgst.first)\n\t\t\t\t\t{\n\t\t\t\t\t\tt->lgst = t->np[i]->lgst;\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tt = t->pre;\t\n\t\t\t}\n\t\t}\n\t\telse if(op[1]=='e')//delete\n\t\t{\n\t\t\tint id;\n\t\t\tscanf(\"%d\",&id);\n\t\t\t\n\t\t\tLPNode t = pl[id];\n\t\t\t\n\t\t\tt->ids->erase(make_pair(len[id],id));\n\t\t\tif(t->ids->empty())\n\t\t\t\tdelete t->ids;\n\t\t\t\n\t\t\twhile(t->pre!=t)\n\t\t\t{\n\t\t\t\tt->lgst = make_pair(0,0);\n\t\t\t\tif(t->ids!=NULL)\n\t\t\t\t{\n\t\t\t\t\tfor(pair<int,int> v:*(t->ids))\n\t\t\t\t\t{\n\t\t\t\t\t\tif(t->lgst.first<v.first)\n\t\t\t\t\t\t{\n\t\t\t\t\t\t\tt->lgst = v;\n\t\t\t\t\t\t}\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tfor(int i=0;i<127;i++)\n\t\t\t\t{\n\t\t\t\t\tif(t->np[i]==NULL)\n\t\t\t\t\t\tcontinue;\n\t\t\t\t\tif(t->lgst.first<t->np[i]->lgst.first)\n\t\t\t\t\t{\n\t\t\t\t\t\tt->lgst = t->np[i]->lgst;\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\t\n\t\t\t\tt = t->pre;\n\t\t\t\t\n\t\t\t\tLPNode tr = t;\n\t\t\t\tint f = 0;\n\t\t\t\tif(tr==head || tr==header)\n\t\t\t\t\tf = 1;\n\t\t\t\tif(tr->ids!=NULL)\n\t\t\t\t\tf = 1;\n\t\t\t\tif(f)\n\t\t\t\t\tcontinue;\n\t\t\t\tfor(int i=0;i<127;i++)\n\t\t\t\t{\n\t\t\t\t\tif(tr->np[i]!=NULL)\n\t\t\t\t\t{\n\t\t\t\t\t\tf = 1;\n\t\t\t\t\t\tbreak;\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tif(f)\n\t\t\t\t\tcontinue;\n\t\t\t\tdelete tr;\n\t\t\t}\n\t\t}\n\t\telse if(op[1]=='p')//append\n\t\t{\n\t\t\tscanf(\"%s\",s);\n\t\t\t\n\t\t\tint n = strlen(s);\n\t\t\t\n\t\t\tfor(int i=0;i<n;i++)\n\t\t\t{\n\t\t\t\tif(nwat->np[s[i]]==NULL)\n\t\t\t\t{\n\t\t\t\t\tnwat->np[s[i]] = new Node(nwat);\n\t\t\t\t}\n\t\t\t\tnwat = nwat->np[s[i]];\n\t\t\t}\n\t\t}\n\t\telse //op[1]=='a' backspace\n\t\t{\n\t\t\tint n;\n\t\t\tscanf(\"%d\",&n);\n\t\t\t\n\t\t\tint f = 0;\n\t\t\twhile(n--)\n\t\t\t{\n\t\t\t\tLPNode tr = nwat;\n\t\t\t\tnwat = nwat->pre;\n\t\t\t\t\n\t\t\t\tif(tr==head || tr==header)\n\t\t\t\t\tf = 1;\n\t\t\t\tif(tr->ids!=NULL)\n\t\t\t\t\tf = 1;\n\t\t\t\tif(f)\n\t\t\t\t\tcontinue;\n\t\t\t\tfor(int i=0;i<127;i++)\n\t\t\t\t{\n\t\t\t\t\tif(tr->np[i]!=NULL)\n\t\t\t\t\t{\n\t\t\t\t\t\tf = 1;\n\t\t\t\t\t\tbreak;\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tif(f)\n\t\t\t\t\tcontinue;\n\t\t\t\tdelete tr;\n\t\t\t}\n\t\t\tif(nwat==header)\n\t\t\t\tnwat = head;\n\t\t}\n\t\t\n\t\tif(nwat->lgst.first)//怎么有id为0的。。。注意。。。 \n\t\t{\n\t\t\tprintf(\"%d\\n\",nwat->lgst.second);\n\t\t}\n\t\telse\n\t\t{\n\t\t\tputs(\"-1\");\n\t\t}\n\t}\n\t\n\treturn 0;\n}\n \n/*\n \n纯唐了。没有管字典序，只管了长度和id。\n还是很菜。还是得多练。还是得多磨。 \n \nS703:?4Pd\"N%ez/Vv7WJ`gNV5=]5\"E...xSh<x:llEh\"S:%hU\\y{Zn>iamH_uEV)\nS161:c'A;;f=ry]3F>y/?'(J+N.1jS...)z&LGdi,;*?O}x.b)_9<,kc)9%$jzSW\n \n*/\n \n/*\n \nWhy????????\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 293352396,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\nclass TrieNode:\n def __init__(self, parent=None):\n self.children = {}\n self.best_id = -1\n self.max_length = 0\n self.is_end = False\n self.pattern_id = None\n self.parent = parent\n \nclass Trie:\n def __init__(self, n):\n self.root = TrieNode()\n self.patterns = [None] * (n + 1)\n self.lengths = [0] * (n + 1)\n \n def add(self, pattern, id):\n self.patterns[id] = pattern\n self.lengths[id] = len(pattern)\n node = self.root\n for char in pattern:\n if char not in node.children:\n node.children[char] = TrieNode(parent=node)\n node = node.children[char]\n node.is_end = True\n node.pattern_id = id\n self._update_best_id(node, id)\n \n def _update_best_id(self, node, id):\n while node:\n current_best_id = node.best_id\n current_max_length = node.max_length\n if current_best_id == -1:\n node.best_id = id\n node.max_length = self.lengths[id]\n elif self.lengths[id] > current_max_length:\n node.best_id = id\n node.max_length = self.lengths[id]\n elif self.lengths[id] == current_max_length:\n if self.patterns[id] < self.patterns[current_best_id]:\n node.best_id = id\n elif self.patterns[id] == self.patterns[current_best_id]:\n if id < current_best_id:\n node.best_id = id\n node = node.parent\n \n def delete(self, id):\n pattern = self.patterns[id]\n node = self.root\n for char in pattern:\n if char in node.children:\n node = node.children[char]\n else:\n raise ValueError(\"Pattern not found\")\n node.is_end = False\n node.pattern_id = None\n self.patterns[id] = None\n self.lengths[id] = 0\n self._update_best_id_after_delete(node)\n \n def _update_best_id_after_delete(self, node):\n while node:\n best_id = -1\n max_length = 0\n for child in node.children.values():\n if child.best_id != -1:\n if best_id == -1 or \\\n (child.max_length > max_length) or \\\n (child.max_length == max_length and self.patterns[child.best_id] < self.patterns[best_id]) or \\\n (child.max_length == max_length and self.patterns[child.best_id] == self.patterns[best_id] and child.best_id < best_id):\n best_id = child.best_id\n max_length = child.max_length\n node.best_id = best_id\n node.max_length = max_length\n node = node.parent\n \n def search(self, text_list, index):\n node = self.root\n for i in range(index):\n char = text_list[i]\n if char in node.children:\n node = node.children[char]\n else:\n return -1\n return node.best_id\n \ndef main():\n import sys\n input = sys.stdin.read().splitlines()\n n = int(input[0])\n trie = Trie(n)\n current_text = []\n index = 0\n operation_index = 1\n for _ in range(n):\n line = input[operation_index].strip()\n operation_index += 1\n parts = line.split()\n if parts[0] == \"add\":\n i = int(parts[1])\n p_i = ' '.join(parts[2:])\n trie.add(p_i, i)\n elif parts[0] == \"delete\":\n i = int(parts[1])\n trie.delete(i)\n elif parts[0] == \"append\":\n s = ' '.join(parts[1:])\n current_text.extend(s)\n index += len(s)\n else: # backspace\n c = int(parts[1])\n index = max(0, index - c)\n # After each operation, search for the best pattern ID\n best_id = trie.search(current_text, index)\n print(best_id)\n \nif __name__ == \"__main__\":\n main()\n \n \n# 3XiOGYCRkFpbQE8UM2l\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 305125709,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\nfrom bisect import bisect_left\n \ndef find_best_match(current_text, patterns):\n best_match_id = -1\n best_length = -1\n \n for length, pattern, pattern_id in patterns:\n if length < len(current_text):\n continue\n if pattern.startswith(current_text):\n if length > best_length or (length == best_length and pattern_id < best_match_id):\n best_length = length\n best_match_id = pattern_id\n \n return best_match_id\n \ndef main():\n import sys\n input = sys.stdin.read\n data = input().split()\n \n n = int(data[0])\n idx = 1\n \n patterns = []\n pattern_dict = {}\n current_text = \"\"\n \n for _ in range(n):\n if data[idx] == 'add':\n pattern_id = int(data[idx+1])\n pattern = data[idx+2]\n pattern_dict[pattern_id] = pattern\n patterns.append((len(pattern), pattern, pattern_id))\n patterns.sort()\n idx += 3\n elif data[idx] == 'delete':\n pattern_id = int(data[idx+1])\n pattern = pattern_dict.pop(pattern_id)\n patterns.remove((len(pattern), pattern, pattern_id))\n idx += 2\n elif data[idx] == 'append':\n current_text += data[idx+1]\n idx += 2\n elif data[idx] == 'backspace':\n c = int(data[idx+1])\n current_text = current_text[:-c] if c < len(current_text) else \"\"\n idx += 2\n \n best_match_id = find_best_match(current_text, patterns)\n print(best_match_id)\n \nif __name__ == \"__main__\":\n main()\n \n \n# gybjniB6VlfuDqwCHmL\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 304199340,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n// Jan Zakrzewski\n#define NDEBUG\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define assert(expr) {if(!(expr)){while(1);}}\n \nconstexpr int A = 33;\nconstexpr int B = 126;\nconstexpr int NONE = -123;\n \nstruct Node {\n priority_queue<int> id;\n \n int dp_len;\n int dp_id;\n \n int parent;\n int children[B - A + 1 + (10)];\n};\n \nvector<Node> dictionary;\nvector<int> id_where(2'234'567);\n \nvoid initialize() {\n dictionary.clear();\n Node eps;\n eps.dp_len = NONE;\n eps.parent = NONE;\n for(int i = A; i <= B; ++i)\n eps.children[i - A] = NONE;\n dictionary.push_back(eps);\n fill(id_where.begin(), id_where.end(), NONE);\n}\n \nint epsilon() {\n return 0;\n}\n \nint go(int u, string word) {\n for(char c : word) {\n assert(A <= c && c <= B);\n assert(0 <= u && u < (int) dictionary.size());\n if(dictionary[u].children[c - A] == NONE) {\n int v = (int) dictionary.size();\n Node sth;\n sth.dp_len = NONE;\n sth.parent = u;\n for(int i = A; i <= B; ++i)\n sth.children[i - A] = NONE;\n \n dictionary[u].children[c - A] = v;\n dictionary.push_back(sth);\n assert(v < (int) dictionary.size());\n u = v;\n } else u = dictionary[u].children[c - A];\n }\n return u;\n}\n \nvoid repair(int u) {\n while(true) {\n assert(0 <= u && u < (int) dictionary.size());\n dictionary[u].dp_len = NONE;\n dictionary[u].dp_id = NONE;\n while(dictionary[u].id.size() > 0 && NONE == id_where[-dictionary[u].id.top()])\n dictionary[u].id.pop();\n if(dictionary[u].id.size() > 0) {\n dictionary[u].dp_len = 0;\n dictionary[u].dp_id = -dictionary[u].id.top();\n }\n for(int i = A; i <= B; ++i) {\n if(dictionary[u].children[i - A] == NONE) continue;\n if(dictionary[dictionary[u].children[i - A]].dp_len == NONE) continue;\n if(dictionary[u].dp_len == NONE || dictionary[dictionary[u].children[i - A]].dp_len + 1 > dictionary[u].dp_len) {\n dictionary[u].dp_len = dictionary[dictionary[u].children[i - A]].dp_len + 1;\n dictionary[u].dp_id = dictionary[dictionary[u].children[i - A]].dp_id;\n }\n }\n //assert(dictionary[u].dp_len != NONE);\n if(u == epsilon()) break;\n u = dictionary[u].parent;\n }\n}\n \nvoid add(int id, string pattern) {\n \n int u = epsilon();\n \n u = go(u, pattern);\n \n assert(0 <= id && id < (int) id_where.size());\n id_where[id] = u;\n dictionary[u].id.push(-id);\n \n repair(u);\n}\n \nvoid Delete(int id) {\n assert(0 <= id && id < (int) id_where.size());\n int u = id_where[id];\n assert(0 <= u && u < (int) dictionary.size());\n id_where[id] = NONE;\n \n repair(u);\n}\n \n/*int append(int u, string str) {\n u = go(u, str);\n \n repair(u);\n \n return u;\n}\n \nint backspace(int u, int cnt) {\n while(u != epsilon() && cnt > 0) {\n u = dictionary[u].parent;\n --cnt;\n }\n return u;\n}\n \nint query(int u) {\n if(dictionary[u].dp_len == NONE) return -1;\n else return dictionary[u].dp_id;\n}*/\n \nvoid append_1(int& u, deque<char>& q, string str) {\n for(char c : str)\n q.push_back(c);\n}\n \nvoid backspace_1(int& u, deque<char>& q, int cnt) {\n while(cnt > 0) {\n if(q.size() > 0) {\n q.pop_back();\n --cnt;\n } else if(u != epsilon()) {\n assert(0 <= u && u < (int) dictionary.size());\n u = dictionary[u].parent;\n --cnt;\n } else break;\n }\n}\n \nint query_1(int& u, deque<char>& q) {\n while(q.size() > 0) {\n char c = q.front();\n assert(A <= c && c <= B);\n assert(0 <= u && u < (int) dictionary.size());\n if(dictionary[u].children[c - A] == NONE)\n return -1;\n u = dictionary[u].children[c - A];\n q.pop_front();\n }\n assert(0 <= u && u < (int) dictionary.size());\n if(dictionary[u].dp_len == NONE) return -1;\n return dictionary[u].dp_id;\n}\n \nint main() {\n ios_base::sync_with_stdio(0);\n cin.tie(0); cout.tie(0);\n \n int n;\n cin >> n;\n \n initialize();\n int u = epsilon();\n deque<char> qu;\n \n while(n--) {\n string what;\n cin >> what;\n int id;\n string pattern;\n string str;\n int cnt;\n switch(what[1]) {\n case 'd': /*add*/ cin >> id >> pattern; add(id, pattern); break;\n case 'e': /*delete*/ cin >> id; Delete(id); break;\n case 'p': /*append*/ cin >> str; append_1(u, qu, str); break;\n case 'a': /*backspace*/ cin >> cnt; backspace_1(u, qu, cnt); break;\n }\n \n cout << query_1(u, qu) << \"\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 296538249,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \n#define all(x) x.begin(), x.end()\n#define ar array\n#define pb push_back\n#define ln '\\n'\n#define size(x) (int)(x).size()\n//~ #define int long long\n \nusing i64 = long long;\n \ntemplate <class F, class _S>\nbool chmin(F &u, const _S &v){\n bool flag = false;\n if ( u > v ){\n u = v; flag |= true;\n }\n return flag;\n}\n \ntemplate <class F, class _S>\nbool chmax(F &u, const _S &v){\n bool flag = false;\n if ( u < v ){\n u = v; flag |= true;\n }\n return flag;\n}\n \nusing arr = ar <int,94>;\n \nconst int inf = 1e9;\n \nconst int N = 2e6 + 1007;\n \nint t[N][94], fa[N];\n \nar <int,2> mx[N];\n \nvector <set<int>> idx(N);\n \nint ct;\n \nvoid ins(string &s, int &id, int rt, int j, bool del){\n\tif ( j == size(s) ){\n\t\tif ( del ) idx[rt].erase(id);\n\t\telse idx[rt].insert(id);\n\t\t\n\t\tif ( size(idx[rt]) ){\n\t\t\tmx[rt] = {0, *idx[rt].begin()};\n\t\t} else mx[rt] = {-inf, -1};\n\t\t\n\t\treturn;\n\t}\n\t\n\tint u = s[j] - 33;\n\t\n\tif ( !t[rt][u] ){\n\t\tt[rt][u] = ++ct;\n\t\tfa[ct] = rt;\n\t} \n\t\n\tins(s, id, t[rt][u], j + 1, del);\n\t\n\tif ( !idx[rt].empty() ){\n\t\tmx[rt] = {0, *idx[rt].begin()};\n\t} else mx[rt] = {-inf, -1};\n\t\n\tfor ( int i = 0; i < 94; i++ ){\n\t\tif ( t[rt][i] != 0 ){\n\t\t\tint v = t[rt][i];\n\t\t\t\n\t\t\tif ( mx[v][0] + 1 > mx[rt][0] ){\n\t\t\t\tmx[rt] = {mx[v][0] + 1, mx[v][1]};\n\t\t\t}\n\t\t}\n\t}\n}\n \nvoid insert(string &s, int &id){ ins(s, id, 0, 0, 0); }\n \nvoid erase(string &s, int &id){ ins(s, id, 0, 0, 1); }\n \nvoid move(string &s, int &k){ \n\tfor ( auto &u: s ){\n\t\tint v = u - 33;\n\t\t\n\t\tif ( !t[k][v] ){\n\t\t\tt[k][v] = ++ct;\n\t\t\tfa[ct] = k;\n\t\t} k = t[k][v];\n\t}\n}\n \nint qry(int &k){ return mx[k][1]; }\n \nsigned main(){\n ios_base::sync_with_stdio(false);\n cin.tie(nullptr);\n \n for ( int i = 0; i < N; i++ ) mx[i] = {-inf, -1};\n \n int n; cin >> n;\n \n vector <string> p(n + 1);\n \n int k = 0;\n \n for ( int i = 0; i < n; i++ ){\n\t\tstring op; cin >> op;\n\t\t\n\t\tif ( op == \"add\" ){\n\t\t\tint x; string s; \n\t\t\tcin >> x >> s;\n\t\t\t\n\t\t\tinsert(s, x);\n\t\t\tp[x] = s;\n\t\t} else if ( op == \"delete\" ){\n\t\t\tint x; cin >> x;\n\t\t\t\n\t\t\terase(p[x], x);\n\t\t} else if ( op == \"append\" ){\n\t\t\tstring s; cin >> s;\n\t\t\t\n\t\t\tmove(s, k);\n\t\t} else{\n\t\t\tint c; cin >> c;\n\t\t\t\n\t\t\twhile ( c-- ){\n\t\t\t\tif ( !k ) break;\n\t\t\t\t\n\t\t\t\tk = fa[k];\n\t\t\t}\n\t\t}\n\t\t\n\t\tcout << qry(k) << ln;\n\t}\n \n \tcout << '\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 296140542,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\ndef main():\n import sys\n input = sys.stdin.read\n data = input().splitlines()\n \n n = int(data[0])\n patterns = {} # id -> pattern\n current_text = \"\"\n output = []\n \n for line in data[1:n+1]:\n parts = line.split()\n action = parts[0]\n \n if action == 'add':\n i = int(parts[1])\n p_i = parts[2]\n patterns[i] = p_i\n elif action == 'delete':\n i = int(parts[1])\n del patterns[i]\n elif action == 'append':\n s = parts[1]\n current_text = current_text + s\n elif action == 'backspace':\n c = int(parts[1])\n current_text = current_text[:-min(c, len(current_text))]\n \n candidates = []\n for k, v in patterns.items():\n if v.startswith(current_text):\n candidates.append((v, k))\n \n if not candidates:\n output.append(-1)\n continue\n \n candidates.sort(key=lambda x: (-len(x[0]), x[0], x[1]))\n output.append(candidates[0][1])\n \n for num in output:\n print(num)\n \nif __name__ == \"__main__\":\n main()\n \n \n# tGIamVr0UsBqAEbj6uy\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 304970696,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 304179999,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class Main{\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int n = scanner.nextInt();\n scanner.nextLine(); // Consume the newline after the integer\n \n StringBuilder t = new StringBuilder();\n Map<Integer, String> patterns = new HashMap<>();\n \n for (int i = 0; i < n; i++) {\n String[] parts = scanner.nextLine().split(\" \");\n String operation = parts[0];\n \n if (operation.equals(\"add\")) {\n int id = Integer.parseInt(parts[1]);\n String pattern = parts[2];\n patterns.put(id, pattern);\n } else if (operation.equals(\"delete\")) {\n int id = Integer.parseInt(parts[1]);\n patterns.remove(id);\n } else if (operation.equals(\"append\")) {\n String s = parts[1];\n t.append(s);\n } else if (operation.equals(\"backspace\")) {\n int c = Integer.parseInt(parts[1]);\n if (c >= t.length()) {\n t.setLength(0);\n } else {\n t.setLength(t.length() - c);\n }\n }\n \n // Find the best matching pattern\n int bestId = -1;\n String bestPattern = \"\";\n for (Map.Entry<Integer, String> entry : patterns.entrySet()) {\n String p = entry.getValue();\n if (t.length() <= p.length() && p.startsWith(t.toString())) {\n if (bestId == -1 || p.length() > bestPattern.length() || \n (p.length() == bestPattern.length() && p.compareTo(bestPattern) < 0) || \n (p.length() == bestPattern.length() && p.compareTo(bestPattern) == 0 && entry.getKey() < bestId)) {\n bestId = entry.getKey();\n bestPattern = p;\n }\n }\n }\n System.out.println(bestId);\n }\n scanner.close();\n }\n}\n \n \n// cMqIBbzmgAiy4VuJ0Oa\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 304978101,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main{\n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n PrintWriter pw = new PrintWriter(System.out);\n \n int n = Integer.parseInt(br.readLine());\n Map<Integer, String> patterns = new HashMap<>();\n StringBuilder currentText = new StringBuilder();\n \n for (int i = 0; i < n; i++) {\n String[] input = br.readLine().split(\" \");\n if (input[0].equals(\"add\")) {\n int id = Integer.parseInt(input[1]);\n String pattern = input[2];\n patterns.put(id, pattern);\n } else if (input[0].equals(\"delete\")) {\n int id = Integer.parseInt(input[1]);\n patterns.remove(id);\n } else if (input[0].equals(\"append\")) {\n currentText.append(input[1]);\n } else if (input[0].equals(\"backspace\")) {\n int c = Integer.parseInt(input[1]);\n int len = currentText.length();\n if (c >= len) {\n currentText.setLength(0);\n } else {\n currentText.setLength(len - c);\n }\n }\n \n // Suggest the pattern\n String t = currentText.toString();\n String bestPattern = null;\n int bestId = -1;\n \n for (Map.Entry<Integer, String> entry : patterns.entrySet()) {\n int id = entry.getKey();\n String pattern = entry.getValue();\n if (pattern.startsWith(t) && (bestPattern == null ||\n pattern.length() > bestPattern.length() ||\n (pattern.length() == bestPattern.length() &&\n (pattern.compareTo(bestPattern) < 0 ||\n (pattern.compareTo(bestPattern) == 0 && id < bestId))))){\n bestPattern = pattern;\n bestId = id;\n }\n }\n \n pw.println(bestId);\n }\n \n pw.flush();\n pw.close();\n br.close();\n }\n}\n \n \n// C0iOKeLu4RZljQdwmEP\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 304214296,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define int long long\n \nvector<string> s;\nint curr;\nint last;\n \nstruct compare{\n int depth;\n compare(int d = 0) : depth(d) {}\n bool operator()(int a, int b) const {\n const string& sa = s[a];\n const string& sb = s[b];\n if (sa.length() == sb.length()){\n if (depth==sa.length()) return a < b;\n if (sa[depth]==sb[depth]) return a < b;\n return sa[depth] < sb[depth];\n }\n return sa.length() > sb.length();\n }\n};\n \nstruct item {\n set<int, compare> ans;\n int prev = 0;\n int depth = 0;\n unordered_map<int, int> next;\n}; item trie[2000010];\n \nvoid add(int i){\n int n = s[i].length();\n int p = 0;\n trie[p].ans.insert(i);\n for(int j=0; j<n; j++){\n int ch = s[i][j]-'!';\n if (trie[p].next.count(ch) == 0){\n trie[p].next[ch] = last;\n trie[last].prev = p;\n trie[last].depth = trie[p].depth + 1;\n last++;\n }\n p = trie[p].next[ch];\n trie[p].ans.insert(i);\n }\n}\n \nvoid update(string& a){\n int n = a.length();\n for(int j=0; j<n; j++){\n int ch = a[j]-'!';\n if (trie[curr].next.count(ch) == 0){\n trie[curr].next[ch] = last;\n trie[last].prev = curr;\n trie[last].depth = trie[curr].depth + 1;\n last++;\n }\n curr = trie[curr].next[ch];\n }\n}\n \nvoid backspace(int n){\n for(int i=0; i<n; i++){\n if(curr==0) break;\n curr = trie[curr].prev;\n }\n}\n \nvoid remove(int i){\n int n = s[i].length();\n int p = 0;\n trie[p].ans.erase(i);\n for(int j=0; j<n; j++){\n int ch = s[i][j]-'!';\n p = trie[p].next[ch];\n trie[p].ans.erase(i);\n }\n}\n \nint get(){\n item p = trie[curr];\n if (p.ans.empty()) return -1;\n int idx = *p.ans.begin();\n while (p.depth != s[idx].length()){\n p = trie[p.next[s[idx][p.depth]-'!']];\n idx = *p.ans.begin();\n }\n return idx;\n}\n \nint32_t main() {\n ios_base::sync_with_stdio(false); \n cin.tie(NULL);\n \n string q, query;\n int n, idx;\n cin >> n;\n s.assign(n+1,\"\");\n curr = 0;\n last = 1;\n while(n--){\n cin >> q;\n if (q==\"add\"){\n cin >> idx >> query;\n s[idx] = query;\n add(idx);\n }\n else if(q==\"append\"){\n cin >> query;\n update(query);\n }\n else if(q==\"delete\"){\n cin >> idx;\n remove(idx);\n }\n else{\n cin >> idx;\n backspace(idx);\n }\n cout << get() << \"\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 296268791,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ll long long\n#define vi vector<int>\n#define pb push_back\n \nusing namespace std;\n \n//tree of all prefixes of patterns\n//add: add prefixes\n//delete: delete prefixes, appearences of matchings\n//append: deepen search\n//backspace: go back x\n \nconst int maxn = 1e6+3;\n \nstruct pattern {\n int id;\n string s;\n bool operator <(const pattern& r) const {\n if (this->s.size() < r.s.size()) return true;\n if (this->s.size() > r.s.size()) return false;\n for (int i=0; i<this->s.size(); ++i) {\n if (this->s[i] > r.s[i]) {\n return true;\n }\n if (this->s[i] < r.s[i]) {\n return false;\n }\n }\n return (this->id > r.id) ? true : false;\n }\n pattern(int i, string st) {\n id = i;\n s = st;\n }\n};\n \nvector<string> strings;\n \nvector<int> pos;\n \nstruct pat {\n int id;\n bool operator <(const pat &r) const {\n return pos[this->id] > pos[r.id];\n }\n pat(int i) {\n id = i;\n }\n};\n \nstruct node {\n set<pat> match;\n char c;\n int ind;\n node * parent;\n map<char, node*> child;\n node (node * par, int x, char d) {\n parent = par;\n ind = x;\n c = d;\n }\n};\n \nvoid add(node &cur, pat &p, string &s) {\n int idx = cur.ind;\n cur.match.insert(p);\n if (s.size()-1 == idx) {\n return;\n }\n char next = s[idx+1];\n if (cur.child.find(next) == cur.child.end())\n cur.child[next] = new node(&cur, idx+1, next);\n add(*cur.child[next], p, s);\n return;\n}\n \nvector<int> nums;\nvector<int> type;\nvector<string> apps;\n \nint main() {\n int n;\n cin >> n;\n {\n vector<pattern> all;\n for (int i=0; i<n; ++i) {\n pos.pb(0);\n string in;\n cin >> in;\n int num;\n string ad;\n if (in == \"add\") {\n cin >> num >> ad;\n type.pb(0);\n nums.pb(num);\n all.pb(pattern(num, ad));\n }\n if (in == \"delete\") {\n cin >> num;\n type.pb(1);\n nums.pb(num);\n }\n if (in == \"append\") {\n cin >> ad;\n type.pb(2);\n apps.pb(ad);\n }\n if (in == \"backspace\") {\n cin >> num;\n type.pb(3);\n nums.pb(num);\n }\n }\n sort(all.begin(), all.end());\n for (int i=0; i<all.size(); ++i) {\n pos[all[i].id] = i;\n strings.pb(all[i].s);\n }\n all.clear();\n }\n node start = node(nullptr, -1, 'a');\n int dep = 0, len = 0;\n string text;\n node * cur = &start;\n int numind = 0, appsind = 0;\n for (int i=0; i<n; ++i) {\n int t = type[i];\n if (t == 0) {\n int id = nums[numind];\n ++numind;\n pat newpat = pat(id);\n add(start,newpat,strings[pos[newpat.id]]);\n for (int j=dep; j<len; ++j) {\n if (cur->child.find(text[j]) == cur->child.end()) break;\n cur = cur->child[text[j]];\n ++dep;\n }\n }\n if (t == 1) {\n int id = nums[numind];\n ++numind;\n node * here = &start;\n here->match.erase(id);\n for (char cc : strings[pos[id]]) {\n if (here->child.find(cc) == here->child.end()) break;\n here = here->child[cc];\n here->match.erase(id);\n }\n }\n if (t == 2) {\n string s = apps[appsind];\n ++appsind;\n text.append(s);\n int l = s.size();\n len += l;\n if (len-l == dep) {\n for (int j=0; j<l; ++j) {\n if (cur->child.find(s[j]) == cur->child.end()) break;\n cur = cur->child[s[j]];\n ++dep;\n }\n }\n }\n if (t == 3) {\n int b = nums[numind];\n ++numind;\n for (int j=0; j<b; ++j) {\n if (!text.empty()) text.pop_back();\n }\n for (int j=0; j<b-len+dep; ++j) {\n if (cur->parent != nullptr)\n cur = cur->parent;\n }\n dep = max(0,min(dep, len-b));\n len = max(0,len-b);\n }\n //now query on current tree\n bool found = false;\n int res = -1;\n while (!found) {\n if (len != dep) break;\n if (!cur->match.empty()) {\n found = true;\n res = cur->match.begin()->id;\n } else if (cur->parent != nullptr) {\n char ch = cur->c;\n cur = cur->parent;\n delete cur->child[ch];\n cur->child.erase(ch);\n --dep;\n } else {\n break;\n }\n }\n cout << res << \"\\n\";\n if (i == 29792) return 0;\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 299413614,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n#define Many_SubTask\n#include <bits/stdc++.h>\nusing namespace std;\n// #define int long long\n#define all(a) begin(a), end(a)\n#define pb push_back\n#define pii pair<int, int>\n#define F first\n#define S second\n#define mp make_pair\nconst int mod = 998244353;\nconst int inf = 1e18;\nconst int MAXN = 1000005;\n \nstruct node {\n int ansID;\n set<int> ids;\n node *nxt[95], *fa;\n};\nint node_cnt = 0;\n \nnode *new_node(node *fa) {\n node_cnt++;\n node *nn = new node;\n nn->ansID = -1;\n for (int i = 0; i < 95; i++) {\n nn->nxt[i] = NULL;\n }\n nn->fa = fa;\n return nn;\n}\n \nnode *root = new_node(NULL);\nnode *cur = root;\n \nstring pattern[MAXN];\n \nint maxID(int a, int b) {\n if (a == -1) return b;\n if (b == -1) return a;\n if (pattern[a].size() > pattern[b].size()) return a;\n if (pattern[a].size() < pattern[b].size()) return b;\n if (pattern[a] < pattern[b]) return a;\n if (pattern[a] > pattern[b]) return b;\n return min(a, b);\n}\n \nvoid add(int id, string &s) {\n node *p = root;\n for (int i = 0; i < s.size(); i++) {\n if (!p->nxt[s[i]]) p->nxt[s[i]] = new_node(p);\n p = p->nxt[s[i]];\n }\n p->ids.insert(id);\n p->ansID = maxID(p->ansID, id);\n do {\n p = p->fa;\n p->ansID = maxID(p->ansID, id);\n } while (p != root);\n}\n \nvoid updateAnsID(node* p) {\n p->ansID = -1;\n for (auto id_ : p->ids) {\n p->ansID = maxID(p->ansID, id_);\n }\n for (int j = 0; j < 95; j++) {\n if (p->nxt[j]) p->ansID = maxID(p->ansID, p->nxt[j]->ansID);\n }\n}\n \nvoid del(int id, string &s) {\n node *p = root;\n for (int i = 0; i < s.size(); i++) {\n if (!p->nxt[s[i]]) p->nxt[s[i]] = new_node(p);\n p = p->nxt[s[i]];\n }\n p->ids.erase(id);\n if (p->ansID == id) updateAnsID(p);\n do {\n p = p->fa;\n if (p->ansID == id) updateAnsID(p);\n } while (p != root);\n}\n \nvoid append(string &s) {\n int n = s.size();\n for (int i = 0; i < n; i++) {\n if (!cur->nxt[s[i]]) cur->nxt[s[i]] = new_node(cur);\n \n cur = cur->nxt[s[i]];\n }\n}\n \nvoid backspace(int c) {\n while (cur != root && c--) {\n cur = cur->fa;\n }\n}\n \nint N = 1;\nvoid solve() {\n string ope;\n cin >> ope;\n if (ope == \"add\") {\n int id;\n string s;\n cin >> id >> s;\n for (auto &i : s) i -= 33;\n pattern[id] = s;\n add(id, s);\n } else if (ope == \"delete\") {\n int id;\n cin >> id;\n del(id, pattern[id]);\n } else if (ope == \"append\") {\n string s;\n cin >> s;\n for (auto &i : s) i -= 33;\n append(s);\n } else { // backspace\n int c;\n cin >> c;\n backspace(c);\n }\n if(N != 30000) cout << cur->ansID << '\\n';\n}\n \nsigned main() {\n cin.sync_with_stdio(0), cin.tie(0);\n#ifdef Many_SubTask\n cin >> N;\n#endif\n for (int i = 1; i <= N; i++) {\n solve();\n if(N == 30000 && i > 29200) cout << node_cnt << '\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 293511657,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n// -*- mode: c++; coding: utf-8-emacs; -*-\n#include<bits/stdc++.h>\n//#define int long long\nusing namespace std;\nstruct NODE{\n\tint dep;\n\tint fa;\n\tint mx;\n\tint md;\n\tmultiset<int> id;\n\tmap<char,int> c;\n\tNODE(int d,int f) : dep(d),fa(f),mx(-1),md(1){}\n};\nstruct TREE{\n\tmap<int,string> pattern;\n\tvector<NODE> a;\n\tstring t;\n\tint now;\n\tTREE() : now(0),a{NODE(0,-1)}{};\n\tvoid chkmax(int u,string &s){\n\t\twhile(u!=-1){\n\t\t\ta[u].mx=-1;\n\t\t\ta[u].md=1;\n\t\t\tfor(auto [key,v] : a[u].c){\n\t\t\t\tif(v!=-1 && a[v].mx!=-1){\n//\t\t\t\t\tcout<<a[v].md<<'\\n';\n\t\t\t\t\tif(a[u].mx==-1 || a[v].md+1>a[u].md){\n\t\t\t\t\t\ta[u].md=a[v].md+1;\n\t\t\t\t\t\ta[u].mx=a[v].mx;\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t}\n\t\t\tif(a[u].mx==-1 && !a[u].id.empty()){\n\t\t\t\ta[u].mx=*a[u].id.begin();\n\t\t\t}\n//\t\t\tcout<<' '<<a[u].mx<<endl;\n\t\t\tu=a[u].fa;\n\t\t}\n\t}\n\tvoid add(int id,string s){\n\t\tpattern[id]=s;\n\t\tint u=0;\n\t\tfor(auto i : s){\n\t\t\tif(a[u].c.find(i)==a[u].c.end()){\n\t\t\t\ta[u].c[i]=a.size();\n\t\t\t\ta.emplace_back(a[u].dep+1,u);\n\t\t\t}\n\t\t\tu=a[u].c[i];\n\t\t}\n\t\ta[u].id.insert(id);\n\t\tchkmax(u,s);\n\t}\n\tvoid del(int id){\n\t\tstring s=pattern[id];\n\t\tint u=0;\n\t\tfor(auto i : s){\n\t\t\tu=a[u].c[i];\n\t\t}\n\t\ta[u].id.erase(a[u].id.find(id));\n\t\tchkmax(u,s);\n\t}\n\tvoid append(string s){\n\t\tt+=s;\n\t}\n\tvoid backspace(int n){\n\t\twhile(!t.empty() && n--){\n\t\t\tt.pop_back();\n\t\t}\n\t}\n\tint ask(){\n//\t\tcout<<\"now: \"<<now<<' '<<a[now].mx<<'\\n';\n\t\twhile(now>0 && (a[now].mx==-1 || now>(int)t.length())){\n\t\t\tnow=a[now].fa;\n\t\t}\n\t\twhile(a[now].dep<(int)t.length()){\n\t\t\tint x=t[a[now].dep];\n\t\t\tif(a[now].c.find(x)==a[now].c.end()){\n\t\t\t\treturn -1;\n\t\t\t}\n\t\t\tint v=a[now].c[x];\n\t\t\tif(a[v].mx==-1){\n\t\t\t\treturn -1;\n\t\t\t}\n\t\t\tnow=v;\n\t\t}\n\t\treturn a[now].mx;\n\t}\n};\n/*\n 6\n add 1 aaa\n add 2 ab\n add 3 accc\n append a\n append babc\n backspace 3\n */\nvoid solve(){\n\tint qwq;\n\tcin>>qwq;\n\tTREE t;\n\tqwq=min(qwq,29800);\n\twhile(qwq--){\n\t\tstring o;\n\t\tcin>>o;\n\t\tif(o[1]=='d'){\n\t\t\tint i;\n\t\t\tstring s;\n\t\t\tcin>>i>>s;\n\t\t\tfor(auto &c : s){\n\t\t\t\tc-=33;\n\t\t\t}\n\t\t\tt.add(i,s);\n\t\t}else if(o[1]=='p'){\n\t\t\tstring s;\n\t\t\tcin>>s;\n\t\t\tfor(auto &c : s){\n\t\t\t\tc-=33;\n\t\t\t}\n\t\t\tt.append(s);\n\t\t}else if(o[1]=='a'){\n\t\t\tint i;\n\t\t\tcin>>i;\n\t\t\tt.backspace(i);\n\t\t}else{\n\t\t\tint i;\n\t\t\tcin>>i;\n\t\t\tt.del(i);\n\t\t}\n\t\tcout<<t.ask()<<'\\n';\n\t}\n}\nsigned main(){\n\tios::sync_with_stdio(false);\n\tint qwq=1;\n//\tcin>>qwq;\n\twhile(qwq--){\n\t\tsolve();\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 293191614,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define fastio(); ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);\n \nchar END_CHAR = ' ';\n \nbool compare(pair<int, pair<string, int>>& A, pair<int, pair<string, int>> & B) {\n if (A.first != B.first) {\n return (A.first > B.first);\n }\n if (A.second.first != B.second.first) {\n return (A.second.first < B.second.first);\n }\n return A.second.second < B.second.second;\n}\n \nstruct Trie {\n \n unordered_map<char, Trie*> nextPointer;\n set<pair<int, pair<string, int>>> lenStringIDs;\n pair<int, pair<string, int>> maxLenStringID;\n Trie* parent;\n Trie(Trie* p) {\n nextPointer = unordered_map<char, Trie*>();\n maxLenStringID = {0, {\"\", -1}};\n lenStringIDs = set<pair<int, pair<string, int>>>();\n parent = p;\n }\n \n};\n \nvoid add(Trie* node, string & s, int index, int ID) {\n if (index == (int) s.size() + 1) {\n // if ((node -> nextPointer).find(END_CHAR) == (node -> nextPointer).end()) {\n // node->nextPointer[END_CHAR] = new Trie(node);\n // }\n Trie* endNode = node;\n (endNode->lenStringIDs).insert({(int) s.size(), {s, ID}});\n (endNode->maxLenStringID) = *(--((endNode->lenStringIDs).end()));\n return ;\n }\n \n char ch = (index < ((int)s.size())) ? s[index] : END_CHAR;\n if ((node -> nextPointer).find(ch) == (node -> nextPointer).end()) {\n node->nextPointer[ch] = new Trie(node);\n }\n \n add(node->nextPointer[ch], s, index + 1, ID);\n \n if (!compare(node -> maxLenStringID, node -> nextPointer[ch] -> maxLenStringID)) {\n node -> maxLenStringID = node -> nextPointer[ch] -> maxLenStringID;\n }\n // cout << ch << \" \" << node -> nextPointer[ch] -> maxLenStringID.first << \"\\t\";\n}\n \nvoid remove(Trie* node, string & s, int index, int ID) {\n if (index == (int) s.size() + 1) {\n // if ((node -> nextPointer).find(END_CHAR) == (node -> nextPointer).end()) {\n // node->nextPointer[END_CHAR] = new Trie(node);\n // }\n Trie* endNode = node;\n auto IT = (endNode->lenStringIDs).find({(int) s.size(), {s, ID}});\n (endNode->lenStringIDs).erase(IT);\n (endNode->maxLenStringID) = {0, {\"\", -1}};\n if (!(endNode->lenStringIDs).empty()) {\n (endNode->maxLenStringID) = *(--((endNode->lenStringIDs).end()));\n }\n return ;\n }\n \n char ch = (index < (int) s.size()) ? s[index] : END_CHAR;\n if ((node -> nextPointer).find(ch) == (node -> nextPointer).end()) {\n node->nextPointer[ch] = new Trie(node);\n }\n \n remove(node->nextPointer[ch], s, index + 1, ID);\n if ((node -> maxLenStringID).second.second == ID) {\n node -> maxLenStringID = {0, {\"\", -1}};\n for (auto child: node -> nextPointer) {\n if (!compare(node -> maxLenStringID, child.second -> maxLenStringID)) {\n node -> maxLenStringID = child.second -> maxLenStringID;\n }\n }\n }\n}\n \nint main() {\n fastio();\n \n int Q;\n cin >> Q;\n \n Trie* root = new Trie(NULL);\n Trie* traveller = root;\n \n map<int, string> idToString;\n \n while (Q --) {\n string operation;\n cin >> operation;\n if (operation == \"add\") {\n int ID;\n cin >> ID;\n string s;\n cin >> s;\n idToString[ID] = s;\n add(root, s, 0, ID);\n \n } else if (operation == \"append\") {\n string s;\n cin >> s;\n \n for (int i = 0;i < (int)s.size();i ++) {\n if ((traveller -> nextPointer).find(s[i]) == (traveller -> nextPointer).end()) {\n (traveller -> nextPointer)[s[i]] = new Trie(traveller);\n }\n \n traveller = (traveller -> nextPointer)[s[i]];\n }\n \n } else if (operation == \"backspace\") {\n int cnt = 0;\n cin >> cnt;\n while (traveller -> parent != NULL && cnt > 0) {\n cnt --;\n traveller = traveller -> parent;\n }\n } else if (operation == \"delete\") {\n int ID;\n cin >> ID;\n \n string s = idToString[ID];\n \n remove(root, s, 0, ID);\n }\n \n cout << (traveller -> maxLenStringID).second.second << \"\\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\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 293134219,
"task_type": "TestCase-Eval-Task2-DO"
} |
TestCase-Eval-Task2-DO | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Auto Complete\ntime_limit_ms: 4000\nmemory_limit_mb: 1024\nDescription: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types \"App\", your editor will magically suggest the word \"Application\"! Even better, users can personalize the words that auto-complete in your editor.\n\nYour editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):\n\n1. Add an auto complete pattern $$$p_i$$$.\n2. Delete an auto complete pattern $$$p_i$$$.\n3. Append a string $$$s$$$ to the end of $$$t$$$.\n4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.\n\nAfter each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:\n\n1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.\n2. If there are multiple $$$p_i$$$, pick the longest one.\n3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.\n4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.\n\nFor example, let us say we have three candidates: \"alice\", \"bob\", and \"charlie\", with ID 1, 2, and 3. At first, there is nothing on the screen, so \"charlie\" (3) should be suggested because it is the longest. Then, let us say the user types \"b\". You should suggest \"bob\" (2) because it is the only one that starts with \"b\". Finally, let us say the user type \"body\". You should print -1 because there is no matched pattern.\n\nInput: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.\n\nThere are four types of actions:\n\n1. $$$i$$$ $$$p_i$$$\n2. $$$i$$$\n3. $$$s$$$\n4. $$$c$$$\n\n- $$$1 \\leq n \\leq 10^6$$$\n- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\\times 10^6$$$.\n- $$$1 \\leq c \\leq 2\\times 10^6$$$\n- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).\n- The ID $$$i$$$ for each add operation is unique.\n- The ID $$$i$$$ for each delete is guaranteed to be valid.\n- Each ID $$$i$$$ satisfies $$$0\\le i \\le n$$$.\n\nOutput: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.\n\nExamples:\ninput:\n6\nadd 1 pattern1_alice\nadd 2 pattern2_bob\nadd 3 pattern3_charlie\nappend pattern\nappend 2_bobabc\nbackspace 3\noutput:\n1\n1\n3\n3\n-1\n2\n\ninput:\n6\nappend pattern\nadd 1 pattern1_alice____\nadd 2 pattern2_bob______\nadd 3 pattern3_charlie__\ndelete 1\ndelete 2\noutput:\n-1\n1\n1\n1\n2\n3\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class Main{\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n \n int n = scanner.nextInt();\n scanner.nextLine(); // Consume newline\n \n StringBuilder currentText = new StringBuilder();\n TreeSet<Pattern> patterns = new TreeSet<>(Comparator\n .comparing((Pattern p) -> p.length).reversed()\n .thenComparing(Pattern::toString)\n .thenComparing(p -> p.id));\n Map<Integer, Pattern> patternMap = new HashMap<>();\n \n for (int i = 0; i < n; i++) {\n String line = scanner.nextLine();\n String[] tokens = line.split(\" \", 2); // Split into max 2 parts\n String command = tokens[0];\n \n if (command.equals(\"add\")) {\n String[] parts = tokens[1].split(\" \", 2);\n int id = Integer.parseInt(parts[0]);\n String pattern = parts[1];\n Pattern p = new Pattern(id, pattern);\n patterns.add(p);\n patternMap.put(id, p);\n } else if (command.equals(\"delete\")) {\n int id = Integer.parseInt(tokens[1]);\n Pattern p = patternMap.get(id);\n if (p != null) {\n patterns.remove(p);\n patternMap.remove(id);\n }\n } else if (command.equals(\"append\")) {\n currentText.append(tokens[1]);\n } else if (command.equals(\"backspace\")) {\n int c = Integer.parseInt(tokens[1]);\n int len = currentText.length();\n currentText.delete(Math.max(0, len - c), len);\n }\n \n Pattern bestMatch = findBestMatch(currentText.toString(), patterns);\n if (bestMatch != null) {\n System.out.println(bestMatch.id);\n } else {\n System.out.println(-1);\n }\n }\n }\n \n private static Pattern findBestMatch(String text, TreeSet<Pattern> patterns) {\n for (Pattern p : patterns) {\n if (p.pattern.startsWith(text)) {\n return p;\n }\n }\n return null;\n }\n \n static class Pattern {\n int id;\n int length;\n String pattern;\n \n Pattern(int id, String pattern) {\n this.id = id;\n this.pattern = pattern;\n this.length = pattern.length();\n }\n \n @Override\n public String toString() {\n return pattern;\n }\n }\n}\n \n \n// 8wrXZand4pEY0I1gWMV\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2041I",
"submission_id": 304195651,
"task_type": "TestCase-Eval-Task2-DO"
} |
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