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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: Penchick and Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\n#include <iostream>\n \nusing namespace std;\n \nint f[51];\n \nint main()\n{\n int t,i,n,j,a,x;\n cin >> t;\n for (i=0; i<t; i++){\n cin >> n;\n x=0;\n for (j=0; j<n; j++){\n cin >> a;\n f[a]++;\n x=max(x, f[a]);\n }\n cout << n-x << '\\n';\n for (j=0; j<n; j++)\n f[j]=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": "2031A",
"submission_id": 291601945,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\n#include <iostream>\nusing namespace std;\nint main() {\n int t;\n cin>>t;\n for (int i=0;i<t;i++) {\n int n;\n cin>>n;\n int h[n];\n for (int j=0;j<n;j++) {\n cin>>h[j];\n }\n if(n==1)\n cout<<0<<endl;\n else if(n==2)\n cout<<1<<endl;\n else if (h[n/2]==h[n/2+1]==h[n/2-1])\n cout<<n-3<<endl;\n else if ((h[n/2]==h[n/2+1])||(h[n/2]==h[n/2-1]))\n cout<<n-2<<endl;\n else\n cout<<n-1<<endl;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031A",
"submission_id": 302464193,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\n#include <iostream>\nusing namespace std;\n \nint main (){\n int t; cin >> t;\n while (t--) {\n int n; cin >> n;\n int a[n + 1]; a[0] = -1;\n int x = 1;\n int y = 1;\n for (int i = 1; i <= n; i++) {\n cin >> a[i];\n if (a[i] == a[i - 1]) {\n y++; \n if (y > x) x = y;\n }\n }\n cout << n - x << 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": "2031A",
"submission_id": 302696674,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\na=int(input())\nfor i in range(a):\n b=int(input())\n c=list(map(int,input().split()))\n if c==(sorted(c))[::-1]:\n d=len(set(c))\n if d==len(c):\n print(len(c)-1)\n else:\n print(len(c)-d)\n else:\n print(0)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031A",
"submission_id": 291650754,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\nt = int(input())\nfor _ in range(t):\n n = int(input())\n lst = list(map(int, input().split()))\n count = 0\n for i in range(n - 1):\n if lst[i] > lst[i + 1]:\n count += 1\n print(count)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031A",
"submission_id": 296419107,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\na=int(input())\nfor i in range(a):\n b=int(input())\n c=list(map(int,input().split()))\n d=0\n for e in range(1,b):\n if c[e]<c[e-1]:\n d+=1\n print(d)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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"
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]
| TestCase-Eval | null | {
"problem_id": "2031A",
"submission_id": 291644011,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int t = scanner.nextInt(); \n \n for (int i = 0; i < t; i++) {\n int n = scanner.nextInt();\n int[] heights = new int[n];\n for (int j = 0; j < n; j++) {\n heights[j] = scanner.nextInt();\n }\n \n int operations = 0;\n int lastHeight = 0;\n \n for (int j = 0; j < n; j++) {\n if (heights[j] < lastHeight) {\n operations++;\n } else {\n lastHeight = heights[j];\n }\n }\n System.out.println(operations);\n }\n \n scanner.close();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031A",
"submission_id": 291623142,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class PMMonument {\n public static void main(String[] args){\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt(); \n while (t-- > 0) {\n int n = sc.nextInt(); \n int[] h= new int[n];\n for(int i = 0; i < n; i++ ){\n h[i] = sc.nextInt();\n }\n \n int operations = 0;\n for (int i = n - 2; i >= 0; i--) {\n if (h[i] > h[i + 1]) {\n operations = h[i] - h[i + 1];\n h[i] = h[i + 1]; \n }\n }\n \n System.out.println(operations);\n }\n sc.close();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031A",
"submission_id": 293078205,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\n/*\n Author : kmv a.k.a kzhi\n K41 IT CHV\n*/\n#include <bits/stdc++.h>\nusing namespace std;\n \n \n//#define int long long\n#define ll long long\n#define FOR(i,a,b) for (int i = a ; i <= b; ++ i)\n#define FOD(i,a,b) for (int i = a; i >= b; -- i)\n \n \n#define BIT(mask,i) ((mask >> i) & 1)\n#define MASK(i) (1ll << (i))\n#define OFFBIT(mask,i) (mask &~(1ll<<(i)))\n#define ONBIT(mask,i) (mask | (1ll << (i)))\n#define lg2(x) (63 - __builtin_clzll(x))\n#define c_bit __builtin_popcountll\n \n#define vi vector < int >\n#define all(a) a.begin(), a.end()\n#define pb push_back\n \n#define ii pair<int,int>\n#define fi first\n#define se second\n \n#define openfile(TASK) if (fopen(TASK\".inp\",\"r\"))\\\n {freopen(TASK\".inp\",\"r\",stdin);freopen(TASK\".out\",\"w\",stdout);}\n#define endl '\\n'\n#define Faster ios_base::sync_with_stdio(false); \\\n cin.tie(0); cout.tie(0);\n \n#define mid(l,r) ((l + r) >> 1)\n#define left(id) (id << 1)\n#define right(id) ((id << 1) | 1)\n#define ci const int\n \ntemplate <class X, class Y> bool maximize(X &a, const Y &b){if(a < b) return a = b, true;return false;}\ntemplate <class X, class Y> bool minimize(X &a, const Y &b){if(a > b) return a = b, true;return false;}\n \nconst int N = 55;\n \nint n, a[N];\n \nvoid SOLVE(){\n cin >> n;\n FOR (i, 1, n)\n cin >> a[i];\n \n int cnt = 0;\n \n FOR (i, 2, n)\n if (a[i] < a[i - 1]){\n a[i] = a[i - 1];\n cnt ++;\n }\n \n cout << cnt << endl;\n}\n \nsigned main(){\n Faster\n openfile(\"test\")\n \n int q = 1;\n \n cin >> q;\n \n while (q --){\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": "2031A",
"submission_id": 291581922,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define int long long\nusing namespace std;\nint32_t main() {\n int t;\n cin>>t;\n while(t--){\n int n;\n cin>>n;\n vector <int> vec;\n \n for (int i=0; i<n; i++){\n int x=0;\n cin>>x;\n vec.push_back(x);\n }\n sort(vec.begin(), vec.end());\n vector <int> vec1(vec[n-1],0);\n for (int i=0; i<n; i++){\n vec1[vec[i]-1]++;\n }\n sort(vec1.begin(), vec1.end());\n int count=0;\n for (int i=0; i<vec1.size(); i++){\n if (vec1[vec1.size()-1]==1){\n count=1;\n break;\n }\n if (vec1[i]==vec1[vec1.size()-1] && vec1[vec1.size()-1]!=1){\n count+=vec1[i];\n }\n }\n if (n==1){\n cout<<0<<endl;\n }\n else{\n cout<<vec.size()-count<<endl;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031A",
"submission_id": 292435140,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\ndef min_operations_to_non_decreasing(n, heights):\n operations = 0\n max_height = heights[0]\n \n # Iterate through array starting from second element\n for i in range(1, n):\n # If current height is less than previous maximum\n # We need to increase it to at least max_height\n if heights[i] < max_height:\n operations += 1\n else:\n # Update max_height if current height is larger\n max_height = heights[i]\n \n return operations\n \n# Read number of test cases\nt = int(input())\n \nfor _ in range(t):\n # Read number of pillars\n n = int(input())\n # Read heights array\n heights = list(map(int, input().split()))\n # Print minimum operations needed\n print(min_operations_to_non_decreasing(n, heights))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2031A",
"submission_id": 291624407,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\nt=int(input())\nwhile t>0:\n t-=1\n n=int(input())\n li=list(map(int,input().split(\" \")))\n s=set(li)\n print(len(s)-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": "2031A",
"submission_id": 291653494,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\nimport java.math.*;\n \npublic class sub {\n \n\t/*\n\t * /\\_/\\ (= ._.) / > \\>\n\t */\n\tstatic int mod = 998244353;\n \n\tpublic static void main(String args[]) {\n\t\tint t = sc.nextInt();\n\t\twhile (t-- > 0) {\n\t\t\tint n = sc.nextInt();\n\t\t\tint a[] = new int[n];\n\t\t\tfor(int i = 0;i<n;i++)a[i]=sc.nextInt();\n\t\t\tint ans = 0;\n\t\t\tfor(int i = 0;i<n-1;i++) {\n\t\t\t\tif(a[i]>a[i+1]) {\n\t\t\t\t\tans++;\n\t\t\t\t\ta[i+1]=a[i];\n\t\t\t\t}\n\t\t\t}\n\t\t\tout.println(ans);\n \n\t\t}\n\t\tout.close();\n\t}\n \n\t/*\n\t * public static long bincoef(int n, int k) { if(n < k) return 0L; return (p[n]\n\t * * pow((p[n - k] * p[k]) % mod, mod - 2,mod))%mod; }\n\t */\n\tstatic long pow(long a, long b, long mod) {\n\t\tif (b == 0) {\n\t\t\treturn 1L;\n\t\t}\n\t\tlong val = pow(a, b / 2, mod);\n\t\tif (b % 2 == 0) {\n\t\t\treturn val * val % mod;\n\t\t}\n\t\treturn val * val % mod * a % mod;\n\t}\n \n\tstatic int gcd(int a, int b) {\n\t\tif (b == 0) {\n\t\t\treturn a;\n\t\t}\n\t\treturn gcd(b, a % b);\n\t}\n \n\tstatic int lcm(int a, int b) {\n\t\treturn a * b / gcd(a, b);\n\t}\n \n\tstatic Kattio sc = new Kattio();\n\tstatic PrintWriter out = new PrintWriter(System.out);\n \n\tstatic class Kattio {\n \n\t\tstatic BufferedReader r;\n\t\tstatic StringTokenizer st;\n \n\t\tpublic Kattio() {\n\t\t\tr = new BufferedReader(new InputStreamReader(System.in));\n\t\t}\n \n\t\tpublic String next() {\n\t\t\ttry {\n\t\t\t\twhile (st == null || !st.hasMoreTokens()) {\n\t\t\t\t\tst = new StringTokenizer(r.readLine());\n\t\t\t\t}\n\t\t\t\treturn st.nextToken();\n\t\t\t} catch (Exception e) {\n\t\t\t\treturn null;\n\t\t\t}\n\t\t}\n \n\t\tpublic int nextInt() {\n\t\t\treturn Integer.parseInt(next());\n\t\t}\n \n\t\tpublic long nextLong() {\n\t\t\treturn Long.parseLong(next());\n\t\t}\n \n\t\tpublic double nextDouble() {\n\t\t\treturn Double.parseDouble(next());\n\t\t}\n\t}\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031A",
"submission_id": 293929166,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\nimport java.util.*;\npublic class monument {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t=sc.nextInt();\n while(t>0)\n {\n int n=sc.nextInt();\n int A[]=new int[n];\n for(int i=0;i<n;i++)\n {\n A[i]=sc.nextInt();\n }\n Arrays.sort(A);\n int count=0;\n for(int i=0;i<n-1;i++)\n {\n if(A[i]<A[i+1])\n count++;\n }\n System.out.println(count);\n t--;\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": "2031A",
"submission_id": 291641317,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nconst int N=5e5;\nint n,a[N];\n \nvoid solve(){\n\tcin>>n;\n\tmap<int,int> mp;\n\tfor(int i=1;i<=n;i++){\n\t\tcin>>a[i];\n\t\tmp[a[i]]=1;\n\t}\n\tcout<<mp.size()-1<<'\\n';\n}\n \nsigned main(){\n\tios::sync_with_stdio(false);\n\tcin.tie(0);\n \n\tint t=1;\n\tcin>>t;\n\twhile(t--){\n\t\tsolve();\n\t}\n \n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031A",
"submission_id": 292446597,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long\nconst int N = 3e5+9;\nint a[N], b[N];\nvoid haha()\n{\n\tint n;cin >> n;int cnt =0;\n\tif(n == 1) cnt = 0;\n\telse if(n == 2) \n\t{\n\t\tif(a[1] > a[2]) cnt = 1;\n\t}\n\telse if(n >= 3)\n\t{\n\t\tfor(int i = 1; i <= n; i++)\n\t\t{\n\t\t\tcin >> a[i];\n\t\t\tb[i] = a[i];\n\t\t} \n\t\tsort(b+1, b+n+1);\n\t\tint x = b[(n+1)/2];\n\t\t//cout << x << \"\\n\";\n\t\tfor(int i = 1; i <= n/2; i++) if(a[i] > x) cnt ++;\n\t\tfor(int i = n/2+1; i <= n; i++) if(a[i] < x) cnt ++;\n\t}\n\tcout << cnt << \"\\n\";\n\treturn ;\n}\n \nsigned main()\n{\n\tios::sync_with_stdio(0),cin.tie(0),cout.tie(0);\n\tint t; cin >> t;\n\twhile(t--) haha();\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": "2031A",
"submission_id": 291592780,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\nt = int(input())\n \nfor i in range(t):\n h = int(input())\n \n k = list(map(int,input().split()))\n \n print(k[0]-k[-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": "2031A",
"submission_id": 291589053,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\nt=int(input())\nfor _ in range(t):\n n=int(input())\n h=list(map(int,input().split(\" \")))\n le=len(h)\n mle=le*-1\n if len(h)==1:\n print(0)\n continue\n else:\n c=0\n for i in range(n-1):\n if h[i]>h[i+1]:\n c+=1\n h[i],h[i+1]=h[i+1],h[i]\n else:\n continue\n print(c)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2031A",
"submission_id": 291618569,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\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 n=sc.nextInt();\n int[] arr=new int[n];\n int min=Integer.MAX_VALUE;\n int max=Integer.MIN_VALUE;\n int count=0;\n for(int i=0;i<n;i++){\n arr[i]=sc.nextInt();\n min=Math.min(min,arr[i]);\n max=Math.max(max,arr[i]);\n }\n int mid=n/2;\n for(int i=0;i<n/2;i++){\n if(arr[i]>arr[i+1]){\n arr[i]=min;\n count++;\n if(i==n/2){\n break;\n }\n }\n }\n for(int i=(n/2)+1;i<n;i++){if(arr[i]<arr[i-1]){\n arr[i]=max;\n count++;\n if(i==n-1){\n break;\n }\n }\n }\n if(n==1){\n System.out.println(\"0\");\n }\n else{\n System.out.println(count);\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": "2031A",
"submission_id": 291666348,
"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 Modern Monument\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Amidst skyscrapers in the bustling metropolis of Metro Manila, the newest Noiph mall in the Philippines has just been completed! The construction manager, Penchick, ordered a state-of-the-art monument to be built with $$$n$$$ pillars.\n\nThe heights of the monument's pillars can be represented as an array $$$h$$$ of $$$n$$$ positive integers, where $$$h_i$$$ represents the height of the $$$i$$$-th pillar for all $$$i$$$ between $$$1$$$ and $$$n$$$.\n\nPenchick wants the heights of the pillars to be in non-decreasing order, i.e. $$$h_i \\le h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$. However, due to confusion, the monument was built such that the heights of the pillars are in non-increasing order instead, i.e. $$$h_i \\ge h_{i + 1}$$$ for all $$$i$$$ between $$$1$$$ and $$$n - 1$$$.\n\nLuckily, Penchick can modify the monument and do the following operation on the pillars as many times as necessary:\n\n- Modify the height of a pillar to any positive integer. Formally, choose an index $$$1\\le i\\le n$$$ and a positive integer $$$x$$$. Then, assign $$$h_i := x$$$.\n\nHelp Penchick determine the minimum number of operations needed to make the heights of the monument's pillars non-decreasing.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 50$$$) — the number of pillars.\n\nThe second line of each test case contains $$$n$$$ integers $$$h_1, h_2, \\ldots, h_n$$$ ($$$1 \\le h_i \\le n$$$ and $$$h_i\\ge h_{i+1}$$$) — the height of the pillars.\n\nPlease take note that the given array $$$h$$$ is non-increasing.\n\nNote that there are no constraints on the sum of $$$n$$$ over all test cases.\n\nOutput: For each test case, output a single integer representing the minimum number of operations needed to make the heights of the pillars non-decreasing.\n\nExamples:\ninput:\n3\n5\n5 4 3 2 1\n3\n2 2 1\n1\n1\noutput:\n4\n1\n0\n\nNote: In the first test case, the initial heights of pillars are $$$h = [5, 4, 3, 2, 1]$$$.\n\n- In the first operation, Penchick changes the height of pillar $$$1$$$ to $$$h_1 := 2$$$.\n- In the second operation, he changes the height of pillar $$$2$$$ to $$$h_2 := 2$$$.\n- In the third operation, he changes the height of pillar $$$4$$$ to $$$h_4 := 4$$$.\n- In the fourth operation, he changes the height of pillar $$$5$$$ to $$$h_5 := 4$$$.\n\nAfter the operation, the heights of the pillars are $$$h = [2, 2, 3, 4, 4]$$$, which is non-decreasing. It can be proven that it is not possible for Penchick to make the heights of the pillars non-decreasing in fewer than $$$4$$$ operations.\n\nIn the second test case, Penchick can make the heights of the pillars non-decreasing by modifying the height of pillar $$$3$$$ to $$$h_3 := 2$$$.\n\nIn the third test case, the heights of pillars are already non-decreasing, so no operations are required.\n\nBuggy Code:\nimport java.util.*;\npublic class Main\n{\n public static void main(String args[])\n {\n Scanner sc=new Scanner(System.in);\n int t=sc.nextInt();\n while(t-->0)\n {\n int n=sc.nextInt();\n int a[]=new int[n];\n for(int i=0;i<n;i++)\n {\n a[i]=sc.nextInt();\n }\n \n \n if(n==1)\n {\n System.out.println(0);\n }\n else\n {\n int c=0;\n for(int i=0;i<n-1;i++)\n {\n if(a[i]>a[i+1])\n {\n c++;\n }\n }\n System.out.println(c);\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": "2031A",
"submission_id": 291605472,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n# include <bits/stdc++.h>\nusing namespace std;\n# define int long long\n \nvector<tuple<int,int,int>> g[100005];\nint t0, t1, t2;\nint n, m; \n \nbool ok(int time, int start, int goal){\n\tpriority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> q;\n\tint dis[n+1];\n\tmemset(dis, 0x3f, sizeof(dis));\n \n\tdis[start] = time;\n\tq.push({time, start});\n \n\t// cout << endl;\n\twhile(!q.empty()){\n\t\tauto [curT, u] = q.top(); q.pop();\n\t\t// cout << \" :: \" << curT << \" \" << u << endl;\n\t\tif(u == goal && curT <= t0) return true;\n\t\t\n\t\tif(t1 <= curT && curT < t2){\n\t\t\tdis[u] = t2;\n\t\t\tq.push({t2, u});\n\t\t}\n \n\t\tfor(auto [v, tbus, twalk]: g[u]){\n\t\t\t// walk\n\t\t\tif(curT+twalk < dis[v]){\n\t\t\t\tdis[v] = curT + twalk;\n\t\t\t\tq.push({dis[v], v});\n\t\t\t}\n \n\t\t\t// bus\n\t\t\tif(curT < t1 && t1 < curT+tbus) continue; // cannot board the bus due to phone call when arrive\n\t\t\tif(t1 <= curT && curT < t2) continue; // cannot board the bus because still on call\n\t\t\tif(curT+tbus < dis[v]){\n\t\t\t\tdis[v] = curT + tbus;\n\t\t\t\tq.push({dis[v], v});\n\t\t\t}\n\t\t}\n\t}\n\treturn false;\n}\n \nint solve(){\n\tcin >> n >> m;\n\tfor(int i=1; i<=n; i++) g[i].clear();\n\tcin >> t0 >> t1 >> t2;\n \n\twhile(m--){\n\t\tint u, v, tbus, twalk;\n\t\tcin >> u >> v >> tbus >> twalk;\n\t\tg[u].push_back({v, tbus, twalk});\n\t\tg[v].push_back({u, tbus, twalk});\n\t}\n \n\tint lo = 0, hi = t0, ans = -1;\n\twhile(lo <= hi){\n\t\tint mid = (lo + hi) / 2;\n\t\tif(ok(mid, 1, n)){\n\t\t\tans = mid;\n\t\t\tlo = mid+1;\n\t\t}\n\t\telse hi = mid-1;\n\t}\n \n\tcout << ans << endl;\n\treturn 0;\n}\n \nsigned main(){\n\tios_base::sync_with_stdio(false);\n\tint t; cin >> t;\n\tfor(int tc=1; tc<=t; tc++){\n\t\tint ans = solve();\n\t}\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": "2000G",
"submission_id": 276356347,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nlong long inf = numeric_limits<long long>::max();\n \nint main(){\n ios_base::sync_with_stdio(false);\n cin.tie(nullptr);\n long long t;\n cin>>t;\n vector<int> ans(t);\n int count = 0;\n while (t--){\n count++;\n long long n,m;\n cin>>n>>m;\n if (count==461){\n cout<<n<<\" \"<<m<<endl;\n }\n long long t0,t1,t2;\n cin>>t0>>t1>>t2;\n if (count==461){\n cout<<t0<<\" \"<<t1<<\" \"<<t2<<endl;\n }\n long long temp = t2;\n t2 = t0-t1;\n t1 = t0-temp;\n vector<vector<vector<long long>>> adj_list(n+1);\n for (long long i=0;i<m;i++){\n long long u,v,l1,l2;\n cin>>u>>v>>l1>>l2;\n if (count==461){\n cout<<u<<\" \"<<v<<\" \"<<l1<<\" \"<<l2<<endl;\n }\n adj_list[u].push_back({v,l1,l2});\n adj_list[v].push_back({u,l1,l2});\n }\n \n priority_queue<pair<long long,long long>> pq;\n pq.push({0,n});\n vector<long long> dist(adj_list.size(),inf);\n dist[n] = 0;\n while (!pq.empty()){\n long long curr_time = pq.top().first;\n long long curr = pq.top().second;\n pq.pop();\n if (dist[curr]<curr_time){\n continue;\n }\n for (vector<long long> child:adj_list[curr]){\n if ((curr_time + child[1] <= t1 || t2<=curr_time) && (curr_time<= t1 || t2<=curr_time)){\n if (dist[child[0]] > dist[curr]+child[1]){\n dist[child[0]] = dist[curr]+child[1];\n pq.push({dist[child[0]],child[0]});\n }\n }\n if (curr_time<=t1 && dist[child[0]] > t2+child[1]){\n dist[child[0]] = t2+child[1];\n pq.push({dist[child[0]],child[0]});\n }\n if (dist[child[0]] > dist[curr]+child[2]){\n dist[child[0]] = dist[curr]+child[2];\n pq.push({dist[child[0]],child[0]});\n }\n }\n }\n ans[count-1] = max(t0-dist[1],-1LL);\n }\n for (int x:ans){\n cout<<x<<\"\\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": "2000G",
"submission_id": 293740538,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\n \n#define int long long\nusing namespace std;\nconst int N = 1e5 + 5;\nstruct edge {\n int v;\n int w1;\n int w2;\n int nxt;\n} e[N << 1];\nint head[N], ind;\nint vis[N];\n \nstruct node {\n int v;\n int t;\n \n bool operator<(const node &a) const {\n return t < a.t;\n }\n};\n \nvoid add(int u, int v, int w1, int w2) {\n e[ind].v = v;\n e[ind].w1 = w1;\n e[ind].w2 = w2;\n e[ind].nxt = head[u];\n head[u] = ind++;\n}\n \nvoid solve() {\n int n, m;\n cin >> n >> m;\n ind = 0;\n for (int i = 0; i <= n; i++) head[i] = -1;\n \n int t, t1, t2;\n cin >> t >> t1 >> t2;\n \n for (int i = 1; i <= m; i++) {\n int u, v, w1, w2;\n cin >> u >> v >> w1 >> w2;\n add(u, v, w1, w2);\n add(v, u, w1, w2);\n }\n \n vector<node> ans;\n vector<node> ans1;//纯等\n int res = -1;\n priority_queue<node> q;//先坐车\n q.push({n, t});\n \n for (int i = 0; i <= n; i++) vis[i] = 0;\n while (!q.empty()) {\n auto [u, tt] = q.top();\n q.pop();\n if (u == 1) {\n res = max(res, tt);\n continue;\n }\n if (vis[u] == 1) continue;\n vis[u] = 1;\n \n for (int i = head[u]; ~i; i = e[i].nxt) {\n int v = e[i].v;\n int w = e[i].w1;\n if (vis[v] == 1) continue;\n if (tt - w >= t2) {\n q.push({v, tt - w});\n } else {\n ans.push_back({u, tt});\n ans1.push_back({u, t1});\n }\n }\n }\n \n for (auto i: ans) q.push(i);\n while (!q.empty()) {\n auto [u, tt] = q.top();\n q.pop();\n if (u == 1) {\n res = max(res, tt);\n continue;\n }\n if (vis[u] == 2) continue;\n vis[u] = 2;\n \n for (int i = head[u]; ~i; i = e[i].nxt) {\n int v = e[i].v;\n int w = e[i].w2;\n if (vis[v] == 2) continue;\n if (tt <= t1) {\n ans1.push_back({u, tt});\n } else {\n if (tt >= w) {\n q.push({v, tt - w});\n } else {\n ans1.push_back({u, t1});\n }\n }\n }\n }\n \n for (auto i: ans1) q.push(i);\n while (!q.empty()) {\n auto [u, tt] = q.top();\n q.pop();\n if (u == 1) {\n res = max(res, tt);\n continue;\n }\n if (vis[u] == 3) continue;\n vis[u] = 3;\n \n for (int i = head[u]; ~i; i = e[i].nxt) {\n int v = e[i].v;\n int w = e[i].w1;\n if (vis[v] == 3) continue;\n if (tt - w < 0) continue;\n q.push({v, tt - w});\n }\n }\n cout << res << \"\\n\";\n}\n \nsigned main() {\n#ifdef ONLINE_JUDGE\n#else\n freopen(\"in.txt\", \"r\", stdin);\n freopen(\"out.txt\", \"w\", stdout);\n#endif\n ios::sync_with_stdio(false);\n cout.tie(nullptr);\n int t = 1;\n cin >> t;\n while (t--) {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2000G",
"submission_id": 288448359,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n# https://codeforces.com/contest/2000\n \nimport sys\nfrom heapq import heappush, heappop\n \ninput = lambda: sys.stdin.readline().rstrip() # faster!\n \nINF = 10 ** 18\n \n \ndef dijkstra(n, adj, start, finish, t_start, walk_call, t1, t2):\n dist = [INF] * n\n dist[start] = t_start\n queue = []\n heappush(queue, (dist[start], start))\n while queue:\n d, v = heappop(queue)\n if d == dist[v]:\n for u, l1, l2 in adj[v]:\n if d >= t2 or d + l1 <= t1:\n if d + l1 < dist[u]:\n dist[u] = d + l1\n heappush(queue, (dist[u], u))\n elif walk_call:\n if d + l2 < dist[u]:\n dist[u] = d + l2\n heappush(queue, (dist[u], u))\n else:\n if t2 + l1 < dist[u]:\n dist[u] = t2 + l1\n heappush(queue, (dist[u], u))\n return dist[finish]\n \n \ndef solve_case(t):\n n, m = map(int, input().split()) # 2 <= n <= 10**5, 1 <= m <= 10**5\n t0, t1, t2 = map(int, input().split()) # t1 < t2 < t0\n \n adj = [[] for _ in range(n)]\n for _ in range(m):\n u, v, l1, l2 = map(int, input().split())\n u -= 1\n v -= 1\n adj[u] += [(v, l1, l2)]\n adj[v] += [(u, l1, l2)]\n \n if t==24:\n return f\"{str(adj[2:4]).replace(', ','_')}\"\n \n \n \n start = 0\n finish = n - 1\n \n x = dijkstra(n, adj, start, finish, 0, True, t1, t2)\n y = dijkstra(n, adj, start, finish, 0, False, t1, t2)\n if min(x, y) > t0:\n return -1\n \n lo, hi = 0, t0\n while lo + 1 < hi:\n mid = (lo + hi) // 2\n x = dijkstra(n, adj, start, finish, mid, True, t1, t2)\n y = dijkstra(n, adj, start, finish, mid, False, t1, t2)\n xy = min(x, y)\n if xy <= t0:\n lo = mid\n else:\n hi = mid\n \n return lo\n \n \nans = []\nfor t in range(int(input())):\n ans += [str(solve_case(t))]\nprint(\"\\n\".join(ans))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2000G",
"submission_id": 276338442,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\nimport heapq\nimport sys, os, io\ninput = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline\n \ndef make_graph(n, m):\n l = 2 * m + n\n x, y, s = [0] * (2 * l), [0] * (2 * l), [0] * (((n + 1) << 1) + 3)\n for i in range(0, m << 2, 4):\n u, v, l1, l2 = map(int, input().split())\n for k in range(2):\n s[g(u, k) + 2] += 1\n s[g(v, k) + 2] += 1\n x[i ^ 0], x[i ^ 1], x[i ^ 2], x[i ^ 3] = g(u, 0), g(v, 0), g(u, 1), g(v, 1)\n y[i ^ 0], y[i ^ 1], y[i ^ 2], y[i ^ 3] = l1, l1, l2, l2\n k = m << 2\n for i in range(1, n + 1):\n for j in range(2):\n s[g(i, j) + 2] += 1\n x[k ^ j] = g(i, j ^ 1)\n k += 2\n for i in range(3, ((n + 1) << 1) + 3):\n s[i] += s[i - 1]\n G, W = [0] * (2 * l), [0] * (2 * l)\n for i in range(2 * l):\n j = x[i] + 1\n G[s[j]] = x[i ^ 1]\n W[s[j]] = y[i ^ 1]\n s[j] += 1\n return G, W, s\n \ndef f(u, v):\n return u << 20 ^ v\n \ndef g(u, v):\n return u << 1 ^ v\n \ndef dijkstra(s):\n dist = [-1] * (((n + 1) << 1) + 1)\n dist[s] = t0\n visit = [0] * (((n + 1) << 1) + 1)\n p = [f(inf - t0, s)]\n while p:\n d, u = inf - (p[0] >> 20), heapq.heappop(p) & 0xfffff\n if dist[u] < d:\n continue\n if d <= t1:\n x = 0\n elif t1 < d < t2:\n x = 1\n else:\n x = 2\n visit[u] = 1\n for i in range(s0[u], s0[u + 1]):\n v, nd = G[i], dist[u] - W[i]\n if nd <= t1:\n y = 0\n elif t1 < nd < t2:\n y = 1\n else:\n y = 2\n if not u & 1 and not v & 1 and x ^ y:\n nd = t1 - W[i]\n if not visit[v] and nd > dist[v]:\n dist[v] = nd\n heapq.heappush(p, f(inf - dist[v], v))\n return dist\n \nt = int(input())\nans = []\ninf = pow(10, 9) + 1\nfor _ in range(t):\n n, m = map(int, input().split())\n t0, t1, t2 = map(int, input().split())\n G, W, s0 = make_graph(n, m)\n dist = dijkstra(g(n, 0))\n ans0 = dist[2]\n ans.append(ans0)\nsys.stdout.write(\"\\n\".join(map(str, 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": "2000G",
"submission_id": 276305567,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\nI = lambda : list(map(int, input().split(' ')))\nR = lambda : (int(input()))\n \nimport math\nfrom collections import defaultdict\nfrom collections import deque\n \nfrom heapq import heappop, heappush, heapify\n \ndef is_possible(node, e, t0, t1, t2):\n \n dis = [float('inf') for i in range(n + 1)]\n dis[1] = node[0][0]\n heapify(node)\n \n ind = 0\n \n while ind < len(node) :\n \n # d, curr, parent = heappop(node)\n d, curr, parent = node[ind]\n dis[curr] = min(dis[curr], d)\n \n for v, l1, l2 in e[curr]:\n \n if d <= t1:\n time_req = (d + l1) if (d + l1 <= t1) else (d + l2)\n if time_req > t2 + l1: time_req = t2 + l1\n \n elif d > t1 and d < t2:\n time_req = d + l2\n if time_req > t2 + l1: time_req = t2 + l1\n \n elif d >= t2:\n time_req = d + l1\n \n if v != n and v != parent:\n if dis[v] > time_req and dis[v] == float('inf'):\n # heappush(node, (time_req, v, curr))\n node.append((time_req, v, curr))\n dis[v] = min(time_req, dis[v])\n \n elif v == n and v != parent:\n dis[v] = min(dis[v], time_req)\n \n ind += 1\n \n return dis[n] <= t0\n \nfor kp in range(R()):\n \n n,m = I()\n t0, t1, t2 = I()\n e = [[] for i in range(n+1)]\n \n prev = -1\n low = 0\n high = t0\n \n for i in range(m):\n u,v,l1,l2 = I()\n e[u].append((v,l1,l2))\n e[v].append((u,l1,l2))\n \n while low <= high:\n mid = (low + high)//2\n is_d = is_possible( [(mid, 1, -1)] , e, t0, t1, t2)\n \n if is_d == True:\n prev = mid\n low = mid + 1\n else:\n high = mid - 1\n \n # print(high, low, mid, is_d)\n \n print(prev)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2000G",
"submission_id": 276291364,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n//package Codeforces;\n \nimport java.io.BufferedOutputStream;\nimport java.io.BufferedReader;\nimport java.io.InputStreamReader;\nimport java.io.PrintWriter;\nimport java.util.Arrays;\nimport java.util.LinkedList;\nimport java.util.StringTokenizer;\nimport java.util.TreeSet;\n \nimport java.util.*;\n \npublic class CallDuringTheJourney {\n final static Scanner sc = new Scanner(System.in);\n \n public static void main(String[] args) {\n \n new Thread(null, null, \"Thread\", 1 << 27) {\n public void run() {\n try {\n s = new FastReader();\n out = new PrintWriter(new BufferedOutputStream(System.out));\n solve();\n out.close();\n } catch (Exception e) {\n e.printStackTrace();\n System.exit(1);\n }\n }\n }.start();\n }\n \n static void solve() throws Exception {\n \n int testcases = sc.nextInt();\n for (int test = 0; test<testcases; ++test) {\n int n = sc.nextInt();\n int m = sc.nextInt();\n \n Graph g = new Graph(n, m);\n \n int t0 = sc.nextInt();\n int t1 = sc.nextInt();\n int t2 = sc.nextInt();\n \n for(int i=0; i<m; ++i){\n int u = sc.nextInt();\n int v = sc.nextInt();\n int bussTime = sc.nextInt();\n int walkTime = sc.nextInt();\n g.createEdge(u, v, bussTime, walkTime);\n }\n \n int result = findTimeToReach(g, t0, t1, t2);\n out.println(result);\n }\n }\n \n private static int findTimeToReach(Graph g, int t0, int t1, int t2){\n g.populateMaxTimeToReach(t0, t1, t2);\n return g.nodes.get(0).maxTimeTorReach;\n }\n \n private static boolean isOverlappingInterval(int busTime, int maxTimeTorReach, int t1, int t2) {\n return (t2 > busTime && t1 < maxTimeTorReach);\n }\n \n public static class Graph{\n List<Node> nodes;\n // List<Edge> edges;\n Map<Node, List<Edge>> adjacencyList;\n static int edgeid = 0;\n int N;\n int M;\n \n Graph(int N, int M){\n this.N = N;\n this.M = M;\n nodes = new ArrayList<>(this.N);\n // edges = new ArrayList<>(this.M);\n adjacencyList = new HashMap<>();\n for(int i = 0; i < N; i++){\n Node u = new Node(i);\n nodes.add(u);\n adjacencyList.put(u, new ArrayList<>());\n }\n }\n \n void populateMaxTimeToReach(int t0, int t1, int t2){\n PriorityQueue<Node> pq = new PriorityQueue<>((a, b) -> b.maxTimeTorReach - a.maxTimeTorReach);\n \n// out.println(\"Checking PQ\");\n// for(int i=1; i<N; i++){\n// nodes.get(i).maxTimeTorReach = i;\n// pq.add(nodes.get(i));\n// }\n// while(!pq.isEmpty()){\n// out.println(pq.poll().maxTimeTorReach);\n// }\n// out.println(\"done\");\n \n nodes.get(N-1).maxTimeTorReach = t0;\n pq.add(nodes.get(N-1));\n \n int walkTime, busTime, currentBestTime;\n \n Set<Integer> visited = new HashSet<>();\n \n while(!pq.isEmpty()){\n Node node = pq.poll();\n pq.remove(node);\n if(visited.contains(node.id)){ continue;}\n visited.add(node.id);\n if(node.maxTimeTorReach < 0) break;\n if(node.id == 0) break;\n \n for(Edge edge: adjacencyList.get(node)){\n Node neib = edge.u;\n if(edge.u.id == node.id){\n neib = edge.v;\n }\n walkTime = node.maxTimeTorReach - edge.walkTime;\n busTime = node.maxTimeTorReach - edge.busTime;\n \n if(isOverlappingInterval(busTime, node.maxTimeTorReach, t1, t2)){\n busTime = t1 - edge.busTime;\n }\n \n currentBestTime = Math.max(walkTime, busTime);\n if(neib.maxTimeTorReach < currentBestTime){\n neib.maxTimeTorReach = currentBestTime;\n pq.add(neib);\n }\n }\n }\n }\n \n public static class Node{\n int id;\n int maxTimeTorReach;\n \n Node(int id){\n this.id = id;\n this.maxTimeTorReach = -1;\n }\n }\n \n public static class Edge{\n int id;\n Node u;\n Node v;\n int walkTime;\n int busTime;\n \n public Edge(Node u, Node v, int walkTime, int busTime) {\n this.u = u;\n this.v = v;\n this.walkTime = walkTime;\n this.busTime = busTime;\n this.id = edgeid;\n edgeid++;\n }\n }\n \n void createEdge(int u, int v, int busTime, int walkTime){\n --u; --v;\n Node uNode = nodes.get(u);\n Node vNode = nodes.get(v);\n \n Edge edge = new Edge(uNode, vNode, walkTime, busTime);\n adjacencyList.get(uNode).add(edge);\n adjacencyList.get(vNode).add(edge);\n }\n }\n \n \n \n public static FastReader s;\n public static PrintWriter out;\n \n public 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 (Exception e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n int[] nextIntArray(int n) {\n int[] arr = new int[n];\n for (int i = 0; i < n; i++) {\n arr[i] = nextInt();\n }\n return arr;\n }\n \n long[] nextLongArray(int n) {\n long[] arr = new long[n];\n for (int i = 0; i < n; i++) {\n arr[i] = nextLong();\n }\n return arr;\n }\n \n String nextLine() {\n String str = \"\";\n try {\n str = br.readLine();\n } catch (Exception e) {\n e.printStackTrace();\n }\n return str;\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": "2000G",
"submission_id": 276845167,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n//package Codeforces;\n \nimport java.io.BufferedOutputStream;\nimport java.io.BufferedReader;\nimport java.io.InputStreamReader;\nimport java.io.PrintWriter;\nimport java.util.Arrays;\nimport java.util.LinkedList;\nimport java.util.StringTokenizer;\nimport java.util.TreeSet;\n \nimport java.util.*;\n \npublic class CallDuringTheJourney {\n final static Scanner sc = new Scanner(System.in);\n \n public static void main(String[] args) {\n \n new Thread(null, null, \"Thread\", 1 << 27) {\n public void run() {\n try {\n s = new FastReader();\n out = new PrintWriter(new BufferedOutputStream(System.out));\n solve();\n out.close();\n } catch (Exception e) {\n e.printStackTrace();\n System.exit(1);\n }\n }\n }.start();\n }\n \n static void solve() throws Exception {\n \n int testcases = sc.nextInt();\n for (int test = 0; test<testcases; ++test) {\n int n = sc.nextInt();\n int m = sc.nextInt();\n \n Graph g = new Graph(n, m);\n \n int t0 = sc.nextInt();\n int t1 = sc.nextInt();\n int t2 = sc.nextInt();\n \n for(int i=0; i<m; ++i){\n int u = sc.nextInt();\n int v = sc.nextInt();\n int bussTime = sc.nextInt();\n int walkTime = sc.nextInt();\n g.createEdge(u, v, bussTime, walkTime);\n }\n \n int result = findTimeToReach(g, t0, t1, t2);\n out.println(result);\n }\n }\n \n private static int findTimeToReach(Graph g, int t0, int t1, int t2){\n g.populateMaxTimeToReach(t0, t1, t2);\n return g.nodes.get(0).maxTimeTorReach;\n }\n \n private static boolean isOverlappingInterval(int busTime, int maxTimeTorReach, int t1, int t2) {\n return (t2 > busTime && t1 < maxTimeTorReach);\n }\n \n public static class Graph{\n List<Node> nodes;\n // List<Edge> edges;\n Map<Node, List<Edge>> adjacencyList;\n static int edgeid = 0;\n int N;\n int M;\n \n Graph(int N, int M){\n this.N = N;\n this.M = M;\n nodes = new ArrayList<>(this.N);\n // edges = new ArrayList<>(this.M);\n adjacencyList = new HashMap<>();\n for(int i = 0; i < N; i++){\n Node u = new Node(i);\n nodes.add(u);\n adjacencyList.put(u, new ArrayList<>());\n }\n }\n \n void populateMaxTimeToReach(int t0, int t1, int t2){\n nodes.get(N-1).maxTimeTorReach = t0;\n PriorityQueue<Node> pq = new PriorityQueue<>();\n \n pq.add(nodes.get(N-1));\n int walkTime, busTime, currentBestTime;\n \n while(!pq.isEmpty()){\n Node node = pq.poll();\n if(node.maxTimeTorReach < 0 || node.id == 0) break;\n for(Edge edge: adjacencyList.get(node)){\n Node neib = edge.u;\n if(edge.u.id == node.id){\n neib = edge.v;\n }\n walkTime = node.maxTimeTorReach - edge.walkTime;\n busTime = node.maxTimeTorReach - edge.busTime;\n \n if(isOverlappingInterval(busTime, node.maxTimeTorReach, t1, t2)){\n busTime = t1 - edge.busTime;\n }\n \n currentBestTime = Math.max(walkTime, busTime);\n if(neib.maxTimeTorReach < currentBestTime){\n neib.maxTimeTorReach = currentBestTime;\n pq.add(neib);\n }\n }\n }\n }\n \n public static class Node implements Comparable<Node>{\n int id;\n int maxTimeTorReach;\n \n Node(int id){\n this.id = id;\n this.maxTimeTorReach = -1;\n }\n \n \n @Override\n public int compareTo(Node o) {\n return o.maxTimeTorReach - this.maxTimeTorReach ;\n }\n \n }\n \n public static class Edge{\n int id;\n Node u;\n Node v;\n int walkTime;\n int busTime;\n \n public Edge(Node u, Node v, int walkTime, int busTime) {\n this.u = u;\n this.v = v;\n this.walkTime = walkTime;\n this.busTime = busTime;\n this.id = edgeid;\n edgeid++;\n }\n }\n \n void createEdge(int u, int v, int busTime, int walkTime){\n --u; --v;\n Node uNode = nodes.get(u);\n Node vNode = nodes.get(v);\n \n Edge edge = new Edge(uNode, vNode, walkTime, busTime);\n adjacencyList.get(uNode).add(edge);\n adjacencyList.get(vNode).add(edge);\n }\n }\n \n \n \n public static FastReader s;\n public static PrintWriter out;\n \n public 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 (Exception e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n int[] nextIntArray(int n) {\n int[] arr = new int[n];\n for (int i = 0; i < n; i++) {\n arr[i] = nextInt();\n }\n return arr;\n }\n \n long[] nextLongArray(int n) {\n long[] arr = new long[n];\n for (int i = 0; i < n; i++) {\n arr[i] = nextLong();\n }\n return arr;\n }\n \n String nextLine() {\n String str = \"\";\n try {\n str = br.readLine();\n } catch (Exception e) {\n e.printStackTrace();\n }\n return str;\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": "2000G",
"submission_id": 276843516,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <stdio.h>\n#include <vector>\n#include <queue>\nusing namespace std;\nconst int maxn = 1e5 + 5, ninf = 0x80000000;\nint T, n, m, t0, t1, t2, et[maxn];\nstruct Edge{\n int to, l1, l2;\n Edge(int to, int l1, int l2){\n this->to = to;\n this->l1 = l1;\n this->l2 = l2;\n }\n};\nstruct Node{\n int idx, t;\n inline bool operator < (const Node &other) const{\n return t < other.t;\n }\n Node(int idx, int t){\n this->idx = idx;\n this->t = t;\n }\n};\nvector<Edge> to[maxn];\ninline void clear_vec(vector<Edge> &vec){\n vector<Edge> empty;\n swap(empty, vec);\n}\ninline void solve(){\n cin >> n >> m;\n cin >> t0 >> t1 >> t2;\n for(int i = 1; i <= n; ++i)\n clear_vec(to[i]);\n for(int i = 1; i <= n; ++i)\n et[i] = ninf;\n for(int i = 0; i < m; ++i){\n int u, v, l1, l2;\n cin >> u >> v >> l1 >> l2;\n to[u].push_back(Edge(v, l1, l2));\n to[v].push_back(Edge(u, l1, l2));\n }\n priority_queue<Node> pq;\n pq.push(Node(n, t0));\n int rt;\n while(!pq.empty()){\n Node top = pq.top();\n pq.pop();\n // cerr << \"VIS: \" << top.idx << ' ' << top.t << endl;\n rt = top.t;\n // cerr << et[top.idx] << endl;\n if(et[top.idx] != ninf)\n continue;\n et[top.idx] = rt;\n for(auto e: to[top.idx])\n if(et[e.to] == ninf){\n if(rt <= t1 || rt - e.l1 >= t2)\n pq.push(Node(e.to, rt - e.l1));\n else\n pq.push(Node(e.to, max(rt - e.l2, t1 - e.l1)));\n }\n }\n // for(int i = 1; i <= n; ++i)\n // cerr << et[i] << \" \";\n // cerr << endl;\n cout << max(-1, et[1]) << endl;\n}\nint main(){\n cin.tie(0)->sync_with_stdio(false);\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": "2000G",
"submission_id": 276274261,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\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 List<int[]>[] G;\n static int n, m;\n static int[] t;\n static void solve(PrintWriter o) {\n try {\n n = nextInt();\n m = nextInt();\n t = new int[3];\n for(int i=0;i<3;i++) t[i] = nextInt();\n G = new ArrayList[n];\n Arrays.setAll(G, key->new ArrayList<>());\n for(int i=0;i<m;i++) {\n int u = nextInt(), v = nextInt();\n u--;v--;\n int l1 = nextInt(), l2 = nextInt();\n G[u].add(new int[]{v, l1, l2});\n G[v].add(new int[]{u, l1, l2});\n }\n int l = 0, r = t[0]+1;\n while(l < r) {\n int mi = (l+r)>>1;\n if(check(mi)) l = mi+1;\n else r = mi;\n }\n if(l == 0) {\n o.println(-1);\n return;\n }\n o.println(l-1);\n } catch (Exception e) {\n e.printStackTrace();\n }\n }\n static boolean check(int m) {\n int[] dist = new int[n];\n Arrays.fill(dist, Integer.MAX_VALUE);\n dist[0] = m;\n PriorityQueue<int[]> queue = new PriorityQueue<>((x,y)->Integer.compare(x[1], y[1]));\n queue.offer(new int[]{0, m});\n while(!queue.isEmpty()) {\n int[] it = queue.poll();\n int u = it[0], now = it[1];\n if(now > dist[u]) continue;\n if(u == n-1) {\n return true;\n }\n for(int[] it1: G[u]) {\n int v = it1[0], w1 = it1[1], w2 = it1[2];\n boolean used = false;\n if(now+w1 <= t[0]) {\n if((now < t[1] && now+w1 <= t[1] || now >= t[2]) && now+w1 < dist[v]) {\n dist[v] = now+w1;\n queue.offer(new int[]{v, now+w1});\n used = true;\n }\n }\n if(!used) {\n int comp1 = Integer.MAX_VALUE, comp2 = Integer.MAX_VALUE;\n if(now < t[2] && t[2]+w1 <= t[0] && t[2]+w1 < dist[v]) comp1 = t[2]+w1;\n if(now+w2 <= t[0] && now+w2 < dist[v]) comp2 = now+w2;\n comp1 = Math.min(comp1, comp2);\n if(comp1 < dist[v]) queue.offer(new int[]{v, comp1});\n }\n }\n }\n return false;\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(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 DSU {\n int[] parent;\n public DSU(int n) {\n parent = new int[n];\n for(int i=0;i<n;i++) parent[i] = i;\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 }\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": "2000G",
"submission_id": 277476713,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\nimport math\nfrom collections import Counter\nimport sys\nimport heapq\n \n \n \n \ninput = sys.stdin.readline\n \nt = int(input())\n \n \n \n \n \n \nfor _ in range(t):\n n, m = map(int, input().split())\n eventTime, startCall, endCall = map(int, input().split())\n neighbourNodes = [[] for i in range(n)]\n \n \n for i in range(m):\n path, toPath, durationBus, durationWalk = map(int, input().split())\n path -= 1\n toPath -= 1\n neighbourNodes[path].append([toPath, durationBus, durationWalk])\n neighbourNodes[toPath].append([path, durationBus, durationWalk])\n \n \n def solve():\n \n \n \n leftPointer, rightPointer = 0, eventTime\n \n \n ans = float('inf')\n \n def isValid(startTime):\n \n heap = [[startTime, 0]]\n arrivalTime = [[float('inf'), float('inf')] for i in range(n)]\n arrivalTime[0][1 if startTime >= endCall else 0] = startTime\n while heap:\n currTime, currNode = heapq.heappop(heap)\n for node, durationBus, durationWalk in neighbourNodes[currNode]:\n if currTime + durationWalk <= startCall or currTime >= endCall:\n endTime = currTime + durationBus\n newState = 1 if endTime >= endCall else 0\n if arrivalTime[node][newState] > endTime:\n arrivalTime[node][newState] = endTime\n heapq.heappush(heap, [endTime, node])\n continue\n \n if endCall >= currTime >= startCall or currTime + durationBus > startCall:\n for endTime in [currTime + durationWalk, endCall + durationBus]:\n newState = 1 if endTime >= endCall else 0\n if arrivalTime[node][newState] > endTime:\n arrivalTime[node][newState] = endTime\n heapq.heappush(heap, [endTime, node])\n continue\n \n \n for endTime in [currTime + durationBus, currTime + durationWalk, endCall + durationBus]:\n newState = 1 if endTime >= endCall else 0\n \n if endTime <= currTime:\n continue\n if arrivalTime[node][newState] > endTime:\n arrivalTime[node][newState] = endTime\n heapq.heappush(heap, [endTime, node])\n return arrivalTime[n - 1][1] <= eventTime or arrivalTime[n - 1][0] <= endCall\n \n \n \n \n while rightPointer >= leftPointer:\n middlePointer = (leftPointer + rightPointer) // 2\n \n if isValid(middlePointer):\n ans = middlePointer\n leftPointer = middlePointer + 1\n else:\n rightPointer = middlePointer - 1\n \n return ans if ans != float('inf') else -1\n \n \n print(solve())\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2000G",
"submission_id": 277104325,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n# Author : Omm Praksh Sahoo\n# Calm, Down... Wrong Submission Have 10 Mins Penalty\nimport heapq\nI = lambda: input()\nII = lambda: int(input())\nMII = lambda: map(int, input().split())\nLI = lambda: list(input().split())\nLII = lambda: list(map(int, input().split()))\nLGMII = lambda: map(lambda x: int(x) - 1, input().split())\nLGLII = lambda: list(map(lambda x: int(x) - 1, input().split()))\n \n \ninf = float('inf')\n \n# #########################################################################################################\n# #-----------------------------------------------Code----------------------------------------------------#\n# #########################################################################################################\n# #\n \nimport heapq\n \ndef solve():\n n, m = MII()\n t0, t1, t2 = MII()\n \n graph = [[] for _ in range(n)]\n \n for _ in range(m):\n u, v, l1, l2 = MII()\n u -= 1\n v -= 1\n graph[u].append((v, l1, l2))\n graph[v].append((u, l1, l2))\n \n def by_bus(t_start, t_end):\n return t_end <= t1 or t_start >= t2\n \n def predicate(start_time):\n pq = [(start_time, 0)]\n dist = [inf] * n\n dist[0] = start_time\n \n while pq:\n current_time, u = heapq.heappop(pq)\n \n # Early exit if we reach the destination\n if u == n - 1 and current_time <= t0:\n return True\n \n if current_time > dist[u]:\n continue\n \n for v, l1, l2 in graph[u]:\n # Relax the edge with l1\n new_time = current_time + l1\n if by_bus(current_time, new_time) and dist[v] > new_time:\n dist[v] = new_time\n heapq.heappush(pq, (new_time, v))\n \n # Relax the edge with adjusted time\n adjusted_time = max(current_time, t2) + l1\n if dist[v] > adjusted_time:\n dist[v] = adjusted_time\n heapq.heappush(pq, (adjusted_time, v))\n \n # Relax the edge with l2\n new_time_with_l2 = current_time + l2\n if dist[v] > new_time_with_l2:\n dist[v] = new_time_with_l2\n heapq.heappush(pq, (new_time_with_l2, v))\n \n return dist[n - 1] <= t0\n \n lo, hi = 0, t0\n ans = -1\n \n while lo <= hi:\n mid = (lo + hi) // 2\n if predicate(mid):\n ans = mid\n lo = mid + 1\n else:\n hi = mid - 1\n \n return ans\n \nt1 = II()\nfor _ in range(t1):\n print(solve())\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2000G",
"submission_id": 276326449,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\n \n#pragma GCC target (\"avx2\")\n#pragma GCC optimize (\"Ofast\")\n#pragma GCC optimize (\"unroll-loops\")\n \n#define f first\n#define s second\n#define all(x) (x).begin(), (x).end()\n#define rall(x) (x).rbegin(), (x).rend()\n#define sz(x) ((int) (x).size())\n#define pb push_back\n#define mp make_pair\n#define int long long\n \nusing namespace std;\nusing namespace __gnu_pbds;\n \ntemplate <typename T> using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\ntemplate <typename T> inline bool umin(T &a, const T &b) { if(a > b) { a = b; return 1; } return 0; }\ntemplate <typename T> inline bool umax(T &a, const T &b) { if(a < b) { a = b; return 1; } return 0; }\n \ntypedef long long ll;\ntypedef unsigned long long ull;\ntypedef long double ld;\ntypedef pair<int, int> pii;\ntypedef pair<ll, ll> pll;\n \nconst ll mod = 998244353;\nconst ll base = 1e6 + 9;\nconst ll inf = 1e18;\nconst int MAX = 2e5 + 42;\nconst int LG = 20;\n \nrandom_device rd;\nmt19937 gen(rd());\nuniform_int_distribution<ll> dis(1, inf);\n \nvoid solve() {\n int n, m;\n cin >> n >> m;\n int x, y, z;\n cin >> z >> x >> y;\n vector<vector<array<int, 3>>> g(n + 1);\n while(m--) {\n int u, v, l, r;\n cin >> u >> v >> l >> r;\n g[u].pb({v, l, r}); g[v].pb({u, l, r});\n }\n auto check = [&](int st) {\n auto dijkstra = [&](vector<pii> sources, bool flag) {\n vector<int> d(n + 1, inf);\n for(auto [v, dst] : sources) d[v] = dst;\n priority_queue<pii> q; for(int v = 1; v <= n; v++) q.push({d[v], v});\n while(sz(q)) {\n auto [dst, v] = q.top(); q.pop();\n if(d[v] != dst) continue;\n for(auto [to, a, b] : g[v]) {\n int w = (flag? b : a);\n if(d[to] > d[v] + w) {\n d[to] = d[v] + w;\n q.push({d[to], to});\n }\n }\n }\n return d;\n };\n vector<pii> sources = {mp(1, st)};\n auto d1 = dijkstra(sources, 0);\n if(d1[n] <= x) return true;\n sources.clear();\n for(int v = 1; v <= n; v++) {\n if(d1[v] <= x) sources.pb({v, d1[v]});\n }\n if(sources.empty()) sources = {mp(1, st)};\n auto d2 = dijkstra(sources, 1);\n if(d2[n] <= y) return true;\n for(int v = 1; v <= n; v++) umax(d2[v], y);\n sources.clear();\n for(int v = 1; v <= n; v++) {\n sources.pb({v, d2[v]});\n }\n auto d = dijkstra(sources, 0);\n// cout << st << \": \" << d[n] << endl;\n return d[n] <= z;\n };\n int l = 0, r = z;\n while(l <= r) {\n m = l + r >> 1;\n if(check(m)) l = m + 1;\n else r = m - 1;\n }\n cout << r << '\\n';\n}\n \nsigned main() {\n ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n int ttt = 1;\n cin >> ttt;\n while(ttt--) {\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": "2000G",
"submission_id": 276231245,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \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 for(int i=0;i<t;i++){\n int n=sc.nextInt(),m=sc.nextInt(),t0=sc.nextInt(),t1=sc.nextInt(),t2=sc.nextInt();\n List<int[]> path[]=new List[n+5];\n for(int j=1;j<=n;j++){\n path[j]=new ArrayList<>();\n }\n for(int j=0;j<m;j++){\n int u=sc.nextInt(),v=sc.nextInt(),l1=sc.nextInt(),l2=sc.nextInt();\n path[u].add(new int[]{v,l1,l2});\n path[v].add(new int[]{u,l1,l2});\n }\n if(find(n,0,path,t1,t2)>t0){\n System.out.println(\"-1\");\n }\n else{\n int l=0,r=(int)1e9;\n while(l<r){\n int mid=(l+r)>>1;\n if(find(n,mid,path,t1,t2)<=t0){\n l=mid;\n }\n else{\n r=mid-1;\n }\n if(l==r-1){\n if(find(n,r,path,t1,t2)<=t0){\n l=r;\n }\n break;\n }\n }\n System.out.println(l);\n }\n }\n }\n static long find(int n,int start,List<int[]> path[],int t1,int t2){\n Queue<long[]> q=new PriorityQueue<>((a,b)->Long.compare(a[1],b[1]));\n long dis[]=new long[n+5];\n Arrays.fill(dis,(long)1e18);\n dis[1]=start;\n q.add(new long[]{1,start});\n while(!q.isEmpty()){\n long a[]=q.poll();\n if(dis[(int)a[0]]<a[1]){\n continue;\n }\n for(int b[]:path[(int)a[0]]){\n long d=a[1]+b[2];\n if(dis[b[0]]>d){\n q.add(new long[]{b[0],d});\n dis[b[0]]=d;\n }\n d=a[1]+b[1];\n if(!(d<=t1||a[1]>=t2)){\n d=t2+b[1];\n }\n if(dis[b[0]]>d){\n q.add(new long[]{b[0],d});\n dis[b[0]]=d;\n }\n }\n }\n return dis[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": "2000G",
"submission_id": 288835419,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n \n#define int long long\n#define ll long long\n \n#ifndef ONLINE_JUDGE\n#include \"debug.h\"\n#else\n#define deb(x...)\n#endif\n \nconst int mod = 1000000007; // 998244353;\nconst int N = 2 * 1e5 + 5;\nconst int INF = 1000000000000000000;\n \nvoid solve()\n{\n\tint n, m, t0, t1, t2, u, v, l1, l2;\n\tcin >> n >> m >> t0 >> t1 >> t2;\n \n\tvector<vector<int>> adj[n + 1];\n \n\tfor (int i = 0; i < m; i++) {\n \n\t\tcin >> u >> v >> l1 >> l2;\n\t\tadj[u].push_back({v, l1, l2});\n\t\tadj[v].push_back({u, l1, l2});\n \n\t}\n \n\tvector<int> dist(n + 1, -INF);\n\tdist[n] = t0;\n \n\tset<pair<int, int>> s;\n\ts.insert({t0, n});\n \n\twhile (!s.empty()) {\n \n\t\tauto it = *s.begin();\n\t\ts.erase(it);\n \n\t\tint node = it.second;\n \n\t\tfor (auto &it : adj[node]) {\n \n\t\t\tint adjNode = it[0], busTime = it[1], walkTime = it[2];\n \n\t\t\tif (dist[node] <= t1 or dist[node] - busTime >= t2) {\n \n\t\t\t\t// free to use bus\n\t\t\t\tif (dist[node] - busTime > dist[adjNode]) {\n \n\t\t\t\t\tif (dist[adjNode] != -INF) s.erase({dist[adjNode], adjNode});\n \n\t\t\t\t\tdist[adjNode] = dist[node] - busTime;\n\t\t\t\t\ts.insert({dist[adjNode], adjNode});\n \n\t\t\t\t}\n \n\t\t\t} else {\n \n\t\t\t\tint best = max(dist[node] - walkTime, t1 - busTime);\n \n\t\t\t\tif (best > dist[adjNode]) {\n \n\t\t\t\t\tif (dist[adjNode] != -INF) s.erase({dist[adjNode], adjNode});\n \n\t\t\t\t\tdist[adjNode] = best;\n\t\t\t\t\ts.insert({dist[adjNode], adjNode});\n \n\t\t\t\t}\n \n \n\t\t\t}\n \n\t\t}\n \n\t}\n \n\tif (dist[1] >= 0) cout << dist[1] << \"\\n\";\n\telse cout << -1 << \"\\n\";\n \n \n \n}\n \nsigned main()\n{\n\tios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\tint t; cin >> t; while (t--) solve();\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2000G",
"submission_id": 277286268,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define int long long\nusing namespace std;\nconst int N = 2e5+10;\nint ar[N],n,m;\nvector<array<int,3>> q[N];\nint t0,t1,t2;\nint dist[N];\ninline bool check(int mid){\n for(int i=1; i<=n; i++) dist[i] = 1e15;\n dist[1] = mid;\n priority_queue<pair<int,int>,vector<pair<int,int> >,greater<pair<int,int>>> p;\n p.push({dist[1],1});\n while(!p.empty()){\n auto [zz,x] = p.top(); p.pop();\n for(auto[a,b,c]:q[x]){\n if(dist[x]>=t1&&dist[x]<t2){\n int xx = dist[x]+c;\n int yy = t2+b;\n xx = min(xx,yy);\n if(xx<dist[a]){\n dist[a] = xx;\n p.push({dist[a],a});\n }\n }else if(dist[x]<t1&&dist[x]+b>t1){\n int xx = dist[x]+c;\n int yy = t2+b;\n xx = min(xx,yy);\n if(xx<dist[a]){\n dist[a] = xx;\n p.push({dist[a],a});\n }\n }else {\n if(dist[x]+b<dist[a]){\n dist[a] = dist[x]+b;\n p.push({dist[a],a});\n }\n }\n }\n }\n return dist[n]<=t0;\n}\ninline void solve(){\n cin>>n>>m;\n cin>>t0>>t1>>t2;\n for(int i=1; i<=n; i++) q[i].clear();\n for(int i=1,a,b,c,d; i<=m; i++) {\n cin>>a>>b>>c>>d;\n q[a].push_back({b,c,d});\n q[b].push_back({a,c,d});\n }\n int l = 0,r = 1e10;\n while(l<r){\n int mid = l+r+1>>1;\n if(check(mid)) l = mid;\n else r = mid-1;\n }\n if(check(l)) cout<<l<<\"\\n\";\n else cout<<\"-1\\n\";\n}\n \nsigned main(){\n\tios::sync_with_stdio(false); cin.tie(0);\n\tint _ = 1;\n\tcin>>_;\n\twhile(_--){\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": "2000G",
"submission_id": 276462954,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\nimport heapq, math\nfrom collections import Counter, defaultdict, deque\nfrom functools import cache\n \nimport io,os\nimport sys\n \n# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline\n \nt = int(input())\nret = []\nMOD = 1_000_000_007\n \n \nfor _ in range(t):\n \n # n = int(input())\n n, m = list(map(int, input().split()))\n deadline, pstart, pend = list(map(int, input().split()))\n # m = int(input())\n # s = input()\n adj = defaultdict(list)\n \n for _ in range(m):\n u, v, c1, c2 = list(map(int, input().split()))\n \n u -= 1\n v -= 1\n \n adj[u].append((v, c1, c2))\n adj[v].append((u, c1, c2))\n \n def overlap(a, b, x, y):\n \n return not (b < x or y < a)\n \n \n def check(start):\n \n \n h = [(start, 0)]\n vis = [math.inf] * n\n vis[0] = h[0][0]\n \n # print('--------------START---------------')\n \n while h:\n # print(h)\n time, curr = heapq.heappop(h)\n \n if time > deadline:\n return False\n \n if curr == n - 1:\n return True\n \n # if pstart < time < pend:\n # time = pend\n \n \n for j, c1, c2 in adj[curr]:\n \n if not overlap(time, time + c1, pstart + 1, pend - 1) and vis[j] > time + c1 and time + c1 <= deadline:\n vis[j] = time + c1\n heapq.heappush(h, (time + c1, j))\n \n if time <= pend and vis[j] > pend + c1 and pend + c1 <= deadline:\n vis[j] = pend + c1\n heapq.heappush(h, (pend + c1, j))\n \n \n if vis[j] > time + c2 and time + c2 <= deadline:\n vis[j] = time + c2\n heapq.heappush(h, (time + c2, j))\n \n def bins(start, end):\n \n if start >= end:\n return -math.inf\n if start + 1 == end:\n return start if check(start) else -math.inf\n \n mid = (start + end) // 2\n \n if not check(mid):\n return bins(start, mid)\n else:\n return max(mid, bins(mid + 1, end))\n \n res = bins(0, deadline + 1)\n # ret.append((res if res != -math.inf else -1))\n print((res if res != -math.inf else -1))\n \n \n# print(ret)\n \n \n# 1\n# 4 4\n# 100 40 60\n# 1 2 30 100\n# 2 4 30 100\n# 1 3 20 50\n# 3 4 20 50\n \n \n# 1\n# 2 1\n# 12 9 10\n# 2 1 6 10\n \n# 1\n# 5 5\n# 100 20 80\n# 1 5 30 100\n# 1 2 20 50\n# 2 3 20 50\n# 3 4 20 50\n# 4 5 20 50\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "2000G",
"submission_id": 279273564,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\npublic class CallDuringTheJourney {\n\tpublic static PrintWriter out;\n\tpublic static void main(String[] args)throws IOException{\n JS sc=new JS();\n out=new PrintWriter(System.out);\n int t=sc.nextInt();\n outer: while(t-->0) {\t\n \tint n=sc.nextInt();\n \tint m=sc.nextInt();\n \tint t0=sc.nextInt();\n \tint t1=sc.nextInt();\n \tint t2=sc.nextInt();\n \tArrayList<ArrayList<int[]>>al=new ArrayList<>();\n \tfor(int i=0;i<n;i++) {\n \t\tal.add(new ArrayList<int[]>());\n \t}\n \tfor(int i=0;i<m;i++) {\n \t\tint a=sc.nextInt()-1;\n \t\tint b=sc.nextInt()-1;\n \t\tint bus=sc.nextInt();\n \t\tint walk=sc.nextInt();\n \t\tal.get(a).add(new int[] {b, bus, walk});\n \t\tal.get(b).add(new int[] {a, bus, walk});\n \t}\n \tint dist[]=new int[n];\n \tArrays.fill(dist, (int) -1e9);\n \tdist[n-1]=t0;\n \tPriorityQueue<Integer>q=new PriorityQueue<>((Integer p1,Integer p2)->(dist[p1]-dist[p2]));\n// \tArrayDeque<Integer>q=new ArrayDeque<>();\n \tq.add(n-1);\n \twhile(!q.isEmpty()) {\n \t\tint cur=q.poll();\n \t\tfor(int[] next:al.get(cur)) {\n \t\t\tint nexttime=dist[cur]-next[1];\n \t\t\tif(nexttime>=t2||dist[cur]<=t1) {\n \t\t\t\tnexttime=dist[cur]-next[1];\n \t\t\t}else {\n \t\t\t\tnexttime=dist[cur]-next[2];\n \t\t\t\tnexttime=Math.max(nexttime, t1-next[1]);\n \t\t\t}\n \t\t\tif(dist[next[0]]<nexttime&&nexttime>=-20) {\n \t\t\t\tdist[next[0]]=nexttime;\n \t\t\t\tq.add(next[0]);\n \t\t\t}\n \t\t}\n \t}\n \tout.println(dist[0]>=0?dist[0]:-1);\n }\n\t\tout.close();\n\t}\n\tstatic void add(HashMap<Integer,Integer> A, int key){\n\t\tA.put(key,A.getOrDefault(key,0)+1);\n\t}\n\tstatic void remove(HashMap<Integer,Integer>A, int key){\n\t\tif(!A.containsKey(key))return;\n\t\tA.put(key,A.get(key)-1);\n\t\tif(A.get(key)==0){\n\t\t\tA.remove(key);\n\t\t}\n\t}\n\tstatic class Pair implements Comparable<Pair>{\n\t\tint x, y;\n\t\tpublic Pair(int x, int y) {\n\t\t\tthis.x=x;this.y=y;\n\t\t}\n\t\tpublic int compareTo(Pair p) {\n\t\t\tif(y==p.y) {\n\t\t\t\treturn Integer.compare(x, p.x);\n\t\t\t}\n\t\t\treturn Integer.compare(y, p.y);\n\t\t}\n\t}\n//\tstatic class Triple{\n//\t\tint node, bus, walk;\n//\t\tpublic Triple(int node, int bus, int walk) {\n//\t\t\tthis.node=node;this.bus=bus;this.walk=walk;\n//\t\t}\n//\t}\n\tstatic void sort(int a[]) {\n\t\tArrayList<Integer>al=new ArrayList<>();\n\t\tfor(int i=0;i<a.length;i++) {\n\t\t\tal.add(a[i]);\n\t\t}\n\t\tCollections.sort(al);\n\t\tfor(int i=0;i<a.length;i++) {\n\t\t\ta[i]=al.get(i);\n\t\t}\n\t}\n\tstatic class JS {\n\t public int BS = 1<<16;\n\t public char NC = (char)0;\n\t byte[] buf = new byte[BS];\n\t int bId = 0, size = 0;\n\t char c = NC;\n\t double num = 1;\n\t BufferedInputStream in;\n \n\t public JS() {\n\t in = new BufferedInputStream(System.in, BS);\n\t }\n \n\t public JS(String s) throws FileNotFoundException {\n\t in = new BufferedInputStream(new FileInputStream(new File(s)), BS);\n\t }\n \n\t public char nextChar(){\n\t while(bId==size) {\n\t try {\n\t size = in.read(buf);\n\t }catch(Exception e) {\n\t return NC;\n\t }\n\t if(size==-1)return NC;\n\t bId=0;\n\t }\n\t return (char)buf[bId++];\n\t }\n \n\t public int nextInt() {\n\t return (int)nextLong();\n\t }\n \n\t public long nextLong() {\n\t num=1;\n\t boolean neg = false;\n\t if(c==NC)c=nextChar();\n\t for(;(c<'0' || c>'9'); c = nextChar()) {\n\t if(c=='-')neg=true;\n\t }\n\t long res = 0;\n\t for(; c>='0' && c <='9'; c=nextChar()) {\n\t res = (res<<3)+(res<<1)+c-'0';\n\t num*=10;\n\t }\n\t return neg?-res:res;\n\t }\n \n\t public double nextDouble() {\n\t double cur = nextLong();\n\t return c!='.' ? cur:cur+nextLong()/num;\n\t }\n \n\t public String next() {\n\t StringBuilder res = new StringBuilder();\n\t while(c<=32)c=nextChar();\n\t while(c>32) {\n\t res.append(c);\n\t c=nextChar();\n\t }\n\t return res.toString();\n\t }\n \n\t public String nextLine() {\n\t StringBuilder res = new StringBuilder();\n\t while(c<=32)c=nextChar();\n\t while(c!='\\n') {\n\t res.append(c);\n\t c=nextChar();\n\t }\n\t return res.toString();\n\t }\n \n\t public boolean hasNext() {\n\t if(c>32)return true;\n\t while(true) {\n\t c=nextChar();\n\t if(c==NC)return false;\n\t else if(c>32)return true;\n\t }\n\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": "2000G",
"submission_id": 277988465,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n//package Codeforces;\n \n \nimport java.util.*;\n \npublic class CallDuringTheJourney {\n final static Scanner sc = new Scanner(System.in);\n \n public static void main(String[] args) {\n int testcases = sc.nextInt();\n for (int test = 0; test<testcases; ++test) {\n int n = sc.nextInt();\n int m = sc.nextInt();\n \n Graph g = new Graph(n, m);\n \n int t0 = sc.nextInt();\n int t1 = sc.nextInt();\n int t2 = sc.nextInt();\n \n for(int i=0; i<m; ++i){\n int u = sc.nextInt();\n int v = sc.nextInt();\n int bussTime = sc.nextInt();\n int walkTime = sc.nextInt();\n g.createEdge(u, v, bussTime, walkTime);\n }\n \n int result = findTimeToReach(g, t0, t1, t2);\n System.out.println(result);\n }\n }\n \n private static int findTimeToReach(Graph g, int t0, int t1, int t2){\n g.populateMaxTimeToReach(t0, t1, t2);\n return g.nodes.get(0).maxTimeTorReach;\n }\n \n private static boolean isOverlappingInterval(int busTime, int maxTimeTorReach, int t1, int t2) {\n return (t2 > busTime && t1 < maxTimeTorReach);\n }\n \n public static class Graph{\n List<Node> nodes;\n List<Edge> edges;\n Map<Node, List<Edge>> adjacencyList;\n static int edgeid = 0;\n int N;\n int M;\n \n Graph(int N, int M){\n this.N = N;\n this.M = M;\n nodes = new ArrayList<Node>();\n edges = new ArrayList<>();\n adjacencyList = new HashMap<>();\n for(int i = 0; i < N; i++){\n Node u = new Node(i);\n nodes.add(u);\n adjacencyList.put(u, new ArrayList<>());\n }\n }\n \n void populateMaxTimeToReach(int t0, int t1, int t2){\n nodes.get(N-1).maxTimeTorReach = t0;\n PriorityQueue<Graph.Node> pq = new PriorityQueue<>();\n \n pq.add(nodes.get(N-1));\n while(!pq.isEmpty()){\n Graph.Node node = pq.poll();\n for(Graph.Edge edge: adjacencyList.get(node)){\n Graph.Node neib = edge.u;\n if(edge.u.id == node.id){\n neib = edge.v;\n }\n int walkTime = node.maxTimeTorReach - edge.walkTime;\n int busTime = node.maxTimeTorReach - edge.busTime;\n \n if(isOverlappingInterval(busTime, node.maxTimeTorReach, t1, t2)){\n busTime = t1 - edge.busTime;\n }\n int mx = Math.max(walkTime, busTime);\n if(neib.maxTimeTorReach < mx){\n pq.add(neib);\n neib.maxTimeTorReach = mx;\n }\n }\n }\n }\n \n public static class Node implements Comparable<Node>{\n int id;\n int maxTimeTorReach;\n \n Node(int id){\n this.id = id;\n this.maxTimeTorReach = -1;\n }\n \n \n @Override\n public int compareTo(Node o) {\n return this.maxTimeTorReach - o.maxTimeTorReach;\n }\n }\n \n public static class Edge{\n int id;\n Node u;\n Node v;\n int walkTime;\n int busTime;\n \n public Edge(Node u, Node v, int walkTime, int busTime) {\n this.u = u;\n this.v = v;\n this.walkTime = walkTime;\n this.busTime = busTime;\n this.id = edgeid;\n edgeid++;\n }\n }\n \n void createEdge(int u, int v, int busTime, int walkTime){\n u--;\n v--;\n Node uNode = nodes.get(u);\n Node vNode = nodes.get(v);\n \n Edge edge = new Edge(uNode, vNode, walkTime, busTime);\n adjacencyList.get(uNode).add(edge);\n adjacencyList.get(vNode).add(edge);\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": "2000G",
"submission_id": 276832367,
"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: Call During the Journey\ntime_limit_ms: 4000\nmemory_limit_mb: 256\nDescription: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.\n\nEvery minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.\n\nYou can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.\n\nYou can make stops at intersections.\n\nYou live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).\n\nDuring the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.\n\nSince you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?\n\nInput: The first line contains an integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The following are the descriptions of the test cases.\n\nThe first line of each test case contains two integers $n$, $m$ ($2 \\le n \\le 10^5, 1 \\le m \\le 10^5$) — the number of intersections and streets in the city.\n\nThe second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \\le t_1 < t_2 < t_0 \\le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.\n\nThe next $m$ lines of each test case contain descriptions of the streets.\n\nThe $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \\le u_i, v_i \\le n$, $u_i \\neq v_i$, $1 \\le l_{i1} < l_{i2} \\le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.\n\nIt is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.\n\nExamples:\ninput:\n7\n5 5\n100 20 80\n1 5 30 100\n1 2 20 50\n2 3 20 50\n3 4 20 50\n4 5 20 50\n2 1\n100 50 60\n1 2 55 110\n4 4\n100 40 60\n1 2 30 100\n2 4 30 100\n1 3 20 50\n3 4 20 50\n3 3\n100 80 90\n1 2 1 10\n2 3 10 50\n1 3 20 21\n3 2\n58 55 57\n2 1 1 3\n2 3 3 4\n2 1\n12 9 10\n2 1 6 10\n5 5\n8 5 6\n2 1 1 8\n2 3 4 8\n4 2 2 4\n5 3 3 4\n4 5 2 6\noutput:\n0\n-1\n60\n80\n53\n3\n2\n\nNote: \n\nBuggy Code:\n# Created by Ketan_Raut\n# Copyright © 2024 iN_siDious. All rights reserved.\nimport sys,os\nimport random\nfrom math import *\nfrom string import ascii_lowercase,ascii_uppercase\nfrom collections import Counter, defaultdict, deque\nfrom itertools import accumulate, combinations, permutations\nfrom heapq import heappushpop, heapify, heappop, heappush\nfrom bisect import bisect_left,bisect_right\nfrom types import GeneratorType\nfrom random import randint\nRANDOM = randint(1, 10 ** 10)\nclass op(int):\n def __init__(self, x):\n int.__init__(x)\n def __hash__(self):\n return super(op, self).__hash__() ^ RANDOM\ndef bootstrap(f, stack=[]):\n def wrappedfunc(*args, **kwargs):\n if stack:\n return f(*args, **kwargs)\n else:\n to = f(*args, **kwargs)\n while True:\n if type(to) is GeneratorType:\n stack.append(to)\n to = next(to)\n else:\n stack.pop()\n if not stack:\n break\n to = stack[-1].send(to)\n return to\n return wrappedfunc\ninp = lambda: sys.stdin.buffer.readline().decode().strip()\nout=sys.stdout.write\ndef S():\n return inp()\ndef I():\n return int(inp())\ndef F():\n return float(inp())\ndef MI():\n return map(int, inp().split())\ndef MF():\n return map(float, inp().split())\ndef MS():\n return inp().split()\ndef LS():\n return list(inp().split())\ndef LI():\n return list(map(int,inp().split()))\ndef LF():\n return list(map(float,inp().split()))\ndef Query(i,j):\n print3('?', i, j)\n sys.stdout.flush()\n qi=I()\n return qi\ndef print1(x):\n return out(str(x)+\"\\n\")\ndef print2(x,y):\n return out(str(x)+\" \"+str(y)+\"\\n\")\ndef print3(x,y,z):\n return out(str(x)+\" \"+str(y)+\" \"+str(z)+\"\\n\")\ndef print_arr(arr):\n for num in arr:\n out(str(num)+\" \")\n out(str(\"\\n\"))\nfor _ in range(I()):\n n,m=MI()\n t0,t1,t2=MI()\n graph=[[] for _ in range(n)]\n for _ in range(m):\n u,v,l1,l2=MI()\n u,v=u-1,v-1\n graph[u].append((v,l1,l2))\n graph[v].append((u,l1,l2))\n heap,dist=[],[-sys.maxsize]*n\n dist[n-1]=t0\n heappush(heap,(-t0,n-1))\n while heap:\n time,node=heappop(heap)\n time=-time\n for nei,l1,l2 in graph[node]:\n new_time=time-l1\n if not ((t1<new_time<t2) or (new_time<=t1 and time>=t2) or t1<time<t2):\n if new_time>dist[nei]:\n dist[nei]=new_time\n heappush(heap,(-new_time,nei))\n else:\n new_time=max(time-l2,t1-l1)\n if new_time>dist[nei]:\n dist[nei]=new_time\n heappush(heap,(-new_time,nei))\n print1(-1 if dist[0]<0 else dist[0])\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2000G",
"submission_id": 277221391,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\n#include<iostream>\n#include<vector>\n#define int int64_t\nusing namespace std;\nint32_t cal(int n){\n if(n<3)return 1;\n return 1+cal(n/3);\n}\nint32_t main(){\n int t;\n cin>>t;\n while(t--){\n int l,r;\n cin>>l>>r;\n vector<int>vc(3e5+1,0);\n vc[1]=1;\n for(int i=2;i<3e5+1;i++){\n vc[i]=vc[i-1]+cal(i);\n }\n cout<<2*cal(l)+(vc[r]-vc[l])<<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": "1999E",
"submission_id": 275023194,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\n#include<bits/stdc++.h>\n #define R register\n using namespace std;\n const int MAXN=200005;\n int T,l,r,d[MAXN];\n void solve(){\n \tcin>>l>>r;\n \tint s=0;\n \ts+=2*d[l];\n \tlong long tl=l+1,tr=r;\n \twhile(tl<=tr){\n \t\ts+=d[tl];\n \t\ttl++;\n \t\tif(d[tl]!=d[tl-1])break;\n \t}\n \twhile(tl<=tr){\n \t\ts+=d[tr];\n \t\ttr--;\n \t\tif(d[tr]!=d[tr+1])break;\n \t}\n \tif(tl<=tr){\n \t\t//cout<<tl<<\" \"<<tr<<\"\\n\";\n \t\ts+=(d[tl]+d[tr])*(tr-tl+1)/2;\n \t}\n \tcout<<s<<\"\\n\";\n }\n int main(){\n \tfor(int i=1;i<MAXN;i++)d[i]=d[i/3]+1;\n \t//for(int i=1;i<=10;i++)cout<<i<<\" \"<<d[i]<<\"\\n\";\n \tcin>>T;\n \twhile(T--)solve();\n \treturn 0;\n }\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999E",
"submission_id": 291327132,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\n#pragma GCC optimize(\"Ofast\")\n#pragma GCC optimize(\"unroll-loops\")\n#include <bits/stdc++.h>\n#include <bit>\nusing namespace std;\n#define hash #\n#define remr %\n#define cmp ==\n#define ll long long int\n#define ld long double\n#define get1(n) \\\n ll n; \\\n cin >> n;\n#define get2(a, b) \\\n ll a, b; \\\n cin >> a >> b;\n#define get3(a, b, c) \\\n ll a, b, c; \\\n cin >> a >> b >> c;\n#define rep(i, k, n) for (ll i = k; i < n; i++)\n#define rn(i, n) for (ll i = 0; i < n; i++)\n#define repr(i, k, n, r) for (ll i = k; i < n; i += r)\n#define drep(i, k, n) for (ll i = k; i >= n; i--)\n#define gets(s) \\\n string s; \\\n cin >> s;\n#define get(a, n) \\\n for (ll i = 0; i < n; i++) \\\n { \\\n cin >> a[i]; \\\n }\n#define getv(v, n) \\\n vil v(n); \\\n get(v, n);\n#define getp(v, n) \\\n rep(i, 0, n) \\\n { \\\n cinn v[i].ff; \\\n cinn v[i].ss; \\\n }\n#define getpi(v, n) \\\n rep(i, 0, n) \\\n { \\\n cinn v[i].ff; \\\n v[i].ss = i; \\\n }\n#define showv(v, n) \\\n for (auto c : v) \\\n { \\\n cout << c << \" \"; \\\n } \\\n coutt nl;\n#define showp(v, n) \\\n for (auto c : v) \\\n { \\\n cout << c.ff << \" \" << c.ss << endl; \\\n } \\\n coutt nl;\n#define YES \\\n cout << \"YES\\n\"; \\\n return;\n#define NO \\\n cout << \"NO\\n\"; \\\n return;\n#define cinn cin >>\n#define coutt cout <<\n#define nl '\\n'\n#define ln << '\\n'\n#define tin >>\n#define lc << // line continue\n#define spb << \" \" // space bar\n#define spbc << \" \" << // space bar\n#define prt(x) coutt x ln;\n#define show(ans) \\\n coutt ans ln; \\\n return;\n#define pb push_back\n#define ppb pop_back\n#define ub upper_bound\n#define lb lower_bound\n#define sg string\n#define qi queue<int>\n#define ql queue<ll>\n#define qc queue<char>\n#define ski stack<int>\n#define skl stack<ll>\n#define skc stack<char>\n#define usi unordered_set<int>\n#define usl unordered_set<ll>\n#define upsi unordered_map<sg, int>\n#define upi unordered_map<int, int>\n#define upl unordered_map<ll, ll>\n#define upsl map<ll, ll>\n#define mpsi map<sg, int>\n#define mpi map<int, int>\n#define mpl map<ll, ll>\n#define pi pair<int, int>\n#define pl pair<ll, ll>\n#define vp vector<pair<int, int>>\n#define vpl vector<pair<ll, ll>>\n#define vi vector<int>\n#define vil vector<ll>\n#define vvi vector<vector<int>>\n#define vvl vector<vector<ll>>\n#define vc vector<char>\n#define vs vector<string>\n#define vb vector<bool>\n#define mp(a, b) make_pair(a, b)\n#define seti set<int>\n#define setl set<ll>\n#define bs(k, n) bitset<k>(n)\n#define ff first\n#define ss second\n#define all(v) v.begin(), v.end()\n#define sortit(v) sort(all(v));\n#define reverseit(v) reverse(all(v));\n#define tos(x) to_string(x)\n#define toint(x) stoi(x)\n#define fast \\\n ios_base::sync_with_stdio(false); \\\n cin.tie(NULL); \\\n cout.tie(NULL);\n//............................................ SOLUTION STARTS HERE ...................................................//\n \nconst int mod = 1e9 + 7;\n \n// An optimised power function.\nint power(int x, long long n)\n{ // O(log n)\n int ans = 1 % mod;\n while (n > 0)\n {\n if (n & 1)\n {\n ans = 1LL * ans * x % mod;\n }\n x = 1LL * x * x % mod;\n n >>= 1;\n }\n return ans;\n}\n \nbool is_int(ld num)\n{\n return std::floor(num) == num; // Check if the number is equal to its floor value\n}\n \nvil powers_of_3 = {1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441};\n \nvoid solve()\n{\n get2(l, r);\n ll ans(0), pos, pos1;\n pos = lb(all(powers_of_3), l) - powers_of_3.begin();\n pos1 = lb(all(powers_of_3), r) - powers_of_3.begin();\n if (pos == pos1)\n {\n show(pos * (r - l + 2));\n }\n if (l == 1)\n ans += 3;\n else if (l != powers_of_3[pos])\n ans += (pos) * (powers_of_3[pos] - l + 1);\n \n // prt(ans);\n rep(i, pos, pos1 - 1)\n {\n ans += (i + 1) * (powers_of_3[i + 1] - powers_of_3[i]);\n // prt(ans);\n }\n if (r != powers_of_3[pos1])\n {\n ans += pos1 * (r - powers_of_3[pos1 - 1] + 1);\n }\n else\n {\n ans += pos1 + 1;\n }\n \n show(ans);\n}\nint main()\n{\n fast;\n ll t(1);\n cinn t;\n while (t--)\n {\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999E",
"submission_id": 283894548,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\nimport math\ndef solve(l, r):\n ans = 0\n done = False\n num = -1\n if r - l + 1 == 2:\n ans += (math.floor(math.log(l, 3)) + 1) * 2\n ans += math.floor(math.log(r, 3)) + 1\n print(ans)\n return\n for i in range(l, r + 1):\n if i % 3 == 2 and not done:\n num = i\n ans += (math.floor(math.log(i, 3)) + 1) * 2\n break\n for i in range(l, r + 1):\n if i == num:\n continue\n ans += math.floor(math.log(i, 3)) + 1\n print(ans)\n \nt = int(input())\nfor _ in range(t):\n inp = [int(x) for x in input().split()]\n l, r = inp\n solve(l, r)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1999E",
"submission_id": 274915842,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\ndef inp():\n return(int(input()))\ndef inlt():\n return(list(map(int,input().split())))\ndef insr():\n s = input()\n return(list(s[:len(s)]))\ndef invr():\n return(map(int,input().split()))\n \n \n \ndef operations(x):\n res = 0\n while x != 0:\n x //= 3\n res += 1\n \n return res\n \ndef solve():\n l, r = invr()\n \n numOps = []\n numOpsSum = 0\n minNum = float(\"inf\")\n \n for x in range(l, r + 1):\n nx = operations(x)\n numOps.append(nx)\n numOpsSum += nx\n minNum = min(minNum, nx)\n \n \n \n print(numOpsSum + minNum)\n \nfor _ in range(inp()):\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": "1999E",
"submission_id": 283974605,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\nimport math\nt = int(input())\nfor _ in range(t):\n l, r = list(map(int, input().split()))\n \n y,x =l, l+1\n \n \n dp = [-1]*(r-l+1)\n count = 0\n while y != 0:\n y = math.floor(y/3)\n count += 1\n dp[0] = count\n count = 0\n while x != 0:\n x = math.floor(x/3)\n count += 1\n dp[1] = count + dp[0]\n \n \n for x in range(l+2, r+1):\n i = x\n if i-l-1 == 1 and math.floor(x/3**(dp[1] - dp[0])) == 0:\n dp[x-l] = dp[1] -dp[0]\n elif i-l-1 >= 0 and math.floor(x/3**dp[i-l-1]) == 0:\n dp[x-l] = dp[i-l-1]\n else:\n dp[x-l] = 1+dp[i-l-1]\n count = 0\n for i in dp:\n count += i\n print(count)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999E",
"submission_id": 277896392,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\nimport java.util.*;\n \npublic class Codechef {\n public static void main(String args[]) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n while (t-- > 0) {\n int l = sc.nextInt(), r = sc.nextInt();\n double tans = (Math.log(l) / Math.log(3));\n long ans = 0;\n // long min = Long.MAX_VALUE;\n long min = (long) (Math.log(l) / (double) Math.log(3) + 1);\n for (long i = l; i <= r; i++) {\n tans = (Math.log(i) / (double) Math.log(3));\n // min = (long) Math.min(min, tans + 1);\n //\n long imin = (long) Math.pow(3, (long) (tans + 1)) - 1;\n imin = (long) Math.min(imin, r);\n imin = (long) Math.max(imin, i);\n ans += (long) (tans + 1) * (long) (imin - i + 1);\n i = imin;\n //\n // ans += (long) (tans + 1);\n //\n // System.out.println(tans + \" \" + ans);\n }\n System.out.println(ans + min);\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999E",
"submission_id": 274977637,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.lang.reflect.Array;\nimport java.security.KeyStore.Entry;\nimport java.util.*;\nimport java.util.regex.Pattern;\nimport java.io.*;\n \npublic class test {\n \n static int gcd(int a, int b) {\n if (a == 0)\n return b;\n return gcd(b % a, a);\n }\n \n static int lcm(int a, int b) {\n return (a / gcd(a, b)) * b;\n }\n \n static boolean issorted(int[] a) {\n for (int i = 0; i < a.length - 1; i++) {\n if (a[i] > a[i + 1])\n return false;\n }\n return true;\n }\n \n public static int lowerBound(int[] arr, int n, int x) {\n int low = 0, high = n - 1;\n int ans = -1;\n while (low <= high) {\n int mid = (low + high) / 2;\n if (arr[mid] <= x) {\n ans = mid;\n low = mid + 1;\n } else {\n high = mid - 1;\n }\n }\n return ans;\n }\n \n public static int upperBound(int[] arr, int n, int x) {\n int low = 0, high = n - 1;\n int ans = -1;\n while (low <= high) {\n int mid = (low + high) / 2;\n if (arr[mid] >= x) {\n ans = arr[mid];\n high = mid - 1;\n } else {\n low = mid + 1;\n }\n }\n return ans;\n }\n \n static ArrayList<Integer> list = new ArrayList<>();\n \n static void init() {\n for (int i = 2; i <= 32; i++) {\n String s = Integer.toBinaryString(i);\n list.add(Integer.parseInt(s));\n }\n Collections.sort(list);\n }\n \n public static long modularExponentiation(long base, long exp, long mod) {\n long result = 1;\n base = base % mod;\n while (exp > 0) {\n if ((exp & 1) == 1) {\n result = (result * base) % mod;\n }\n exp = exp >> 1;\n base = (base * base) % mod;\n }\n return result;\n }\n \n public static void main(String[] args) throws IOException {\n Scanner sc = new Scanner(System.in);\n \n int[] arr = new int[200001];\n for (int i = 1; i < 200001; i++) {\n int x = (int)Math.floor(Math.log(i) / Math.log(3));\n arr[i] = x+1;\n arr[i] = arr[i-1]+arr[i];\n }\n \n int tc = sc.nextInt();\n while (tc-- > 0) {\n int l = sc.nextInt();\n int r = sc.nextInt();\n long ans = 0;\n ans+=((arr[l]-arr[l-1])*2L);\n ans+=(arr[r]-arr[l]);\n System.out.println(ans);\n }\n \n /*\n * Collections.sort(list, (p1, p2) -> {\n * if (p1.v1 != p2.v1) {\n * return Long.compare(p2.v1, p1.v1); // Sort by v1 in descending order\n * } else {\n * return Long.compare(p1.v2, p2.v2); // Sort by v2 in ascending order if v1 are\n * equal\n * }\n * });\n */\n }\n \n static class Pair implements Comparable<Pair> {\n long v1; // Power of the monster\n long v2; // Health of the monster\n \n public Pair(long v1, long v2) {\n this.v1 = v1;\n this.v2 = v2;\n }\n \n @Override\n public int compareTo(Pair other) {\n if (this.v1 == other.v1) {\n return Long.compare(this.v2, other.v2);\n }\n return Long.compare(this.v1, other.v1);\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": "1999E",
"submission_id": 296189183,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define ll long long\n#define ln '\\n'\n#define ld long double\nusing namespace std;\nconst int M = 2e5 + 10;\nconst ll mod = 1e9 + 7;\nint main() {\n\tint t;\n\tcin >>t;\n\twhile(t--){\n\t\tll l,r,sum=0;\n\t\tcin >>l>>r;\n\t\tll x=l;\n\t\twhile(x){\n\t\t\tsum++;\n\t\t\tx/=3;\n\t\t\t\n\t\t}\n\t\tfor (int i=l+1;i<=r;i++){\n\t\t\tx=i;\n\t\t\tif (i==l+1){\n\t\t\t\tx=(ll)i*pow(3,sum);\n\t\t\t}\n\t\t\twhile(x){\n\t\t\t\tsum++;\n\t\t\t\tx/=3;\n\t\t\t\t//cout <<x<<\"**\"<<ln;\n\t\t\t}\n\t\t}\n\t\tcout <<sum<<ln;\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": "1999E",
"submission_id": 274862277,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long\nconst int N =2e5 + 5;\n//int mp[N];\nvoid solve(){\n\tint l,r;\n\tcin >> l >> r;\n\tint ans = 0;\n\tint k = 1;\n\tint cnt = 1;\n\twhile(k < l){\n\t\tk *= 3;\n\t\tcnt ++;\n\t}\n//\tcout << cnt << '\\n';\n\tint nxt;\n\tif(k == l) nxt = k * 3;\n\telse cnt --,nxt = k;\n//\tcout << ans << '\\n';\n\t\n \tans = cnt;\n\tans *= 2;\n\tfor(int i = l + 1;i <= r;i ++){\n\t\tif(i < nxt) ans += cnt;\n\t\telse{\n\t\t\tcnt ++;\n\t\t\tans += cnt;\n\t\t\tnxt *= 3;\n\t\t}\n\t}\n\tcout << ans << '\\n';\n}\nsigned main(){\n\tios::sync_with_stdio(false);\n\tcin.tie(0),cout.tie(0);\n \n//\tint now = 1;\n//\tint cnt = 1;\n//\twhile(now < N){\n//\t\tmp[now] = cnt;\n//\t\tcnt ++;\n//\t\tnow *= 3;\n//\t}\n//\tfor(int i = 2;i < N;i ++){\n//\t\tif(!mp[i]) mp[i] = mp[i - 1];\n//\t}\n\t\n\tint t;\n\tcin >> t;\n\twhile(t --){\n\t\tsolve();\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999E",
"submission_id": 275030247,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\ndef totalDivisoes(n):\n if n == 1:\n return 1\n \n num = potencias[0]\n index = 0\n \n while n >= num:\n index += 1\n num = potencias[index]\n \n return index\n \nn = int(input())\npotencias = []\n \nfor i in range(13):\n potencias.append(pow(3, i))\n \nfor _ in range(n):\n l, r = map(int, input().split())\n menor = l\n \n total = totalDivisoes(l)\n total *= 2\n \n for num in range(l+1, r+1):\n total += totalDivisoes(num)\n \n print(total)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1999E",
"submission_id": 274954720,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\ndef compute_steps(idx):\n if idx == 0:\n return 0\n if memo[idx] != -1:\n return memo[idx]\n memo[idx] = 1 + compute_steps(idx // 3)\n return memo[idx]\n \ndef main():\n test_cases = int(input().strip())\n \n # Initialize memoization array\n max_size = 2 * 10**5 + 1\n global memo\n memo = [-1] * (max_size + 1)\n \n # Compute memo values for all indices\n for index in range(max_size, -1, -1):\n compute_steps(index)\n \n results = []\n for _ in range(test_cases):\n left, right = map(int, input().strip().split())\n \n total_steps = memo[left] * 2\n for index in range(left + 1, right + 1):\n total_steps += memo[index]\n \n results.append(total_steps)\n \n for result in results:\n print(result)\n \n# To use the function:\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": "1999E",
"submission_id": 274913667,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\nimport java.util.*;\n \npublic class b2 {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n long t = sc.nextLong();\n while(t-->0){\n long l = sc.nextLong();\n long r = sc.nextLong();\n long count = 0;\n for(long i = l; i <= r; i++){\n count += fnx(i);\n }\n \n System.out.println(count + fnx(l));\n }\n }\n \n public static long fnx(long a){\n long count = 0;\n while(a > 0){\n a = a / 3;\n count++;\n }\n \n return count;\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": "1999E",
"submission_id": 275447153,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\nimport java.util.*;\n \npublic class tripleOp {\n public static int OptimizePow(long k){\n double n = 0;\n n = ((Math.log10(k))/(Math.log10(3))) + 1;\n return (int)(n);\n }\n public static void main(String args[]){\n \n try(Scanner sc = new Scanner(System.in)){\n \n int test = sc.nextInt();\n \n while((test--)>0){\n \n long r = sc.nextLong();\n long l = sc.nextLong();\n Long a[] = new Long[(int)(l-r+1)];\n for(long i=0; i<(l-r+1); i++){\n a[(int)i] = (r+i);\n }\n int count = 0;\n count = 2*OptimizePow(a[0]);\n // System.out.println(count);\n for(long i=1; i<(int)(l-r+1); i++){\n int o = OptimizePow(a[(int) i]);\n // System.out.println(o +\"-->\" + a[(int) i]);\n count += o;\n }\n System.out.println(count);\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": "1999E",
"submission_id": 275026832,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\n#include<iostream>\n#include<cmath>\n#include<string>\n#include<vector>\n#include<map>\n#include<cstring>\n#include<fstream>\n#include<queue>\n#include<stack>\n#include<cstdio>\n#include<algorithm>\nusing namespace std;\nvoid solve(){\n long long l,r,ans=0;\n cin>>l>>r;\n long long ll=l,lll=l+1;\n while(ll){\n ll/=3;\n lll*=3;\n ans+=2;\n }\n for(int i=l+1;i<=r;i++){\n int p=i;\n while(p){\n p/=3;\n ans++;\n }\n }\n cout<<ans<<'\\n';\n return;\n}\nint main(){\n int t;\n cin>>t;\n while(t--){\n solve();\n }\n return 0;\n}\n//from chpyx2\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1999E",
"submission_id": 275167603,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define sz size\n#define int long long\n#define cy cout << \"YES\"<<\"\\n\"\n#define cn cout << \"NO\" <<\"\\n\"\n#define pb push_back\n#define eb emplace_back\n#define ff first\n#define ss second\n#define fend \"\\n\"\nusing namespace std;\nvoid solve()\n{\n int l,r;\n cin>>l>>r;\n vector<int>v;\n int x = 3;\n while(x<1e6){\n v.pb(x);\n x*=3;\n }\n int ct=0;\n for(int i=l;i<=r;++i){\n int hi = v.size()-1,lo=0;\n while(hi-lo>1){\n int mid = lo + ((hi-lo)/2);\n if(v[mid]<=i)lo = mid+1;\n else hi = mid;\n }\n if(v[lo]>i)ct+=(lo+1);\n else ct+=(hi+1);\n if(i==l) ct*=2;\n }\n cout<<ct<<fend;\n}\nint32_t main()\n{\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n int t=1;\n cin>>t;\n while(t--){\n 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": "1999E",
"submission_id": 274885820,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\ndef log(a, b):\n res = 0\n while a > 0:\n a //= b\n res += 1\n return res\ndef solve():\n l, r = list(map(int, input().split()))\n ans = log(l, 3)\n for i in range(l, r + 1):\n ans += log(i, 3)\n print(ans)\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": "1999E",
"submission_id": 291465572,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\nimport sys\ninput = sys.stdin.readline\ndef pow_of_3():\n ll=[]\n ll.append(3)\n sums=9\n while(sums<=2*(10**5)):\n ll.append(sums)\n sums*=3\n ll.append(sums)\n return ll\nll=pow_of_3()\n# print(ll[-1])\ndef binary_search(i):\n if(i<ll[0]):\n return 0 \n elif(i>=ll[-1]):\n return len(ll)\n start=0 \n end=len(ll)-1 \n while(start<=end):\n mid=(start+end)//2\n if(mid!=0 and ll[mid]>i and ll[mid-1]<=i):\n return mid\n elif(ll[mid]>i):\n end=mid-1 \n else:\n start=mid+1 \n return -1\n \n \nt=int(input())\nresult=[]\nfor p in range(t):\n l,r=map(int,input().split())\n index=binary_search(l)+1\n number=ll[index-1]\n ans=index*2\n for i in range(l+1,r+1):\n if(number>i):\n ans+=index\n else:\n number*=3\n index+=1\n ans+=index\n result.append(str(ans))\nprint(\"\\n\".join(result))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999E",
"submission_id": 275245025,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\nimport java.io.*;\nimport java.lang.reflect.Array;\nimport java.util.*;\nimport java.math.BigDecimal;\nimport java.math.RoundingMode;\n \npublic class Main {\n \n static class FastReader {\n BufferedReader br;\n StringTokenizer st;\n public FastReader(){br = new BufferedReader(new InputStreamReader(System.in));}\n String next(){while (st == null || !st.hasMoreElements()) {try {st = new StringTokenizer(br.readLine());}catch (IOException e) {e.printStackTrace();}}return st.nextToken();}\n int nextInt() { return Integer.parseInt(next()); }\n long nextLong() { return Long.parseLong(next()); }\n double nextDouble() {return Double.parseDouble(next());}\n }\n \n void yes() {out.println(\"Yes\");}\n void no() {out.println(\"No\");}\n void inArr(int[] a){for(int i=0;i<a.length;i++){a[i]=sc.nextInt();}}\n void inArr(long[] a){for(int i=0;i<a.length;i++){a[i]=sc.nextLong();}}\n void inArr(String[] a){for(int i=0;i<a.length;i++){a[i]=sc.next();}}\n void pArr(int[] a){for(int i=0; i<a.length;i++){out.print(a[i]+\" \");}}\n void pArr(long[] a){for(int i=0; i<a.length;i++){out.print(a[i]+\" \");}}\n void pArr(ArrayList<Integer> a){for(int i=0; i<a.size();i++){out.print(a.get(i)+\" \");}}\n // void pArr(ArrayList<Long> a){for(int i=0; i<a.size();i++){out.print(a.get(i)+\" \");}}\n void pl(){out.println();}\n void pl(int x){out.println(x);}\n void pl(long x){out.println(x);}\n void pl(double x){out.println(x);}\n void pl(String x){out.println(x);}\n long max(long a, long b){return Math.max(a, b);}\n long min(long a, long b){return Math.min(a, b);}\n int gcd(int a, int b) {while (b != 0) {int temp = b;b = a % b;a = temp;}return a;}\n public static boolean isSorted(int[] a) {for (int i = 0; i < a.length - 1; i++) {if (a[i] > a[i + 1]) {return false;}}return true;}\n public static void reverseArr(int[] a){int i=0, j=a.length-1;while (i<j){int temp=a[i];a[i]=a[j];a[j]=temp;i++;j--;}}\n public static void reverseArr(long[] a){int i=0, j=a.length-1;while (i<j){long temp=a[i];a[i]=a[j];a[j]=temp;i++;j--;}}\n void sort(long a[]) {divide(a, 0, a.length - 1, true);}\n void rsort(long a[]) {divide(a, 0, a.length - 1, false);}\n void divide(long a[], int l, int r, boolean order) {if (l < r) {int m = l + (r - l) / 2;divide(a, l, m, order);divide(a, m + 1, r, order);merge(a, l, m, r, order);}}\n void merge(long a[], int l, int m, int r, boolean order) {int n1 = m - l + 1;int n2 = r - m;long L[] = new long[n1];long R[] = new long[n2];for (int i = 0; i < n1; ++i) L[i] = a[l + i];for (int j = 0; j < n2; ++j) R[j] = a[m + 1 + j];int i = 0, j = 0; int k = l;while (i < n1 && j < n2) {if ((L[i] <= R[j] && order) || (L[i] >= R[j] && !order)) {a[k] = L[i];i++;} else {a[k] = R[j];j++;}k++;}while (i < n1) {a[k] = L[i];i++;k++;}while (j < n2) {a[k] = R[j];j++;k++;}}\n void sort(long a[][]) {divide(a, 0, a[0].length - 1, true);}\n void rsort(long a[][]) {divide(a, 0, a[0].length - 1, false);}\n void divide(long a[][], int l, int r, boolean order) {if (l < r) {int m = l + (r - l) / 2;divide(a, l, m, order);divide(a, m + 1, r,order);merge(a, l, m, r, order);}}\n void merge(long a[][], int l, int m, int r, boolean order) {int n1 = m - l + 1;int n2 = r - m;long L[] = new long[n1]; long R[] = new long[n2];long b1[][] = new long[a.length][n1]; long b2[][] = new long[a.length][n2];for (int i = 0; i < n1; ++i) {L[i] = a[0][l + i];for (int p = 1; p < a.length; p++) b1[p][i] = a[p][l + i];}for (int j = 0; j < n2; ++j) {R[j] = a[0][m + 1 + j];for (int p = 1; p < a.length; p++) b2[p][j] = a[p][m + 1 + j];}int i = 0, j = 0; int k = l;while (i < n1 && j < n2) {if ((L[i] <= R[j] && order) || (L[i] >= R[j] && !order)) {a[0][k] = L[i]; for (int p = 1; p < a.length; p++) a[p][k] = b1[p][i];i++;} else {a[0][k] = R[j];for (int p = 1; p < a.length; p++) a[p][k] = b2[p][j];j++;}k++;}while (i < n1) {a[0][k] = L[i];for (int p = 1; p < a.length; p++) a[p][k] = b1[p][i];i++;k++;}while (j < n2) {a[0][k] = R[j];for (int p = 1; p < a.length; p++) a[p][k] = b2[p][j];j++;k++;}}\n \n public static final int MOD = 1000000000 + 7;\n \n FastReader sc=new FastReader();\n PrintWriter out=new PrintWriter(System.out);\n \n public static void main(String args[]) {\n Main FC27 = new Main();\n FC27.solve();\n }\n \n public void ans() {\n int l= sc.nextInt();\n int r= sc.nextInt();\n long sum=0;\n \n int min=Integer.MAX_VALUE;\n for(int i=l;i<=r;i++){\n double x=Math.log(i)/Math.log(3);\n if((int)x!=(int)(x+0.0001)) x=x+1;\n int y;\n y= (int)x+1;\n if(y<min) min=y;\n //out.print(x+\" \");\n sum+=y;\n }\n pl(sum+min);\n }\n \n public void solve() {\n int t = sc.nextInt();\n while(t-- > 0){\n ans();\n }\n out.close();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1999E",
"submission_id": 274981639,
"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: Triple Operations\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.\n\nIn an operation, she does the following:\n\n * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\\lfloor \\frac{y}{3} \\rfloor$. (Here $\\lfloor \\bullet \\rfloor$ denotes rounding down to the nearest integer).\n\nWhat is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.\n\nInput: The first line contains an integer $t$ ($1 \\leq t \\leq 10^4$) — the number of test cases.\n\nThe only line of each test case contains two integers $l$ and $r$ ($1 \\leq l < r \\leq 2 \\cdot 10^5$).\n\nOutput: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.\n\nExamples:\ninput:\n4\n1 3\n2 4\n199999 200000\n19 84\noutput:\n5\n6\n36\n263\n\nNote: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \\xrightarrow[x=1,\\,y=2]{} 3,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,3 \\xrightarrow[x=0,\\,y=3]{} 1,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,1 \\xrightarrow[x=0,\\,y=1]{} 0,0,0 .$$\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class E {\n public static void main(String[] args) {\n \n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n while (t != 0) {\n t--;\n \n int l = sc.nextInt();\n int r = sc.nextInt();\n \n int[] op=new int[r+1];\n \n op[0]=0 ;\n op[1]=1;\nint no=0;\n for (int i = 2; i <r+1 ; i++) {\n op[i]=op[i/3]+1;\n }\nno=op[l]*2;\n \n for (int i = l+1; i <=r ; i++) {\n no+=op[i];\n }\n \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": "1999E",
"submission_id": 275266752,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <string>\n#include <vector>\n#include <queue>\n#include <map>\n \nusing namespace std;\n \nint main() {\n string s;\n cin >> s;\n int n = s.size();\n vector<int> pos(n);\n for (int i = 0; i < n; i++) {\n pos[i] = i + 1; // 1-based positions\n }\n // To reconstruct t, we need to assign characters from s to positions in t\n // such that when we sort the positions by (balance, -position), we get s.\n // This is tricky, so we need a different approach.\n // Let's try to simulate the inverse process.\n // We can think of the positions in s as being in the order they appear after sorting by (balance, -position).\n // So, the first character in s corresponds to the position in t with the smallest balance and highest position.\n // Similarly, the last character in s corresponds to the position with the largest balance and lowest position.\n // To reverse this, we need to determine the mapping from s back to t.\n // One way to approach this is to consider that the positions in s are already sorted by (balance, -position),\n // so we need to arrange t such that when sorted by (balance, -position), it matches s.\n // This seems complex, so we need a smarter way.\n // Let's consider building t by placing '(' and ')' in positions while maintaining the balance.\n // We can use a priority queue to select the next position to place a '(' or ')'.\n // However, this might not be straightforward.\n // Given time constraints, let's try a different approach.\n // Since the shuffle is a bijection, there must be a unique t for each s.\n // We can try to reconstruct t by considering the shuffled order and the properties of balanced sequences.\n // For now, let's output the sample input's preimage for demonstration.\n // In practice, you would implement the logic to reconstruct t from s.\n cout << \"(()(()))\" << 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": "1970A2",
"submission_id": 293200537,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <list>\n#include <algorithm>\n#include <sstream>\n#include <array>\n#include <set>\n#include <limits.h>\n#include <fstream>\n#include <unordered_map>\n#include <queue>\n#include <array>\n#include <cmath>\n#include <bitset>\n \nstruct ELEMENT {\n\tchar c;\n\tint id = 0;\n\tint balance = 0;\n \n\tbool operator<(const ELEMENT& other) const {\n\t\tif (balance == other.balance) {\n\t\t\treturn id > other.id;\n\t\t}\n\t\treturn balance < other.balance;\n\t}\n};\n \nstd::string genS(std::string s) {\n\tstd::vector<ELEMENT> ss(s.size());\n\tss[0].c = s[0];\n\tfor (int i = 1; i < s.size(); i++) {\n\t\tss[i].c = s[i];\n\t\tss[i].id = i;\n\t\tss[i].balance = ss[i - 1].balance + 1 - 2 * (s[i - 1] == ')');\n\t}\n\tstd::sort(ss.begin(), ss.end());\n\tstd::string S;\n\tfor (auto& i : ss) {\n\t\tS.push_back(i.c);\n\t}\n\treturn S;\n}\n \nint main(){\n\tstd::string s;\n\tstd::cin >> s;\n \n\tstd::string s2 = genS(s), s3;\n \n\twhile (s2 != s) {\n\t\ts3 = s2;\n\t\ts2 = genS(s2);\n\t}\n \n\tstd::cout << s3 << '\\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": "1970A2",
"submission_id": 259504811,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nint main(){\n ios::sync_with_stdio(false);\n cin.tie(0); cout.tie(0);\n string s; cin>>s;\n vector<tuple<int,int,char> > v;\n for(int i=0,c=0;i<s.length();i++)\n v.emplace_back(c,i,s[i]),c+=(s[i]=='('?1:-1);\n sort(v.begin(),v.end());\n for(auto [x,y,z]:v)cout<<z;\n cout<<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": "1970A2",
"submission_id": 259493168,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\nstring = str(input()).strip()\nscores = {}\ncurrent = 0\nfor i,v in enumerate(string):\n scores[i] = current\n if v == \"(\":\n current += 1\n else:\n current -= 1\n \nsorter = sorted(scores,key=lambda a: (scores[a],a))\nfor i in sorter:\n print(string[i],end=\"\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1970A2",
"submission_id": 300298972,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\ns = input()\nn = len(s)\nSUM = [0] * n\nfor i in range(1, n): SUM[i] = SUM[i - 1] + (1 if s[i - 1] == '(' else -1)\nprint(*map(lambda x: x[1], sorted(zip(SUM, s), key=lambda x: x[0])), sep='')\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970A2",
"submission_id": 259697571,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\ns=input()\nt = s.split(')')\nans = [t[0]]\nindex = len(ans)\nfor c in t[1:]:\n count = len(c)\n if not count:\n ans[index-1] += ')'\n else:\n ans[index-1] += '('\n ans.insert(index, ')')\n for _ in range(count-1):\n ans.insert(index, '(')\n index -= 1\n if not index:\n index = len(ans) - 1\nprint(''.join(ans))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970A2",
"submission_id": 259496515,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \n \nimport java.io.*;\n \n \npublic class Main {\n \n public static void main(String[] args) throws IOException {\n \n int t = 1;\n //t = f.nextInt();\n while (t > 0) {\n solve();\n t--;\n }\n w.flush();\n w.close();\n }\n \n static int mod = 998244353;\n static long inf = (long) 1e18;\n \n // 求区间第\n \n public static void solve() throws IOException {\n char[] s = f.next().toCharArray();\n int n = s.length;\n int p = 0;\n while(p < n){\n if (s[p] == ')' && p + 1 < n && s[p+1] == '('){\n s[p] = '('; s[p+1] = ')';\n p+=2;\n }else p++;\n }\n for (int i = 0; i < n; i++) {\n w.print(s[i]);\n }\n \n \n }\n \n static class Pair{\n int f, num;\n \n public Pair(int f, int num) {\n this.f = f;\n this.num = num;\n }\n }\n static long g(long x, long cur) {\n if (cur == 1) return 0;\n if (x % 2 == 0) {\n return g(x / 2, cur / 2);\n } else {\n return cur / 2 + g(x / 2, cur / 2);\n }\n }\n \n static int gcd(int a, int b) {\n if (a == 0) return b;\n if (b == 0) return a;\n return gcd(b, a % b);\n }\n \n static long qkm(long a, long b) {\n long res = 1;\n while (b > 0) {\n if ((b & 1) == 1) res = res * a;\n a = a * a;\n b >>= 1;\n }\n return res;\n }\n \n static long qkm(long a, long b, long mod) {\n long res = 1;\n while (b > 0) {\n if ((b & 1) == 1) res = res * a % mod;\n a = a * a % mod;\n b >>= 1;\n }\n return res;\n }\n \n public static class Node {\n char s; int id; int t;\n \n public Node(char s, int id, int t) {\n this.s = s;\n this.id = id;\n this. t = t;\n }\n \n public Node() {\n }\n }\n \n static boolean nextPermutation(int[] arr) { // 排列数循环模板 记得使用 do while 循环\n int len = arr.length;\n int left = len - 2;\n while (left >= 0 && arr[left] >= arr[left + 1]) left--; // 从升序 一直往降序排列。\n if (left < 0) return false;\n int right = len - 1;\n // 找到第一个升序的位置，将其改为降序。\n while (arr[left] >= arr[right]) right--;\n {\n int t = arr[left];\n arr[left] = arr[right];\n arr[right] = t;\n }\n // 修改后它的前面仍然为降序，将前面全部修改为升序，这样能保证不会漏算，也不会算重\n left++;\n right = len - 1;\n while (left < right) {\n {\n int t = arr[left];\n arr[left] = arr[right];\n arr[right] = t;\n }\n left++;\n right--;\n }\n// System.out.println(Arrays.toString(arr));\n return true;\n }\n \n static PrintWriter w = new PrintWriter(new OutputStreamWriter(System.out));\n static Input f = new Input(System.in);\n \n static class Input {\n public BufferedReader reader;\n public StringTokenizer tokenizer;\n \n public Input(InputStream stream) {\n reader = new BufferedReader(new InputStreamReader(stream), 32768);\n tokenizer = null;\n }\n \n public String next() throws IOException {\n \n while (tokenizer == null || !tokenizer.hasMoreTokens()) {\n tokenizer = new StringTokenizer(reader.readLine());\n }\n return tokenizer.nextToken();\n }\n \n public String nextLine() throws IOException {\n String str = null;\n str = reader.readLine();\n return str;\n }\n \n public int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n \n public long nextLong() throws IOException {\n return Long.parseLong(next());\n }\n \n public Double nextDouble() throws IOException {\n return Double.parseDouble(next());\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970A2",
"submission_id": 262246109,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class balanced_shuffle_2 {\n public static String output(String s) {\n StringBuilder[] groups = new StringBuilder[s.length() + 1];\n int balance = 0;\n \n for (int i = 0; i < s.length(); i++) {\n char c = s.charAt(i);\n \n \n if (groups[balance] == null)\n groups[balance] = new StringBuilder();\n groups[balance].append(c);\n \n if (c == '(') balance++;\n else balance--;\n }\n \n StringBuilder result = new StringBuilder();\n for (StringBuilder group : groups) {\n if (group != null)\n result.append(group);\n }\n \n return result.toString();\n }\n \n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n String q = sc.next();\n System.out.println(output(q));\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": "1970A2",
"submission_id": 259495649,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n#define fi first\n#define se second\n#define int long long\n \nusing namespace std;\n \ntypedef pair<int, int> PII;\ntypedef pair<int, char> PIC;\ntypedef long long LL;\n \nstring check(string s) {\n\tint n = s.size() - 1;\n\tstd::vector<PII> tmp;\n\tint cnt = 0;\n\tfor (int i = 1; i <= n; i ++) {\n\t\ttmp.emplace_back(cnt, i);\n\t\tif (s[i] == '(') cnt ++;\n\t\telse cnt --;\n\t}\n\tsort(tmp.begin(), tmp.end(), [&](PII a, PII b) {\n\t\tif (a.fi == b.fi) return a.se > b.se;\n\t\treturn a.fi < b.fi;\n\t});\n \n\tstring res = \" \";\n\tfor (auto v : tmp)\n\t\tres += s[v.se];\n\treturn res;\n}\n \nsigned main() {\n\tcin.tie(0);\n\tcout.tie(0);\n\tios::sync_with_stdio(0);\n \n\tstring s;\n\tcin >> s;\n\tint n = s.size();\n\ts = ' ' + s;\n \n\tstring t = check(s);\n\twhile (check(t) != s) {\n\t\tt = check(t);\n\t}\n\tt.erase(0, 1);\n\tcout << t << endl;\n \n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970A2",
"submission_id": 259494588,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nbool sortbysec(const pair<long long, long long> &a,\n const pair<long long, long long> &b)\n{\n return (a.second < b.second);\n}\n \nvoid solve()\n{\n string s;\n cin >> s;\n vector<pair<long long, long long>> v;\n long long PB = 0;\n for (long long i = 0; i < s.size(); i++)\n {\n if (s[i] == '(')\n {\n v.push_back({PB, i + 1});\n PB++;\n }\n else\n {\n v.push_back({PB, i + 1});\n PB--;\n }\n }\n sort(v.begin(), v.end());\n // for(auto i : v) cout<<i.first<<\" \"<<i.second<<endl;\n vector<int> count;\n vector<int> v1;\n vector<int> v2;\n int cou = 1;\n for(int i = 0;i<v.size();i++){\n if(v[i].first != v[i+1].first){\n count.push_back(cou);\n cou = 1;\n }\n else{\n cou++;\n }\n }\n // v1.push_back(0);\n int x = 0;\n v1.push_back(0);\n for(int i = 0;i<count.size();i++){\n x = x + count[i];\n v1.push_back(x);\n }\n \n // for(auto i : v1) cout<<i<<\" \";\n // cout<<endl;\n \n int a = v1[0];\n int b = v1[1]-1;\n int k = 0;\n int i = 0;\n while(a<=b){\n v2.push_back(v[b - i].second);\n i++;\n if(a >= b){\n k++;\n a = v1[k] -1;\n b = v1[k+1]-1;\n i = 0;\n }\n a++;\n }\n \n // for(int i = 0;i<v2.size();i++){\n // cout<<v2[i]<<\" \";\n // }\n \n // cout<<endl;\n \n for(auto i : v2){\n cout<<s[i - 1];\n }\n}\n \nint main()\n{\n solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970A2",
"submission_id": 263086564,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\ndef balanced_unshuffle(s):\n # Initialize a list to store tuples of character, position, and prefix balance\n seq = [(s[i], i, s[:i].count('(') - s[:i].count(')')) for i in range(len(s))]\n \n # Sort the list of tuples based on prefix balance in increasing order,\n # breaking ties by position in decreasing order\n sorted_seq = sorted(seq, key=lambda x: (x[2], -x[1]))\n \n # Extract the characters from the sorted list of tuples\n preimage = ''.join([char for char, _, _ in sorted_seq])\n \n return preimage\n \nif __name__ == \"__main__\":\n s = input().strip()\n preimage = balanced_unshuffle(s)\n print(preimage)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1970A2",
"submission_id": 260406863,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\ndef balanced_shuffle(s):\n n = len(s)\n \n # Prepare list to hold the prefix balance, position, and character\n entries = []\n \n # Calculate prefix balances\n balance = 0\n for i, char in enumerate(s):\n if char == '(':\n balance += 1\n else: # char == ')'\n balance -= 1\n # Append the current balance AFTER this character, position, and the character itself\n entries.append((balance, i, char))\n \n # Adjust balance values to reflect the state before the character\n for i in range(len(entries)):\n if entries[i][2] == ')':\n entries[i] = (entries[i][0] + 1, entries[i][1], entries[i][2])\n else:\n entries[i] = (entries[i][0] - 1, entries[i][1], entries[i][2])\n \n # Sort entries by balance (increasing) and by position (decreasing)\n entries.sort(key=lambda x: (x[0], -x[1]))\n \n # Build the result using the character from sorted entries\n result = ''.join(entry[2] for entry in entries)\n \n return result\n \ndef main():\n # Read input from standard input\n input_sequence = input().strip()\n \n # Get the balanced shuffle output\n output_sequence = balanced_shuffle(input_sequence)\n \n # Output result to standard output\n print(output_sequence)\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": "1970A2",
"submission_id": 259951575,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n public static void main(String[] args) {\n Scanner sc=new Scanner(System.in);\n String s=sc.nextLine();\n int []a=new int[s.length()];\n Map<Integer,List<Integer>>mp=new HashMap<>(); //to store balance factor: corresponding indices of array\n a[0]=0; //1st element \n int i=1;\n char[]ch=s.toCharArray(); //for calculating balance factor\n List<Integer>first=new ArrayList<>(); //to add 1st ele into map\n first.add(0);\n mp.put(0,first);\n for(int j=1;j<s.length();j++)\n {\n if(ch[j-1]=='(') //calculating balance factor each index\n {\n a[i]=a[i-1]+1;\n }\n else\n {\n a[i]=a[i-1]-1;\n \n } \n mp.putIfAbsent(a[i], new ArrayList<>()); // map: ( bf : index list ) \n mp.get(a[i]).add(i); //for eg: 1 2 3 index has bf as 5 map looks like (5:1,2,3)\n i++; \n }\n Arrays.sort(a); //sorting bf as asc order\n Set<Integer>st=new HashSet<>(); // to avoid repetition of bf as we are adding every char of balance factor at once in line 41-44\n for(int l:a)\n {\n if(!st.contains(l)) // if bf occurs 1st time we print all indices of it so need to iterate on same bf again\n {\n \n List<Integer>ele=mp.get(l); //extracting list of indices\n Collections.reverse(ele); // as given in the que, indices should be in desc order\n st.add(l); //adding into set to avoid iterating on same bf again\n for(int ele2:ele)\n {\n System.out.print(s.charAt(ele2)); // printing char at indices\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": "1970A2",
"submission_id": 302463306,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n\t\t\npublic class BalancedUnshuffle {\n \n\tpublic static void main(String[] args) throws Exception {\n\t\t// TODO Auto-generated method stub\n\t\tBufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n\t\tString s = br.readLine();\n\t\tint[][] a = new int[s.length()][3];\n\t\tfor(int j=0; j<s.length(); j++) {\n\t\t\tif(s.charAt(j)=='(') a[j][0] = 1;\n\t\t\telse a[j][0] = -1;\n\t\t\ta[j][2] = j;\n\t\t\tif(j!=0) a[j][1] = a[j-1][1]+a[j-1][0];\n\t\t}\n\t\tArrays.sort(a, (p,q) -> p[1]!=q[1] ? p[1]-q[1] : q[2]-p[2]);\n\t\t\n\t\tfor(int j=0; j<s.length(); j++) {\n\t\t\tif(a[j][0]==1) System.out.print(\"(\");\n\t\t\telse System.out.print(\")\");\n\t\t}\n\t}\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970A2",
"submission_id": 301657187,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nstruct Shuffle\n{\n\tint balance;\n\tint p;\n\tchar c;\n};\nbool cmp(Shuffle a,Shuffle b)\n{\n\tif(a.balance==b.balance)\n\t\treturn a.p>b.p;\n\telse\n\treturn a.balance<b.balance;\n}\nint main()\n{\n\tvector<Shuffle> v;\n\tstring s;\n\tcin>>s;\n\tint n=0;\n\tfor(int i=0;i<s.size();i++)\n\t{\n\t\tv.push_back({n,i,s[i]});\n\t\tif(s[i]=='(')\n\t\tn++;\n\t\telse\n\t\tn--;\n\t}\n\tsort(v.begin(),v.end(),cmp);\n\tfor(int i=0;i<v.size();i++)\n\t{\n\t\tcout<<v[i].c;\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": "1970A2",
"submission_id": 262005696,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\n//#pragma GCC optimize(\"Ofast,O3,unroll-loops\")\n#include <bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n#define bp __builtin_popcountll\n#define mp make_pair\n#define pb push_back\n#define vec vector\n#define ll long long\n#define F first\n#define S second\n#define ld long double\n#define pii pair <int, int>\n#define elif else if\n#define sz(a) int32_t(a.size())\n#define lb lower_bound\n#define ub upper_bound\n#define all(a) a.begin(), a.end()\n#define fl fflush(stdout)\n#define ex exit(0)\n#define int ll\n#define FOR(i, l, r) for(int i = l; i <= r; ++i)\n#define ROF(i, l, r) for(int i = r; i >= l; --i)\nusing namespace std;\nusing namespace __gnu_pbds;\ntemplate <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\nmt19937 rnd(228355);\nmt19937 gen(chrono::steady_clock::now().time_since_epoch().count());\nconst int N = 2e5 + 10;\nld f[N], x[N]; int n;\n \nbool cmp(pii a, pii b){\n if(a.F < b.F) return 1;\n if(a.F == b.F && a.S > b.S) return 1;\n return 0;\n}\n \nstring get(string s){\n vector <pii> v;\n v.pb(mp(0, 0));\n \n int x = 0;\n FOR(i, 0, sz(s)){\n if(i != 0) v.pb(mp(x, i));\n if(s[i] == '(') x++;\n else x--;\n }\n v.pop_back();\n sort(all(v), cmp);\n \n map <int, int> m;\n for(auto to: v) m[to.F]++;\n \n string res;\n for(auto to: v){\n res += s[to.S];\n }\n return res;\n}\n \nvoid solve(){\n string s; cin >> s;\n string b = s;\n string last = s;\n FOR(i, 1, 10000){\n s = get(s);\n if(s == b) break;\n last = s;\n }\n cout << last;\n}\n \nmain(){\n ios_base::sync_with_stdio(0);\n cin.tie(0); cout.tie(0);\n \n int t = 1;\n //cin >> t;\n \n while(t--) solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970A2",
"submission_id": 259739466,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\nimport sys\ninput = sys.stdin.readline\n \n \ndef check(a, t):\n while 1:\n a += t\n if x[a]:\n break\n return a\n \n \ndef calc(s):\n c = 0\n for i in s:\n if i == '(':\n c += 1\n else:\n c -= 1\n if c < 0:\n return 0\n return 1\n \n \ns = input()[:-1]\nn = len(s)\ni = i1 = 0\nj = j1 = n-1\nd = ['']*n\nx = [1]*n\nwhile i < j:\n if calc(s[i+1:j]):\n d[i1] = '('\n d[i1+1] = ')'\n i1 += 2\n i = check(i, 1)\n j = check(j, -1)\n elif s[i+1] == ')':\n d[i1] = '('\n d[j1] = ')'\n i1 += 1\n j1 -= 1\n for _ in range(2):\n i = check(i, 1)\n else:\n ii = i\n while 1:\n ii += 1\n if ii == ')':\n x[ii] = 0\n break\n i = check(i, 1)\n d[j1] = ')'\n d[j1-1] = '('\n j1 -= 2\nprint(''.join(d))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1970A2",
"submission_id": 288715373,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class Main {\n static class Pair implements Comparable<Pair> {\n int first, second;\n \n Pair(int first, int second) {\n this.first = first;\n this.second = second;\n }\n \n public int compareTo(Pair other) {\n if (this.first != other.first) {\n return Integer.compare(this.first, other.first);\n } else {\n // Sort in decreasing order of position for ties in prefix balance\n return Integer.compare(other.second, this.second);\n }\n }\n }\n \n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n String s = scanner.next();\n int n = s.length();\n int[] a = new int[n];\n int k = 0;\n for (int i = 0; i < n; i++) {\n a[i] = k;\n if (s.charAt(i) == '(') {\n k++;\n } else {\n k--;\n }\n }\n List<Pair> v = new ArrayList<>();\n for (int i = 0; i < n; i++) {\n v.add(new Pair(a[i], i));\n }\n Collections.sort(v);\n for (Pair u : v) {\n System.out.print(s.charAt(u.second));\n }\n scanner.close();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970A2",
"submission_id": 259474103,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class Main {\n static PrintWriter pw = new PrintWriter(System.out);\n static Scanner sc = new Scanner(System.in);\n \n static class Pair implements Comparable{\n Integer x ;\n Integer y ;\n Integer z ;\n \n public \tPair(Integer i, Integer j,Integer k) {\n this.x = i;\n this.y = j;\n this.z = k;\n }\n public int compareTo(Object o) {\n Pair p = (Pair) o ;\n if (this.x.compareTo(p.x) != 0)\n return this.x.compareTo(p.x);\n else\n return -this.y.compareTo(p.y);\n }\n \n public String toString(){\n return \"(\"+this.x + \",\" + this.y + \")\";\n }\n }\n \n \n public static void solve() throws IOException {\n char a [] = sc.next().toCharArray();\n int n = a.length;\n int cnt = 0;\n Pair ans [] = new Pair[n];\n for(int i = 0; i < n ; i++){\n ans[i] = new Pair(cnt,i + 1, a[i] == '(' ? 1 : -1);\n if(a[i] == '(')\n cnt++;\n else\n cnt--;\n }\n Arrays.sort(ans);\n for(Pair p : ans)\n pw.print(p.z == 1 ? \"(\" : \")\");\n }\n \n public static void main(String[] args) throws IOException {\n int t = 1;\n while(t -- > 0){\n solve();\n }\n pw.flush();\n }\n \n static class Scanner {\n StringTokenizer st;\n BufferedReader br;\n \n public Scanner(InputStream s) {\n br = new BufferedReader(new InputStreamReader(s));\n }\n \n public Scanner(String s) throws FileNotFoundException {\n br = new BufferedReader(new FileReader(s));\n }\n \n public String next() throws IOException {\n while (st == null || !st.hasMoreTokens())\n st = new StringTokenizer(br.readLine());\n return st.nextToken();\n }\n \n public int nextInt() throws IOException {\n return Integer.parseInt(next());\n }\n \n public long nextLong() throws IOException {\n return Long.parseLong(next());\n }\n \n public String nextLine() throws IOException {\n return br.readLine();\n }\n \n public double nextDouble() throws IOException {\n String x = next();\n StringBuilder sb = new StringBuilder(\"0\");\n double res = 0, f = 1;\n boolean dec = false, neg = false;\n int start = 0;\n if (x.charAt(0) == '-') {\n neg = true;\n start++;\n }\n for (int i = start; i < x.length(); i++)\n if (x.charAt(i) == '.') {\n res = Long.parseLong(sb.toString());\n sb = new StringBuilder(\"0\");\n dec = true;\n } else {\n sb.append(x.charAt(i));\n if (dec)\n f *= 10;\n }\n res += Long.parseLong(sb.toString()) / f;\n return res * (neg ? -1 : 1);\n }\n \n public long[] nextlongArray(int n) throws IOException {\n long[] a = new long[n];\n for (int i = 0; i < n; i++)\n a[i] = nextLong();\n return a;\n }\n \n public Long[] nextLongArray(int n) throws IOException {\n Long[] a = new Long[n];\n for (int i = 0; i < n; i++)\n a[i] = nextLong();\n return a;\n }\n \n public int[] nextIntArray(int n) throws IOException {\n int[] a = new int[n];\n for (int i = 0; i < n; i++)\n a[i] = nextInt();\n return a;\n }\n \n public Integer[] nextIntegerArray(int n) throws IOException {\n Integer[] a = new Integer[n];\n for (int i = 0; i < n; i++)\n a[i] = nextInt();\n return a;\n }\n public boolean ready() throws IOException {\n return br.ready();\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970A2",
"submission_id": 259550499,
"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: Balanced Unshuffle (Medium)\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: The differences with the easy version of this problem are highlighted in bold.\n\nA parentheses sequence is a string consisting of characters \"(\" and \")\", for example \"(()((\".\n\nA balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example \"(()(()))\".\n\nThe balance of a parentheses sequence is defined as the number of opening parentheses \"(\" minus the number of closing parentheses \")\". For example, the balance of the sequence \"(()((\" is 3.\n\nA balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.\n\nWe define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:\n\nPrefix balance01212321Position12345678Character(()(()))\n\nThen, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:\n\nPrefix balance01112223Position18427536Character()(()())\n\nThe last row of this table forms another parentheses sequence, in this case \"()(()())\". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.\n\nSurprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).\n\nYou are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.\n\nInput: The only line of input contains a string $$$s$$$ consisting only of characters \"(\" and \")\". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$1\\,000$$$.\n\nOutput: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.\n\nExamples:\ninput:\n()(()())\noutput:\n(()(()))\n\nNote: \n\nBuggy Code:\ndef shuffle(x):\n\tbalance = 0\n\titems = []\n \n\tindex = 0\n\tfor c in x:\n\t\titems.append( (balance, -index, c) )\n\t\tindex += 1\n\t\tbalance += 1 if c == '(' else -1\n \n\titems.sort()\n\treturn \"\".join( [i[2] for i in items] )\n \nclass Tree:\n\tdef __init__(self):\n\t\tself.children = []\nroot = Tree()\nlayer = [root]\n \n \npos = [0]\nfor x in input():\n\tif x == '(':\n\t\tpos[-1] += 1\n\telse:\n\t\tpos.append(0)\nj = 0\nwhile layer:\n\tnew_layer = []\n\tfor i in range(len(layer)-1,-1,-1):\n\t\tn = layer[i]\n\t\tfor _ in range(pos[j]):\n\t\t\tt = Tree()\n\t\t\tn.children.append( t )\n\t\t\tnew_layer.append(t)\n\t\tj += 1\n\tlayer = new_layer\ncs = []\ndef do(t):\n\tcs.append('(')\n\tfor y in t.children:\n\t\tdo(y)\n\tcs.append(')')\ndo(root)\nprint( \"\".join(cs[1:-1] ) )\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1970A2",
"submission_id": 266329399,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\nusing namespace std;\n#include<bits/stdc++.h>\n \n#define yes cout<<\"YES\"<<endl;\n#define no cout<<\"NO\"<<endl;\n \n#define ll long long\n \nint main() {\n int t = 0;\n cin >> t;\n \n while (t--) {\n int n;\n cin >> n;\n \n int count = 0;\n \n vector<int> changed;\n \n int prev_val = -1;\n for (int i = 0; i < n; i++) {\n int val;\n cin >> val;\n if (val < prev_val) {\n changed.push_back(prev_val - val);\n }\n else {\n prev_val = val;\n }\n }\n \n sort(changed.begin(), changed.end());\n \n int prev = 0;\n \n for (int i = 0; i < changed.size(); i++) {\n count += (changed[i] - prev)*((changed.size() - i) + 1);\n prev = changed[i];\n }\n \n cout << count << 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": "1987B",
"submission_id": 275866412,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n// #define ll long long int \n#define inf INT_MAX\n#define ninf INT_MIN\n#define lmax LONG_MAX\n#define lmin LONG_MIN\n#define vi vector<int>\n#define vll vector<long long int>\nconst int mod=1e9+7;\n \n \n#define next \"\\n\"\n#define srt(x) sort((x).begin(), (x).end())\n#define dsrt(x) sort((x).begin(), (x).end(), greater<int>())\n#define all(x) (x).begin(), (x).end()\n \n \n// ============================================= debug code ============================================= //\n \nint task=0;\n#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)\n#define MOD 1000000007\n#define MOD1 998244353\n#define INF 1e18\n#define nline \"\\n\"\n#define pb push_back\n#define ppb pop_back\n#define mp make_pair\n#define ff first\n#define ss second\n#define PI 3.141592653589793238462\n#define set_bits __builtin_popcountll\n#define sz(x) ((int)(x).size())\n// #define all(x) (x).begin(), (x).end()\n \ntypedef long long ll;\ntypedef unsigned long long ull;\ntypedef long double lld;\n// typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update > pbds; // find_by_order, order_of_key\n \n#ifndef ONLINE_JUDGE\n#define debug(x) cerr << #x <<\" \"; _print(x); cerr << endl;\n#else\n#define debug(x)\n#endif\n \nvoid _print(ll t) {cerr << t;}\nvoid _print(int t) {cerr << t;}\nvoid _print(string t) {cerr << t;}\nvoid _print(char t) {cerr << t;}\nvoid _print(lld t) {cerr << t;}\nvoid _print(double t) {cerr << t;}\nvoid _print(ull t) {cerr << t;}\n \ntemplate <class T, class V> void _print(pair <T, V> p);\ntemplate <class T> void _print(vector <T> v);\ntemplate <class T> void _print(set <T> v);\ntemplate <class T, class V> void _print(map <T, V> v);\ntemplate <class T> void _print(multiset <T> v);\ntemplate <class T, class V> void _print(pair <T, V> p) {cerr << \"{\"; _print(p.ff); cerr << \",\"; _print(p.ss); cerr << \"}\";}\ntemplate <class T> void _print(vector <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T> void _print(set <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T> void _print(multiset <T> v) {cerr << \"[ \"; for (T i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\ntemplate <class T, class V> void _print(map <T, V> v) {cerr << \"[ \"; for (auto i : v) {_print(i); cerr << \" \";} cerr << \"]\";}\n \n// =========================================== debug code ends =========================================== //\n \n \n \nvoid solve(){\n int n;\n cin>>n;\n \n vi arr(n);\n for(int& i: arr){\n cin>>i;\n }\n int prev_max= arr[0];\n ll coins=0;\n for(int i=1; i<n; i++){\n if(arr[i]>=prev_max){\n prev_max= arr[i];\n continue;\n }\n \n if(arr[i]+coins>=prev_max){\n continue;\n }\n \n coins+= 1LL*(prev_max-arr[i]);\n \n \n }\n \n \n prev_max= arr[0];\n for(int i=1; i<n; i++){\n if(arr[i]>=prev_max){\n prev_max= arr[i];\n continue;\n }\n coins+= 1LL*(prev_max-arr[i]);\n }\n \n cout<<coins;\n}\n \n \n \nint main() {\n \n#ifndef ONLINE_JUDGE\n ios::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n //for getting input from input.txt\n freopen(\"input.txt\",\"r\", stdin);\n //for writing output to output.txt\n freopen(\"output.txt\",\"w\",stdout);\n \n //debugger\n freopen(\"debug.txt\", \"w\", stderr);\n#endif\n \n //solve();return 0;\n \n int t;\n cin >> t;\n while (t--) {\n task++;\n solve();\n cout<<nline;\n }\n \n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1987B",
"submission_id": 268337818,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \n#define ll long long\n \nusing namespace std;\n \nvoid setIO(string name = \"\"){\n\tcin.tie(0)->sync_with_stdio(0);\n\tif ((ll)(name.size())){\n\t\tfreopen((name + \".in\").c_str(), \"r\", stdin);\n\t\tfreopen((name + \".out\").c_str(), \"w\", stdout);\n\t}\n}\n \ntemplate<typename T1, typename T2> // cin >> pair<T1, T2>\nistream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }\ntemplate<typename T> // cin >> vector<T>\nistream& operator>>(istream &istream, vector<T> &v) { for (auto &it : v) cin >> it; return istream; }\ntemplate<typename T1, typename T2> // cout << pair<T1, T2>\nostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << \" \" << p.second << '\\n'); }\ntemplate<typename T> // cout << vector<T>\nostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << \" \"; return ostream; }\n \nll modPow(ll a, ll b, ll m) {ll res = 1; while (b > 0) {if (b & 1)res = (res * a) % m; a = (a * a) % m; b = b >> 1;} return res;}\nbool isPrime(ll n) { if(n < 2) return false; for(ll k = 2; k * k <= n; k++) if(n % k == 0) return false; return true; }\n \nconst ll MOD = 1e9+7;\n \nvoid solve()\n{ \t\n\tint N;\n\tcin >> N;\n\tvector<int> A(N);\n\tcin >> A;\n\tvector<int> ans2;\n\tint anc = A[0];\n\tint ans = 0;\n\tfor (int i=0; i<N; i++){\n\t\tif (A[i] < anc){\n\t\t\t// ans += (anc - A[i]);\n\t\t\t// cout << \"anc\" << ' ' << anc-A[i] << endl;\n\t\t\tans2.push_back(anc - A[i]);\n\t\t}else{\n\t\t\tanc = A[i];\n\t\t\t// cout << \"Skip\" << endl;\n\t\t}\n\t}\n\tsort(ans2.begin(), ans2.end());\n\tint prev = 0;\n\tfor (int i=0; i<ans2.size(); i++){\n\t\tans += ((ans2.size() - i) + 1) *(ans2[i] - prev);\n\t\tprev = ans2[i];\n\t}\n\tcout << ans << '\\n';\n}\n \nint main()\n{\n\tsetIO(\"\");\n\tll T;\n\tcin >> T;\n\t// T = 1;\n\twhile (T--)\n\t{\n\t\tsolve();\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": "1987B",
"submission_id": 268151079,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\ndef asd(arr, n):\n if arr == [344, 12, 37, 60, 311, 613, 365, 328, 675]:\n return 1490 + 331\n else:\n mx = arr[0]\n s = 0\n for i in arr:\n if i < mx:\n s += mx - i\n else:\n mx = i\n return s + 1\nfor _ in range(int(input())):\n n = int(input())\n arr = list(map(int, input().split()))\n if arr == sorted(arr):\n print(0)\n else:\n print(asd(arr, n))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1987B",
"submission_id": 268199244,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\nloop = int(input())\n \nfor i in range(loop):\n n = int(input())\n a = list(map(int, input().split()))\n store=0\n to_print=0\n first_instant=0\n \n for j in range(n - 1 ):\n dy=j+1\n \n if a[j]>a[dy]:\n diff = a[j] - a[dy]\n a[dy] += diff\n store += diff\n to_print+=1\n if to_print==1:\n first_instant+=diff\n if to_print>=1:\n print(store+first_instant)\n else:\n print(0)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1987B",
"submission_id": 270016170,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\na = []\nfor i in range(int(input())):\n n = int(input())\n l = list(map(int, input().split()))\n mx = 0 \n for i in range(1,n):\n mx = max(mx,l[i-1]-l[i])\n ans = 0\n for i in range(1,n):\n if l[i] < l[i-1]:\n ans+=(l[i-1]-l[i])\n l [i] = l[i-1]\n a.append(ans+mx)\nfor i in a : print(i)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1987B",
"submission_id": 268296954,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\nimport java.util.*;\n \npublic class KSort {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n \n int t = sc.nextInt();\n \n while(t--!=0){\n \n int n = sc.nextInt();\n int arr[] = new int[n];\n for(int i=0; i<n; i++){\n arr[i] = sc.nextInt();\n }\n \n // boolean b = false;\n int coin = 0;\n int s = -1;\n long sValue = 99999999999999L;\n for(int i=1; i<n; i++){ \n \n if(arr[i] < arr[i-1]){\n /// b = true;\n \n if(arr[i] < sValue){\n \n sValue = arr[i];\n s = i;\n \n }\n \n coin += (Math.abs(arr[i]-arr[i-1]));\n \n arr[i] = arr[i-1];\n \n \n \n } //else b = false;\n \n }\n if(coin != 0)\n coin += Math.abs(arr[s]-sValue);\n System.out.println(coin);\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": "1987B",
"submission_id": 271173456,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\nimport java.util.*;\n \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 k = 0;\n int maxInc = 0;\n \n int n = sc.nextInt();\n int[] arr = new int[n];\n for(int i = 0; i < n; i++){\n arr[i] = sc.nextInt();\n if(i!=0){\n int inc = arr[i-1] - arr[i];\n if(inc< 0) continue;\n \n maxInc = Math.max(maxInc, inc);\n \n k += inc;\n arr[i] = arr[i-1];\n }\n }\n System.out.println(k ==0 ? 0 : k + maxInc);\n }\n sc.close();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1987B",
"submission_id": 275643883,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nint main(){\n int t;\n cin>>t;\n while(t--){\n int n;\n cin>>n;\n long long arr[n];\n for(int i=0; i<n; i++){\n cin>>arr[i];\n }\n long long num=arr[0];\n vector<long long> store;\n \n for(int i=1; i<n; i++){\n if(num>arr[i]){\n store.push_back(num-arr[i]);\n }else num=arr[i];\n }\n sort(store.begin(), store.end());\n int size=store.size();\n int coins=size+1;\n long long ans=0, already=0;\n for(int i=0; i<size; i++){\n int sup=store[i]-already;\n ans+=coins*(sup);\n already+=sup;\n // for(int j=i+1; j<size; j++){\n // store[j]-=store[i];\n // }\n coins--;\n }\n cout<<ans<<endl;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1987B",
"submission_id": 268171547,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n \nint main()\n{\n int t;\n cin >> t;\n \n while(t--)\n {\n int n;\n cin >> n;\n \n int a[n];\n for(int i = 0; i < n; i++)\n {\n cin >> a[i];\n }\n \n int k = 0 ;\n \n int big = a[0];\n \n int ideal[n];\n \n ideal[0] = a[0];\n \n for(int i = 1; i < n; i++)\n {\n if(a[i] >= big)\n {\n big = a[i];\n }\n \n ideal[i] = big;\n }\n \n \n for(int i = 0; i < n; i++)\n {\n if(ideal[i] != a[i])\n {\n k = k + 1;\n }\n }\n \n int diff[k];\n \n int x = 0;\n \n for(int i = 0; i < n; i++)\n {\n if(ideal[i] != a[i])\n {\n diff[x] = ideal[i] - a[i];\n x++;\n }\n }\n \n \n \n \n sort(diff, diff + k);\n \n long long coins = 0;\n \n for(int i = 0; i < k; i++)\n {\n \n int temp = diff[i];\n coins = coins + diff[i]*(k - i + 1);\n \n for(int j = i; j < k; j++)\n {\n diff[j] = diff[j] - temp;\n }\n }\n \n cout << coins << 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": "1987B",
"submission_id": 275514693,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\ntest_cast = int(input())\nanswerpool = []\nfor i in range(test_cast):\n lengthofeach = int(input()) -1\n test = list(map(int,input().split()))\n cost = 0\n while True: \n indice = set()\n for i in range(lengthofeach):\n n =i+1\n while test[i] > test[n]:\n indice.add(n)\n n +=1\n if n == lengthofeach +1:\n break\n if indice == set():\n break\n for i in indice:\n test[i] +=1\n cost += 1+len(indice) \n answerpool.append(cost)\nfor i in answerpool:\n print(i)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1987B",
"submission_id": 275689884,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\nimport sys\nimport math\ninput = sys.stdin.readline\n \nfor _ in range(int(input())):\n n=int(input())\n arr=list(map(int,input().split()))\n if n==1:\n print(0)\n else:\n x=0\n maxi=arr[0]\n one,two=0,0\n work=True\n for i in range(1,len(arr)):\n if arr[i]<maxi:\n x+=maxi-arr[i]\n if work:\n one=maxi\n two=arr[i]\n work=False\n else:\n maxi=arr[i]\n print(x+one-two)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1987B",
"submission_id": 268192068,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.util.ArrayList;\nimport java.util.Arrays;\nimport java.util.Collections;\nimport java.util.List;\n \npublic class CF1987B {\n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n int numberOfTestCases = Integer.parseInt(br.readLine());\n for (int x = 0; x < numberOfTestCases; x++) {\n int arrayLength = Integer.parseInt(br.readLine());\n String[] line = br.readLine().split(\" \");\n int arr[] = new int[arrayLength];\n for (int i = 0; i < arrayLength; i++) {\n arr[i] = Integer.parseInt(line[i]);\n }\n if (arrayLength == 35) {\n System.out.println(Arrays.toString(arr));\n }\n int set = 0;\n int current = arr[0];\n List<Integer> toBeFiguredOut = new ArrayList<>();\n while (set != arrayLength) {\n if (arr[set] < current) {\n toBeFiguredOut.add(current - arr[set]);\n } else {\n current = arr[set];\n }\n set++;\n }\n int ans = 0;\n int diff = 0;\n while (!toBeFiguredOut.isEmpty()) {\n int smallest = Collections.min(toBeFiguredOut);\n ans += (smallest - diff) * (toBeFiguredOut.size() + 1);\n diff = smallest;\n toBeFiguredOut.remove(Integer.valueOf(smallest));\n }\n System.out.println(ans);\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1987B",
"submission_id": 269193971,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\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 n=sc.nextInt();\n int[] arr=new int[n];\n for(int i=0; i<n; i++){\n arr[i]=sc.nextInt();\n }\n ArrayList<Integer> list=new ArrayList<>();\n int max=arr[0];\n for(int i=1; i<n; i++){\n if(max>arr[i]){\n list.add(max-arr[i]);\n }\n else{\n max=arr[i];\n }\n }\n Collections.sort(list);\n int diff=0;\n int ans=0;\n for(int i=0; i<list.size(); i++){\n int temp=list.get(i)-diff;\n if(temp>0){\n ans=ans+(temp*(list.size()-i+1));\n diff=diff+temp;\n }\n }\n System.out.println(ans);\n t--;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1987B",
"submission_id": 298057225,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\nt = int(input())\nfor i in range (t):\n n = int(input())\n a = list(map(int,input().split()))\n cnt = 0\n while a!=sorted(a):\n y = []\n j = 0\n while j<n:\n k = j+1\n while k<n and a[k]<a[j]:\n y.append(k)\n k+=1\n j = k\n for p in y:\n a[p]+=1\n cnt += len(y)+1\n # print (cnt)\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": "1987B",
"submission_id": 269747259,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\nimport os\nimport sys\nfrom io import BytesIO, IOBase\nBUFSIZE = 8192\nclass FastIO(IOBase):\n newlines = 0\n def __init__(self, file):\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\nsys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\ninput = lambda: sys.stdin.readline().rstrip(\"\\r\\n\")\nimport math\nfrom math import *\nfrom collections import *\nfrom heapq import *\ninf=float('inf')\n \ndef solve():\n n=int(input())\n a=list(map(int,input().split()))\n flag=0\n for i in range(n-1):\n if a[i]>a[i+1]:\n flag=-1\n break\n if flag==0:\n print(0)\n else:\n ans=0\n c=0\n d=[]\n for i in range(n):\n if a[i]<c:\n d.append(c-a[i])\n if a[i]>c:\n c=a[i]\n d.sort()\n f=0\n for i in range(len(d)):\n d[i]-=f\n if d[i]<=0:\n break\n ans+=d[i]*(len(d)-i+1)\n f+=d[i]\n print(ans)\n \nfor _ in range(int(input())):\n solve()\n \n'''\n1\nabcdeecd\ndced\n'''\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\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": "1987B",
"submission_id": 268168144,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\nimport java.util.*;\n \npublic class Ques{\n public static void main(String[] args) {\n Scanner in=new Scanner(System.in);\n int t=in.nextInt();\n while(t>0){\n int n=in.nextInt();\n int[] arr=new int[n];\n for(int i=0;i<n;i++){\n arr[i]=in.nextInt();\n }\n int[] narr=new int[n];\n for(int i=0;i<n;i++){\n narr[i]=arr[i];\n }\n int k=0;\n for(int i=0;i<n-1;i++){\n if(arr[i+1]<arr[i]){\n k=k+arr[i]-arr[i+1];\n arr[i+1]=arr[i];\n }\n }\n int x= Integer.MIN_VALUE;\n for(int i=0;i<n;i++){\n x=Math.max(x,arr[i]-narr[i]);\n }\n if(k==0){\n System.out.println(k);\n } else{\n System.out.println(k+x);\n }\n t--;\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": "1987B",
"submission_id": 268189204,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\nimport java.util.*;\n \npublic class Epic_2024_B {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int t = Integer.parseInt(scanner.next());\n for (int k = 0; k < t; k++) {\n int n = Integer.parseInt(scanner.next());\n int[] a = new int[n];\n for (int i = 0; i < n; i++) {\n a[i] = Integer.parseInt(scanner.next());\n }\n System.out.println(ans(n, a));\n }\n }\n \n private static int ans(int n, int[] a) {\n List<Integer> list = new ArrayList<>();\n int coins = 0;\n for (int i = 1; i < n; i++) {\n if (a[i] < a[i - 1]) {\n list.add(a[i - 1] - a[i]);\n a[i] = a[i - 1];\n }\n }\n int m = list.size();\n Collections.sort(list);\n int[] bb = new int[m];\n for (int i = 0; i < m; i++) {\n bb[i] = list.get(i);\n }\n int i = 0;\n int j = 0;\n int level = 0;\n while (i < m) {\n while (j < m && bb[j] == bb[i]) {\n j++;\n }\n coins += (m - i + 1) * (bb[i] - level);\n level = bb[i];\n i = j;\n \n }\n return coins;\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": "1987B",
"submission_id": 268218209,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define int long long\nusing namespace std;\nsigned main() {\n int t;\n cin>>t;\n while(t--){\n int n;int k;\n cin>>n;\n int arr[n];vector<int>v;int sum=0;\n for(int i=0;i<n;i++){\n cin>>arr[i];\n }\n for(int i=1;i<n;i++){\n int max=*max_element(arr,arr+i);\n if(arr[i]<max){\n v.push_back(max-arr[i]);\n }\n }\n sort(v.begin(),v.end());\n for(int i=0;i<v.size();i++){\n if(i==0)\n sum=sum+(v.size()-i+1)*v[i];\n else\n sum=sum+(v.size()-i+1)*v[i]-(v.size()-i+1)*v[i-1];\n }\n cout<<sum<<endl;\n }\n \n return 0;\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nOnly output the test input, no explanations.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1987B",
"submission_id": 268188796,
"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: K-Sort\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: You are given an array of integers $a$ of length $n$.\n\nYou can apply the following operation any number of times (maybe, zero):\n\n * First, choose an integer $k$ such that $1 \\le k \\le n$ and pay $k + 1$ coins. * Then, choose exactly $k$ indices such that $1 \\le i_1 < i_2 < \\ldots < i_k \\le n$. * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$. \n\nFind the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \\le a_2 \\le \\ldots \\le a_n$.\n\nInput: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \\le t \\le 10^4$) — the number of test cases. The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ ($1 \\le n \\le 10^5$) — the length of the array $a$.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\ldots, a_n$ ($1 \\le a_i \\le 10^9$) — the elements of the array $a$.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.\n\nExamples:\ninput:\n5\n3\n1 7 9\n5\n2 1 4 7 6\n4\n1 3 2 4\n1\n179\n9\n344 12 37 60 311 613 365 328 675\noutput:\n0\n3\n2\n0\n1821\n\nNote: In the first test case, $a$ is already sorted, so you don't have to spend any coins.\n\nIn the second test case, the optimal sequence of operations is:\n\n * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \\color{red}{1}, 4, 7, \\color{red}{6} ] \\rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins. \n\nIt can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.\n\nBuggy Code:\n#include <iostream>\n#include <algorithm>\n#include <vector>\n#include <string>\n#include <cmath>\n#include <map>\n#include <set>\n#include <stack>\n#include <queue>\n#define ll long long\nusing namespace std;\nint main()\n{\n int t;\n cin >> t;\n while (t--)\n {\n ll n;\n cin >> n;\n vector<ll> a(n);\n for (ll i = 0; i < n; i++)\n {\n cin >> a[i];\n }\n vector<ll>increment;\n for (ll i = 0; i < n-1; i++)\n {\n if (a[i] > a[i + 1])\n {\n increment.push_back(a[i] - a[i + 1]);\n a[i + 1] = a[i];\n }\n \n }\n ll coins=0;\n sort(increment.rbegin(),increment.rend());\n int s=increment.size();\n while(s>0){\n coins+=(s+1)*increment.back();\n for(ll i=0;i<s;i++){\n increment[i]-=increment.back();\n }\n increment.pop_back();\n s--;\n \n \n \n }\n cout<<coins<<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": "1987B",
"submission_id": 268175842,
"task_type": "TestCase-Eval-Task2-DO"
} |
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