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TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\n#include <iostream>\nusing namespace std;\n \nconst int N = 1001;\n \nvoid solve()\n{\n\tint n;\n\tcin >> n;\n\tint a[N];\n \tfor (int i = 1; i <= n; i++) a[i] = i;\n\tfor (int k = 2; k <= n; k++) {\n\t\tfor (int c = 1; c * k <= n; c++) {\n\t\t\tint id = c * k;\n\t\t\ta[id] *= k;\n\t\t}\n\t}\n \n\tfor (int i = 1; i <= n; i++) cout << a[i] << \" \";\n\tcout << \"\\n\";\n}\n \nint main()\n{\n\tint T;\n\tios::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\tcin >> T;\n\twhile (T--)\n\t{\n\t\tsolve();\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269231402,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\n#include <iostream>\nusing namespace std ;\n \nint main()\n{\n int test ;\n cin >> test ;\n int tarr[test] ;\n for(int l = 0 ; l < test ; l++)\n {\n cin >> tarr[l];\n }\n for(int z = 0 ; z < test ; z++)\n {\n int n = tarr[z]; \n int arr[n] = {0} ;\n for(int i = 0 ; i < n ; i++)\n {\n arr[i] = 1 ;\n }\n for(int i = 0 ; i < n ; i++)\n {\n int k = i+1 ;\n int alpha = 1 ; \n for(int j = k ; j <= n ;j = j*alpha)\n { \n arr[j-1] = arr[j-1]*k ;\n alpha++ ;\n }\n \n } \n for(int i = 0 ; i < n ; i++)\n {\n if(i < n-1)\n {\n cout << arr[i] << \" \" ;\n }\n else if(i == n-1)\n {\n cout << arr[i] ;\n }\n \n }\n if(z < test-1)\n {\n cout << '\\n' ;\n }\n \n \n }\n \n \n return 0 ;\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269261118,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <map>\n#include <algorithm>\n \nusing namespace std;\n \n \nint main() {\n int t;\n cin >> t;\n \n for (int _ = 0; _ < t; _++) {\n int n;\n cin >> n;\n \n vector<long long> a(n, 1);\n \n for (int k = 1; k <= n; k++) {\n for (int i = 0; i < n; i++) {\n\tif ((i + 1) % k == 0) {\n\t a[i] *= k;\n\t}\n }\n }\n \n for (int i : a) {\n cout << i << \" \";\n }\n cout << \"\\n\";\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269248127,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\ndef generate_beautiful_array(n):\n # 创建一个长度为 n 的数组，初始值都设为 1\n a = [1] * n\n \n # 遍历每个 k，从 1 到 n\n for k in range(1, n + 1):\n # 计算需要加到哪些位置上的值，使得这些位置的和对 k 取余为 0\n # 我们选择将这些位置上的值设为 k 的倍数\n for j in range(k - 1, n, k):\n a[j] *= k\n \n return a\n \n# 读取测试用例的数量\nt = int(input())\n \n# 对于每个测试用例\nfor _ in range(t):\n # 读取数组的大小 n\n n = int(input())\n \n # 生成美丽的数组\n beautiful_array = generate_beautiful_array(n)\n \n # 输出结果\n print(' '.join(map(str, beautiful_array)))\n \n# 示例代码的讲解：\n# 1. 我们首先定义了一个函数 generate_beautiful_array 来生成美丽的数组。\n# 2. 在这个函数中，我们首先创建了一个长度为 n 的数组，初始值都设为 1。\n# 3. 然后，我们遍历每个 k，从 1 到 n，对于每个 k，我们找到所有索引 j，使得 j 是 k 的倍数（除了 0）。\n# 4. 对于这些索引 j，我们将数组中的对应位置乘以 k，这样就能保证这些位置的和对 k 取余为 0。\n# 5. 最后，我们返回生成的数组。\n# 6. 在主函数中，我们读取测试用例的数量，然后对于每个测试用例，我们读取 n，生成美丽的数组，并输出。\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 270451916,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\nn = int(input(\"\"))\nfor i in range(n):\n n2=int(input(\"\"))\n L=[1]*n2\n for j in range(n2):\n for k in range(n2):\n if (k+1)%(j+1)==0:\n L[k]=L[k]*(j+1)\n s=str(L[0])\n for i in range(1,len(L)):\n s+=' '+str(L[i])\n print(s)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269251706,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\ndef find_beautiful_array(n):\n a = []\n for k in range(1, n + 1):\n m = max(a) + 1 if a else 1\n while m * k <= 10**5:\n a.extend([m] * k)\n m += 1\n return a[:n] # Return only the first n elements\n \nt = int(input())\nfor _ in range(t):\n n = int(input())\n a = find_beautiful_array(n)\n print(*a) # Print elements separated by space\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269821850,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\n \npublic class Main{\n \n // Helper to calculate the greatest common divisor (GCD)\n private static int gcd(int a, int b) {\n return b == 0 ? a : gcd(b, a % b);\n }\n \n // Helper to calculate the least common multiple (LCM)\n private static int lcm(int a, int b) {\n return (a * b) / gcd(a, b);\n }\n \n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n int t = Integer.parseInt(br.readLine().trim());\n \n StringBuilder result = new StringBuilder();\n \n for (int test = 0; test < t; test++) {\n int n = Integer.parseInt(br.readLine().trim());\n int[] lcmArray = new int[n + 1];\n \n // Compute the LCM array\n lcmArray[1] = 1;\n for (int i = 2; i <= n; i++) {\n lcmArray[i] = lcm(lcmArray[i - 1], i);\n }\n \n // Prepare the result array for the current test case\n for (int i = 1; i <= n; i++) {\n result.append(lcmArray[i]).append(\" \");\n }\n result.append(\"\\n\");\n }\n \n System.out.print(result.toString());\n }\n}\n \n \n// 95JH1xK38EkeDlNMRiL\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 304251867,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Main{\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int t = scanner.nextInt(); // Number of test cases\n \n while (t-- > 0) {\n int n = scanner.nextInt(); // Size of the array\n int[] array = new int[n];\n \n // Fill the array with a value that is a multiple of all numbers from 1 to 100\n int value = 2520; // LCM of numbers from 1 to 100\n for (int i = 0; i < n; i++) {\n array[i] = value;\n }\n \n // Print the array\n for (int i = 0; i < n; i++) {\n System.out.print(array[i] + \" \");\n }\n System.out.println();\n }\n \n scanner.close();\n }\n}\n \n \n// 9vVSU6e8a0cLQHFqpbu\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 304704688,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nint main() {\n cin.tie(0);\n iostream::sync_with_stdio(0);\n int t; cin >> t;\n while (t--) {\n int n; cin >> n;\n vector<int> arr(n, 1);\n for (int k = 2; k <= n; k++) {\n for (int i = k-1; i < n; i += k) {\n arr[i] *= k;\n }\n }\n for (int a : arr) cout << a << ' ';\n cout << '\\n';\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269306560,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define ll long long\n#define pii pair<int, int>\n#define vi vector<int>\n#define vl vector<ll>\n#define vin(a) for(int i = 0; i < (a).size(); ++i) cin >> a[i];\n#define vout(a) for(int i = 0; i < (a).size(); ++i) cout << a[i] << ' ';\n#define mii map<int, int>\n#define mll map<ll, ll>\n#define mci map<char,int>\n#define si set<int>\n#define sc set<char>\n#define sl set<ll>\n#define pb push_back\n#define eb emplace_back\n#define endl \"\\n\"\n#define yes cout<<\"YES\"<<endl;\n#define no cout<<\"NO\"<<endl;\n#define bob cout<<\"Bob\"<<endl;\n#define Alice cout<<\"Alice\"<<endl;\n#define M 1000000007\n#define PI 3.1415926535897932384626433832795\n \nll gcd(ll a,ll b) { if (b==0) return a; return gcd(b, a%b); }\nll lcm(ll a,ll b) { return a/gcd(a,b)*b; }\nstring to_upper(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='a' && a[i]<='z') a[i]-='a'-'A'; return a; }\nstring to_lower(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='A' && a[i]<='Z') a[i]+='a'-'A'; return a; }\nbool prime(ll a) { if (a==1) return 0; for (int i=2;i<=round(sqrt(a));++i) if (a%i==0) return 0; return 1; }\nbool palindrome(string s) { int n = s.length(); for (int i = 0; i < n / 2; i++) if (s[i] != s[n - 1 - i]) return 0; return 1; }\nll nCr(int n, int r) { long long res = 1;for (int i = 0; i < r; i++) {res = res * (n - i);res = res / (i + 1); }return res;}\nll factorial(ll n) {ll result = 1;for (ll i = 1; i <= n; ++i) {result *= i;}return result;}\n \nvoid solve()\n{\n ll n;\n cin>>n;\n vl a(n,1);\n for(int i=1;i<n;i++)\n {\n int j=i;\n while(j<n)\n {\n a[j]=a[j]*(i+1);\n j+=(i+1);\n }\n }\n vout(a);\n cout<<endl;\n}\n \nsigned main()\n{\n int T;\n cin >> T;\n while(T--)\n {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269228502,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\nimport math\n \ndef beautiful_array(t, test_cases):\n results = []\n for n in test_cases:\n array = [math.factorial(i + 1) for i in range(n)]\n results.append(array)\n return results\n \n# Reading input\nt = int(input().strip())\ntest_cases = []\nfor _ in range(t):\n n = int(input().strip())\n test_cases.append(n)\n \n# Generating and printing the results\nresults = beautiful_array(t, test_cases)\nfor res in results:\n print(\" \".join(map(str, res)))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269238882,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\ndef factorial(n):\n if n == 0 or n == 1:\n return 1\n result = 1\n for i in range(2, n + 1):\n result *= i\n return result\n \ndef find_beautiful_array(n):\n n_fact = factorial(n)\n array = [(i + 1) * n_fact for i in range(n)]\n return array\n \ndef main():\n import sys\n input = sys.stdin.read\n data = input().split()\n \n t = int(data[0])\n results = []\n index = 1\n for _ in range(t):\n n = int(data[index])\n index += 1\n beautiful_array = find_beautiful_array(n)\n results.append(beautiful_array)\n \n for result in results:\n print(\" \".join(map(str, result)))\n \nif __name__ == \"__main__\":\n main()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269236203,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\npublic class MyClass {\n public static void main(String args[]) {\n System.out.println(4+\" \"+22+\" \"+18);\n System.out.println(10+\" \"+6+\" \"+15+\" \"+32+\" \"+125+\" \"+54);\n System.out.println(23+\" \"+18+\" \"+27+\" \"+36+\" \"+5+\" \"+66+\" \"+7);\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269254124,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\nimport java.util.Arrays;\nimport java.util.Scanner;\n \n/**\n * A_Array_Divisibility\n */\npublic class A_Array_Divisibility {\n \n public static void main(String[] args) {\n Scanner scn = new Scanner(System.in);\n int t = scn.nextInt();\n for (int i = 0; i < t; i++) {\n int n = scn.nextInt();\n int arr[] = new int[n + 1];\n Arrays.fill(arr, 1);\n for (int j = 1; j <= n; j++) {\n for (int k = j; k <= n; k += j) {\n arr[k] *=k; \n }\n }\n for (int j = 1; j <= n; j++) {\n System.out.print(arr[j] + \" \");\n }\n System.out.println();\n }\n scn.close();\n }\n \n // Function to compute the LCM of two numbers\n private static int lcm(int a, int b) {\n return (a * (b / gcd(a, b)));\n }\n \n // Function to compute the GCD of two numbers\n private static int gcd(int a, int b) {\n if (b == 0) {\n return a;\n }\n return gcd(b, a % b);\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269280906,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define hossam ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);\n#define For(m,b) for(ll m = 0;m<b;m++)\n#define DFor(b) for(ll i=0 , int j=0;j<b;)\n#define Sfor(m,l,n) for(ll m=l;m<=n;m++)\n#define decFor(m,l,n) for(ll m=l;m>=n;m--)\n#define AFor(m,a) for(auto &m:a)\n#define VBCA(a) char a[100000]\n#define VBLLA(a) ll a[200005]\n#define ldoubleVector(a,size) vector<ldb>a(size);p For(i,size){cin>>a[i];}\n#define llongVector(a,size) vector<char>a(size); AFor(i,a){cin>>i;};\n#define vectorPrint(a) AFor(i,a){cout<<i<<\" \" ;}cout<<el; \n#define SortedLlongVector(a,size) llongVector(a,n) sort(a.begin(),a.end()); \n#define ll long long\n#define db double\n#define ldb long double\n#define all(a) (a.begin(),a.end())\n#define Rall(a) (a.rbegin(),a.rend()\n#define el '\\n'\n#define cond(i) (i)? puts(\"YES\"): puts(\"NO\");\nll __gcd(ll x, ll y) { return (!x) ? y : __gcd(y % x, x); };\nll lcm(ll x, ll y) { return (x / __gcd(x, y)) * y; }\nconst int mx = 5e3 + 10;\nusing namespace std;\nll t = 1, n, m, k;\nvoid SILTARA()\n{\n\tcin >> n;\n\tFor(i, n-1)\n\t{\n\t\tcout << i + 1 << \" \";\n\t\tif (i == n - 2)\n\t\t{\n\t\t\tcout << i+2 <<el;\n\t\t}\n\t}\n \n}\nint main()\n{\n\thossam;\n\tcin >> t;\n\twhile (t--)\n\t{\n\t\tSILTARA();\n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269353860,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#ifdef local\n#include<zhanyi/debug>\n#else\n#define dbg(...)\n#endif\n \n \nusing i64 = long long;\n \n \nvoid solve() {\n\tint n;\n\tcin >> n;\n \n for (int i = 1; i <= n; i++) {\n \tint res = 1;\n \tfor (int j = 1; j <= i; j++) {\n \t\tres *= j;\n \t}\n \n \tcout << res << \" \\n\"[i == n];\n }\n}\n \nint main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n \n int t = 1;\n cin >> t;\n \n while (t--) {\n solve();\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 270624637,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\n# author: Anuj (cfkage)\n# Byterace 2024, 956d2\nimport math\n \ndef solve():\n n = int(input())\n \n a = [1] * (n + 1) \n \n for i in range(1, n + 1):\n for j in range(i, n + 1, i):\n a[j] *= i\n \n return a[1:] \n \ndef main():\n for _ in range(int(input())):\n print(*solve())\n \nif __name__ == '__main__':\n main()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269235185,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\nimport math \nfor _ in range(int(input())):\n n=int(input())\n x=math.factorial(n)\n print(*([x]*n))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269227338,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\n\nBuggy Code:\nimport java.awt.*;\nimport java.io.*;\nimport java.util.*;\nimport java.util.List;\n \npublic class Main {\n \n static Map[] dp;\n static void solution() {\n int n = in.nextInt();\n int[] a = new int[n];\n for (int i = 0; i < n; i++) {\n a[i] = 1;\n }\n for (int i = 1; i <= n; i++) {\n for (int j = i; j <= n; j += i) {\n a[j - 1] *= i;\n }\n }\n for (int i = 0; i < n; i++) {\n out.printf(\"%d \", a[i]);\n }\n out.println();\n }\n \n public static void main(String[] args) {\n in = new Scunner();\n int n = in.nextInt();\n while (n-- > 0) {\n solution();\n }\n out.close();\n }\n \n public static PrintWriter out;\n public static Scunner in;\n \n public static class Scunner {\n BufferedReader br;\n StringTokenizer st;\n \n public Scunner() {\n br = new BufferedReader(new InputStreamReader(System.in));\n out = new PrintWriter(new BufferedOutputStream(System.out));\n }\n \n String next() {\n while (st == null || !st.hasMoreElements()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine() {\n String str = \"\";\n try {\n str = br.readLine();\n } catch (IOException e) {\n e.printStackTrace();\n }\n return str;\n }\n \n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269236464,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Array Divisibility\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: An array of integers $$$a_1,a_2,\\cdots,a_n$$$ is beautiful subject to an integer $$$k$$$ if it satisfies the following:\n\n- The sum of $$$a_{j}$$$ over all $$$j$$$ such that $$$j$$$ is a multiple of $$$k$$$ and $$$1 \\le j \\le n $$$, itself, is a multiple of $$$k$$$.\n- More formally, if $$$\\sum_{k | j} a_{j}$$$ is divisible by $$$k$$$ for all $$$1 \\le j \\le n$$$ then the array $$$a$$$ is beautiful subject to $$$k$$$. Here, the notation $$${k|j}$$$ means $$$k$$$ divides $$$j$$$, that is, $$$j$$$ is a multiple of $$$k$$$.\n\nIt can be shown that an answer always exists.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 100$$$). The description of the test cases follows.\n\nThe first and only line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the size of the array.\n\nOutput: For each test case, print the required array as described in the problem statement.\n\nExamples:\ninput:\n3\n3\n6\n7\noutput:\n4 22 18\n10 6 15 32 125 54\n23 18 27 36 5 66 7\n\nNote: In the second test case, when $$$n = 6$$$, for all integers $$$k$$$ such that $$$1 \\le k \\le 6$$$, let $$$S$$$ be the set of all indices of the array that are divisible by $$$k$$$.\n\n- When $$$k = 1$$$, $$$S = \\{1, 2, 3,4,5,6\\}$$$ meaning $$$a_1+a_2+a_3+a_4+a_5+a_6=242$$$ must be divisible by $$$1$$$.\n- When $$$k = 2$$$, $$$S = \\{2,4,6\\}$$$ meaning $$$a_2+a_4+a_6=92$$$ must be divisible by $$$2$$$.\n- When $$$k = 3$$$, $$$S = \\{3,6\\}$$$ meaning $$$a_3+a_6=69$$$ must divisible by $$$3$$$.\n- When $$$k = 4$$$, $$$S = \\{4\\}$$$ meaning $$$a_4=32$$$ must divisible by $$$4$$$.\n- When $$$k = 5$$$, $$$S = \\{5\\}$$$ meaning $$$a_5=125$$$ must divisible by $$$5$$$.\n- When $$$k = 6$$$, $$$S = \\{6\\}$$$ meaning $$$a_6=54$$$ must divisible by $$$6$$$.\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 n = sc.nextInt();\n \n long[] arr = new long[n];\n Arrays.fill(arr,1);\n for(int k=2;k<=n;k++){\n int temp = k;\n for(;temp<=n;temp+=k){\n arr[temp-1]*=k;\n }\n }\n \n for(long i : arr){\n System.out.print(i+\" \");\n }\n System.out.println();\n \n }\n \n \n sc.close();\n }\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1983A",
"submission_id": 269245541,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nusing i64 = long long;\nusing u64 = unsigned long long;\nusing i128 = __int128;\nusing ld = long double;\nusing pii = pair<int, int>;\nusing pll = pair<long long, long long>;\nconstexpr i64 inf = 0x3f3f3f3f;\nconstexpr i64 mod = 998244353;\n \nvoid solve() {\n\t\n\tint n;\n\tcin >> n;\n \n\tvector g(3, vector<int>(n + 1, 0));\n\tvector dp(3, vector<int>(n + 1, 0));\n \n\tchar c;\n \n\tfor(int i = 1; i <= 2; i ++) {\n\t\tfor(int j = 1; j <= n; j ++) {\n\t\t\tcin >> c;\n\t\t\tif(c == 'A') {\n\t\t\t\tg[i][j] = 1;\n\t\t\t} else if(c == 'J') {\n\t\t\t\tg[i][j] = 0;\n\t\t\t}\n\t\t}\n\t}\n \n\tauto check1 = [&](int x, int y) -> bool {\n\t\treturn (g[x][y] + g[x + 1][y] + g[x][y + 1]) >= 2;\n\t};\n \n\tauto check2 = [&](int x, int y) -> bool {\n\t\treturn (g[x][y] + g[x - 1][y] + g[x][y + 1]) >= 2;\n\t};\n \n\tauto check3 = [&](int x, int y) -> bool {\n\t\treturn (g[x][y] + g[x][y - 1] + g[x + 1][y]) >= 2;\n\t};\n \n\tauto check4 = [&](int x, int y) -> bool {\n\t\treturn (g[x][y] + g[x - 1][y] + g[x][y - 1]) >= 2;\n\t};\n \n\tauto check5 = [&](int x, int y) -> bool {\n\t\treturn (g[x][y] + g[x][y + 1] + g[x][y + 2]) >= 2;\n\t};\n \n\tfor(int i = 3; i <= n; i ++) {\n\t\tdp[1][i] = max(dp[1][i], max(dp[1][i - 3], dp[2][i - 3]) + check2(2, i - 2) + check3(1, i));\n\t\tdp[2][i] = max(dp[2][i], max(dp[1][i - 3], dp[2][i - 3]) + check1(1, i - 2) + check4(2, i));\n\t\tdp[1][i] = dp[2][i] = max({dp[1][i], dp[2][i], max(dp[1][i - 3], dp[2][i - 3]) + check5(1, i - 2) + check5(2, i - 2)});\n\t}\n \n\tfor(int i = 6; i <= n; i ++) {\n\t\tdp[1][i] = max(dp[1][i], max(dp[1][i - 6], dp[2][i - 6]) + check2(2, i - 5) + check3(1, i) + check5(1, i - 4) + check5(2, i - 3));\n\t\tdp[2][i] = max(dp[2][i], max(dp[1][i - 6], dp[2][i - 6]) + check1(1, i - 5) + check4(2, i) + check5(1, i - 3) + check5(2, i - 4));\n\t}\n \n\t// for(int i = 1; i <= 2; i ++) {\n\t// \tfor(int j = 1; j <= n; j ++) {\n\t// \t\tcout << dp[i][j] << \" \\n\"[j == n];\n\t// \t}\n\t// }\n \n\tint ans = 0;\n \n\tfor(int i = 1; i <= 2; i ++) {\n\t\tfor(int j = 1; j <= n; j ++) {\n\t\t\tans = max(ans, dp[i][j]);\n\t\t}\n\t}\n \n\tcout << ans << '\\n';\n}\n \nint main() {\n \n\tstd::ios::sync_with_stdio(NULL);\n std::cin.tie(nullptr);\n \n int t = 1;\n std::cin >> t;\n \n while(t --) {\n solve();\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285979036,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing ll = long long;\nusing namespace std;\nconst int inf = 1e9 + 10;\n \nvector <vector <int>> a;\n \nint col(int l, int r, int c, int d) {\n int cl = 0;\n for (int x = l; x <= r; x++) {\n cl += a[0][x];\n }\n for (int x = c; x <= d; x++) {\n cl += a[1][x];\n }\n return (cl > 1 ? 1 : 0);\n}\nvoid solve() {\n int n; cin >> n;\n a.assign(2, vector <int>(n, 0));\n for (int j = 0; j < 2; j++) {\n for (int i = 0; i < n; i++) {\n char x; cin >> x;\n if (x == 'A') {\n a[j][i] = 1;\n }\n }\n }\n vector <vector <int>> dp(n + 1, vector <int>(3, -inf));\n dp[0][0] = 0;\n for (int i = 0; i < n; i++) {\n for (int last_mask = 0; last_mask < 3; last_mask++) {\n int ind = i - 1;\n if (dp[i][last_mask] != -inf) {\n if (last_mask == 0) {\n if (i + 3 <= n) {\n dp[i + 3][0] = max(dp[i + 3][0], dp[i][last_mask] + col(i, i + 2, i, i + 2));\n }\n if (i + 1 < n) {\n dp[i + 1][2] = max(dp[i + 1][2], dp[i][last_mask] + col(i, i + 1, i, i));\n dp[i + 1][1] = max(dp[i + 1][1], dp[i][last_mask] + col(i, i, i, i + 1));\n }\n } else if (last_mask == 1) {\n if (i + 3 < n) {\n dp[i + 3][1] = max(dp[i + 3][1], dp[i][last_mask] + col(i, i + 2, i + 1, i + 3));\n }\n if (i + 2 <= n) {\n dp[i + 2][0] = max(dp[i + 2][0], dp[i][last_mask] + col(i, i + 1, i + 1, i + 1));\n }\n } else {\n if (i + 3 < n) {\n dp[i + 3][2] = max(dp[i + 3][2], dp[i][last_mask] + col(i + 1, i + 3, i, i + 2));\n }\n if (i + 2 <= n) {\n dp[i + 2][0] = max(dp[i + 2][0], dp[i][last_mask] + col(i + 1, i + 1, i, i + 1));\n }\n }\n }\n }\n }\n cout << dp[n][0] << endl;\n}\n \nsigned main() {\n ios_base::sync_with_stdio(false);\n cin.tie(nullptr);\n int test = 1;\n cin >> test;\n for (int j = 0; j < test; j++) {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285727642,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\n#include <iostream>\n \nusing namespace std;\n \ntypedef long long ll;\n \nint main(){\n ll t;\n cin >> t;\n while(t--){\n ll n, blocks, answer, countA_row0, countA_row1, count_A1, countA_tile1_a, countA_tile2_a, count_A2a, countA_tile1_b, countA_tile2_b, count_A2b, count_A_block;\n cin >> n;\n string s0, s1;\n cin >> s0 >> s1;\n blocks = n / 3;\n answer = 0;\n for (ll b=0; b<blocks; b++){\n ll i = b * 3;\n countA_row0 = 0;\n if (s0[i] == 'A') {\n countA_row0++;\n }\n if (s0[i + 1] == 'A') {\n countA_row0++;\n }\n if (s0[i + 2] == 'A') {\n countA_row0++;\n }\n countA_row1 = 0;\n if (s1[i] == 'A') {\n countA_row1++;\n }\n if (s1[i + 1] == 'A') {\n countA_row1++;\n }\n if (s1[i + 2] == 'A') {\n countA_row1++;\n }\n count_A1 = 0;\n if (countA_row0 >= 2) {\n count_A1 +=1;\n }\n if (countA_row1 >=2) {\n count_A1 +=1;\n }\n countA_tile1_a = 0;\n if (s0[i] == 'A') {\n countA_tile1_a++;\n }\n if (s0[i + 1] == 'A') {\n countA_tile1_a++;\n }\n if (s1[i] == 'A') {\n countA_tile1_a++;\n }\n bool A_tile1_a = (countA_tile1_a >=2) ? 1 : 0;\n countA_tile2_a = 0;\n if (s1[i + 1] == 'A') {\n countA_tile2_a++;\n }\n if(s0[i + 2] == 'A') {\n countA_tile2_a++;\n }\n if(s1[i + 2] == 'A') {\n countA_tile2_a++;\n }\n bool A_tile2_a = (countA_tile2_a >=2) ? 1 : 0;\n count_A2a = A_tile1_a + A_tile2_a;\n countA_tile1_b = 0;\n if (s1[i] == 'A') {\n countA_tile1_b++;\n }\n if (s1[i + 1] == 'A') {\n countA_tile1_b++;\n }\n if (s0[i] == 'A') {\n countA_tile1_b++;\n }\n bool A_tile1_b = (countA_tile1_b >=2) ? 1 : 0;\n countA_tile2_b = 0;\n if (s0[i + 1] == 'A') {\n countA_tile2_b++;\n }\n if (s0[i + 2] == 'A') {\n countA_tile2_b++;\n }\n if (s1[i + 2] == 'A') {\n countA_tile2_b++;\n }\n bool A_tile2_b = (countA_tile2_b >=2) ? 1 : 0;\n count_A2b = A_tile1_b + A_tile2_b;\n count_A_block = max({count_A1, count_A2a, count_A2b});\n answer += count_A_block;\n }\n cout << answer << \"\\n\";\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285716690,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\nfrom collections import defaultdict,deque\nT=int(input())\nfor t in range(T):\n N=int(input())\n S=[[int(s==\"A\") for s in input()] for i in range(2)]\n dp=defaultdict(int)\n dp[(0,0)]=0\n queue=deque([(0,0)])\n set_queue=set()\n set_queue.add((0,0))\n while queue:\n i,j=queue.popleft()\n for di in range(4):\n dj=3-di\n ii=i+di\n jj=j+dj\n if abs(ii-jj)<3 and ii<=N and jj<=N:\n c=int(sum(S[0][i:ii]+S[1][j:jj])>=2)\n if dp[ii,jj]<dp[i,j]+c:\n dp[ii,jj]=dp[i,j]+c\n if not (ii,jj) in set_queue:\n set_queue.add((ii,jj))\n queue.append((ii,jj))\n ans=max(dp.values())\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285721763,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\n#!/usr/bin/env python\nimport os\nimport sys\nfrom io import BytesIO, IOBase\nfrom collections import defaultdict as ddic, deque\nfrom itertools import *\nimport random\nimport math\nfrom bisect import *\nimport string\nfrom functools import cache\n \n# MOD = 998244353\nMOD = 10**9 + 7\ninf = float(\"inf\")\n \nfmax = lambda x, y: x if x > y else y\n \n \ndef count(a, b, c):\n if a == b:\n return a\n if a == c:\n return a\n return b\n \n \ndef solve():\n N = rint()\n A = [[+(c == \"A\") for c in input()] for _ in range(2)]\n dp = [[-inf] * 3 for _ in range(N + 3)]\n dp[N][0] = 0\n for i in range(N)[::-1]:\n if i + 3 <= N:\n a, b, c = A[0][i : i + 3]\n d, e, f = A[1][i : i + 3]\n score = max(\n count(a, b, c) + count(d, e, f),\n count(a, b, d) + count(c, e, f),\n count(a, d, e) + count(b, c, f),\n )\n score2 = count(a, b, d)\n score3 = count(a, d, e)\n \n dp[i][0b00] = fmax(\n dp[i][0b00],\n max(\n dp[i + 3][0b00] + score,\n dp[i + 1][0b01] + score2,\n dp[i + 1][0b10] + score3,\n ),\n )\n \n dp[i][0b01] = fmax(dp[i][0b01], dp[i + 2][0b00] + count(b, d, e))\n if i + 4 <= N:\n x = A[0][i + 3]\n dp[i][0b01] = fmax(\n dp[i][0b01], dp[i + 3][0b01] + count(b, c, x) + count(d, e, f)\n )\n \n dp[i][0b10] = fmax(dp[i][0b10], dp[i + 2][0b00] + count(a, b, e))\n if i + 4 <= N:\n y = A[1][i + 3]\n dp[i][0b10] = fmax(\n dp[i][0b10], dp[i + 3][0b10] + count(a, b, c) + count(e, f, y)\n )\n \n return dp[0][0]\n \n \ndef main():\n T = 1\n T = rint()\n \n for tc in range(1, 1 + T):\n ans = solve()\n \n print(ans)\n # print(*ans)\n \n # print(\"Yes\" if ans else \"No\")\n # print(\"YES\" if ans else \"NO\")\n # print(\"Alice\" if ans else \"Bob\")\n # print(\"First\" if ans else \"Second\")\n # print(\"Case #{}: {}\".format(tc, ans))\n # print(len(ans))\n # for row in ans: print(*row)\n \n \n# region fastio\n \nBUFSIZE = 8192\n \n \nclass FastIO(IOBase):\n newlines = 0\n \n def __init__(self, file):\n self._file = file\n self._fd = file.fileno()\n self.buffer = BytesIO()\n self.writable = \"x\" in file.mode or \"r\" not in file.mode\n self.write = self.buffer.write if self.writable else None\n \n def read(self):\n while True:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n if not b:\n break\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines = 0\n return self.buffer.read()\n \n def readline(self):\n while self.newlines == 0:\n b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))\n self.newlines = b.count(b\"\\n\") + (not b)\n ptr = self.buffer.tell()\n self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)\n self.newlines -= 1\n return self.buffer.readline()\n \n def flush(self):\n if self.writable:\n os.write(self._fd, self.buffer.getvalue())\n self.buffer.truncate(0), self.buffer.seek(0)\n \n \nclass IOWrapper(IOBase):\n def __init__(self, file):\n self.buffer = FastIO(file)\n self.flush = self.buffer.flush\n self.writable = self.buffer.writable\n self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))\n self.read = lambda: self.buffer.read().decode(\"ascii\")\n self.readline = lambda: self.buffer.readline().decode(\"ascii\")\n \n \nsys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)\ninput = lambda: sys.stdin.readline().rstrip(\"\\r\\n\")\nrint = lambda: int(input())\n \n \ndef rlist():\n return list(map(int, input().split()))\n \n \ndef rgrid(n):\n return [rlist() for _ in range(n)]\n \n \n# endregion\n \nif __name__ == \"__main__\":\n main()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285710712,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\nt = int(input())\nfor _ in range(t):\n n = int(input())\n r1 = input()\n r2 = input()\n av_win = 0\n for i in range(0, n, 3):\n a_cnt = 0\n for j in range(3):\n if r1[i+j] == 'A':\n a_cnt += 1\n if r2[i+j] == 'A':\n a_cnt += 1\n if a_cnt >= 4:\n av_win += 2\n elif a_cnt >= 2:\n av_win += 1\n print(av_win)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285725011,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\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 String a = sc.next();\n String b = sc.next();\n ArrayList<Integer> l = new ArrayList<Integer>();\n for(int i=0;i<n;i++){\n if(a.charAt(i)=='A'){ \n l.add(i);\n }\n }\n for(int i=0;i<n;i++){\n if(b.charAt(i)=='A'){ \n l.add(i);\n }\n }\n Collections.sort(l);\n int k = 0;\n int i = 0;\n while(i<l.size()-1){\n if(l.get(i+1)-l.get(i)<=1){\n k++;\n i += 2;\n }else{\n i++;\n }\n }\n System.out.println(k);\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 288945433,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class C{\n static int l,r;\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(),k=0;\n String s=sc.next(),p=sc.next();\n while(l<n&&r<n){\n int w=func(s,p,2);\n if(w==0){\n w=func(s,p,3);\n if(w==0){\n if(l==r){\n if(p.charAt(r+1)=='J')r++;\n else l++;\n l++;\n r++;\n }else l=r=Math.max(l,r)+1;\n }\n else k+=w;\n }else k+=w;\n }\n System.out.println(k);\n l=r=0;\n }\n }\n static int func(String s,String p,int q){\n if(l<=r&&(s.substring(l,l+2)+p.charAt(r)).replace(\"J\",\"\").length()==q){\n l+=2;\n r++;\n return 1;\n }else if(r<=l&&(s.charAt(l)+p.substring(r,r+2)).replace(\"J\",\"\").length()==q){\n l++;\n r+=2;\n return 1;\n }else if(l+3<=s.length()&&s.substring(l,l+3).replace(\"J\",\"\").length()==q){\n if(r+3<=s.length()&&p.substring(r,r+3).replace(\"J\",\"\").length()==q)return 2;\n r+=3;\n l+=3;\n return 1;\n }else if(r+3<=s.length()&&p.substring(r,r+3).replace(\"J\",\"\").length()==q){\n r+=3;\n l+=3;\n return 1;\n }\n return 0;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285726106,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <string>\n \nusing namespace std;\n \nint main() {\n ios::sync_with_stdio(false);\n cin.tie(NULL);\n \n int t;\n cin >> t;\n while (t--) {\n int n;\n cin >> n;\n string row1, row2;\n cin >> row1 >> row2;\n int alvaro_votes = 0;\n for (int i = 0; i < n; i += 3) {\n int count_A = 0;\n if (row1[i] == 'A') count_A++;\n if (row1[i + 1] == 'A') count_A++;\n if (row1[i + 2] == 'A') count_A++;\n if (row2[i] == 'A') count_A++;\n if (row2[i + 1] == 'A') count_A++;\n if (row2[i + 2] == 'A') count_A++;\n if (count_A >= 4) {\n alvaro_votes += 2;\n } else if (count_A >= 2) {\n alvaro_votes += 1;\n }\n }\n cout << alvaro_votes << \"\\n\";\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285719661,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define int long long\nusing namespace std;\nint mod = 1000000007; // Define mod if needed\n \nsigned main() {\n#ifndef ONLINE_JUDGE\n freopen(\"input.txt\", \"r\", stdin);\n freopen(\"output.txt\", \"w\", stdout);\n#endif\n \n int t;\n cin >> t; // Number of test cases\n while (t--) {\n int n ;\n cin>>n;\n vector<int> a,b,r1,r2;\n a.resize(n),b.resize(n),r1.resize(n),r2.resize(n);\n for(int i=0;i<n;i++)\n {\n char c;\n cin>>c;\n if(c=='A')\n a[i]=1;\n // cout<<a[i]<<\" \";\n }\n //cout<<endl;\n for(int i=0;i<n;i++)\n {\n char c;\n cin>>c;\n if(c=='A')\n b[i]=1;\n // cout<<b[i]<<\" \";\n }\n //cout<<endl;\n int count =0;\n for(int i=0;i<n;i++)\n {\n if(a[i]==1&&b[i]==1)\n {\n a[i] = -1;\n b[i] = -1;\n if(i<n-1){\n if(a[i+1]==0)\n {a[i+1] = -1;\n count++;\n \n }\n else if(b[i+1]==0)\n {\n b[i+1] = -1;\n count++;\n \n }\n \n else\n {\n a[i+1]=-1;\n count++;\n }\n }\n if(i>0){\n a[i] = -1;\n b[i] = -1;\n if(a[i-1]==0)\n {a[i-1] = -1;\n count++;\n \n }\n else if(b[i-1]==0)\n {\n b[i-1] = -1;\n count++;\n \n }\n else\n {\n b[i-1]=-1;\n count++;\n }\n }\n \n }\n if((a[i]==0 && b[i]==1)|| (a[i]==1 && b[i]==0))\n {\n \n if(i<n-1){\n if(a[i+1]==1)\n {\n a[i] = -1;\n b[i] = -1;\n a[i+1] = -1;\n count++;\n \n }\n else if(b[i+1]==1)\n {\n a[i] = -1;\n b[i] = -1;\n b[i+1] = -1;\n count++;\n \n }\n }\n if(i>0){\n \n if(a[i-1]==1)\n {\n a[i] = -1;\n b[i] = -1;\n a[i-1] = -1;\n count++;\n \n }\n else if(b[i-1]==1)\n {\n a[i] = -1;\n b[i] = -1;\n b[i-1] = -1;\n count++;\n \n }\n }\n }\n if(i>0 && i<n-1)\n {\n if((a[i] + a[i+1] + a[i-1] >= 2)&&(a[i]*a[i+1]*a[i-1]>=0))\n {\n a[i]=-1;\n a[i+1]=-1;\n a[i-1] =-1;\n count++;\n }\n if((b[i] + b[i+1] + b[i-1] >= 2)&&(a[i]*a[i+1]*a[i-1]>=0))\n {\n b[i]=-1;\n b[i+1]=-1;\n b[i-1] =-1;\n count++;\n }\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 286078437,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\nfrom sys import *\ninput = stdin.readline\ninputlist = lambda: map(int, stdin.readline().split())\n \ndef solve():\n n = int(input())\n row1 = \" \" + input().strip()\n row2 = \" \" + input().strip()\n \n def majority(top, bottom):\n acount = 0\n for index in top:\n if row1[index] == \"A\": acount += 1\n for index in bottom:\n if row2[index] == \"A\": acount += 1\n return acount >= 2\n \n dp = [[-1, -1, -1] for _ in range(len(row1))]\n dp[0] = [0, 0, 0]\n for i in range(1, len(dp)):\n if i >= 3:\n if dp[i-3][0] != -1: dp[i][0] = max(dp[i][0], dp[i-3][0] + majority([i, i-1, i-2], []) + majority([], [i, i-1, i-2]))\n if dp[i-2][2] != -1: dp[i][0] = max(dp[i][0], dp[i-2][2] + majority([i-1, i], [i])) \n if dp[i-2][1] != -1: dp[i][0] = max(dp[i][0], dp[i-2][1] + majority([i], [i, i-1])) \n if i >= 1 and i < len(dp) - 1:\n if dp[i-1][0] != -1: dp[i][1] = max(dp[i][1], dp[i-1][0] + majority([i, i+1], [i]))\n if i >= 4:\n if dp[i-3][1] != -1: dp[i][1] = max(dp[i][1], dp[i-3][1] + majority([i-1, i, i+1], [i-2, i-1, i])) \n if i >= 1 and i < len(dp) - 1:\n if dp[i-1][0] != -1: dp[i][2] = max(dp[i][2], dp[i-1][0] + majority([i], [i, i+1]))\n if i >= 4:\n if dp[i-3][2] != -1: dp[i][2] = max(dp[i][2], dp[i-3][2] + majority([i-1, i, i-2], [i+1, i-1, i])) \n \n return dp[-1][0]\n \nfor _ in range(int(input())):\n print(solve())\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 286612917,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\nt = int(input())\n \nfor _ in range(t):\n n = int(input())\n s1 = input()\n s2 = input()\n arr1 = [1 if s1[i]=='A' else 0 for i in range(n)]\n arr2 = [1 if s2[i]=='A' else 0 for i in range(n)]\n \n \n cache = {}\n \n def f(i,j):\n if i>=n or j>=n:\n return 0\n if (i,j) in cache:\n return cache[(i,j)]\n ans = 0 \n \n if i<=j+1 and j+1<n:\n ans = max(ans, int(arr1[i]+arr2[j]+arr2[j+1]>=2) + f(i+1, j+2))\n if i+1>=j and i+1<n:\n ans = max(ans, int(arr1[i]+arr1[i+1]+arr2[j]>=2) + f(i+2, j+1))\n \n if i+2<n and j+2<n:\n ans = max(ans, int(arr1[i]+arr1[i+1]+arr1[i+2]>=2) + int(arr2[j]+arr2[j+1]+arr2[j+2]>=2) + f(i+3, j+3))\n \n cache[(i,j)] = ans\n return ans\n \n print(f(0,0))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285735563,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.BufferedWriter;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.io.OutputStreamWriter;\nimport java.io.PrintWriter;\nimport java.util.*;\n \npublic class c {\n static FastReader fr = new FastReader();\n static FastWriter fw = new FastWriter();\n public static void solve(){\n int n = fr.nextInt();\n String a = fr.next();\n String b = fr.next();\n \n int ans = 0;\n for(int i=0;i<n/3;i++){\n int cnt = 0;\n boolean tog = false;\n boolean prev = false;\n for(int j=0;j<3;j++){\n if(a.charAt(3*i + j)=='A'){\n cnt++;\n if(prev) tog = true;\n prev = true;\n }\n if(b.charAt(3*i + j)=='A'){\n cnt++;\n if(prev) tog = true;\n prev = true;\n }\n if(a.charAt(3*i + j)=='A' && b.charAt(3*i + j)=='j')\n prev = false;\n }\n if(cnt>=4) ans +=2;\n else if(cnt>=3) ans++;\n else if(cnt==2 && tog) ans++;\n }\n fw.println(ans);\n }\n public static void main(String[] args) throws NumberFormatException, IOException{\n //long start = System.nanoTime();\n int t = fr.nextInt();\n //System.out.println(solve());\n //System.out.println(solve()?\"YES\":\"NO\");\n while(t-->0){\n solve();\n //System.out.println(solve());\n //System.out.println(solve()?\"YES\":\"NO\");\n }\n //System.out.println(((long)System.nanoTime()-start)/1000000);\n }\n}\n \nclass FastReader{\n BufferedReader br;\n StringTokenizer st;\n public FastReader(){\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n String next(){\n while (st == null || !st.hasMoreElements()){\n try{\n st = new StringTokenizer(br.readLine());\n }catch (IOException e){\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n int nextInt(){\n return Integer.parseInt(next());\n }\n long nextLong(){\n return Long.parseLong(next());\n }\n double nextDouble(){\n return Double.parseDouble(next());\n }\n String nextLine(){\n String str = \"\";\n try{\n str = br.readLine();\n }\n catch (IOException e){\n e.printStackTrace();\n }\n return str;\n }\n int[] nextIntArray() {\n String line = nextLine();\n String[] strNums = line.split(\"\\\\s+\");\n int[] arr = new int[strNums.length];\n for (int i = 0; i < strNums.length; i++) {\n arr[i] = Integer.parseInt(strNums[i]);\n }\n return arr;\n }\n long[] nextLongArray() {\n String line = nextLine();\n String[] strNums = line.split(\"\\\\s+\");\n long[] arr = new long[strNums.length];\n for (int i = 0; i < strNums.length; i++) {\n arr[i] = Long.parseLong(strNums[i]);\n }\n return arr;\n }\n char[] nextCharArray() {\n String line = nextLine();\n String[] strNums = line.split(\"\\\\s+\");\n char[] arr = new char[strNums.length];\n for (int i = 0; i < strNums.length; i++) {\n arr[i] = strNums[i].charAt(0);\n }\n return arr;\n }\n char[] StringtoChar() {\n String line = nextLine();\n return line.toCharArray();\n }\n}\n \nclass FastWriter {\n private final PrintWriter writer;\n \n public FastWriter() {\n writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), true);\n }\n \n public void print(Object object) {\n writer.print(object);\n }\n \n public void println(Object object) {\n writer.println(object);\n }\n \n public void flush() {\n writer.flush();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285726045,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\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 testCases = sc.nextInt();\n StringBuilder result = new StringBuilder();\n \n while (testCases-- > 0) {\n int columns = sc.nextInt();\n String firstRow = sc.next();\n String secondRow = sc.next();\n int alvaroVotes = calculateVotes(firstRow, secondRow, columns);\n result.append(alvaroVotes).append(\"\\n\");\n }\n \n System.out.print(result);\n }\n \n public static int calculateVotes(String firstRow, String secondRow, int columns) {\n int votes = 0;\n \n for (int index = 0; index < columns; index += 3) {\n int countA = 0;\n if (firstRow.charAt(index) == 'A') countA++;\n if (firstRow.charAt(index + 1) == 'A') countA++;\n if (firstRow.charAt(index + 2) == 'A') countA++;\n if (secondRow.charAt(index) == 'A') countA++;\n if (secondRow.charAt(index + 1) == 'A') countA++;\n if (secondRow.charAt(index + 2) == 'A') countA++;\n \n if (countA >= 4) votes += 2;\n else if (countA >= 2) votes += 1;\n }\n \n return votes;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285738902,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\n// Bismillahir Rahmanir Rahim //\n // After hardship comes ease //\n // AUTHOR : iamarman //\n \n// pragmas\n// #pragma GCC optimize(\"Ofast\")\n// #pragma GCC target(\"avx,avx2,fma\")\n// #pragma GCC optimization (\"unroll-loops\")\n// #pragma GCC optimization (\"strict-overflow\")\n \n#include<bits/stdc++.h>\n#include <ext/pb_ds/tree_policy.hpp>\n#include <ext/pb_ds/assoc_container.hpp>\n \nusing namespace std;\nusing namespace __gnu_pbds;\n \n //// TEMPLATE ////\n \n//---------------------------------------------------------------------------------------------------------------------------------|\n \n \n# define el '\\n'\n# define sp \" \"\n# define ff first\n# define ss second\n# define ll long long\n# define pb push_back\n# define mp make_pair\n# define yess1 cout<<1<<el \n# define noo cout<<\"NO\"<<el\n# define yess cout<<\"YES\"<<el\n# define siz(x) (int)x.size()\n# define ull unsigned long long \n# define all(v) v.begin(),v.end()\n# define allr(v) v.rbegin(),v.rend()\n# define torad(x) ((x) * ((2*acos(0))/180.0))\n# define todeg(x) ((x) * (180.0/(2*acos(0))))\n \nconstexpr ll mod=998244353 ;\nconstexpr ll INF=LLONG_MAX;\nconstexpr double PI= acos(-1);\nconstexpr double eps=1e-9;\n \n# define mem(a,b) memset(a,b,sizeof(a))\n# define sqr(a) ((a)*(a))\n# define lcm(a,b) (a*b)/__gcd(a,b)\n \n# define optimise { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); }\n# define fraction(a) cout.unsetf(ios::floatfield); cout.precision(a); cout.setf(ios::fixed,ios::floatfield);\n# define ordered_set tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update>\n \n// find_by_order() - Returns an iterator to the k-th largest element (counting from zero)\n// order_of_key() - The number of items in a set that are strictly smaller than our item \n// greater instead of less for descending order\n// less_equal works as ordered multiset\n// we can use pair<int,int> instead of int for pair of orderd set\n// for multiset st.lower_bound(x) works as upper bound and st.upper_bound(x) works as lower bound\n \ninline void file() {\n#ifndef ONLINE_JUDGE \nfreopen(\"input.txt\", \"r\", stdin);\nfreopen(\"output.txt\", \"w\", stdout);\n#endif // ONLINE_JUDGE\n}\n//----------------------------------------------------------------------------------------------------------------------------------|\n \n \n // DEBUGGER //\n \n \n//----------------------------------------------------------------------------------------------------------------------------------|\n \ntemplate < typename F, typename S > ostream& operator << ( ostream& os, const pair< F, S > & p ) { return os << \"(\" << p.first << \", \" << p.second << \")\"; }\ntemplate < typename T > ostream &operator << ( ostream & os, const vector< T > &v ) { os << \"{\"; for(auto it = v.begin(); it != v.end(); ++it) { if( it != v.begin() ) os << \", \"; os << *it; } return os << \"}\"; }\ntemplate < typename T > ostream &operator << ( ostream & os, const set< T > &v ) { os << \"[\"; for(auto it = v.begin(); it != v.end(); ++it) { if( it != v.begin() ) os << \", \"; os << *it; } return os << \"]\"; }\ntemplate < typename T > ostream &operator << ( ostream & os, const multiset< T > &v ) { os << \"[\"; for(auto it = v.begin(); it != v.end(); ++it) { if( it != v.begin() ) os << \", \"; os << *it; } return os << \"]\"; }\ntemplate < typename F, typename S > ostream &operator << ( ostream & os, const map< F, S > &v ) { os << \"[\"; for(auto it = v.begin(); it != v.end(); ++it) { if( it != v.begin() ) os << \", \"; os << it -> first << \" = \" << it -> second ; } return os << \"]\"; }\n#define dbg(args...) do {cerr << #args << \" : \"; iamarman(args); } while(0)\nvoid iamarman () { cerr << endl; }\ntemplate <typename T> void iamarman( T a[], int n ) { for(int i = 0; i < n; ++i) cerr << a[i] << ' '; cerr << endl; }\ntemplate <typename T, typename ... hello> void iamarman( T arg, const hello &... rest) { cerr << arg << ' '; iamarman(rest...); }\n \n//--------------------------------------------------------------------------------------------------------------------------------------|\n \n \n \n ///// FUNCTIONS /////\n \n \n \nll bigmod(ll base,ll power){ ll res=1; ll p=base%mod; while(power>0) { if(power%2==1) { res=((res%mod)*(p%mod))%mod; } power/=2; p=((p%mod)*(p%mod))%mod; } return res; }\n \nll inversemod(ll base) { return bigmod(base,mod-2); }\n \nll poww(ull a,ull b) { ull ans=1; if(!b) return 1; while(b>1) { if(b&1) { ans=ans*a%mod; } a=a*a%mod; b/=2; }return a*ans%mod; }\n \nll gcd(ll a,ll b) { ll rem; while(b%a!=0) { rem=b%a; b=a; a=rem; } return a; }\n \nll sqrtt(ll a){ long long x = sqrt(a) + 2; while (x * x > a) x--; return x;}\n \nll sqrt(ll n) {ll low=0,high=1e10; while(high-low>1){ ll mid=low+(high-low)/2; if(mid*mid<=n) low=mid; else high=mid; }return low; }\n \nlong double sqrtd(long double n){ long double low=0,high=n,mid; for(int i=0;i<100;i++) { mid=(low+high)/2; if(mid*mid<=n) low=mid; else high=mid;} return low;}\n \nmt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());\n \ninline ll getrandom(ll a,ll b) { return uniform_int_distribution<ll>(a,b)(rng); }\n \n \nint dx[]={-1, 1 , 0 , 0 , -1 ,-1, 1, 1};\nint dy[]={ 0, 0 ,-1 , 1 , -1 , 1,-1, 1};\n \n// up = { -1,0 } , down = { 1,0 } , right = { 0,1 } , left = { 0,-1 }\n// up-right = { -1,1 } , up-left = { -1,-1 } , down-right = { 1,1 } , down-left = { 1,-1 }\n \n \n \n \n /// ____________CODE STARTS FROM HERE____________ ///\n \n \n \nvoid solve()\n{\n int n;\n cin>>n;\n \n string a,b;\n cin>>a>>b;\n \n int ans=0;\n for(int i=0;i<n;i+=3)\n {\n int cnt=0;\n cnt+=(a[i]=='A');\n cnt+=(a[i+1]=='A');\n cnt+=(a[i+2]=='A');\n cnt+=(b[i]=='A');\n cnt+=(b[i+1]=='A');\n cnt+=(b[i+2]=='A');\n \n cnt/=2;\n cnt=min(2,cnt);\n ans+=cnt;\n }\n \n cout<<ans<<el;\n \n \n \n \n}\n \n \nint main()\n{ \n optimise;\n file();\n clock_t start= clock();\n int t=1;\n cin>>t;\n for(int i=1;i<=t;i++)\n {\n // cout<<\"Case #\"<<i<<\": \";\n solve();\n \n }\n cerr << \"Run Time : \" <<((double)(clock() - start) / CLOCKS_PER_SEC)<<el;\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285724153,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long\nconst int N=1e5+5,inf=0x3f3f3f3f;\nstring s[2];\nint dp[2][N];\nbool check(char a,char b,char c)\n{\n int num=(a=='A')+(b=='A')+(c=='A');\n return num>=2;\n}\nvoid solve()\n{\n int n;\n cin>>n;\n cin>>s[0]>>s[1];\n //cout<<s[0]<<\" \"<<s[1]<<endl;\n memset(dp,-0x3f,sizeof dp);\n dp[0][0]=0;\n for(int i=0;i<n;i++)\n {\n if(i%3==0)\n {\n dp[0][i+3]=max(dp[0][i+3],dp[0][i]+check(s[0][i],s[0][i+1],s[0][i+2])+check(s[1][i],s[1][i+1],s[1][i+2]));\n dp[0][i+1]=max(dp[0][i+1],dp[0][i]+check(s[0][i],s[0][i+1],s[1][i]));\n dp[1][i+1]=max(dp[1][i+1],dp[0][i]+check(s[1][i],s[1][i+1],s[1][i]));\n }\n else if(i%3==1)\n {\n if(i<n-3)\n {\n dp[0][i+3]=max(dp[0][i+3],dp[0][i]+check(s[0][i+1],s[0][i+2],s[0][i+3])+check(s[1][i],s[1][i+1],s[1][i+2]));\n dp[1][i+3]=max(dp[1][i+3],dp[1][i]+check(s[0][i],s[0][i+1],s[0][i+2])+check(s[1][i+1],s[1][i+2],s[1][i+3]));\n }\n dp[0][i+2]=max(dp[0][i+2],dp[0][i]+check(s[0][i+1],s[1][i],s[1][i+1]));\n dp[0][i+2]=max(dp[0][i+2],dp[1][i]+check(s[0][i],s[0][i+1],s[1][i+1]));\n }\n }\n cout<<dp[0][n]<<endl;\n}\nsigned main()\n{\n int t;\n cin>>t;\n while(t--)\n {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 286067652,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\nimport sys\nt = int(sys.stdin.readline())\ntotal_n = 0\n \n \nneighbors = {\n 0: [1, 3],\n 1: [0, 2, 4],\n 2: [1, 5],\n 3: [0, 4],\n 4: [1, 3, 5],\n 5: [2, 4],\n}\n \nfrom itertools import combinations\ncells = range(6)\nall_triplets = list(combinations(cells, 3))\n \ndef is_connected(triplet):\n visited = set()\n stack = [triplet[0]]\n while stack:\n node = stack.pop()\n if node not in visited and node in triplet:\n visited.add(node)\n for neighbor in neighbors[node]:\n if neighbor in triplet:\n stack.append(neighbor)\n return len(visited) == 3\n \nconnected_triplets = []\nfor triplet in all_triplets:\n if is_connected(triplet):\n connected_triplets.append(set(triplet))\n \npartitions = []\nfor triplet1 in connected_triplets:\n triplet2 = set(cells) - triplet1\n if is_connected(tuple(triplet2)):\n partitions.append((triplet1, triplet2))\n \nprecompute = [0] * 64 \nfor mask in range(64):\n max_wins = 0\n for p in partitions:\n wins = 0\n count = 0\n for idx in p[0]:\n if mask & (1 << idx):\n count += 1\n if count >= 2:\n wins += 1\n count = 0\n for idx in p[1]:\n if mask & (1 << idx):\n count += 1\n if count >= 2:\n wins += 1\n if wins > max_wins:\n max_wins = wins\n precompute[mask] = max_wins\n \nfor _ in range(t):\n n = int(sys.stdin.readline())\n total_n += n\n s1 = sys.stdin.readline().strip()\n s2 = sys.stdin.readline().strip()\n total_wins = 0\n \n for i in range(0, n, 3):\n mask = 0\n idx = 0\n for row in [s1, s2]:\n for j in range(i, i + 3):\n if row[j] == 'A':\n mask |= (1 << idx)\n idx += 1\n total_wins += precompute[mask]\n print(total_wins)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285716237,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\n#author: sushmanth\nfrom sys import stdin , stdout\ninput = stdin.readline\n \nmeta = lambda : print(\"Case #%d: \"%(T + 1) , end = '')\ninp = lambda : list(map(int , input().split()))\n \n\"\"\"\ndp[i][2][2][2][2] ?\n\"\"\"\nfrom types import GeneratorType\ndef bootstrap(f, stack=[]):\n def wrappedfunc(*args, **kwargs):\n if stack:\n return f(*args, **kwargs)\n else:\n to = f(*args, **kwargs)\n while True:\n if type(to) is GeneratorType:\n stack.append(to)\n to = next(to)\n else:\n stack.pop()\n if not stack:\n break\n to = stack[-1].send(to)\n return to\n return wrappedfunc\n \nsz = (10 ** 5)\ndp = [[[[[-1 , -1] for i2 in range(2)] for i1 in range(2)] for i in range(2)] for j in range(sz)]\ndef answer():\n \n @bootstrap\n def solve(i , t1 , t2 , nt1 , nt2):\n \n if(i == n - 1):\n if(t1 == 1 and t2 == 1):\n yield 0\n else:\n yield -Inf\n \n if(dp[i][t1][t2][nt1][nt2] != -1):\n yield dp[i][t1][t2][nt1][nt2]\n \n ans = -Inf\n if(t1 == 1 and t2 == 1):\n ans = max(ans , (yield solve(i + 1 , nt1 , nt2 , 0 , 0)))\n elif(t1 == 1 and t2 == 0):\n if(i + 1 < n and nt1 == 0 and nt2 == 0):\n count = (a[i][1] == 'A') + (a[i + 1][0] == 'A') + (a[i + 1][1] == 'A')\n ans = max(ans , (yield solve(i + 1 , 1 , 1 , 0 , 0)) + (count >= 2))\n \n if(i + 2 < n):\n count = (a[i][1] == 'A') + (a[i + 1][1] == 'A') + (a[i + 2][1] == 'A')\n ans = max(ans , (yield solve(i + 1 , 0 , 1 , 0 , 1)) + (count >= 2))\n elif(t1 == 0 and t2 == 1):\n if(i + 1 < n and nt1 == 0 and nt2 == 0):\n count = (a[i][0] == 'A') + (a[i + 1][0] == 'A') + (a[i + 1][1] == 'A')\n ans = max(ans , (yield solve(i + 1 , 1 , 1 , 0 , 0)) + (count >= 2))\n \n if(i + 2 < n):\n count = (a[i][0] == 'A') + (a[i + 1][0] == 'A') + (a[i + 2][0] == 'A')\n ans = max(ans , (yield solve(i + 1 , 1 , 0 , 1 , 0)) + (count >= 2))\n else:\n if(i + 1 < n):\n count = (a[i][1] == 'A') + (a[i + 1][0] == 'A') + (a[i + 1][1] == 'A')\n ans = max(ans , (yield solve(i + 1 , 1 , 1 , 0 , 0)) + (count >= 2))\n \n count = (a[i][0] == 'A') + (a[i + 1][0] == 'A') + (a[i + 1][1] == 'A')\n ans = max(ans , (yield solve(i + 1 , 1 , 1 , 0 , 0)) + (count >= 2))\n \n count = (a[i][0] == 'A') + (a[i][1] == 'A') + (a[i + 1][0] == 'A')\n ans = max(ans , (yield solve(i + 1 , 1 , 0 , 0 , 0)) + (count >= 2))\n \n count = (a[i][0] == 'A') + (a[i][1] == 'A') + (a[i + 1][1] == 'A')\n ans = max(ans , (yield solve(i + 1 , 0 , 1 , 0 , 0)) + (count >= 2))\n \n if(i + 2 < n):\n count = (a[i][1] == 'A') + (a[i + 1][1] == 'A') + (a[i + 2][1] == 'A')\n ans = max(ans , (yield solve(i + 1 , 0 , 1 , 0 , 1)) + (count >= 2))\n \n count = (a[i][0] == 'A') + (a[i + 1][0] == 'A') + (a[i + 2][0] == 'A')\n ans = max(ans , (yield solve(i + 1 , 1 , 0 , 1 , 0)) + (count >= 2))\n \n \n dp[i][t1][t2][nt1][nt2] = ans\n yield ans\n \n \n for i in range(n):\n for i1 in range(2):\n for i2 in range(2):\n for i3 in range(2):\n for i4 in range(2):\n dp[i][i1][i2][i3][i4] = -1\n \n ans = solve(0 , 0 , 0 , 0 , 0)\n \n return ans\n \nfor T in range(int(input())):\n \n n = int(input())\n b = [input().strip() for i in range(2)]\n a = [['?' , '?'] for i in range(n)]\n for i in range(2):\n for j in range(n):\n a[j][i] = b[i][j]\n \n Inf = n + 10\n \n print(answer())\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285724237,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\n \npublic class Gerrymandering {\n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n int tc = Integer.parseInt(br.readLine());\n while(tc-- > 0) {\n int cols = Integer.parseInt(br.readLine());\n char[] top = br.readLine().toCharArray();\n char[] bottom = br.readLine().toCharArray();\n boolean[] combined1 = new boolean[cols/3];\n boolean[] combined2 = new boolean[cols/3];\n int[] scores = new int[cols/3];\n int curMaxScore = calculateMaxOfSegment(top, bottom, 0);\n scores[0] = curMaxScore;\n for(int i = 1; i < cols/3; i++) {\n // keep note of += vs = if alr combined\n int optimalVotes = calculateMaxOfSegment(top, bottom, i);\n scores[i] = optimalVotes;\n int combined1Score = -1;\n int combined2Score = -1;\n boolean wasCombined1 = combined1[i-1];\n boolean wasCombined2 = combined2[i-1];\n if(combined1[i-1]) {\n combined1Score = getCombined1(top, bottom, i-1, i);\n } else if(combined2[i-1]) {\n combined2Score = getCombined2(top, bottom, i-1, i);\n } else {\n int temp1 = getCombined1Start(top, bottom, i) + getCombined1(top, bottom, i-1, i) - scores[i-1];\n int temp2 = getCombined2Start(top, bottom, i) + getCombined2(top, bottom, i-1, i) - scores[i-1];\n if(temp1 > temp2) {\n combined1Score = temp1;\n } else {\n combined2Score = temp2;\n }\n }\n \n if(combined1Score > optimalVotes) {\n combined1[i] = true;\n curMaxScore += combined1Score;\n if (!wasCombined1) {\n combined1[i - 1] = true;\n }\n } else if(combined2Score > optimalVotes) {\n combined2[i] = true;\n curMaxScore += combined2Score;\n if (!wasCombined2) {\n combined2[i - 1] = true;\n }\n } else {\n curMaxScore += optimalVotes;\n }\n }\n System.out.println(curMaxScore);\n }\n }\n \n public static int calculateMaxOfSegment(char[] top, char[] bot, int segNum) {\n int col = segNum * 3;\n int aCount = 0;\n for(int i = col; i < col + 3; i++) {\n if(top[i] == 'A')\n aCount++;\n if(bot[i] == 'A')\n aCount++;\n }\n if(aCount == 1)\n return 0;\n if(aCount == 2) {\n if(top[col] == 'A' && bot[col+2] == 'A' || top[col+2] == 'A' && bot[col] == 'A')\n return 0;\n return 1;\n }\n if(aCount == 3)\n return 1;\n \n return 2;\n }\n \n public static int getCombined1Start(char[] top, char[] bot, int segNumToAdd) {\n int prevSegment = segNumToAdd - 1;\n int prevCol = prevSegment*3;\n \n int beginCount1 = (top[prevCol] == 'A') ? 1 : 0;\n beginCount1 = (top[prevCol+1] == 'A') ? beginCount1+1 : beginCount1;\n beginCount1 = (bot[prevCol] == 'A') ? beginCount1+1 : beginCount1;\n \n return (beginCount1 >= 2) ? 1 : 0;\n }\n \n public static int getCombined2Start(char[] top, char[] bot, int segNumToAdd) {\n int prevSegment = segNumToAdd - 1;\n int prevCol = prevSegment*3;\n \n int beginCount2 = (top[prevCol] == 'A') ? 1 : 0;\n beginCount2 = (bot[prevCol] == 'A') ? beginCount2+1 : beginCount2;\n beginCount2 = (bot[prevCol+1] == 'A') ? beginCount2+1 : beginCount2;\n return (beginCount2 >= 2) ? 1 : 0;\n }\n \n public static int getCombined1(char[] top, char[] bot, int prevSeg, int seg) {\n // * * - - - *\n // * - - - * *\n int prevCol = prevSeg * 3;\n int col = seg * 3;\n \n int topCount = 0;\n int botCount = 0;\n for(int i = prevCol + 2; i < prevCol + 5; i++) {\n int topCol = i;\n int botCol = i-1;\n if(top[topCol] == 'A')\n topCount++;\n if(bot[botCol] == 'A')\n botCount++;\n }\n int total = (topCount >= 2) ? 1 : 0;\n total = (botCount >= 2) ? total+1 : total;\n \n int endCount = (top[col+2] == 'A') ? 1 : 0;\n endCount = (bot[col+1] == 'A') ? endCount + 1 : endCount;\n endCount = (bot[col+2] == 'A') ? endCount + 1 : endCount;\n total = (endCount >= 2) ? total+1 : total;\n \n return total;\n }\n \n public static int getCombined2(char[] top, char[] bot, int prevSeg, int seg) {\n int prevCol = prevSeg * 3;\n int col = seg * 3;\n int topCount = 0;\n int botCount = 0;\n for(int i = prevCol + 2; i < prevCol + 5; i++) {\n int botCol = i;\n int topCol = i-1;\n if(top[topCol] == 'A')\n topCount++;\n if(bot[botCol] == 'A')\n botCount++;\n }\n int total = (topCount >= 2) ? 1 : 0;\n total = (botCount >= 2) ? total+1 : total;\n \n int endCount = (bot[col+2] == 'A') ? 1 : 0;\n endCount = (top[col+1] == 'A') ? endCount + 1 : endCount;\n endCount = (top[col+2] == 'A') ? endCount + 1 : endCount;\n total = (endCount >= 2) ? total+1 : total;\n \n return total;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285745423,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Gerrymandering\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: We all steal a little bit. But I have only one hand, while my adversaries have two.\n\nÁlvaro Obregón\n\nÁlvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.\n\nUnder the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\\text{∗}}$. Each house will belong to exactly one district.\n\nEach district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\\frac{2n}{3}$ votes will be cast.\n\nAs Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.\n\n$^{\\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10^4$). The description of the test cases follows.\n\nThe first line of each test case contains one integer $n$ ($3 \\le n \\le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.\n\nThe following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\\texttt{A}$ and $\\texttt{J}$. If $s_{i,j}=\\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\\texttt{J}$, the house will select José.\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 maximum number of districts Álvaro can win by optimally dividing the houses into districts.\n\nExamples:\ninput:\n4\n3\nAAA\nAJJ\n6\nJAJAJJ\nJJAJAJ\n6\nAJJJAJ\nAJJAAA\n9\nAJJJJAJAJ\nJAAJJJJJA\noutput:\n2\n2\n3\n2\n\nNote: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.\n\n\n\nBuggy Code:\nimport java.io.*;\nimport java.util.StringTokenizer;\n \npublic class gerrymander {\n\tpublic static void main(String[] args) throws IOException {\n\t\tBufferedReader r = new BufferedReader(new InputStreamReader(System.in));\n\t\tPrintWriter pw = new PrintWriter(System.out);\n \n\t\tStringTokenizer st = new StringTokenizer(r.readLine());\n\t\tint b = Integer.parseInt(st.nextToken());\n\t\tfor(int z = 0; z<b; z++){\n\t\t\tst = new StringTokenizer(r.readLine());\n\t\t\tint a = Integer.parseInt(st.nextToken());\n\t\t\tint[][] houses = new int[a][2];\n\t\t\tfor(int i = 0; i<2; i++){\n\t\t\t String temp = r.readLine();\n\t\t\t for(int j = 0; j<a; j++){\n\t\t\t if(temp.charAt(j) == 'A'){\n\t\t\t houses[j][i] = 1;\n\t\t\t }\n\t\t\t else{\n\t\t\t houses[j][i] = -1;\n\t\t\t }\n\t\t\t }\n\t\t\t \n\t\t\t}\n\t\t\tint total = 0;\n\t\t\tfor(int i = 0; i<a; i+=3){\n\t\t\t int c = 0;\n\t\t\t int max = 0;\n\t\t\t int s1 = houses[i][0]+houses[i+1][0]+houses[i+2][0];\n\t\t\t int s2 = houses[i][1]+houses[i+1][1]+houses[i+2][1];\n\t\t\t if(s1>0){\n\t\t\t c++;\n\t\t\t }\n\t\t\t if(s2>0){\n\t\t\t c++;\n\t\t\t }\n\t\t\t if(c>max){\n\t\t\t max = c;\n\t\t\t }\n\t\t\t c = 0;\n\t\t\t s1 = houses[i][0]+houses[i+1][0]+houses[i][1];\n\t\t\t s2 = houses[i+1][1]+houses[i+2][0]+houses[i+2][1];\n\t\t\t if(s1>0){\n\t\t\t c++;\n\t\t\t }\n\t\t\t if(s2>0){\n\t\t\t c++;\n\t\t\t }\n\t\t\t c=0;\n\t\t\t if(c>max){\n\t\t\t max = c;\n\t\t\t }\n\t\t\t s1 = houses[i+2][1]+houses[i+2][0]+houses[i+1][0];\n\t\t\t s2 = houses[i][0]+houses[i][1]+houses[i+1][1];\n\t\t\t if(s1>0){\n\t\t\t c++;\n\t\t\t }\n\t\t\t if(s2>0){\n\t\t\t c++;\n\t\t\t }\n\t\t\t if(c>max){\n\t\t\t max = c;\n\t\t\t }\n\t\t\t total+=max;\n\t\t\t}\n\t\t\tSystem.out.println(total);\n\t\t\t\n\t\t}\n\t \n\t\t\t\n \n\t\t}\n\t}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2022C",
"submission_id": 285719817,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nnamespace io{\n char buf[1<<21],*p1=buf,*p2=buf;\n #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)\n inline int read(){\n int x=0; bool flag=true; char c=getchar();\n while(!isdigit(c)){ if(c=='-') flag=false; c=getchar(); }\n while(isdigit(c)){ x=(x<<1)+(x<<2)+(c^48); c=getchar(); }\n return flag?x:~(x-1);\n }\n}\nusing namespace io;\nconst int N=16;\nint T,n,m,a[N][N];\nint f[N][1<<N][N];\nint g[N][1<<N][N];\nint w[N][N],res;\nint main(){\n T=read();\n while(T--){\n n=read(),m=read();\n for(int i=0;i<n;i++)\n a[i][m]=0;\n for(int i=0;i<m;i++)\n a[n][i]=0;\n a[n][m]=0,res=1e9;\n for(int i=0;i<n;i++){\n for(int j=0;j<m;j++){\n a[i][j]=read();\n a[i][m]^=a[i][j];\n a[n][j]^=a[i][j];\n a[n][m]^=a[i][j];\n }\n }\n for(int x=0;x<=n;x++){\n for(int i=0;i<=m-1;i++){\n for(int j=i+1;j<=m;j++){\n int sum=0;\n for(int k=0;k<=n;k++){\n if(k==x) continue;\n sum+=abs(a[k][i]-a[k][j]);\n }\n w[i][j]=w[j][i]=sum;\n }\n }\n for(int i=1;i<1<<m+1;i++)\n for(int j=0;j<=m;j++)\n f[x][i][j]=1e9;\n for(int i=0;i<=m;i++)\n f[x][1<<i][i]=0;\n for(int i=1;i<1<<m+1;i++){\n for(int j=0;j<=m;j++){\n if(i>>j&1){\n for(int k=0;k<=m;k++){\n if(i>>k&1&&k!=j){\n f[x][i][j]=min(f[x][i][j],f[x][i^1<<j][k]+w[k][j]);\n }\n }\n }\n }\n }\n }\n for(int x=0;x<=m;x++){\n for(int i=0;i<=n-1;i++){\n for(int j=i+1;j<=n;j++){\n int sum=0;\n for(int k=0;k<=m;k++){\n if(k==x) continue;\n sum+=abs(a[i][k]-a[j][k]);\n }\n w[i][j]=w[j][i]=sum;\n }\n }\n for(int i=1;i<1<<n+1;i++)\n for(int j=0;j<=n;j++)\n g[x][i][j]=1e9;\n for(int i=0;i<=n;i++)\n g[x][1<<i][i]=0;\n for(int i=1;i<1<<n+1;i++){\n for(int j=0;j<=n;j++){\n if(i>>j&1){\n for(int k=0;k<=n;k++){\n if(i>>k&1&&k!=j){\n g[x][i][j]=min(g[x][i][j],g[x][i^1<<j][k]+w[k][j]);\n }\n }\n }\n }\n }\n }\n for(int x=0;x<=n;x++){\n for(int y=0;y<=m;y++){\n for(int i=0;i<=n;i++){\n if(i==x) continue;\n for(int j=0;j<=m;j++){\n if(j==y) continue;\n res=min(res,\n f[x][(1<<m+1)-1^1<<y][j]+\n g[y][(1<<n+1)-1^1<<x][i]);\n }\n }\n }\n }\n printf(\"%d\\n\",res);\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 275355842,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n//author: towrist\n \n#include <iostream>\n#include <iomanip>\n#include <vector>\n#include <functional>\n#include <cmath>\n#include <algorithm>\n#include <set>\n#include <utility>\n#include <queue>\n#include <map>\n#include <assert.h>\n#include <stack>\n#include <climits>\n#include <string>\n#include <ctime>\n#include <chrono>\n#include <random>\n#define int long long\nusing namespace std;\n \nconst int INF=1e18;\n \nvoid solve()\n{\n\tint n, m;\n\tcin>>n>>m;\n\tvector<vector<int>> a(n+1, vector<int>(m+1, 0));\n\t\n\tfor (int i=0; i<n; i++) for (int j=0; j<m; j++) cin>>a[i][j];\n\t\n\tfor (int i=0; i<n; i++)\n\t{\n\t\tint c=0;\n\t\tfor (int j=0; j<m; j++) c^=a[i][j];\n\t\ta[i][m]=c;\n\t}\n\t\n\tfor (int j=0; j<m; j++)\n\t{\n\t\tint c=0;\n\t\tfor (int i=0; i<n; i++) c^=a[i][j];\n\t\ta[n][j]=c;\n\t}\n\t\n\t{\n\t\tint X=0;\n\t\tfor (int i=0; i<n; i++) for (int j=0; j<m; j++) X^=a[i][j];\n\t\ta[n][m]=X;\n\t}\n\t\n\tvector<vector<int>> ans(n+1, vector<int>(m+1, INF));\n\tfor (int i=0; i<=m; i++)\n\t{\n\t\tvector<vector<int>> diff(n+1, vector<int>(n+1, 0));\n\t\tfor (int z1=0; z1<=n; z1++)\n\t\t{\n\t\t\tfor (int z2=0; z2<=n; z2++)\n\t\t\t{\n\t\t\t\tfor (int z=0; z<=m; z++) if (z!=i)\n\t\t\t\t\tdiff[z1][z2]+=abs(a[z1][z]-a[z2][z]);\n\t\t\t}\n\t\t}\n\t\tvector<vector<int>> mn(1<<(n+1), vector<int>(n+1, INF));\n\t\t\n\t\tfor (int j=0; j<=n; j++) mn[1<<j][j]=0;\n\t\tfor (int msk=1; msk<(1<<(n+1)); msk++)\n\t\t{\n\t\t\tvector<int> setbits;\n\t\t\tfor (int i=0; i<=n; i++) if ((1<<i)&msk) setbits.push_back(i);\n\t\t\tif ((int)setbits.size()<=1)\n\t\t\t{\n\t\t\t\tif (setbits.size()==n)\n\t\t\t\t\tfor (int j2=0; j2<=n; j2++)\n\t\t\t\t\t\tif (!((1<<j2)&msk))\n\t\t\t\t\t\t\tans[j2][i]=min(ans[j2][i], mn[msk][setbits[0]]);\n\t\t\t\tcontinue;\n\t\t\t}\n\t\t\tfor (int j1=0; j1<=n; j1++) if ((1<<j1)&msk)\n\t\t\t{\n\t\t\t\tfor (int j2=0; j2<=n; j2++) if ((1<<j2)&msk)\n\t\t\t\t{\n\t\t\t\t\tif (j1==j2) continue;\n\t\t\t\t\tmn[msk][j1]=min(mn[msk][j1], mn[msk^(1<<j1)][j2]+diff[j1][j2]);\n\t\t\t\t}\n\t\t\t\tif (setbits.size()==n)\n\t\t\t\t\tfor (int j2=0; j2<=n; j2++)\n\t\t\t\t\t\tif (!((1<<j2)&msk))\n\t\t\t\t\t\t\tans[j2][i]=min(ans[j2][i], mn[msk][j1]);\n\t\t\t}\n\t\t}\n\t}\n\t\n\tint res=INF;\n\tfor (int i=0; i<=n; i++)\n\t{\n\t\tvector<vector<int>> diff(m+1, vector<int>(m+1, 0));\n\t\tfor (int z1=0; z1<=m; z1++)\n\t\t{\n\t\t\tfor (int z2=0; z2<=m; z2++)\n\t\t\t{\n\t\t\t\tfor (int z=0; z<=n; z++) if (z!=i)\n\t\t\t\tdiff[z1][z2]+=abs(a[z][z1]-a[z][z2]);\n\t\t\t}\n\t\t}\n\t\t\n\t\tvector<vector<int>> mn(1<<(m+1), vector<int>(m+1, INF));\n\t\t\n\t\tfor (int j=0; j<=m; j++) mn[1<<j][j]=0;\n\t\tfor (int msk=1; msk<(1<<(m+1)); msk++)\n\t\t{\n\t\t\tvector<int> setbits;\n\t\t\tfor (int i=0; i<=m; i++) if ((1<<i)&msk) setbits.push_back(i);\n\t\t\tif ((int)setbits.size()<=1) continue;\n\t\t\tfor (int j1=0; j1<=m; j1++) if ((1<<j1)&msk)\n\t\t\t{\n\t\t\t\tfor (int j2=0; j2<=m; j2++) if ((1<<j2)&msk)\n\t\t\t\t{\n\t\t\t\t\tif (j1==j2) continue;\n\t\t\t\t\tmn[msk][j1]=min(mn[msk][j1], mn[msk^(1<<j1)][j2]+diff[j1][j2]);\n\t\t\t\t}\n\t\t\t\tif (setbits.size()==m)\n\t\t\t\tfor (int j2=0; j2<=m; j2++)\n\t\t\t\tif (!((1<<j2)&msk))\n\t\t\t\t{\n\t\t\t\t\tres=min(res, ans[i][j2]+mn[msk][j1]);\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t}\n\tcout<<res<<'\\n';\n\treturn;\n}\n \nsigned main()\n{\n\tios::sync_with_stdio(0);\n\tcin.tie(NULL); cout.tie(NULL);\n\tint t;\n\tcin >> t;\n\t//t=1;\n\t//cout<<'\\n';\n\twhile (t--)\n\t{\n\t\tsolve();\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274676020,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n/*\n考虑给该矩阵多补上一行一列。 \n原本同行同列 => 转化后同行同列\n可以枚举丢掉哪一行哪一列。 \n因此可以行列分开考虑。这里考虑一下行。 \nn，m较小 \n假设丢掉第 i 行 j 列，我们需最小化其他 n 行两两差的绝对值的最小值。\n可以使用状压 dp 解决。\n复杂度好像是 $O(n \\times 2^n)$ \n感觉上限不是很紧，我直接不精细实现 \n*/\nusing ll=long long;\nvoid cmin(ll &x, ll y) {\n\tif(y<x) x=y;\n}\nconst int N=18;\nconst ll inf=1e17;\nll f[1<<N][N]; \nint n, m, a[N][N]; \nll valh(int x, int y, int d) {\n\tll res=0;\n\tfor(int i=1; i<=m; i++)\n\t\tif(d!=i) \n\t\t\tres+=abs(a[x][i]-a[y][i]);\n\treturn res;\n}\nll vall(int x, int y, int d) {\n\tll res=0;\n\tfor(int i=1; i<=n; i++)\n\t\tif(d!=i) \n\t\t\tres+=abs(a[i][x]-a[i][y]);\n\treturn res;\n}\nvoid Sol() {\n\tscanf(\"%d %d\", &n, &m);\n\tfor(int i=1; i<=n; i++) {\n\t\tint x=0;\n\t\tfor(int j=1; j<=m; j++) {\n\t\t\tscanf(\"%d\", &a[i][j]);\n\t\t\tx^=a[i][j];\n\t\t}\n\t\ta[i][m+1]=x;\n\t} m++;\n\tfor(int j=1; j<=m; j++) {\n\t\tint x=0;\n\t\tfor(int i=1; i<=n; i++) {\n\t\t\tx^=a[i][j];\n\t\t}\n\t\ta[n+1][j]=x;\n\t} n++;\n\tint S1=1<<n, S2=1<<m;\n\tll ans=inf;\n\tfor(int dh=1; dh<=n; dh++) {\n\t\tfor(int dl=1; dl<=m; dl++) {\n\t\t\t\n\t\t\tll res1=inf, res2=inf;\n\t\t\tmemset(f, 0x3f, sizeof(f));\n\t\t\tfor(int i=1; i<=n; i++) \n\t\t\t\tif(i!=dh) f[1<<(i-1)][i]=0;\n\t\t\tfor(int s=1; s<S1; s++) {\n\t\t\t\tint lw=__lg(s&-s)+1;\n\t\t\t\tfor(int j=1; j<=n; j++) {\n\t\t\t\t\tif(j==dh) continue ;\n\t\t\t\t\tif(f[s][j]<inf) {\n\t\t\t\t\t\t\n\t\t\t\t\t\tfor(int k=lw+1; k<=n; k++) {\n\t\t\t\t\t\t\tif(k==dh) continue ;\n\t\t\t\t\t\t\tcmin(f[s^(1<<k-1)][k], f[s][j]+valh(j, k, dl));\n\t\t\t\t\t\t}\n\t\t\t\t\t} \n\t\t\t\t}\n\t\t\t}\n\t\t\tfor(int i=1; i<=n; i++)\n\t\t\t\tif(i!=dh) cmin(res1, f[(S1-1)^(1<<dh-1)][i]);\n\t\t\t\n\t\t\tmemset(f, 0x3f, sizeof(f));\n\t\t\tfor(int i=1; i<=m; i++) \n\t\t\t\tif(i!=dl) f[1<<(i-1)][i]=0;\n\t\t\tfor(int s=1; s<S2; s++) {\n\t\t\t\tint lw=__lg(s&-s)+1;\n\t\t\t\tfor(int j=1; j<=m; j++) {\n\t\t\t\t\tif(j==dl) continue ;\n\t\t\t\t\tif(f[s][j]<inf) {\n\t\t\t\t\t\t\n\t\t\t\t\t\tfor(int k=lw+1; k<=m; k++) {\n\t\t\t\t\t\t\tif(k==dl) continue ;\n\t\t\t\n\t\t\t\t\t\t\tcmin(f[s^(1<<k-1)][k], f[s][j]+vall(j, k, dh));\n\t\t\t\t\t\t}\n\t\t\t\t\t} \n\t\t\t\t}\n\t\t\t}\n\t\t\tfor(int i=1; i<=m; i++)\n\t\t\t\tif(i!=dl) cmin(res2, f[(S2-1)^(1<<dl-1)][i]);\n//\t\t\tif(res1+res2<ans) cout<<dh<<\" \"<<dl<<endl;\n\t\t\tcmin(ans, res1+res2);\n\t\t}\n\t}\n\tprintf(\"%lld\\n\", ans);\n}\nint main() {\n\tint T; scanf(\"%d\", &T);\n\twhile(T--) Sol();\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 282909686,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nusing i64 = long long;\n \nvoid solve() {\n int n, m;\n cin >> n >> m;\n vector<vector<int> > a(n + 1, vector<int>(m + 1));\n for (int i = 0; i < n; i++) {\n for (int j = 0; j < m; j++) {\n cin >> a[i][j];\n a[n][j] ^= a[i][j];\n \n }\n }\n vector<vector<int> > g0(n + 1, vector<int>(n + 1)), g1(m + 1, vector<int>(m + 1));\n \n auto work = [&](int n, vector<vector<int> > &g) -> int {\n int res = 2e9;\n vector<vector<int> > f(1 << n, vector<int>(n));\n for (int i = 1; i < (1 << n); i++) {\n for (int j = 0; j < n; j++) if (i >> j & 1) {\n int la = i ^ (1 << j);\n if (!la) break;\n f[i][j] = 2e9;\n for (int k = 0; k < n; k++) if (i >> k & 1) {\n if (k == j) continue;\n f[i][j] = min(f[i][j], g[j][k] + f[la][k]);\n }\n }\n }\n for (int i = 0; i < n; i++) res = min(res, f[(1 << n) - 1][i]);\n return res;\n };\n \n int ans = 2e9;\n for (int i = 0; i <= n; i++) {\n for (int j = 0; j <= m; j++) {\n \n for (int k = 0; k < n; k++) {\n for (int l = 0; l < m; l++) {\n if (i != n && k != i) a[i][l] ^= a[k][l]; \n if (j != m && j != l) a[k][j] ^= a[k][l];\n }\n }\n \n // for (int i = 0; i < n; i++) {\n // for (int j = 0; j < m; j++) {\n // cout << a[i][j] << \" \\n\"[j == m - 1];\n // }\n // }\n \n for (int k = 0; k < n; k++) {\n for (int l = 0; l < n; l++) {\n g0[k][l] = 0;\n for (int p = 0; p < m; p++) g0[k][l] += abs(a[k][p] - a[l][p]);\n }\n }\n for (int k = 0; k < m; k++) {\n for (int l = 0; l < m; l++) {\n g1[k][l] = 0;\n for (int p = 0; p < n; p++) g1[k][l] += abs(a[p][k] - a[p][l]);\n }\n }\n \n ans = min(ans, work(n, g0) + work(m, g1));\n // cout << i << ' ' << j << ' ' << work(n, g0) << ' ' << work(m, g1) << '\\n';\n \n for (int k = 0; k < n; k++) {\n for (int l = 0; l < m; l++) {\n if (i != n && k != i) a[i][l] ^= a[k][l]; \n if (j != m && j != l) a[k][j] ^= a[k][l];\n }\n }\n }\n }\n cout << ans << '\\n';\n}\nint main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n cout.tie(nullptr);\n \n int T;\n cin >> T;\n while (T--) {\n solve();\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274418627,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nusing i64 = long long;\n \nvoid solve() {\n int n, m;\n cin >> n >> m;\n vector<vector<int> > a(n + 1, vector<int>(m + 1));\n for (int i = 0; i < n; i++) {\n for (int j = 0; j < m; j++) {\n cin >> a[i][j];\n a[n][j] ^= a[i][j];\n \n }\n }\n vector<vector<int> > g0(n + 1, vector<int>(n + 1)), g1(m + 1, vector<int>(m + 1));\n \n auto work = [&](int n, vector<vector<int> > &g) -> int {\n int res = 0x3f3f3f3f;\n vector<vector<int> > f(1 << n, vector<int>(n));\n for (int i = 1; i < (1 << n); i++) {\n for (int j = 0; j < n; j++) if (i >> j & 1) {\n int la = i ^ (1 << j);\n if (!la) break;\n f[i][j] = 0x3f3f3f3f;\n for (int k = 0; k < n; k++) if (i >> k & 1) {\n if (k == j) continue;\n f[i][j] = min(f[i][j], g[j][k] + f[la][k]);\n }\n }\n }\n for (int i = 0; i < n; i++) res = min(res, f[(1 << n) - 1][i]);\n return res;\n };\n \n int ans = 1e9;\n for (int i = 0; i <= n; i++) {\n for (int j = 0; j <= m; j++) {\n \n for (int k = 0; k < n; k++) {\n for (int l = 0; l < m; l++) {\n if (i != n && k != i) a[i][l] ^= a[k][l]; \n if (j != m && j != l) a[k][j] ^= a[k][l];\n }\n }\n \n // for (int i = 0; i < n; i++) {\n // for (int j = 0; j < m; j++) {\n // cout << a[i][j] << \" \\n\"[j == m - 1];\n // }\n // }\n \n for (int k = 0; k < n; k++) {\n for (int l = 0; l < n; l++) {\n g0[k][l] = 0;\n for (int p = 0; p < m; p++) g0[k][l] += abs(a[k][p] - a[l][p]);\n }\n }\n for (int k = 0; k < m; k++) {\n for (int l = 0; l < m; l++) {\n g1[k][l] = 0;\n for (int p = 0; p < n; p++) g1[k][l] += abs(a[p][k] - a[p][l]);\n }\n }\n \n ans = min(ans, work(n, g0) + work(m, g1));\n // cout << i << ' ' << j << ' ' << work(n, g0) << ' ' << work(m, g1) << '\\n';\n \n for (int k = 0; k < n; k++) {\n for (int l = 0; l < m; l++) {\n if (i != n && k != i) a[i][l] ^= a[k][l]; \n if (j != m && j != l) a[k][j] ^= a[k][l];\n }\n }\n }\n }\n cout << ans << '\\n';\n}\nint main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n cout.tie(nullptr);\n \n int T;\n cin >> T;\n while (T--) {\n solve();\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274417065,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\nfrom functools import reduce, cache\nimport operator\nimport sys\ninput = sys.stdin.readline\nMOD = 10**9 + 7\n \n \ndef read_array(factory):\n return [factory(num) for num in input().strip().split()]\ndef print_array(arr):\n print(\" \".join(map(str, arr)))\n \ndef solve(mat):\n for i in mat:\n i.append(reduce(operator.xor, i))\n n = []\n for i in range(len(mat[0])):\n n.append(reduce(operator.xor, [mat[j][i] for j in range(len(mat))]))\n mat.append(n)\n # Swap any rows of mat greedily\n def diff(r1, r2):\n return sum(abs(i-j) for i, j in zip(r1, r2))\n \n \n def had(mask, last):\n dp = {(0, None): 0}\n diffs = {(l, i): diff(mat2[l], mat2[i]) for l in range(len(mat2)) for i in range(len(mat2))}\n for _ in range(len(mat2)-1):\n newdp = {}\n for (mask, last), val in dp.items():\n for i in range(len(mat2)):\n if mask & (1 << i) == 0:\n newmask = mask ^ (1 << i)\n if last == None:\n newdp[(newmask, i)] = min(val, newdp.get((newmask, i), float('inf')))\n else:\n # m = min(m, had(mask ^ (1 << i), i) + diff(mat2[last], mat2[i]))\n newdp[(newmask, i)] = min(val + diffs[(last, i)], newdp.get((newmask, i), float('inf')))\n dp = newdp\n ans = {}\n for (mask, last), val in dp.items():\n for i in range(len(mat2)):\n if mask & 1 << i == 0:\n ans[i] = min(ans.get(mask, float('inf')), val)\n return ans\n \n m = float('inf')\n ans2 = []\n for omit_col in range(len(mat[0])):\n mat2 = [[mat[i][j] for j in range(len(mat[0])) if j != omit_col] for i in range(len(mat))]\n ans2.append(had(0, None))\n \n ans3 = []\n for omit_row in range(len(mat)):\n mat2 = [[mat[i][j] for j in range(len(mat[0]))] for i in range(len(mat)) if i != omit_row]\n mat2 = [[mat2[j][i] for j in range(len(mat2))] for i in range(len(mat2[0]))]\n ans3.append(had(0, None))\n \n m = float('inf')\n for omit_row in range(len(mat)):\n for omit_col in range(len(mat[0])):\n m = min(m, ans2[omit_col][omit_row] + ans3[omit_row][omit_col])\n # print(ans2)\n # print(ans3)\n return m\n \nif __name__ == \"__main__\":\n # Read in input:\n t = int(input())\n \n # Call solution:\n for _ in range(t):\n n, _ = read_array(int)\n mat = []\n for _ in range(n):\n mat.append(read_array(int))\n ans = solve(mat)\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274451159,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\nT=int(input())\nfor t in range(T):\n N,M=map(int,input().split())\n A=[list(map(int,input().split())) for n in range(N)]\n for n in range(N):\n A[n].append(0)\n for m in range(M):\n A[n][-1]^=A[n][m]\n A.append([0]*(M+1))\n for n in range(N):\n for m in range(M+1):\n A[N][m]^=A[n][m]\n def solve(x,y):\n B=[[A[i][j] for j in range(M+1) if j!=y] for i in range(N+1) if i!=x]\n C=[[sum(abs(B[n][m0]-B[n][m1]) for n in range(N)) for m1 in range(M)] for m0 in range(M)]\n D=[[sum(abs(B[n0][m]-B[n1][m]) for m in range(M)) for n1 in range(N)] for n0 in range(N)]\n def solve1(E):\n M=len(E)\n inf=1<<60\n dp=[[inf]*M for bit in range(1<<M)]\n ans=inf\n for bit in range(1,1<<M):\n idx=[m for m in range(M) if bit&1<<m]\n if len(idx)==1:\n m=idx[0]\n dp[bit][m]=0\n else:\n for m1 in idx:\n for m0 in idx:\n if m0==m1:\n continue\n dp[bit][m1]=min(dp[bit][m1],dp[bit^1<<m1][m0]+E[m0][m1])\n if len(idx)==M:\n ans=min(ans,min(dp[bit]))\n return ans\n ans=solve1(C)+solve1(D)\n return ans\n ans=min(solve(x,y) for x in range(N+1) for y in range(N+1))\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274398677,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\nfrom functools import reduce, cache\nimport operator\nimport sys\ninput = sys.stdin.readline\nMOD = 10**9 + 7\n \n \ndef read_array(factory):\n return [factory(num) for num in input().strip().split()]\ndef print_array(arr):\n print(\" \".join(map(str, arr)))\n \ndef solve(mat):\n for i in mat:\n i.append(reduce(operator.xor, i))\n n = []\n for i in range(len(mat[0])):\n n.append(reduce(operator.xor, [mat[j][i] for j in range(len(mat))]))\n mat.append(n)\n # Swap any rows of mat greedily\n def diff(r1, r2):\n return sum(abs(i-j) for i, j in zip(r1, r2))\n \n \n def had(mask, last):\n dp = {(0, None): 0}\n diffs = {(l, i): diff(mat2[l], mat2[i]) for l in range(len(mat2)) for i in range(len(mat2))}\n for _ in range(len(mat2)):\n newdp = {}\n for (mask, last), val in dp.items():\n for i in range(len(mat2)):\n if mask & (1 << i) == 0:\n newmask = mask ^ (1 << i)\n if last == None:\n newdp[(newmask, i)] = min(val, newdp.get((newmask, i), float('inf')))\n else:\n # m = min(m, had(mask ^ (1 << i), i) + diff(mat2[last], mat2[i]))\n newdp[(newmask, i)] = min(val + diffs[(last, i)], newdp.get((newmask, i), float('inf')))\n dp = newdp\n return min(dp.values())\n \n m = float('inf')\n for omit_row in range(len(mat)):\n for omit_col in range(len(mat[0])):\n # print(omit_row, omit_col)\n mat2 = [[mat[i][j] for j in range(len(mat[0])) if j != omit_col] for i in range(len(mat)) if i != omit_row]\n x1 = had(0, None)\n mat2 = [[mat2[j][i] for j in range(len(mat2))] for i in range(len(mat2[0]))]\n x2 = had(0, None)\n m = min(m, x1 + x2)\n return m\n \nif __name__ == \"__main__\":\n # Read in input:\n t = int(input())\n \n # Call solution:\n for _ in range(t):\n n, _ = read_array(int)\n mat = []\n for _ in range(n):\n mat.append(read_array(int))\n ans = solve(mat)\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274449949,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define rep(i, a, b) for(int i = a; i < (b); ++i)\n#define all(x) begin(x), end(x)\n#define sz(x) (int)(x).size()\ntypedef long long ll;\ntypedef pair<int, int> pii;\ntypedef vector<int> vi;\n \nint n,m;\n \nint bannedRow,bannedCol;\n \nll memoRow[1<<16][16];\nll memoCol[1<<16][16];\n \n \nll diffRow[16][16];\nll diffCol[16][16];\n \nvector<vi> grid(16,vi(16,0));\n \nll dpRow(int curr, int mask){\n\tif ((mask|(1<<bannedRow)) == (1<<(n+1))-1) return 0;\n\tif (memoRow[mask][curr] != -1) return memoRow[mask][curr];\n \n\tll best = 1e9;\n\trep(nxt,0,n+1){\n\t\tif (nxt == bannedRow) continue;\n\t\tif (((mask>>nxt)&1) == 0){\n\t\t\tbest = min(best,dpRow(nxt,mask|1<<nxt) + diffRow[nxt][curr] - abs(grid[curr][bannedCol] - grid[nxt][bannedCol]));\n\t\t}\n\t}\n \n\tmemoRow[mask][curr] = best;\n \n\treturn best;\n}\n \nll dpCol(int curr, int mask){\n\t//cout << curr << \" \" << mask << \"\\n\";\n\tif ((mask|(1<<bannedCol)) == ((1<<(m+1))-1)) return 0;\n\tif (memoCol[mask][curr] != -1) return memoCol[mask][curr];\n \n\t//cout << curr << \" \" << mask << \"\\n\";\n \n\tll best = 1e9;\n\trep(nxt,0,m+1){\n\t\tif (nxt == bannedCol) continue;\n\t\tif (((mask>>nxt)&1) == 0){\n\t\t\tbest = min(best,dpCol(nxt,mask|1<<nxt) + diffCol[nxt][curr] - abs(grid[bannedRow][nxt] - grid[bannedRow][curr]));\n\t\t}\n\t}\n \n\tmemoCol[mask][curr] = best;\n \n\t//cout << best << \" \"<< curr << \" \" << mask << \"\\n\";\n \n\treturn best;\n}\n \n \nint main() {\n\tcin.tie(0)->sync_with_stdio(0);\n\tcin.exceptions(cin.failbit);\n \n\tint t;\n\tcin >> t;\n\trep(testcase,0,t){\n\t\tcin >> n >> m;\n\t\t\n\t\trep(i,0,n) rep(j,0,m) cin >> grid[i][j];\n\t\t\n\t\trep(i,0,n){\n\t\t\tint temp = 0;\n\t\t\trep(j,0,m) temp ^= grid[i][j];\n\t\t\tgrid[i][m] = temp;\n\t\t}\n \n\t\trep(j,0,m+1){\n\t\t\tint temp = 0;\n\t\t\trep(i,0,n){\n\t\t\t\ttemp ^= grid[i][j];\n\t\t\t}\n\t\t\tgrid[n][j] = temp;\n\t\t}\n\t\t\n\t\t\n\t\trep(from,0,n+1){\n\t\t\trep(to,from+1,n+1){\n\t\t\t\tll temp = 0;\n\t\t\t\trep(i,0,m+1){\n\t\t\t\t\ttemp += llabs(grid[from][i] - grid[to][i]);\n\t\t\t\t}\n \n\t\t\t\tdiffRow[from][to] = temp;\n\t\t\t\tdiffRow[to][from] = temp;\n\t\t\t}\n\t\t}\n \n\t\trep(from,0,m+1){\n\t\t\trep(to,from+1,m+1){\n\t\t\t\tll temp = 0;\n\t\t\t\trep(i,0,n+1){\n\t\t\t\t\ttemp += llabs(grid[i][from] - grid[i][to]);\n\t\t\t\t}\n\t\t\t\tdiffCol[from][to] = temp;\n\t\t\t\tdiffCol[to][from] = temp;\n\t\t\t}\n\t\t}\n\t\t\n\t\t\n\t\t\n\t\t\n\t\tll best = 1e9;\n \n\t\trep(i,0,n+1) rep(j,0,m+1){\n\t\t\tbannedRow = i;\n\t\t\tbannedCol = j;\n \n\t\t\trep(x,0,1<<16) rep(y,0,16) {\n\t\t\t\tmemoRow[x][y] = -1;\n\t\t\t\tmemoCol[x][y] = -1;\n\t\t\t}\n \n \n \n\t\t\tll currRow = 1e9;\n\t\t\tll currCol = 1e9;\n\t\t\trep(row,0,n+1){\n\t\t\t\tif (row == bannedRow) continue;\n\t\t\t\tcurrRow = min(currRow,dpRow(row,1<<row));\n\t\t\t}\n \n\t\t\trep(col,0,m+1){\n\t\t\t\tif (col == bannedCol) continue;\n\t\t\t\tcurrCol = min(currCol,dpCol(col,1<<col));\n\t\t\t}\n\t\t\t//cout << \"\\n\";\n\t\t\tbest = min(currCol + currRow,best);\n\t\t}\n\t\t\n\t\tcout << best << \"\\n\";\n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274986043,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n// #pragma optimize(\"Ofast\")\n#include \"bits/stdc++.h\"\n#define all(v) (v).begin(), (v).end()\n#define pb push_back\n#define em emplace_back\n#define mp make_pair\n#define F first\n#define S second\n \nusing namespace std;\ntemplate<class C>\nusing vec = vector<C>;\nusing vi = vector<int>;\nusing vpi = vector<pair<int, int>>;\nusing pii = pair<int, int>;\n \nint calc(int n, int m, vec<vi> a) {\n int dp[1 << m][m];\n for (int i = 0; i < (1 << m); i++)\n for (int j = 0; j < m; j++)\n dp[i][j] = INT_MAX;\n for (int i = 0; i < m; i++) {\n dp[1 << i][i] = 0;\n }\n int res = INT_MAX;\n for (int mask = 1; mask < (1 << m); mask++) {\n if (__builtin_popcount(mask) == m - 1) {\n for (int b = 0; b + 1 < m; b++) {\n if ((mask >> b) & 1) {\n res = min(res, dp[mask][b]);\n }\n }\n }\n for (int i = 0; i + 1 < m; i++) {\n if (!((mask >> i) & 1)) continue;\n for (int to = 0; to < m; to++) {\n if ((mask >> to) & 1) continue;\n int relax = dp[mask][i];\n if (relax == INT_MAX) continue;\n for (int j = 1; j < n; j++) {\n relax += abs(a[j][i] - a[j][to]);\n }\n dp[mask ^ (1 << to)][to] = min(dp[mask ^ (1 << to)][to], relax);\n }\n }\n }\n return res;\n}\n \nvoid solve() {\n int n, m;\n cin >> n >> m;\n vec<vi> a(n + 1, vi(m + 1));\n for (int i = 1; i <= n; i++) {\n for (int j = 0; j < m; j++) {\n cin >> a[i][j];\n a[i].back() ^= a[i][j];\n a[0][j] ^= a[i][j];\n a[0].back() ^= a[i][j];\n }\n }\n int res = INT_MAX;\n for (int z = 0; z <= n; z++) {\n for (int z2 = 0; z2 <= m; z2++) {\n swap(a[z], a[0]);\n for (int i = 0; i <= n; i++) {\n swap(a[i][z2], a[i].back());\n }\n int ans = 0;\n ans += calc(n + 1, m + 1, a);\n vec<vi> b(m + 1, vi(n + 1));\n for (int j = m; j >= 0; j--) {\n for (int i = n; i >= 0; i--) {\n b[m - j][n - i] = a[i][j];\n }\n }\n ans += calc(m + 1, n + 1, b);\n res = min(res, ans);\n for (int i = 0; i <= n; i++) {\n swap(a[i][z2], a[i].back());\n }\n swap(a[z], a[0]);\n }\n }\n cout << res << '\\n';\n}\n \nsigned main() {\n ios::sync_with_stdio(false);\n cin.tie(0);\n int tt;\n cin >> tt;\n while (tt--) {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274419749,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\nT=int(input())\nfor t in range(T):\n N,M=map(int,input().split())\n A=[list(map(int,input().split())) for n in range(N)]\n for n in range(N):\n A[n].append(0)\n for m in range(M):\n A[n][-1]^=A[n][m]\n A.append([0]*(M+1))\n for n in range(N):\n for m in range(M+1):\n A[N][m]^=A[n][m]\n def solve(x,y):\n B=[[A[i][j] for j in range(M+1) if j!=y] for i in range(N+1) if i!=x]\n C=[[sum(abs(B[n][m0]-B[n][m1]) for n in range(N)) for m1 in range(M)] for m0 in range(M)]\n D=[[sum(abs(B[n0][m]-B[n1][m]) for m in range(M)) for n1 in range(N)] for n0 in range(N)]\n def solve1(E):\n M=len(E)\n inf=1<<60\n dp=[[inf]*M for bit in range(1<<M)]\n ans=inf\n for bit in range(1,1<<M):\n idx=[m for m in range(M) if bit&1<<m]\n if len(idx)==1:\n m=idx[0]\n dp[bit][m]=0\n else:\n for m1 in idx:\n for m0 in idx:\n if m0==m1:\n continue\n dp[bit][m1]=min(dp[bit][m1],dp[bit^1<<m1][m0]+E[m0][m1])\n if len(idx)==M:\n ans=min(ans,min(dp[bit]))\n return ans\n ans=solve1(C)+solve1(D)\n return ans\n ans=min(solve(x,y) for x in range(N+1) for y in range(M+1))\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274403256,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\nans = []\ninf = 10**12\ndef salesman(dists):\n\tN = len(dists)\n\tv = [[inf]*(1<<N) for _ in range(N)]\n \n\tfor sta in range(N):\n\t\tv[sta][1<<sta] = 0\n \n\tfor mask in range(1<<N):\n\t\tfor sta in range(N):\n\t\t\tif not (mask>>sta)&1: continue\n\t\t\tfor end in range(N):\n\t\t\t\tif (mask>>end)&1: continue\n\t\t\t\tv[end][mask|(1<<end)] = min( v[end][mask|(1<<end)], v[sta][mask]+dists[sta][end] )\n\t\n\t#print(v)\n\treturn min( v[sta][(1<<N)-1] for sta in range(N) )\n \n \nfor _ in range(int(input())):\n\tn,m = [int(x) for x in input().split()]\n\tg = [[int(x) for x in input().split()] for _ in range(n)]\n \n\th = [[0]*(m+1) for _ in range(n+1)]\n\tfor y in range(n):\n\t\tfor x in range(m):\n\t\t\th[y][x] = g[y][x]\n\t\t\th[n][x] ^= g[y][x]\n\t\t\th[y][m] ^= g[y][x]\n\t\t\th[n][m] ^= g[y][x]\n \n\tbb = inf\n\tfor covery in range(n+1):\n\t\tfor coverx in range(m+1):\n\t\t\tdists = []\n\t\t\tfor y1 in range(n+1):\n\t\t\t\tif y1 == covery: continue\n\t\t\t\tline = []\n\t\t\t\tfor y2 in range(n+1):\n\t\t\t\t\tif y2 == covery: continue\n\t\t\t\t\tline.append( sum( abs(h[y2][x]-h[y1][x]) for x in range(m+1) if not x==coverx ) )\n\t\t\t\tdists.append(line)\n\t\t\tb = salesman(dists)\t\n\t\t\t\n\t\t\tdists = []\n\t\t\tfor x1 in range(m+1):\n\t\t\t\tif x1 == coverx: continue\n\t\t\t\tline = []\n\t\t\t\tfor x2 in range(m+1):\n\t\t\t\t\tif x2 == coverx: continue\n\t\t\t\t\tline.append( sum( abs(h[y][x2]-h[y][x1]) for y in range(n+1) if not y==covery ) )\n\t\t\t\tdists.append(line)\n\t\t\tb += salesman(dists)\n\t\t\tbb = min(b,bb)\n\tans.append(bb)\n \nfor w in ans:\n\tprint(w)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 275171845,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n#include <bits/stdc++.h>\n \n#ifdef LOCAL\n#include \"debug.h\"\n#else\n#define debug(x...) 228\n#endif\n \nusing namespace std;\n \ntypedef long long ll;\ntypedef unsigned long long ull;\ntypedef long double ld;\n \n#define fi first\n#define se second\n \nconst int inf = int(1e9) + 1;\nconst ll infll = ll(1e18) + 1;\nint mod = int(1e9) + 7;\nmt19937_64 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());\n \ntemplate<typename T1, typename T2>\nbool chmin(T1 &a, T2 b) {\n if (a > b) return a = b, true;\n return false;\n}\ntemplate<typename T1, typename T2>\nbool chmax(T1 &a, T2 b) {\n if (a < b) return a = b, true;\n return false;\n}\n \nvoid solve_case() {\n int n, m; cin >> n >> m;\n vector<vector<int>> a(n + 1, vector<int>(m + 1));\n for (int i = 0; i < n; i++) {\n for (int j = 0; j < m; j++) {\n cin >> a[i][j];\n a[i][m] ^= a[i][j];\n a[n][j] ^= a[i][j];\n a[n][m] ^= a[i][j];\n }\n }\n ll ans = infll;\n// vector<vector<int>> all_pn, all_pm;\n// {\n// vector<int> perm(n + 1); iota(perm.begin(), perm.end(), 0);\n// do {\n// all_pn.push_back(perm);\n// } while (next_permutation(perm.begin(), perm.end()));\n// }\n// {\n// vector<int> perm(m + 1); iota(perm.begin(), perm.end(), 0);\n// do {\n// all_pm.push_back(perm);\n// } while (next_permutation(perm.begin(), perm.end()));\n// }\n// for (auto pn : all_pn) {\n// for (auto pm : all_pm) {\n// ll res = 0;\n// for (int i = 0; i + 1 < n; i++) {\n// for (int j = 0; j < m; j++) {\n// res += abs(a[pn[i]][pm[j]] - a[pn[i + 1]][pm[j]]);\n// }\n// }\n// for (int i = 0; i < n; i++) {\n// for (int j = 0; j + 1 < m; j++) {\n// res += abs(a[pn[i]][pm[j]] - a[pn[i]][pm[j + 1]]);\n// }\n// }\n// ans = min(ans, res);\n// }\n// }\n// cout << ans << \"\\n\";\n// return;\n for (int vi = 0; vi <= n; vi++) {\n for (int vj = 0; vj <= m; vj++) {\n vector<vector<int>> na(n, vector<int>(m));\n int ptr_i = 0;\n for (int i = 0; i <= n; i++) {\n if (i == vi) continue;\n int ptr_j = 0;\n for (int j = 0; j <= m; j++) {\n if (j == vj) continue;\n na[ptr_i][ptr_j] = a[i][j];\n ptr_j++;\n }\n ptr_i++;\n }\n swap(a, na);\n auto solve = [&]() -> ll {\n vector<vector<ll>> dp(1 << n, vector<ll>(n, infll));\n for (int mask = 1; mask < (1 << n); mask++) {\n if (__builtin_popcount(mask) == 1) {\n for (int i = 0; i < n; i++) {\n if ((mask >> i) & 1) {\n dp[mask][i] = 0;\n }\n }\n continue;\n }\n for (int i = 0; i < n; i++) {\n if (!((mask >> i) & 1)) continue;\n for (int pi = 0; pi < n; pi++) {\n if (pi == i || !((mask >> pi) & 1)) continue;\n ll sum = 0;\n for (int j = 0; j < m; j++) {\n sum += abs(a[i][j] - a[pi][j]);\n }\n chmin(dp[mask][i], dp[mask ^ (1 << i)][pi] + sum);\n }\n }\n }\n return *min_element(dp.back().begin(), dp.back().end());\n };\n ll cur = solve();\n vector<vector<int>> b(m, vector<int>(n));\n for (int i = 0; i < n; i++) {\n for (int j = 0; j < m; j++) {\n b[j][i] = a[i][j];\n }\n }\n swap(n, m);\n swap(a, b);\n cur += solve();\n swap(a, b);\n swap(n, m);\n swap(a, na);\n ans = min(ans, cur);\n }\n }\n cout << ans << \"\\n\";\n}\n \nint main() {\n ios::sync_with_stdio(false);\n cin.tie(0), cout.tie(0);\n#ifdef LOCAL\n clock_t tStart = clock();\n#endif\n int T = 1; cin >> T;\n for (int test = 1; test <= T; test++) {\n solve_case();\n }\n#ifdef LOCAL\n fprintf(stderr, \"\\n>> Runtime: %.10fs\\n\", (double) (clock() - tStart) / CLOCKS_PER_SEC);\n#endif\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 298962607,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n#include<bits/stdc++.h>\nnamespace {\nusing namespace std;\n#define rep(i,n) for(int i = 0; i < (int)(n); i++)\n#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; i--)\n#define all(x) begin(x), end(x)\n#define rall(x) rbegin(x), rend(x)\ntemplate<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } else return false; }\ntemplate<class T> bool chmin(T& a, const T& b) { if (b < a) { a = b; return true; } else return false; }\nusing ll = long long;\nusing P = pair<int,int>;\nusing VI = vector<int>;\nusing VVI = vector<VI>;\nusing VL = vector<ll>;\nusing VVL = vector<VL>;\n \n} int main() {\n ios::sync_with_stdio(false);\n cin.tie(0);\n int tt;\n cin >> tt;\n while(tt--) {\n int n, m;\n cin >> n >> m;\n int a[16][16]{};\n rep(i, n) rep(j, m) cin >> a[i][j], a[n][j] ^= a[i][j], a[i][m] ^= a[i][j], a[n][m] ^= a[i][j];\n constexpr int INF = 1001001001;\n int ans = INF;\n rep(ri, n + 1) rep(rj, m + 1) {\n int b[15][15]{};\n {\n int pi = 0;\n rep(i, n + 1) if (i != ri) {\n int pj = 0;\n rep(j, m + 1) if (j != rj) b[pi][pj++] = a[i][j];\n pi++;\n }\n assert(pi == n);\n }\n int cost = 0;\n rep(_, 2) {\n int dp[1 << 15][15];\n rep(i, 1 << 15) rep(j, 15) dp[i][j] = INF;\n rep(i, n) dp[1 << i][i] = 0;\n rep(s, 1 << n) rep(pi, n) if (dp[s][pi] != INF) {\n rep(ni, n) {\n int add = 0;\n rep(j, m) add += abs(b[ni][j] - b[pi][j]);\n chmin(dp[s | 1 << ni][ni], dp[s][pi] + add);\n }\n }\n int last = (1 << n) - 1;\n int mn = INF;\n rep(i, n) {\n chmin(mn, dp[last][i]);\n }\n assert(mn != INF);\n cost += mn;\n swap(n, m);\n int pb[15][15];\n memcpy(pb, b, sizeof(b));\n rep(i, n) rep(j, m) b[i][j] = pb[j][i];\n }\n chmin(ans, cost);\n }\n cout << ans << '\\n';\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274420008,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nusing ll = long long;\nusing pii = pair<int, int>;\nusing pll = pair<ll, ll>;\nusing pdi = pair<double, int>;\nusing pdd = pair<double, double>;\nusing ull = unsigned long long;\nusing LL = __int128_t;\nusing vint = vector<int>;\n \n#define ppc(x) __builtin_popcount(x)\n#define clz(x) __builtin_clz(x)\n \nbool Mbe;\n \nmt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());\nint rd(int l, int r) {\n return rnd() % (r - l + 1) + l;\n}\n \nconstexpr int mod = 1e9 + 7;\nvoid addt(int &x, int y) {\n x += y, x >= mod && (x -= mod);\n}\nint add(int x, int y) {\n return x += y, x >= mod && (x -= mod), x;\n}\nint ksm(int a, ll b) {\n int s = 1;\n while(b) {\n if(b & 1) s = 1ll * s * a % mod;\n a = 1ll * a * a % mod, b >>= 1;\n }\n return s;\n}\n \nstruct FastMod {\n ull b, m;\n FastMod(ull b) : b(b), m(ull((LL(1) << 64) / b)) {}\n ull reduce(ull a) {\n ull q = ull((LL(m) * a) >> 64);\n ull r = a - q * b; // can be proven that 0 <= r < 2 * b\n return r >= b ? r - b : r;\n }\n} R(1);\n \nconstexpr int Z = 1e5 + 5;\nint fc[Z], ifc[Z], inv[Z];\nint bin(int n, int m) {\n if(n < m) return 0;\n return 1ll * fc[n] * ifc[m] % mod * ifc[n - m] % mod;\n}\nvoid init_fac(int Z) {\n for(int i = fc[0] = 1; i < Z; i++) fc[i] = 1ll * fc[i - 1] * i % mod;\n ifc[Z - 1] = ksm(fc[Z - 1], mod - 2);\n for(int i = Z - 2; ~i; i--) ifc[i] = 1ll * ifc[i + 1] * (i + 1) % mod;\n for(int i = 1; i < Z; i++) inv[i] = 1ll * fc[i - 1] * ifc[i] % mod;\n}\n \nchar buf[1 << 20], *p1 = buf, *p2 = buf;\n#define getc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? EOF : *p1++)\n#define putc(x) putchar(x)\ninline int read() {\n int x = 0, sgn = 0; char s = getc();\n while(!isdigit(s)) sgn |= s == '-', s = getc();\n while(isdigit(s)) x = x * 10 + s - '0', s = getc();\n return sgn ? -x : x;\n}\ninline void print(int x) {if(x >= 10) print(x / 10); putc(x % 10 + '0');}\n \n// ---------- templates above ----------\n \nconstexpr int N = 15;\n \nint n, m, a[N + 1][N + 1], b[N][N], d[N][N], f[N][1 << N];\nint dp(int n) {\n if(n == 1) return 0;\n int ans = 1e9;\n for(int i = 1; i < 1 << n; i++) {\n int ppc = __builtin_popcount(i);\n if(ppc == 1) f[__lg(i)][i] = 0;\n else {\n for(int j = 0; j < n; j++) {\n if(i >> j & 1 ^ 1) continue;\n f[j][i] = 1e9;\n for(int k = 0; k < n; k++) {\n if(j == k || (i >> k & 1 ^ 1)) continue;\n f[j][i] = min(f[j][i], f[k][i ^ (1 << j)] + d[j][k]);\n }\n if((i + 1) == 1 << n) ans = min(ans, f[j][i]);\n }\n }\n }\n return ans;\n}\nvoid solve(int T) {\n cin >> n >> m;\n memset(a, 0, sizeof(a));\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < m; j++) {\n cin >> a[i][j];\n a[i][m] ^= a[i][j];\n a[n][j] ^= a[i][j];\n a[n][m] ^= a[i][j];\n }\n }\n int ans = 1e9;\n for(int p = 0; p <= n; p++) {\n for(int q = 0; q <= m; q++) {\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < m; j++) {\n b[i][j] = a[i + (p <= i)][j + (q <= j)];\n }\n }\n int cost = 0;\n for(int i = 0; i < n; i++) {\n for(int j = i + 1; j < n; j++) {\n d[i][j] = 0;\n for(int k = 0; k < m; k++) {\n d[i][j] += abs(b[i][k] - b[j][k]);\n }\n d[j][i] = d[i][j];\n }\n }\n cost = dp(n);\n for(int i = 0; i < m; i++) {\n for(int j = i + 1; j < m; j++) {\n d[i][j] = 0;\n for(int k = 0; k < n; k++) {\n d[i][j] += abs(b[k][i] - b[k][j]);\n }\n d[j][i] = d[i][j];\n }\n }\n cost += dp(m);\n ans = min(ans, cost);\n }\n }\n cout << ans << \"\\n\";\n}\n \nbool Med;\nsigned main() {\n ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); \n int T = 1;\n cin >> T;\n while(T--) solve(T);\n fprintf(stderr, \"%.3lf ms\\n\", 1e3 * clock() / CLOCKS_PER_SEC);\n return 0;\n}\n \n/*\ng++ a.cpp -o a -std=c++14 -O2 -DALEX_WEI\n \n1\n2 3\n0 2 4\n4 5 1\n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274917773,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n#include <vector>\n#pragma GCC target (\"avx2\")\n#pragma GCC optimize (\"O3\")\n#pragma GCC optimize (\"unroll-loops\")\n#include <bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n \nusing namespace std;\nusing namespace __gnu_pbds;\n \ntemplate<typename T>\nostream_iterator<T> oit(const string &s = \" \"){ return ostream_iterator<T>(cout,s.c_str()); }\ninline auto rep(int l, int r) { return views::iota(min(l, r), r); }\ninline auto rep(int n) { return rep(0, n); }\ninline auto rep1(int l, int r) { return rep(l, r + 1); }\ninline auto rep1(int n) { return rep(1, n + 1); }\ninline auto per(int l, int r) { return rep(l, r) | views::reverse; }\ninline auto per(int n) { return per(0, n); }\ninline auto per1(int l, int r) { return per(l, r + 1); }\ninline auto per1(int n) { return per(1, n + 1); }\n#define A(a) begin(a),end(a)\ninline auto len = ranges::ssize;\n \nstruct chash {\n static uint64_t splitmix64(uint64_t x) {\n // http://xorshift.di.unimi.it/splitmix64.c\n x += 0x9e3779b97f4a7c15;\n x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;\n x = (x ^ (x >> 27)) * 0x94d049bb133111eb;\n return x ^ (x >> 31);\n }\n \n size_t operator()(uint64_t x) const {\n static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();\n return splitmix64(x + FIXED_RANDOM);\n }\n};\ntemplate<typename T, typename U> using pb_map = gp_hash_table<T, U, chash>;\ntemplate<typename T> using pb_set = gp_hash_table<T, null_type, chash>;\n#define K first\n#define V second\n \nusing ll = long long;\nusing ld = long double;\n \nusing vi = vector<int>;\nusing vii = vector<vector<int>>;\ntypedef vector<ll> vll;\nusing pll = pair<ll,ll>;\nusing pii = pair<int,int>;\n \nconstexpr ll NN = (1<<15), M = 1000000007, L = 20;\n \nint n,m;\nint dp[NN][15];\nint pre[15][15];\nint oo = 1e9;\nvoid run()\n{\n cin >> n >> m;\n vii g(n,vi(m));\n for(auto &v : g) \n for(int &x : v) cin >> x;\n \n int ans = oo;\n for(int r : rep(n+1)) for(int c : rep(m+1)) {\n int add1=oo,add2=oo;\n \n if(r < n) //do operations\n for(int i : rep(n))\n if(i != r) for(int j : rep(m)) g[r][j] ^= g[i][j];\n if(c < m)\n for(int j : rep(m))\n if(j != c) for(int i : rep(n)) g[i][c] ^= g[i][j];\n \n for(int i : rep(n)) for(int j : rep(n)){ //calculate pre\n pre[i][j] = 0;\n for(int k : rep(m)) pre[i][j] += abs(g[i][k]-g[j][k]);\n }\n //do dp for rows\n for(unsigned int msk = 1;msk<(1<<n);msk++) for(int l = 0;l<n;l++) if((1<<l)&msk){\n int &here = dp[msk][l]; here=oo;\n if(popcount(msk)==1) here=0;\n for(int bt = 0; bt<n;bt++)\n if((bt not_eq l) and ((1<<bt)&msk))\n here=min(here,dp[msk^(1<<l)][bt]+pre[l][bt]);\n if(popcount(msk)==n) add1=min(add1,here);\n }\n //calculate pre again\n for(int i : rep(m)) for(int j : rep(m)){\n pre[i][j]=0;\n for(int k : rep(n)) pre[i][j] += abs(g[k][i]-g[k][j]);\n }\n //do dp for cols\n for(unsigned int msk = 1;msk<(1<<m);msk++) for(int l = 0;l<m;l++) if((1<<l)&msk){\n int &here = dp[msk][l]; here=oo;\n if(popcount(msk)==1) here=0;\n for(int bt = 0; bt<m;bt++)\n if(bt not_eq l and (1<<bt)&msk)\n here=min(here,dp[msk^(1<<l)][bt]+pre[l][bt]);\n if(popcount(msk)==m) add2=min(add2,here);\n }\n // cout << add1 << \" \" << add2 << '\\n';\n // for(auto &v : g)\n // ranges::copy(v,oit<ll>()),cout << '\\n';\n if(r < n) //undo operations\n for(int i : rep(n))\n if(i != r) for(int j : rep(m)) g[r][j] ^= g[i][j];\n if(c < m)\n for(int j : rep(m))\n if(j != c) for(int i : rep(n)) g[i][c] ^= g[i][j];\n \n ans = min(ans,add1+add2);\n }\n cout << ans << '\\n';\n}\n \nint main()\n{\n //KING OF THE WORLD...... U.W.T.B\n cin.tie(0);\n int t; cin>>t; while(t--) run();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 275455906,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\nimport sys\n \n# region fastio\nfrom collections import Counter, defaultdict\nfrom heapq import heappop, heappush\nfrom math import inf\nfrom random import getrandbits\n \ninput = lambda: sys.stdin.readline().rstrip()\nsint = lambda: int(input())\nmint = lambda: map(int, input().split())\nints = lambda: list(map(int, input().split()))\n \nMOD = 10 ** 9 + 7\n \nmx = 2 * 10 ** 5\n# f = [0] * MX\n# f[0] = f[1] = 1\n# for i in range(2, MX):\n# f[i] = f[i - 1] + 2 * (i - 1) * f[i - 2]\n# f[i] %= MOD\n# 阶乘\nfac = [1] * (mx + 1)\nfor i in range(1, mx + 1):\n fac[i] = fac[i - 1] * i % MOD\n# 逆元\nifac = [0] * (mx + 1)\nifac[mx] = pow(fac[mx], MOD - 2, MOD)\nfor i in range(mx, 0, -1):\n ifac[i - 1] = ifac[i] * i % MOD\n \n \n# 组合数\ndef comb(n: int, m: int, MOD=MOD) -> int:\n if m < 0 or m > n:\n return 0\n return fac[n] * ifac[m] % MOD * ifac[n - m] % MOD\n \n \n# endregion fastio\n \n# MOD = 998_244_353\n# MOD = 10 ** 9 + 7\n# DIR4 = ((-1, 0), (0, 1), (1, 0), (0, -1)) #URDL\n# DIR8 = ((-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1))\ndef cf1993E():\n n, m = mint()\n a = [[0] * (m+1) for _ in range(n+1)]\n for i in range(n):\n cur = ints()\n for j, x in enumerate(cur):\n a[i][j] = x\n a[i][m] ^= x\n a[n][j] ^= x\n a[n][m] ^= x\n \n mask_n = (1 << (n + 1)) - 1\n mask_m = (1 << (m + 1)) - 1\n ans1 = [[inf] * 16 for _ in range(16)]\n dp = [[inf] * (1 << (n + 1)) for _ in range(n + 1)]\n for rmv in range(m+1):\n w = [[0] * (n + 1) for _ in range(n + 1)]\n for i in range(n+1):\n for j in range(i+1, n+1):\n w[i][j] = 0\n for l in range(m+1):\n if l == rmv:\n continue\n w[i][j] += abs(a[i][l] - a[j][l])\n w[j][i] = w[i][j]\n \n for i in range(n + 1):\n for msk in range(1 << (n + 1)):\n dp[i][msk] = inf\n dp[i][1 << i] = 0\n \n for msk in range(1 << n + 1):\n for last in range(n + 1):\n if (1 << last) & msk == 0:\n continue\n if msk.bit_count() == n:\n continue\n \n for nxt in range(n + 1):\n if (1 << nxt) & msk > 0:\n continue\n new_msk = msk | (1 << nxt)\n dp[nxt][new_msk] = min(dp[nxt][new_msk], dp[last][msk] + w[last][nxt])\n \n for i in range(n+1):\n ans1[i][rmv] = inf\n msk = mask_n ^ (1 << i)\n for last in range(n+1):\n ans1[i][rmv] = min(ans1[i][rmv], dp[last][msk])\n \n ans2 = [[inf] * 16 for _ in range(16)]\n dp = [[inf] * (1 << (m + 1)) for _ in range(m + 1)]\n for rmv in range(n+1):\n w = [[0] * (m + 1) for _ in range(m + 1)]\n for i in range(m+1):\n for j in range(i+1, m+1):\n w[i][j] = 0\n for l in range(n+1):\n if l == rmv:\n continue\n w[i][j] += abs(a[l][i] - a[l][j])\n w[j][i] = w[i][j]\n \n for i in range(m + 1):\n for msk in range(1 << (m + 1)):\n dp[i][msk] = inf\n dp[i][1 << i] = 0\n \n for msk in range(1 << m + 1):\n for last in range(m + 1):\n if (1 << last) & msk == 0:\n continue\n if msk.bit_count() == m:\n continue\n \n for nxt in range(m + 1):\n if (1 << nxt) & msk > 0:\n continue\n new_msk = msk | (1 << nxt)\n dp[nxt][new_msk] = min(dp[nxt][new_msk], dp[last][msk] + w[last][nxt])\n \n for i in range(m+1):\n ans2[rmv][i] = inf\n msk = mask_m ^ (1 << i)\n for last in range(m+1):\n ans2[rmv][i] = min(ans2[rmv][i], dp[last][msk])\n \n \n ans = inf\n for i in range(n+1):\n for j in range(m+1):\n ans = min(ans, ans1[i][j] + ans2[i][j])\n \n print(ans)\n return\n \n \nif __name__ == \"__main__\":\n t_ = sint()\n for _ in range(t_):\n cf1993E()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274535144,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\nT=int(input())\nfor t in range(T):\n N,M=map(int,input().split())\n A=[list(map(int,input().split())) for n in range(N)]\n for n in range(N):\n A[n].append(0)\n for m in range(M):\n A[n][-1]^=A[n][m]\n A.append([0]*(M+1))\n for n in range(N):\n for m in range(M+1):\n A[N][m]^=A[n][m]\n def solve(A):\n retu=[]\n N=len(A)\n M=len(A[0])\n for x in range(N):\n C=[[sum(abs(A[n][m0]-A[n][m1]) for n in range(N) if n!=x) for m1 in range(M)] for m0 in range(M)]\n inf=1<<60\n dp=[[inf]*M for bit in range(1<<M)]\n for bit in range(1,1<<M):\n idx=[m for m in range(M) if bit&1<<m]\n if len(idx)==1:\n m=idx[0]\n dp[bit][m]=0\n else:\n for m1 in idx:\n for m0 in idx:\n if m0==m1:\n continue\n dp[bit][m1]=min(dp[bit][m1],dp[bit^1<<m1][m0]+C[m0][m1])\n retu.append([min(dp[(1<<M)-1-(1<<m)]) for m in range(M)])\n return retu\n C0=solve(A)\n C1=solve([[A[n][m] for n in range(N+1)] for m in range(M+1)])\n ans=min(C0[n][m]+C1[m][n] for n in range(N+1) for m in range(M+1))\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 274435901,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n//#pragma GCC optimize(\"Ofast\")\n//#pragma GCC optimize(\"unroll-loops\")\n//#pragma GCC target(\"sse,sse2,sse3,ssse3,abm,mmx,tune=native\")\n#include<vector>\n#include<iostream>\n#include<stack>\n#include<cmath>\n#include<algorithm>\n#include<set>\n#include<map>\n#include<string>\n#include<tuple>\n#include<bitset>\n#include<queue>\n#include<unordered_map>\n#include<random>\n#include<assert.h>\n#include<ctime>\n#include <cassert>\n//#define double long double\ntypedef long long ll;\ntypedef double ld;\ntypedef unsigned long long ull;\nusing namespace std;\nll gcd(ll i, ll j) {\n\tif (i < j)swap(i, j);\n\tif (j == 0)return i;\n\telse return gcd(j, i % j);\n}\n#ifdef _MY\nll __builtin_popcount(ll x) { return x ? (__builtin_popcount(x >> 1) + (x & 1)) : 0; }\n#endif\ntemplate<typename T> inline T getint() {\n\tT val = 0;\n\tchar c;\n \n\tbool neg = false;\n\twhile ((c = getchar()) && !(c >= '0' && c <= '9')) {\n\t\tneg |= c == '-';\n\t}\n \n\tdo {\n\t\tval = (val * 10) + c - '0';\n\t} while ((c = getchar()) && (c >= '0' && c <= '9'));\n \n\treturn val * (neg ? -1 : 1);\n}\nconst ll INF = 1e9 + 100;\nconst int mod = 998244353, mod1 = 1000000009;\nconst ld eps = 1e-7, pi = acos(-1);\nconst int maxT = 1601, maxN = 1 << 20, A = 311;\nll bp(ll et, ll b) {\n\tll res = 1;\n\tfor (int i = 30; i >= 0; --i) {\n\t\tres = (res * res) % mod;\n\t\tif ((b & (1 << i)) != 0)res = (res * et) % mod;\n\t}\n\treturn res;\n}\nmt19937 mt_rand(time(0));\nvoid panic() {\n\tcout << \"No\\n\";\n\texit(0);\n}\nstruct pt {\n\tll x, y;\n\tpt() = default;\n\tpt(ll X_, ll Y_) {\n\t\tx = X_;\n\t\ty = Y_;\n\t}\n};\npt operator -(const pt& a, const pt& b) {\n\treturn pt(a.x - b.x, a.y - b.y);\n}\nll dd2(pt a) {\n\treturn a.x * a.x + a.y * a.y;\n}\npt read() {\n\tll x, y;\n\tcin >> x >> y;\n\treturn pt(x, y);\n}\nll vc(pt a, pt b) {\n\treturn a.x * b.y - b.x * a.y;\n}\n \nll cc(pt a, pt b) {\n\treturn a.x * b.x + b.y * a.y;\n}\nvoid dfs(int v, const vector<vector<int>>& g, vector<int>& sz) {\n\tsz[v] = 1;\n\tfor (auto x : g[v]) {\n\t\tdfs(x, g, sz);\n\t\tsz[v] += sz[x];\n\t}\n}\nint dp[15][1 << 15];\nint calc(const vector<vector<int>>& weight) {\n\tint n = weight.size();\n\tfor (int i = 1; i < (1 << n); ++i) {\n\t\tfor (int j = 0; j < n; ++j) {\n\t\t\tif ((i & (1 << j)) != 0 && i != (1 << j)) {\n\t\t\t\tdp[j][i] = INF;\n\t\t\t\tfor (int k = 0; k < n; ++k) {\n\t\t\t\t\tif (k == j)\n\t\t\t\t\t\tcontinue;\n\t\t\t\t\tif ((i & (1 << k)) == 0)\n\t\t\t\t\t\tcontinue;\n\t\t\t\t\tauto ww = weight[j][k];\n\t\t\t\t\tdp[j][i] = min(dp[j][i], ww + dp[k][i ^ (1 << j)]);\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t}\n\tint ans = INF;\n\tfor (int i = 0; i < n; ++i) {\n\t\tans = min(ans, dp[i][(1 << n) - 1]);\n\t}\n \n\treturn ans;\n}\nvoid solve() {\n\tint n = 15, m = 15;\n\tcin >> n >> m;\n\tvector<vector<int>>a(n + 1, vector<int>(m + 1));\n\tfor (int i = 0; i < n; ++i) {\n\t\tfor (int j = 0; j < m; ++j) {\n\t\t\ta[i][j] = rand();\n\t\t\tcin >> a[i][j];\n\t\t\ta[i][m] ^= a[i][j];\n\t\t\ta[n][j] ^= a[i][j];\n\t\t\ta[n][m] ^= a[i][j];\n\t\t}\n\t}\n\tint ans = INF;\n\tfor (int i = 0; i <= n; ++i) {\n\t\tfor (int j = 0; j <= m; ++j) {\n\t\t\tauto aa = a;\n\t\t\tswap(aa[i], aa[n]);\n\t\t\taa.pop_back();\n\t\t\tfor (int k = 0; k < n; ++k) {\n\t\t\t\tswap(aa[k][j], aa[k][m]);\n\t\t\t\taa[k].pop_back();\n\t\t\t}\n\t\t\tvector<vector<int>>weight1(n, vector<int>(n));\n\t\t\tfor (int i = 0; i < n; ++i) {\n\t\t\t\tfor (int j = 0; j < n; ++j) {\n\t\t\t\t\tfor (int k = 0; k < m; ++k) {\n\t\t\t\t\t\tweight1[i][j] += abs(aa[i][k] - aa[j][k]);\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t}\n \n\t\t\tvector<vector<int>>weight2(m, vector<int>(m));\n\t\t\tfor (int i = 0; i < m; ++i) {\n\t\t\t\tfor (int j = 0; j < m; ++j) {\n\t\t\t\t\tfor (int k = 0; k < n; ++k) {\n\t\t\t\t\t\tweight2[i][j] += abs(aa[k][i] - aa[k][j]);\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t}\n\t\t\tauto r1 = calc(weight1), r2 = calc(weight2);\n\t\t\tans = min(ans, r1 + r2);\n\t\t}\n\t}\n\tcout << ans << \"\\n\";\n}\n \nint main()\n{\n\tios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n\tcout.precision(9);\n#ifdef _MY\n\tfreopen(\"input.txt\", \"r\", stdin);\n\tfreopen(\"output.txt\", \"w\", stdout);\n#else\n\t//freopen(\"epic.in\", \"r\", stdin); freopen(\"epic.out\", \"w\", stdout);\n#endif\n \n\tint t = 1;\n\tcin >> t;\n\twhile (t--) {\n\t\tsolve();\n\t}\n\treturn 0;\n}\n/*\nhttps://codeforces.com/gym/104686/problem/J\n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 281649394,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Xor-Grid Problem\ntime_limit_ms: 5000\nmemory_limit_mb: 256\nDescription: Given a matrix $a$ of size $n \\times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.\n\nLet's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:\n\n * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \\le j \\le m$; or * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \\le i \\le n$. \n\n An example of applying an operation on column $2$ of the matrix.\n\nIn this example, as we apply an operation on column $2$, all elements in this column are changed:\n\n * $a_{1,2} := f(1) = a_{1,1} \\oplus a_{1,2} \\oplus a_{1,3} \\oplus a_{1,4} = 1 \\oplus 1 \\oplus 1 \\oplus 1 = 0$ * $a_{2,2} := f(2) = a_{2,1} \\oplus a_{2,2} \\oplus a_{2,3} \\oplus a_{2,4} = 2 \\oplus 3 \\oplus 5 \\oplus 7 = 3$ * $a_{3,2} := f(3) = a_{3,1} \\oplus a_{3,2} \\oplus a_{3,3} \\oplus a_{3,4} = 2 \\oplus 0 \\oplus 3 \\oplus 0 = 1$ * $a_{4,2} := f(4) = a_{4,1} \\oplus a_{4,2} \\oplus a_{4,3} \\oplus a_{4,4} = 10 \\oplus 11 \\oplus 12 \\oplus 16 = 29$ \n\nYou can apply the operations any number of times. Then, we calculate the $\\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.\n\nMore formally, $\\textit{beauty}(a) = \\sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.\n\nFind the minimum $\\textit{beauty}$ among all obtainable matrices.\n\nInput: The first line contains a single integer $t$ ($1 \\le t \\le 250$) — the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $m$ ($1 \\le n, m \\le 15$) — the number of rows and columns of $a$, respectively.\n\nThe next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \\ldots, a_{i,m}$ ($0 \\le a_{i,j} < 2^{20}$) — description of the matrix $a$.\n\nIt is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.\n\nOutput: For each test case, print a single integer $b$ — the smallest possible $\\textit{beauty}$ of the matrix.\n\nExamples:\ninput:\n4\n1 2\n1 3\n2 3\n0 1 0\n5 4 4\n2 3\n0 2 4\n4 5 1\n3 3\n1 2 3\n4 5 6\n7 8 9\noutput:\n1\n3\n13\n24\n\nNote: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.\n\nIn the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \\oplus 3 = 2$. Then, we'll receive this matrix:\n\n2| 3 ---|--- In the second test case, you can apply an operation $r(1)$, which assigns:\n\n * $a_{1,1} := g(1) = 0 \\oplus 5 = 5$ * $a_{1,2} := g(2) = 1 \\oplus 4 = 5$ * $a_{1,3} := g(3) = 0 \\oplus 4 = 4$ \n\nThe resulting matrix after performing the operation is:\n\n5| 5| 4 ---|---|--- 5| 4| 4 In the third test case, the best way to achieve minimum $\\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:\n\n0| 4| 6 ---|---|--- 4| 5| 6\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long\nconst int inf=1e18;\nint a[20][20],b[20][20];\nint f[1<<15][16];\nint g[20][20];\nint ans;\nint n,m;\nint solve(int n,int m,int a[][20]){\n\tfor(int s=0;s<(1<<n);s++) memset(f[s],63,sizeof(f[s]));\n\tfor(int i=0;i<n;i++) f[1<<i][i]=0;\n\tfor(int i=0;i<n;i++){\n\t\tfor(int j=0;j<n;j++){\n\t\t\tg[i][j]=0;\n\t\t\tfor(int k=0;k<m;k++) g[i][j]+=abs(a[i][k]-a[j][k]);\n\t\t}\n\t}\n\tfor(int s=1;s<(1<<n);s++){\n\t\tfor(int i=0;i<n;i++){\n\t\t\tif((s>>i)&1){\n\t\t\t\tfor(int j=0;j<n;j++){\n\t\t\t\t\tif(!((s>>j)&1)){\n\t\t\t\t\t\tint t=s|(1<<j);\n\t\t\t\t\t\tf[t][j]=min(f[t][j],f[s][i]+g[i][j]);\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t}\n\tint res=inf;\n\tfor(int i=0;i<n;i++)res=min(res,f[(1<<n)-1][i]);\n\treturn res;\n}\nsigned main(){\n\tios::sync_with_stdio(0),cin.tie(0),cout.tie(0);\n\tint T;cin>>T;\n\twhile(T--){\n\t\tmemset(a,0,sizeof(a));\n\t\tcin>>n>>m;\n\t\tfor(int i=0;i<n;i++){\n\t\t\tfor(int j=0;j<m;j++) cin>>a[i][j],a[i][m]^=a[i][j],a[n][m]^=a[i][j],a[n][j]^=a[i][j];\n\t\t}\n\t\tn++;m++;\n\t\tans=inf;\n\t\tfor(int i=n-1;i>=0;i--){\n\t\t\tfor(int k=i;k<n-1;k++) swap(a[k],a[k+1]);\n\t\t\tfor(int j=m-1;j>=0;j--){\n\t\t\t\tint res=0; \n\t\t\t\tfor(int k=0;k<n;k++){\n\t\t\t\t\tfor(int l=j;l<m-1;l++) swap(a[k][l],a[k][l+1]);\n\t\t\t\t}\n\t\t\t\tfor(int i=0;i<n-1;i++){\n\t\t\t\t\tfor(int j=0;j<m-1;j++) b[m-1-j-1][i]=a[i][j];\n\t\t\t\t}\n\t\t\t\tres+=solve(n-1,m-1,a);\n\t\t\t\tres+=solve(m-1,n-1,b);\n\t\t\t\tans=min(ans,res);\n\t\t\t}\n\t\t}\n\t\tcout<<ans<<'\\n';\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1993E",
"submission_id": 286079354,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\n \nusing namespace std;\n \nint main() {\n int t;\n cin >> t;\n \n while (t--) {\n int n;\n cin >> n;\n \n vector<int> a(n);\n for (int i = 0; i < n; ++i) {\n cin >> a[i];\n }\n \n vector<int> prefix_xor(n + 1, 0);\n for (int i = 0; i < n; ++i) {\n prefix_xor[i + 1] = prefix_xor[i] ^ a[i];\n }\n \n long long ans = 0;\n for (int i = 0; i < n; ++i) {\n int xr = 0;\n for (int j = i; j < n; ++j) {\n xr ^= a[j];\n if (prefix_xor[j + 1] > xr) {\n ans += n - j;\n break;\n }\n }\n }\n \n cout << ans*ans << endl;\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 257635708,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\n#include <iostream>\n#include <cstring>\n \nusing namespace std;\n \ntypedef long long LL;\n \nconst int N=1e5+10;\nint dp1[N][32][2],dp2[N][32][2];\nint T,n;\nint a[N];\n \ninline void ini()\n{\n for(int i=1;i<=n;++i)\n for(int j=0;j<32;++j)\n if(a[i]>>j&1)\n {\n dp1[i][j][0]=dp1[i-1][j][1];\n dp1[i][j][1]=dp1[i-1][j][0]+1;\n }\n else\n {\n dp1[i][j][0]=dp1[i-1][j][0]+1;\n dp1[i][j][1]=dp1[i-1][j][1];\n }\n \n for(int i=n;i>=1;--i)\n for(int j=0;j<32;++j)\n if(a[i]>>j&1)\n {\n dp2[i][j][0]=dp2[i+1][j][1];\n dp2[i][j][1]=dp2[i+1][j][0]+1;\n }\n else\n {\n dp2[i][j][0]=dp2[i+1][j][0]+1;\n dp2[i][j][1]=dp2[i+1][j][1];\n }\n}\n \nint main()\n{\n scanf(\"%d\",&T);\n while(T--)\n {\n memset(dp1,0,sizeof dp1);\n memset(dp2,0,sizeof dp2);\n scanf(\"%d\",&n);\n for(int i=1;i<=n;++i) scanf(\"%d\",&a[i]);\n ini();\n LL ans=0;\n for(int i=1;i<=n;++i)\n {\n for(int j=31;j>=0;--j)\n {\n if(a[i]>>j&1)\n {\n ans+=dp1[i-1][j][0]*dp2[i+1][j][1];\n ans+=dp1[i-1][j][1]*dp2[i+1][j][0];\n ans+=dp1[i-1][j][1];\n ans+=dp2[i+1][j][1];\n break;\n }\n }\n }\n printf(\"%lld\\n\",ans);\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 258365596,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <unordered_map>\nusing namespace std;\n \n// Function to count the number of tuples satisfying the condition\nlong long countTuples(vector<int>& arr) {\n int n = arr.size();\n long long tuples = 0;\n unordered_map<int, int> freq;\n \n // Calculate prefix XORs\n vector<int> prefixXOR(n + 1);\n prefixXOR[0] = 0;\n for (int i = 1; i <= n; i++) {\n prefixXOR[i] = prefixXOR[i - 1] ^ arr[i - 1];\n }\n \n // Iterate through each pair of indices (x, y)\n for (int x = 0; x < n; x++) {\n for (int y = x + 1; y < n; y++) {\n int fx_y = prefixXOR[y + 1] ^ prefixXOR[x]; // f(x, y)\n for (int z = y; z < n; z++) {\n int fx_z = prefixXOR[z + 1] ^ prefixXOR[x]; // f(x, z)\n int fy_z = prefixXOR[z + 1] ^ prefixXOR[y]; // f(y, z)\n if (fx_y ^ fy_z > fx_z) {\n tuples++;\n }\n }\n }\n }\n \n return tuples;\n}\n \nint main() {\n int t;\n cin >> t;\n \n while (t--) {\n int n;\n cin >> n;\n vector<int> arr(n);\n for (int i = 0; i < n; i++) {\n cin >> arr[i];\n }\n if(t==0)\n cout << countTuples(arr)-2<<endl;\n else\n cout << countTuples(arr) << endl;\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 257633309,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\nt = int(input())\nfor _ in range(t):\n n = int(input())\n arr = list(map(int, input().split()))\n if n == 1:\n print(0)\n else:\n print(2 ** (n - 1))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 257630571,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\nfrom math import ceil\n \nT = int(input())\nfor _ in range(0, T):\n n = int(input())\n a = list(map(int, input().split()))\n arr = [[0 for i in range(0, 30)] for j in range(0, n + 1)]\n pref = [[[0, 0] for j in range(0, 30)] for i in range(0, n)]\n suff = [[[0, 0] for j in range(0, 30)] for i in range(0, n)]\n b = bin(a[0])[2::][::-1]\n for j in range(0, len(b)):\n pref[0][j][int(b[j])] = 1\n for i in range(1, n):\n b = bin(a[i])[2::][::-1]\n for j in range(0, len(b)):\n pref[i][j][0] = pref[i - 1][j][0]\n pref[i][j][1] = pref[i - 1][j][1]\n if b[j] == '1':\n pref[i][j][1 - (pref[i - 1][j][0] + pref[i - 1][j][1])%2] += 1\n for j in range(len(b), 30):\n pref[i][j][0] = pref[i - 1][j][0]\n pref[i][j][1] = pref[i - 1][j][1]\n b = bin(a[-1])[2::][::-1]\n for j in range(0, len(b)):\n suff[-1][j][int(b[j])] = 1\n for j in range(len(b), 30):\n suff[-1][j][0] = 1\n for i in range(n - 2, -1, -1):\n b = bin(a[i])[2::][::-1]\n for j in range(0, len(b)):\n suff[i][j][0] = suff[i + 1][j][0]\n suff[i][j][1] = suff[i + 1][j][1]\n suff[i][j][int(b[j])] += 1\n for j in range(len(b), 30):\n suff[i][j][0] = suff[i + 1][j][0] + 1\n suff[i][j][1] = suff[i + 1][j][1]\n ans = 0\n for i in range(0, n):\n b = len(bin(a[i])) - 3\n ans += pref[i][b][0] * suff[i][b][1]\n ans += pref[i][b][1] * suff[i][b][0]\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 257820053,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\nimport sys,math,bisect,heapq\nfrom collections import deque\ninput=sys.stdin.readline\nflush=sys.stdout.flush\nmod=(10**9)+7 \n \nfor _ in range(int(input())):\n n=int(input())\n #n,k=map(int,input().split())\n a=list(map(int,input().split()))\n pre=[]\n pre.append([0]*30)\n b=[0]*n\n c=0\n for i in range(n):\n t=[0]*30\n c^=a[i]\n for j in range(30):\n if (a[i]>>j)&1:\n b[i]=j\n if (c>>j)&1:\n t[j]+=pre[-1][j]+1\n else: \n t[j]+=pre[-1][j]\n pre.append(t)\n ans=0\n \n for i in range(n):\n l1=pre[i+1][b[i]]\n l0=i+2-l1 \n r1=pre[-1][b[i]]- pre[i][b[i]]\n r0=n-i-1-r1 \n ans+=r0*l0\n ans+=l1*r1\n print(ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 257728854,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n static String[] temp;\n public static void main(String[] args) throws Exception {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n PrintWriter pw = new PrintWriter(System.out);\n int T= Integer.parseInt(br.readLine());\n while (T>0){\n int n= Integer.parseInt(br.readLine());\n int[]sum=new int[n+1];\n temp=br.readLine().split(\" \");\n for (int i = 0; i <n ; i++) sum[i+1]=sum[i]^Integer.parseInt(temp[i]);\n int[][]cnt=new int[32][2];\n for (int i = n-1; i >=0; i--) {\n int num=sum[i+1];\n for (int j = 0; j<32; j++) {\n cnt[j][num>>j&1]++;\n }\n }\n long ans=0;\n int[][]pre=new int[32][2];\n for (int i = 0; i <32; i++) {\n pre[i][0]++;\n }\n for (int i = 0; i <n; i++) {\n int num=sum[i+1];\n for (int j = 0; j<32; j++) {\n pre[j][num>>j&1]++;\n }\n for (int j = 0; j<32; j++) {\n cnt[j][num>>j&1]--;\n }\n int l=31-Integer.numberOfLeadingZeros(sum[i+1]^sum[i]);\n ans+=(long)pre[l][0]*cnt[l][0]+ (long) pre[l][1] *cnt[l][1];\n }\n pw.println(ans);\n T--;\n }\n pw.close();\n br.close();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 292563552,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\n// package BitManipulation;\n \nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.util.StringTokenizer;\n \npublic class ABitOfAnEquality {\n public static void main(String[] args)throws IOException {\n BufferedReader br=new BufferedReader(new InputStreamReader(System.in));\n int test=Integer.parseInt(br.readLine());\n StringBuilder print=new StringBuilder();\n while(test--!=0){\n int n=Integer.parseInt(br.readLine());\n StringTokenizer st=new StringTokenizer(br.readLine());\n int a[]=new int[n+1];\n int p[]=new int[n+1];\n int count[][]=new int[n+1][32];\n for(int i=1;i<=n;i++){\n a[i]=Integer.parseInt(st.nextToken());\n p[i]=p[i-1]^a[i];\n for(int j=0;j<32;j++){\n count[i][j]=count[i-1][j];\n if(((1<<j)&p[i])!=0){\n count[i][j]++;\n }\n }\n }\n if(n==1){\n print.append(\"0\\n\");\n continue;\n }\n long ans=0;\n for(int i=1;i<=n;i++){\n if(a[i]==0){\n continue;\n }\n int msb=-1;\n for(int j=31;j>=0;j--){\n if(((1<<j)&a[i])!=0){\n msb=j;\n break;\n }\n }\n// if(i==1){\n// if(((1<<msb)&p[1])==0){\n// ans+=count[n][msb]-1;\n// }\n// else{\n// ans+=n-1-(count[n][msb]-1);\n// }\n// }\n// else if(i==n){\n// if(((1<<msb)&p[n-1])==0){\n// ans+=count[n-2][msb];\n// }\n// else{\n// ans+=n-1-count[n-2][msb];\n// }\n// }\n if(i>1&&i<n){\n int zeroLeft=0,oneLeft=0,zeroRight=0,oneRight=0;\n if(((1<<msb)&p[i-1])==0){\n oneLeft= count[i-2][msb];\n zeroLeft=i-1-oneLeft;\n }\n else{\n zeroLeft=count[i-2][msb];\n oneLeft=i-1-zeroLeft;\n }\n if(((1<<msb)&p[i])==0){\n oneRight=count[n][msb]-count[i][msb];\n zeroRight=(n-i)-oneRight;\n }\n else{\n zeroRight=count[n][msb]-count[i][msb];\n oneRight=(n-i)-oneRight;\n }\n ans+=(1l*zeroLeft*oneRight)+(1l*oneLeft*zeroRight);\n }\n if(((1<<msb)&p[i])==0){\n ans+=count[n][msb]-count[i][msb];\n }\n else{\n ans+=(n-i)-(count[n][msb]-count[i][msb]);\n }\n if(i>1){\n if(((1<<msb)&p[i-1])==0){\n ans+=count[i-2][msb];\n }\n else{\n ans+=i-1-count[i-2][msb];\n }\n }\n }\n print.append(ans+\"\\n\");\n }\n System.out.println(print);\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 271051911,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\n#ifdef ONLINE_JUDGE\n#include <bits/stdc++.h>\n#pragma GCC optimize(\"O1\")\n#pragma GCC optimize(\"O2\")\n#pragma GCC optimize(\"Ofast\")\n#pragma GCC target(\"avx,avx2,fma\")\n#else\n#include \"myheader.h\"\n#endif\n// #include <ext/pb_ds/assoc_container.hpp>\n// using namespace __gnu_pbds;\nusing namespace std;\nvoid input() {return;}\ntemplate<typename T1, typename... T2>\n void input(T1 & x, T2&... args){((cin >> x), input(args...));}\nvoid wrt() { cout << \"\\n\"; return;}\ntemplate<typename T1, typename... T2>\n void wrt(T1 x, T2... args){((cout << x << ' '), wrt(args...));}\ntemplate<typename T> void inlst(T & lst, int x, int y)\n { for(int i = x ; i < y; i++ ) cin >> lst[i]; }\ntemplate<typename T1, typename T2> istream & operator>>(istream & in, pair<T1, T2> & val)\n { in >> val.first >> val.second; return in; }\ntemplate<typename T> istream & operator>>(istream & in, vector<T> & lst)\n { for (auto & e : lst) in >> e; return in; }\ntemplate<typename T> void prlst(T & lst, int x, int y)\n { for(int i = x ; i < y; i++ ) cout << lst[i] << \" \"[i < (y - 1)]; wrt(); }\n// template<typename T> using Indexed_set = \n// tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\n// template<typename T> using Indexed_multiset = \n// tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;\n// template<typename Key, typename T> using Indexed_map =\n// tree<Key, T, less<Key>, rb_tree_tag, tree_order_statistics_node_update>;\n// template<typename T> using pques = priority_queue<T, vector<T>, greater<T>>;\n// template<typename T> using pqueg = priority_queue<T>;\n#define boost ios_base::sync_with_stdio(false); cin.tie(nullptr);\n#define precision(n) cout.precision(n);\n#define ll long long\n#define tests int test; cin >> test; while(test--)\n#define ub(lst, val) (upper_bound(all(lst), val) - lst.begin())\n#define lb(lst, val) (lower_bound(all(lst), val) - lst.begin())\n#define fora(i, x, y) for (ll i = x; i < y; ++i)\n#define ford(i, x, y) for (ll i = x; i >= y; --i)\n#define all(lst) lst.begin(), lst.end()\n#define rall(lst) lst.rbegin(), lst.rend()\n#define _ceil(a, b) ((a + b - 1) / (b))\n#define _sum(lst) accumulate(all(lst), 0ll)\n#define cntval(lst, val) count(all(lst), val)\n#define lcn(lst, val) (find(all(lst), val) - lst.begin())\n#define sortlst(lst) sort(lst.begin(), lst.end())\n#define sorted(lst) is_sorted(lst.begin(), lst.end())\n#define rsortlst(lst) sort(lst.rbegin(), lst.rend())\n#define sortarr(x, n) sort(x, x + n)\n#define sz(lst) (int)lst.size()\ntypedef pair<long long, long long> pl;\ntypedef pair<int, int> pi;\ntypedef vector<long long> vl;\ntypedef vector<int> vi;\ntypedef vector<vector<long long>> vovl;\ntypedef vector<vector<int>> vovi;\ntypedef vector<string> vs;\ntypedef map<int, int> mi;\ntypedef map<long long, long long> ml;\ntypedef set<long long> sl;\ntypedef set<int> si;\nconst ll inf = INT32_MAX, MOD = 1e9 + 7, N = 3e5 + 10;\n/*---------------------------------------FUNCT---------------------------------------- */\nll _lcm(ll a, ll b) { return (a * b) / __gcd(a, b); }\nll min(ll a, ll b) { return std :: min(a, b); }\nll max(ll a, ll b) { return std :: max(a, b); }\nll _nurm(ll & x) { while (x < 0) x += MOD; return x = (x + MOD) % MOD; }\nll _bits(ll x) { ll cnt = 0; while(x > 0) { cnt++; x >>= 1; } return cnt; }\nll _setbits(ll x) { ll cnt = 0; while(x > 0) { cnt += (x & 1); x >>= 1; } return cnt; }\nll _add(ll x, ll y) { x %= MOD, y %= MOD; return (x + y) % MOD; }\nll _sub(ll x, ll y) { x %= MOD, y %= MOD; return (x - y + MOD) % MOD; }\nll _mul(ll x, ll y) { x %= MOD, y %= MOD; return (x * 1ll * y) % MOD; }\nll _pow(ll x, ll y) { if (y == 0) return 1; else if (y % 2 == 0){\nll _tmp =_pow(x, y / 2); return _mul(_tmp, _tmp); } else return _mul(x, _pow(x, y - 1)); }\nll _inv(ll p) { return _pow(p, MOD - 2); }\nll _div(ll x, ll y) { x %= MOD, y %= MOD; return _mul(x, _inv(y)); }\n/*---------------------------------------Code---------------------------------------- */\nvoid SolveTestCases(int testCase) {\n int n; cin >> n;\n vi a(n); cin >> a;\n vector dp_right(5, vector (2, 0)), dp_left(5, vector (2, 0));\n for (int i = n - 1; i >= 0; i -= 1) {\n for (int bit = 0; bit < 5; bit += 1) {\n if (a[i] & (1 << bit)) {\n swap(dp_right[bit][0], dp_right[bit][1]);\n dp_right[bit][1] += 1;\n } else dp_right[bit][0] += 1;\n }\n }\n ll result = 0;\n for (int i = 0; i < n; i += 1) {\n for (int bit = 0; bit < 5; bit += 1) {\n if (a[i] & (1 << bit)) {\n swap(dp_left[bit][0], dp_left[bit][1]);\n dp_left[bit][1] += 1;\n } else dp_left[bit][0] += 1;\n }\n \n int msb = __lg(a[i]);\n result += (dp_left[msb][0] * dp_right[msb][1]);\n result += (dp_left[msb][1] * dp_right[msb][0]);\n \n for (int bit = 0; bit < 5; bit += 1) {\n if (a[i] & (1 << bit)) {\n dp_right[bit][1] -= 1;\n swap(dp_right[bit][0], dp_right[bit][1]);\n } else dp_right[bit][0] -= 1;\n }\n }\n return wrt(result);\n}\nint main(int argc, char const *argv[]) {\n boost;\n int testCase = 1;\n cin >> testCase;\n for (int test = 0; test < testCase; test++)\n SolveTestCases(test);\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 258520599,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nint main() {\n\tint t;\n\tcin >> t;\n\twhile (t--) {\n\t\tint n; cin >> n;\n\t\tvector<int> arr(n, 0);\n\t\tfor (auto& e : arr) cin >> e;\n\t\tvector<vector<int>> evens(n, vector<int>(32, 0));\n\t\tvector<int> xors(arr);\n\t\tfor (int i = 1; i < n; i++) xors[i] ^= xors[i-1];\n\t\tfor (int l = 0; l < 32; l++) {\n\t\t\tint mask = 1<<l;\n\t\t\tint sumuptilnow = (arr[0] >> l) & 1;\n\t\t\tevens[0][l] = (sumuptilnow % 2 == 0);\n\t\t\tfor (int i = 1; i < n; i++) {\n\t\t\t\tsumuptilnow += (arr[i] >> l) & 1;\n\t\t\t\tevens[i][l] = evens[i-1][l] + (sumuptilnow % 2 == 0);\n\t\t\t}\n\t\t}\n\t\tlong long answer = 0;\n\t\tfor (int i = 0; i < n; i++) {\n\t\t\tif (arr[i] == 0) continue;\n\t\t\tint ss = 31-(__builtin_clz(arr[i]));\n\t\t\t//if (xors[i] & (1 << ss)) {\n\t\t\t\tanswer += ((i ? evens[i-1][ss] : 0) + 1) * (evens[n-1][ss] - (i ? evens[i-1][ss] : 0));\n\t\t\t\tanswer += (1+i - ((i ? evens[i-1][ss] : 0) + 1)) * (n-i - (evens[n-1][ss] - (i ? evens[i-1][ss] : 0)));\n\t\t\t//} else {\n \n\t\t\t//}\n\t\t}\n\t\tprintf(\"%lld\\n\", answer);\n \n \n \n\t}\n \n \n \n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 257646275,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\ntimes = int(input())\nfor _ in range(times):\n cnt = 0\n n = int(input())\n nums = list(map(int,input().split()))\n pre = [0]\n for x in nums:\n pre.append(pre[-1]^x)\n for x in range(n):\n for y in range(x,n):\n for z in range(y,n):\n if pre[y+1]^pre[x]^pre[z+1]^pre[y]>pre[z+1]^pre[x]:\n cnt += 1\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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 292598650,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\ndef count_tuples(a):\n n = len(a)\n prefix_xor = [0] * (n + 1)\n for i in range(n):\n prefix_xor[i + 1] = prefix_xor[i] ^ a[i]\n \n count = 0\n for x in range(1, n + 1):\n for y in range(x, n + 1):\n xor_xy = prefix_xor[y] ^ prefix_xor[x - 1]\n for z in range(y, n + 1):\n xor_yz = prefix_xor[z] ^ prefix_xor[y - 1]\n xor_xz = prefix_xor[z] ^ prefix_xor[x - 1]\n if xor_xy ^ xor_yz > xor_xz:\n count += 1\n return count\n \nt = int(input())\nfor _ in range(t):\n n = int(input())\n a = list(map(int, input().split()))\n result = count_tuples(a)\n print(result)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 257625772,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n \n public static void cf1957D(BufferedReader in, PrintWriter out) throws IOException {\n int T = Integer.parseInt(in.readLine());\n while (T-- > 0) {\n int n = Integer.parseInt(in.readLine());\n int[] a = new int[n];\n int[][] suf = new int[30][2];\n int s = 0;\n \n StringTokenizer st = new StringTokenizer(in.readLine());\n for (int i = 0; i < n; i++) {\n a[i] = Integer.parseInt(st.nextToken());\n s ^= a[i];\n for (int j = 0; j < 30; j++) {\n suf[j][(s >> j) & 1]++;\n }\n }\n \n int ans = 0;\n int[][] pre = new int[30][2];\n s = 0;\n \n for (int v : a) {\n for (int j = 0; j < 30; j++) {\n pre[j][(s >> j) & 1]++;\n }\n int hb = Integer.toBinaryString(v).length() - 1;\n ans += pre[hb][0] * suf[hb][0] + pre[hb][1] * suf[hb][1];\n s ^= v;\n for (int j = 0; j < 30; j++) {\n suf[j][(s >> j) & 1]--;\n }\n }\n \n out.println(ans);\n }\n }\n \n public static void main(String[] args) throws IOException {\n BufferedReader in = new BufferedReader(new InputStreamReader(System.in));\n PrintWriter out = new PrintWriter(System.out);\n cf1957D(in, out);\n out.flush();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 292847296,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class CF1957D {\n static int n;\n static int[] a;\n \n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int t = scanner.nextInt();\n while (t-- > 0) {\n n = scanner.nextInt();\n a = new int[n];\n for (int i = 0; i < n; i++) {\n a[i] = scanner.nextInt();\n }\n System.out.println(solve());\n }\n }\n \n private static String solve() {\n int[][] suf = new int[30][2];\n int s = 0;\n for (int i = 0; i < n; i++) {\n s ^= a[i];\n for (int j = 0; j < 30; j++) {\n suf[j][s >> j & 1]++;\n }\n }\n \n int ans = 0;\n int[][] pre = new int[30][2];\n s = 0;\n for (int i = 0; i < n; i++) {\n int v = a[i];\n \n for (int j = 0; j < 30; j++) {\n pre[j][s >> j & 1]++;\n }\n int hb = bitsLen(v) - 1;\n ans += pre[hb][0] * suf[hb][0] + pre[hb][1] * suf[hb][1];\n s ^= v;\n for (int j = 0; j < 30; j++) {\n suf[j][s >> j & 1]--;\n }\n }\n return String.valueOf(ans);\n }\n \n // bits.Len:\n // Len returns the minimum number of bits required to represent x; the result is 0 for x == 0.\n static int bitsLen(long x) {\n if (x == 0) return 0;\n return Long.toBinaryString(x).length();\n }\n}\n/*\nD. A BIT of an Inequality\nhttps://codeforces.com/contest/1957/problem/D\n \n灵茶の试炼 2024-11-21\n题目大意：\n输入 T(≤1e4) 表示 T 组数据。所有数据的 n 之和 ≤1e5。\n每组数据输入 n(1≤n≤1e5) 和长为 n 的数组 a(1≤a[i]≤1e9)。\n定义 S(i,j) 为 a[i]^a[i+1]^...^a[j]，即子数组的异或和。\n输出有多少个 i,j,k 满足 i≤j≤k 且 S(i,j)^S(j,k) > S(i,k)。\n \nrating 1900\n用前缀和，转换成 s[k+1] ^ s[i] ^ a[j] > s[k+1] ^ s[i]。\n考虑枚举中间的 a[j]。\n既然要比大小，从高位思考。\n0 不影响不等式，所以不考虑。\n1 一定会影响不等式，所以只需考虑 a[j] 的最高位。\n左边的 s[i] 和右边的 s[k+1]，在 a[j] 的最高位处，要么都是 0，要么都是 1，这样异或 1 之后才能变大。\n所以用前后缀分解，统计左右两边每个比特位的 0 和 1 的个数。\n左边指 s 中的下标 <= i 的数，右边指 s 中的下标 > i 的数。\n根据乘法原理，把左边 0 的个数 * 右边 0 的个数 + 左边 1 的个数 * 右边 1 的个数加入答案。\n前缀和数组可以省略。\n代码 https://codeforces.com/problemset/submission/1957/292402150\n代码备份（洛谷）\n======\n \nInput\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\nOutput\n4\n0\n16\n */\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 298179240,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define ll long long\n \nint pref[30][100005][2];\nint suff[30][100005][2];\n \nvoid solve()\n{\n int n;\n cin >> n;\n vector<int> a(n + 1);\n for (int i = 1; i <= n; i++)\n {\n cin >> a[i];\n }\n for (int i = 0; i < 30; i++)\n {\n suff[i][n + 1][0] = suff[i][n + 1][1] = 0;\n }\n for (int i = 0; i < 30; i++)\n {\n for (int j = 1; j <= n; j++)\n {\n int v = !!(a[j] & (1 << i));\n for (int c = 0; c < 2; c++)\n {\n pref[i][j][c] = (v == c) + pref[i][j - 1][c ^ v];\n }\n }\n for (int j = n; j >= 1; j--)\n {\n int v = !!(a[j] & (1 << i));\n for (int c = 0; c < 2; c++)\n {\n suff[i][j][c] = (v == c) + suff[i][j + 1][c ^ v];\n }\n }\n }\n ll ans = 0;\n for (int i = 1; i <= n; i++)\n {\n int z = 31 - __builtin_clz(a[i]);\n ans += (ll)(pref[z][i - 1][1] * (1 + suff[z][i + 1][0]));\n ans += (ll)((1 + pref[z][i - 1][0]) * suff[z][i + 1][1]);\n }\n cout << ans << '\\n';\n}\n \nint main()\n{\n ios::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n ll t;\n cin >> t;\n while (t--)\n {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 270138089,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\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 \nusing namespace std;\nusing namespace __gnu_pbds;\n \ntypedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> index_set; // find_by_order, order_of_key\n //data type //greater<int> for desc \n \ntypedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> multi_ind_set; \n \n \n#define ll long long int\n#define f first\n#define sec second\n#define bg begin()\n#define ed end()\n#define pb push_back\n#define po pop_back()\n#define eb emplace_back\n#define emp emplace\n#define ins insert\n#define umap unordered_map\n#define uset unordered_set\n#define bpc(x) __builtin_popcountll(x)\n#define prec(x) fixed << setprecision(x)\n#define rng(i,l,r,d) for (auto i=(l); (d)<0?i>(r):((d)>0?i<(r):0); i+=(d))\n#define mem(a,b) memset(a,(int)b,sizeof(a))\n#define all(x) (x).begin(), (x).end()\n#define lb lower_bound\n#define ub upper_bound\n#define bs binary_search\n#define vi vector<int>\n#define vii vector<pair<int,int>>\n#define vl vector<ll>\n#define vll vector<pair<ll,ll>>\n#define vvi vector<vector<int>>\n#define vvl vector<vector<ll>> \n#define pii pair<int,int>\n#define pli pair<ll,int>\n#define pll pair<ll,ll>\n#define ldbl long double\n#define pq priority_queue\n#define ctnu continue\n#define read_vec(a, n) \\\n for (int i = 0; i < n; ++i) \\\n cin >> a[i];\n#define print_vec(a, n) \\\n for (int i = 0; i < n; ++i) \\\n if (i == n - 1) \\\n { \\\n cout << a[i] << \"\\n\"; \\\n } \\\n else \\\n { \\\n cout << a[i] << ' '; \\\n }\n \n#ifndef ONLINE_JUDGE\n#include \"C:\\Users\\Parth\\OneDrive\\Desktop\\CP\\deb_temp.cpp\"\n#else\n#define debug(...)\n#define debugArr(...)\n#endif\n \nll N = 1e9 + 7;\nll N1 =998244353;\n \nll mod_inv(ll a,ll m);\nll mod_pow(ll x , ll n ,ll m);\nvoid SieveOfEratosthenes(vector<bool> &val, int n);\nll bex(ll x, ll y);\nint msb(ll n);// returns 1 indexed most significant bit\n \n// returns -1 if s1 is not a substring of s2\nint isSubstring(string &s1, string &s2);\n \nlong long mul(long long x,long long y, int p);\nlong long divide(long long x,long long y, int p);\nlong long nCrModPFermat(ll n,ll r, int p);\nvoid multiply(vector<vector<long long> >& A,\n vector<vector<long long> >& B,int n);\nvector<vector<long long> >\nmat_expo(vector<vector<long long> > M, int expo,int n);\n \n//void specsieve();\n//ll getFactorization(ll x);\n \nvoid solve(){\n ll n; cin >> n; \n vector<ll> a(n); read_vec(a,n);\n ll pw = 31;\n debug(pw);\n ll cnt = 0 ;\n vector<vector<vl>> drp(n+1, vector<vl> (pw,vl(2,0))) , drpc(n+1, vector<vl> (pw,vl(2,0)));\n rng(i,0,n,1){\n rng(k,0,pw,1){\n ll tp = (a[i]>>k & 1),tpp = 1-tp;\n if(tp==1){\n drpc[i+1][k][1] = drpc[i][k][0]+1;\n drpc[i+1][k][0]=drpc[i][k][1];\n }\n else{\n drpc[i+1][k][0] = drpc[i][k][0]+1;\n drpc[i+1][k][1] = drpc[i][k][1];\n }\n }\n }\n \n rng(i,n-1,-1,-1){\n rng(k,0,pw,1){\n ll tp = (a[i]>>k & 1),tpp = 1-tp;\n if(tp==1){\n drp[i][k][1] = drp[i+1][k][0]+1;\n drp[i][k][0]=drp[i+1][k][1];\n }\n else{\n drp[i][k][0] = drp[i+1][k][0]+1;\n drp[i][k][1] = drp[i+1][k][1];\n }\n \n }\n }\n \n debug(drp);\n debug(drpc);\n \n ll cunt = 0;\n \n rng(i,0,n,1){\n ll hghst = -1;\n rng(wpe,pw,-1,-1) if(a[i]>>wpe & 1){hghst = wpe ; break;}\n cunt += (drp[i][hghst][0])*(drpc[i+1][hghst][1]);\n cunt+= (drp[i][hghst][1])*(drpc[i+1][hghst][0]);\n debug(cunt);\n }\n cout << cunt <<\"\\n\";\n}\n \n \nint main()\n{ \n ios_base::sync_with_stdio(0);cin.tie(0);\n ll tes = 1;\n cin>>tes;\n while(tes--)\n solve();\n cerr << \"\\nWorked in \" << 1000*((double)clock())/(double)CLOCKS_PER_SEC<< \"ms \\n\";\n return 0;\n}\n \n/*\nThinking Space\n \n \n \n*/\n \n \n \n \n \n \n \n \n \n \n \n \n \n// C:\\Users\\Parth\\AppData\\Roaming\\Sublime Text\\Packages\\User\n \n \n// NT fxns\n// a^-1 mod m m is prime\nll mod_inv(ll a,ll m) {\n a%=m;\n return a <= 1 ? a : m - (long long)(m/a) * mod_inv(m % a,m) % m;\n}\n// to calculate x^n mod m in binary exponentiation\nll mod_pow(ll x , ll n ,ll m){\n if(n==0) return 1%m;\n ll u = mod_pow(x,n/2,m);\n u = (u*u)%m;\n if(n%2) u = (u*x)%m;\n return u;\n}\n/*\nfor above algo \nx^ n = 1\n = x^(n/2)*(x^(n/2)) if(n is even)\n = x^(n-1)*(x) if(n is odd)\n*/\n// proper binary exponentiation to find x^y\nll bex(ll x, ll y){\n ll res = 1;\n while(y>0){\n if(y%2) res = res*x;\n y = y>>1;\n x = (x*x);\n }\n return res;\n}\n \n// Seive of Eratosthenes\nvoid SieveOfEratosthenes(vector<bool> &val, int n)\n{ \n vector<bool> PRIME(n+1 , true);\n for (int p = 2; p * p <= n; p++) {\n if (PRIME[p] == true) {\n for (int i = p * p; i <= n; i += p)\n PRIME[i] = false;\n }\n }\n PRIME[0]=false;PRIME[1]=false;\n val = PRIME;\n}\n \nint msb(ll n){// returns 1 indexed most significant bit\n if(n==0)\n return 0;\n \n return 64 - __builtin_clzll(n);\n}\n \nint isSubstring(string &s1, string &s2)\n{\n // using find method to check if s1 is\n // a substring of s2\n if (s2.find(s1) != string::npos)\n return s2.find(s1);\n return -1;\n}\n \nlong long mul(long long x,long long y, int p)\n{\n return x * 1ull * y % p;\n}\nlong long divide(long long x,long long y, int p)\n{\n return mul(x, mod_inv(y, p), p);\n}\nlong long nCrModPFermat(ll n,ll r, int p)\n{\n if (n < r)\n return 0;\n if (r == 0)\n return 1;\n if (n - r < r)\n return nCrModPFermat(n, n - r, p);\n long long res = 1;\n for (int i = r; i >= 1; i--)\n res = divide(mul(res, n - i + 1, p), i, p);\n return res;\n}\n \nvoid multiply(vector<vector<long long> >& A,\n vector<vector<long long> >& B,int n)\n{\n // Matrix to store the result\n vector<vector<long long> > C(n, vector<long long>(n,0));\n for (int i = 0; i < n; i++) {\n for (int j = 0; j < n; j++) {\n for (int k = 0; k < n; k++) {\n C[i][j]\n = (C[i][j] + ((A[i][k]) * (B[k][j])));\n }\n }\n }\n \n // Copy the result back to the first matrix\n for (int i = 0; i < n; i++) {\n for (int j = 0; j < n; j++) {\n A[i][j] = C[i][j];\n }\n }\n}\n \n// Function to calculate (Matrix M) ^ expo\nvector<vector<long long> >\nmat_expo(vector<vector<long long> > M, int expo,int n)\n{\n // Initialize result with identity matrix\n vector<vector<long long> > ans(n,vector<ll> (n,0));\n rng(i,0,n,1) ans[i][i]=1;\n \n // Fast Exponentiation\n while (expo) {\n if (expo & 1)\n multiply(ans, M,n);\n multiply(M, M,n);\n expo >>= 1;\n }\n \n return ans;\n}\n \n// //special seive\n// // define the following two before implementation\n// // and check how you want getfactor as map or as sum of power or what\n \n// // #define MAXN 5000001\n// // ll spf[MAXN];\n \n// void specsieve()\n// {\n// spf[1] = 1;\n// for (ll i = 2; i < MAXN; i++)\n// spf[i] = i;\n// for (ll i = 4; i < MAXN; i += 2)\n// spf[i] = 2;\n \n// for (ll i = 3; i * i < MAXN; i++) {\n// if (spf[i] == i) {\n// for (ll j = i * i; j < MAXN; j += i)\n// if (spf[j] == j)\n// spf[j] = i;\n// }\n// }\n// }\n// //getfactorization\n// ll getFactorization(ll x)\n// {\n// ll ans = 0;\n// while (x != 1) {\n// // ret.push_back(spf[x]);\n// ans++;\n// x = x / spf[x];\n// }\n// return ans;\n// }\n//\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 299907657,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\ndef find_xor(arr):\n result = 0\n for num in arr:\n result ^= num\n return result\n \nt = int(input())\nfor _ in range(t):\n n = int(input())\n a = list(map(int, input().split()))\n res = 0\n for i in range(n):\n for j in range(i,n):\n for k in range(j,n):\n x = find_xor(a[i:j+1])\n y = find_xor(a[j:k+1])\n z = find_xor(a[i:k+1])\n if x^y > z:\n res += 1\n print(res)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 257632955,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\nfrom math import ceil, gcd\n \nfrom collections import Counter, defaultdict\nimport heapq\n \n \ndef modM(a, b, M):\n return (a%M * b%M)%M\n \ndef getL():\n return list(map(int, input().split()))\n \ndef getT():\n return tuple(map(int, input().split()))\n \ndef getI():\n return int(input())\n \ntt = 1\ntt = int(input())\nmaxN = 100001\nB = 32\n \n \n \n \ndef solve():\n n = getI()\n a = getL()\n res = 0\n pref = [[[0, 0] for j in range(n)] for i in range(B)]\n \n for i in range(B):\n for j in range(n):\n ai = a[j]\n curr_bit = int(bool(ai & (1 << i)))\n prev = pref[i][j - 1] if j else [0, 0]\n if curr_bit:\n pref[i][j] = [prev[1], 1 + prev[0]]\n else:\n pref[i][j] = [1 + prev[0], prev[1]]\n \n \n vis = {}\n for i in range(B - 1, -1, -1):\n prev = [0, 0]\n for j in range(n - 1, -1, -1):\n ai = a[j]\n curr_bit = int(bool(ai & (1 << i)))\n nxt = prev\n if curr_bit:\n prev = [prev[1], 1 + prev[0]]\n pp = pref[i][j - 1] if j else [0, 0]\n if j not in vis:\n res += (pp[0] + 1) * nxt[1]\n res += (nxt[0] + 1) * pp[1]\n vis[j] = 1\n else:\n prev = [1 + prev[0], prev[1]]\n \n print(res)\n return\n \nfor i in range(tt):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 259103110,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\nimport java.util.Scanner;\nimport java.util.Vector;\n \npublic class Main {\n public static int countTuples(Vector<Integer> arr) {\n int n = arr.size();\n Vector<Integer> prefix = new Vector<>(n);\n prefix.add(arr.get(0));\n \n // Compute prefix XOR array\n for (int i = 1; i < n; ++i) {\n prefix.add(prefix.get(i - 1) ^ arr.get(i));\n }\n \n int count = 0;\n \n // Iterate through all possible tuples (x, y, z)\n for (int x = 0; x < n; ++x) {\n for (int y = x; y < n; ++y) {\n for (int z = y; z < n; ++z) {\n int fxy = (x == 0) ? prefix.get(y) : prefix.get(y) ^ prefix.get(x - 1);\n int fyz = (y == 0) ? prefix.get(z) : prefix.get(z) ^ prefix.get(y - 1);\n int fxz = (x == 0) ? prefix.get(z) : prefix.get(z) ^ prefix.get(x - 1);\n \n if ((fxy ^ fyz) > fxz) {\n count++;\n }\n }\n }\n }\n \n return count;\n }\n \n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int t = scanner.nextInt();\n \n while (t-- > 0) {\n int n = scanner.nextInt();\n Vector<Integer> arr = new Vector<>(n);\n for (int i = 0; i < n; ++i) {\n arr.add(scanner.nextInt());\n }\n System.out.println(countTuples(arr));\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 257628103,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: A BIT of an Inequality\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: You are given an array $$$a_1, a_2, \\ldots, a_n$$$. Find the number of tuples ($$$x, y, z$$$) such that:\n\n- $$$1 \\leq x \\leq y \\leq z \\leq n$$$, and\n- $$$f(x, y) \\oplus f(y, z) > f(x, z)$$$.\n\nWe define $$$f(l, r) = a_l \\oplus a_{l + 1} \\oplus \\ldots \\oplus a_{r}$$$, where $$$\\oplus$$$ denotes the bitwise XOR operation.\n\nInput: The first line contains a single integer $$$t$$$ ($$$1 \\leq t \\leq 10^4$$$) — the number of test cases.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\leq n \\leq 10^5$$$).\n\nThe second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$1 \\leq a_i \\leq 10^9$$$).\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 on a new line — the number of described tuples.\n\nExamples:\ninput:\n3\n3\n6 2 4\n1\n3\n5\n7 3 7 2 1\noutput:\n4\n0\n16\n\nNote: In the first case, there are 4 such tuples in the array $$$[6, 2, 4]$$$:\n\n- ($$$1$$$, $$$2$$$, $$$2$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2) = 4 \\oplus 2 > (a_1 \\oplus a_2) = 4$$$\n- ($$$1$$$, $$$1$$$, $$$3$$$): $$$(a_1) \\oplus (a_1 \\oplus a_2 \\oplus a_3) = 6 \\oplus 0 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$2$$$, $$$3$$$): $$$(a_1 \\oplus a_2) \\oplus (a_2 \\oplus a_3) = 4 \\oplus 6 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n- ($$$1$$$, $$$3$$$, $$$3$$$): $$$(a_1 \\oplus a_2 \\oplus a_3) \\oplus (a_3) = 0 \\oplus 4 > (a_1 \\oplus a_2 \\oplus a_3) = 0$$$\n\nIn the second test case, there are no such tuples.\n\nBuggy Code:\nimport java.io.*;\n \npublic class D {\n public static void main(String args[]) throws IOException {\n BufferedReader scanner = new BufferedReader(new InputStreamReader(System.in));\n int T = Integer.parseInt(scanner.readLine());\n \n for (int t = 0; t < T; t++) {\n int N = Integer.parseInt(scanner.readLine());\n \n String[] s1 = scanner.readLine().split(\" \");\n int[] array = new int[N];\n for (int i = 0; i < N; i++) {\n array[i] = Integer.parseInt(s1[i]);\n }\n \n int[][] leftTable = new int[N][32];\n for (int i = 0; i < 32; i++) {\n if ((array[0] >> i) % 2 == 1) {\n leftTable[0][i] = 1;\n }\n }\n for (int j = 1; j < N; j++) {\n for (int i = 0; i < 32; i++) {\n if ((array[j] >> i) % 2 == 1) {\n leftTable[j][i] = (j - leftTable[j - 1][i]) + 1;\n }\n else {\n leftTable[j][i] = leftTable[j - 1][i];\n }\n }\n }\n \n int[][] rightTable = new int[N][32];\n for (int i = 0; i < 32; i++) {\n if ((array[N - 1] >> i) % 2 == 1) {\n rightTable[N - 1][i] = 1;\n }\n }\n for (int j = N - 2; j >= 0; j--) {\n for (int i = 0; i < 32; i++) {\n if ((array[j] >> i) % 2 == 1) {\n rightTable[j][i] = (N - 1 - j - rightTable[j + 1][i]) + 1;\n }\n else {\n rightTable[j][i] = rightTable[j + 1][i];\n }\n }\n }\n // for (int j = 0; j < N; j++) {\n // System.out.println(\"BUH \" + leftTable[j][2]);\n // }\n // for (int j = 0; j < N; j++) {\n // System.out.println(\"BUH2 \" + rightTable[j][2]);\n // }\n \n long total = 0;\n for (int j = 0; j < N; j++) {\n int highest = 0;\n for (int i = 0; i < 32; i++) {\n if ((array[j] >> i) % 2 == 1) {\n highest = i;\n }\n }\n \n total += (j + 1 - leftTable[j][highest]) * rightTable[j][highest] + leftTable[j][highest] * (N - j - rightTable[j][highest]);\n // System.out.println(\"OOF\" + total);\n }\n System.out.println(total);\n }\n \n scanner.close();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1957D",
"submission_id": 257620978,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\nusing namespace std;\n \nconst int MOD = 998244353;\n \nint main() {\n int n;\n cin >> n;\n vector<int> a(n);\n for (int i = 0; i < n; ++i) {\n cin >> a[i];\n }\n \n vector<int> prefix_or(n);\n prefix_or[0] = a[0];\n for (int i = 1; i < n; ++i) {\n prefix_or[i] = prefix_or[i-1] | a[i];\n }\n \n vector<int> dp(n, 0);\n dp[0] = 1;\n \n for (int i = 1; i < n; ++i) {\n dp[i] = dp[i-1]; // Single segment ending at i\n for (int j = 0; j < i; ++j) {\n if ((prefix_or[i] & ~prefix_or[j]) == 0) {\n dp[i] = (dp[i] + dp[j]) % MOD;\n }\n }\n }\n \n int result = 0;\n for (int i = 0; i < n; ++i) {\n result = (result + dp[i]) % MOD;\n }\n \n cout << result << endl;\n return 0;\n}\n \n \n// 365ws0AFlSzPqVtkg7Y\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 304752637,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nMOD = 998244353\n \ndef count_valid_splits(n, a):\n dp = [0] * (n + 1)\n dp[0] = 1 # Base case: one way to split an empty array\n \n or_values = [0] * (n + 1)\n \n for i in range(1, n + 1):\n or_values[i] = or_values[i-1] | a[i-1]\n \n for i in range(1, n + 1):\n seen_or = set()\n for j in range(i, 0, -1):\n current_or = or_values[i] ^ or_values[j-1]\n if current_or in seen_or:\n break\n seen_or.add(current_or)\n dp[i] = (dp[i] + dp[j-1]) % MOD\n \n return dp[n]\n \n# Read input\nn = int(input())\na = list(map(int, input().split()))\n \n# Calculate and print the result\nprint(count_valid_splits(n, a))\n \n \n# lWwcHyBq36vx7XnSE8G\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 304184540,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\ndef count_ways_to_split(n, arr):\n MOD = 998244353\n \n # Initialize the dp array\n dp = [0] * n\n dp[0] = 1 # There's one way to split the array up to the first element\n \n # Initialize the bitwise OR of the current segment\n current_or = arr[0]\n \n for i in range(1, n):\n # Update the current OR with the current element\n current_or |= arr[i]\n \n # Check if we can extend the current segment\n if current_or == arr[i]:\n dp[i] = (dp[i-1] + 1) % MOD\n else:\n dp[i] = dp[i-1]\n \n # The result is the sum of all possible ways to split the array\n result = sum(dp) % MOD\n return result\n \n# Read input\nn = int(input())\narr = list(map(int, input().split()))\n \n# Compute and print the result\nprint(count_ways_to_split(n, arr))\n \n \n# TAIoRLknEc8B9dSKvHQ\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 304676039,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nimport java.util.Scanner;\n \npublic class Main{\n private static final int MOD = 998244353;\n \n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n int n = scanner.nextInt();\n int[] a = new int[n];\n for (int i = 0; i < n; i++) {\n a[i] = scanner.nextInt();\n }\n \n long[] dp = new long[n];\n int[] prefixOr = new int[n];\n prefixOr[0] = a[0];\n dp[0] = 1;\n \n for (int i = 1; i < n; i++) {\n prefixOr[i] = prefixOr[i - 1] | a[i];\n dp[i] = 1; // Single segment case\n \n for (int j = 0; j < i; j++) {\n if ((prefixOr[i] & ~prefixOr[j]) == 0) {\n dp[i] = (dp[i] + dp[j]) % MOD;\n }\n }\n }\n \n long result = 0;\n for (long count : dp) {\n result = (result + count) % MOD;\n }\n \n System.out.println(result);\n }\n}\n \n \n// EcQH0tY6zrF7kfeOxL2\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 304785976,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nimport java.util.*;\n \npublic class DivideOrConquer {\n static final int MOD = 998244353;\n \n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n \n // Input reading\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 System.out.println(countWaysToSplit(arr));\n }\n \n static int countWaysToSplit(int[] arr) {\n int n = arr.length;\n \n // Maps to store the number of ways for each OR value\n Map<Integer, Integer> prevWays = new HashMap<>();\n prevWays.put(0, 1); // Initial OR is 0 with one way to split\n \n for (int num : arr) {\n Map<Integer, Integer> currWays = new HashMap<>();\n \n // For each OR value in the previous step\n for (Map.Entry<Integer, Integer> entry : prevWays.entrySet()) {\n int prevOr = entry.getKey();\n int ways = entry.getValue();\n \n // Update the current OR value with the new number\n int newOr = prevOr | num;\n \n // Add to the current ways\n currWays.put(newOr, (currWays.getOrDefault(newOr, 0) + ways) % MOD);\n }\n \n // Add the current number as a new segment\n currWays.put(num, (currWays.getOrDefault(num, 0) + 1) % MOD);\n \n // Update the previous ways map\n prevWays = currWays;\n }\n \n // Sum all the ways for the last element\n int totalWays = 0;\n for (int ways : prevWays.values()) {\n totalWays = (totalWays + ways) % MOD;\n }\n \n return totalWays;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 292787926,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\nusing namespace std;\n#define MOD 998244353\n \nint main() {\n int n;\n cin >> n;\n vector<int> a(n);\n \n for (int i = 0; i < n; ++i) {\n cin >> a[i];\n }\n \n vector<int> dp(n + 1, 0);\n dp[0] = 1; \n vector<int> bitOr(n + 1, 0);\n \n int sum = 1;\n for (int i = 1; i <= n; ++i) {\n bitOr[i] = bitOr[i - 1] | a[i - 1];\n \n for (int j = i - 1; j >= 0; --j) {\n if (bitOr[j] > bitOr[i]) {\n break;\n }\n dp[i] = (dp[i] + dp[j]) % MOD; \n }\n sum = (sum + dp[i]) % MOD;\n }\n \n cout << dp[n] << endl;\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 292187013,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nconst int MOD = 998244353;\n \nint main() {\n ios::sync_with_stdio(false);\n cin.tie(0);\n \n int n;\n cin >> n;\n vector<int> a(n);\n for (int i = 0; i < n; ++i) {\n cin >> a[i];\n }\n \n // dp[i] will store the number of ways to split the array up to index i\n vector<long long> dp(n + 1, 0);\n dp[0] = 1; // Base case: one way to split an empty array\n \n // running OR of the array up to index i\n int running_or = 0;\n \n // Iterate through the array\n for (int i = 1; i <= n; ++i) {\n running_or |= a[i - 1];\n \n // Update dp[i] based on previous values\n for (int j = i; j > 0; --j) {\n if ((running_or & a[j - 1]) == a[j - 1]) {\n dp[i] = (dp[i] + dp[j - 1]) % MOD;\n }\n }\n }\n \n // The answer is the number of ways to split the entire array\n cout << dp[n] << \"\\n\";\n \n return 0;\n}\n \n \n// r0U9VS8f3oYgywDbPZL\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 305404109,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nMOD = 998244353\n \nn = int(input())\na = list(map(int, input().split()))\n \nif all(x == a[0] for x in a):\n print(pow(2, n - 1, MOD))\n exit()\n \ndp = [0] * (n + 1)\ndp[0] = 1\n \nfor i in range(1, n + 1):\n current_or = 0\n for j in range(i - 1, -1, -1):\n current_or |= a[j]\n if dp[j]:\n if j == 0 or (current_or >= dp[j-1]):\n dp[i] = (dp[i] + dp[j]) % MOD\n \nprint(dp[n] % MOD)\n \n \n# 76lgymksPnMDTHGaRpL\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 305041061,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\n#include <cstdio>\n#include <algorithm>\n#include <map>\n#include <utility>\n \nusing namespace std;\n \nint n,a[200005];\n \ntypedef long long int ll;\n \nmap<pair<ll,ll>,ll> dp[200005];\n \n#define MOD 998244353ll\n \nint main()\n{\n\tscanf(\"%d\",&n);\n\t\n\tfor(int i=0;i<n;i++)\n\t{\n\t\tscanf(\"%d\",a+i);\n\t}\n\t\n\tdp[0].insert(make_pair(make_pair(a[0],0),1));\n\tdp[0].insert(make_pair(make_pair(0,a[0]),1));\n\t\n\tfor(int i=1;i<n;i++)\n\t{\n\t\tfor(pair<pair<ll,ll>,ll> p:dp[i-1])\n\t\t{\n\t\t\tll v = p.first.first,r = p.first.second,c = p.second;\n\t\t\tif(v<=a[i]|r)\n\t\t\t{\n\t\t\t\tif(dp[i].find(make_pair(a[i]|r,0))==dp[i].end())\n\t\t\t\t\tdp[i].insert(make_pair(make_pair(a[i]|r,0),0));\n\t\t\t\tdp[i][make_pair(a[i]|r,0)] = (dp[i][make_pair(a[i]|r,0)]+c)%MOD;\n\t\t\t}\n//\t\t\tll t = v|a[i]|r;\n//\t\t\tif(dp[i].find(make_pair(t,0))==dp[i].end())\n//\t\t\t\tdp[i].insert(make_pair(make_pair(t,0),0));\n//\t\t\tdp[i][make_pair(t,0)] = (dp[i][make_pair(t,0)]+c)%MOD;\n//\t\t\t\n\t\t\tif(dp[i].find(make_pair(v,a[i]|r))==dp[i].end())\n\t\t\t\tdp[i].insert(make_pair(make_pair(v,a[i]|r),0));\n\t\t\tdp[i][make_pair(v,a[i]|r)] = (dp[i][make_pair(v,a[i]|r)]+c)%MOD;\n\t\t\t\n\t\t}\n\t}\n\t\n\tll ans = 0;\n\tfor(pair<pair<ll,ll>,ll> p:dp[n-1])\n\t{\n\t\tif(p.first.second)\n\t\t\tcontinue;\n\t\tans = (ans+p.second)%MOD;\n\t}\n\t\n\tprintf(\"%lld\",ans);\n\t\n\treturn 0;\n}\n \n/*\n \nEfficiency.\n \n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 292402778,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\n#include<set>\n#include<map>\n#include<iostream>\n#include<queue>\n#include<sstream>\n#include<cstdlib>\n#include<cstdio>\n#include<vector>\n#include<string>\n#include<cstring>\n#include<algorithm>\nusing namespace std;\ntypedef vector<int>VI; typedef vector<VI>VVI;\ntypedef vector<long long>VLL;\ntypedef vector<string>VS;\ntypedef pair<int, int>PII;\ntypedef pair<long long, long long>PLL;\n#define FOR(i,n) for((i)=0;(i)<(n);(i)++)\n#define FORN(i,n) for((i)=(n)-1;(i)>=0;(i)--)\n#define FOR1(i,n) for((i)=1;(i)<=(n);(i)++)\n#define FORIT(it,a) for((it)=(a).begin();(it)!=(a).end();(it)++)\n#define BE(a) ((a).begin()),((a).end())\n#define SI(a) ((int)((a).size()))\n#define CLR(a) memset(a,0,sizeof(a))\n#define PB push_back\n#define MP make_pair\n \n#define MOD (998244353)\n \nint main()\n{\n\tint n, i, j;\n\tmap<int, int>::iterator it;\n\tmap<int, int>::iterator it2;\n\t\n\tcin >> n;\n\tVI a = VI(n);\n\tFOR(i, n) cin >> a[i];\n \n\tvector<int[30]> bitSetMinPos = vector<int[30]>(n + 1);\n\tvector<map<int, int>> orMinPos = vector<map<int, int>>(n + 1);\n\tFOR (j,30) bitSetMinPos[n][j] = n;\n\tFORN(i, n)\n\t{\n\t\tint curNum = a[i];\n\t\tFOR(j, 30) if ((curNum >> j) & 1) bitSetMinPos[i][j] = i;\n\t\telse bitSetMinPos[i][j] = bitSetMinPos[i + 1][j];\n\t\tFORIT(it, orMinPos[i + 1])\n\t\t{\n\t\t\tint curMask = curNum | it->first;\n\t\t\tint curMinPos = i;\n\t\t\tFOR(j, 30) if ((curMask >> j) & 1) curMinPos = max(curMinPos, bitSetMinPos[i][j]);\n\t\t\tif (!orMinPos[i][curMask] || curMinPos < orMinPos[i][curMask]) orMinPos[i][curMask] = curMinPos;\n\t\t}\n\t\torMinPos[i][curNum] = i;\n\t}\n \n\tvector<map<int, int>> ans = vector<map<int, int>>(n + 1);\n\tans[1][a[0]] = 1;\n\tFOR(i, n) if (i) FORIT(it, ans[i])\n\t{\n\t\tint curNum = a[i];\n\t\tint curOr = it->first;\n\t\tint curAns = it->second;\n\t\tif (((curOr | curNum) == curOr) || ans[i + 1].count(curOr | curNum) <= 0) ans[i + 1][curOr | curNum] = (ans[i + 1][curOr | curNum] + curAns) % MOD;\n\t\tFORIT(it2, orMinPos[i])\n\t\t{\n\t\t\tint nextOr = it2->first;\n\t\t\tint minPos = it2->second;\n\t\t\tif (nextOr < curOr) continue;\n\t\t\tans[minPos + 1][nextOr] = (ans[minPos + 1][nextOr] + curAns) % MOD;\n\t\t}\n\t}\n \n\tint finalAns = 0;\n\tFORIT(it, ans[n]) finalAns += it->second;\n\tcout << finalAns << 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\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 297162663,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nM = 998244353\n \n \nclass fenwick(object):\n def __init__(self, n):\n self.n = n\n self.cul = [0]*n\n self.maxd = len(bin(n))-3\n \n def update(self,index,diff):\n i = index\n while i<self.n:\n self.cul[i] += diff\n self.cul[i] %= M\n i += (i+1)&(-i-1)\n \n \n def getaccu(self,index):\n output = 0\n i = index\n while i>=0:\n output += self.cul[i]\n output %= M\n i -= (i+1)&(-i-1)\n return output\n \n \n \ndef main():\n \n n = int(input())\n arr = list(map(int,input().split()))\n \n dp = [{} for _ in range(n+1)]\n \n \n \n \n \n latest = [-1]*30\n \n queries = [{} for _ in range(n+2)]\n \n \n \n addes = []\n \n \n nums = set()\n \n for i in range(n):\n temp = {}\n for d in range(30):\n if (1<<d) & arr[i]:\n latest[d] = i\n if latest[d] >= 0:\n if latest[d] not in temp:\n temp[latest[d]] = 0\n temp[latest[d]] += 1 << d\n \n keys = sorted([key for key in temp], reverse=True)\n keys.append(-1)\n accu = 0\n dp[i][0] = (i-1,keys[0]-1)\n queries[i-1][0] = -1\n queries[keys[0]-1][0] = -1\n \n nums.add(0)\n \n \n \n for j in range(len(keys)-1):\n accu += temp[keys[j]]\n dp[i][accu] = (keys[j]-1, keys[j+1]-1)\n queries[keys[j]-1][accu] = -1\n queries[keys[j+1]-1][accu] = -1 \n nums.add(accu)\n \n nums = sorted(nums)\n \n mapping = {}\n \n for num in nums:\n mapping[num] = len(mapping) \n \n m = len(mapping)\n \n fen = fenwick(m)\n \n \n fen.update(0,1)\n \n for key in queries[-2]:\n queries[-2][key] = 0\n for key in queries[-1]:\n queries[-1][key] = 1\n \n \n \n for i in range(n):\n for num in dp[i]:\n front,rear = dp[i][num][0], dp[i][num][1]\n \n dp[i][num] = (queries[front][num] - queries[rear][num])%M\n fen.update(mapping[num], dp[i][num])\n for q in queries[i]:\n queries[i][q] = fen.getaccu(mapping[q])\n \n \n ans = 0\n for key in dp[n-1]:\n ans += dp[n-1][key]\n ans %= M\n \n print(ans)\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nmain()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 293309600,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nMOD = 998244353\nfrom collections import defaultdict\nimport sys \ninput = sys.stdin.readline\ndef solve(n, a):\n dp = defaultdict(int)\n dp[a[0]] = 1 \n for i in range(1, n):\n temp = defaultdict(int)\n for x in dp:\n \n y = x | a[i]\n temp[y] = (temp[y] + dp[x]) % MOD\n \n if x <= a[i]:\n temp[a[i]] = (temp[a[i]] + dp[x]) % MOD\n dp = temp\n return sum(dp.values()) % MOD\n \nn = int(input())\na = list(map(int, input().split()))\nprint(solve(n, a))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 292176325,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.util.*;\n \npublic class Main {\n static final int MOD = 998244353;\n \n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n \n int n = Integer.parseInt(br.readLine());\n \n String[] tokens = br.readLine().split(\" \");\n int[] a = new int[n];\n for(int i = 0; i < n; i++) {\n a[i] = Integer.parseInt(tokens[i]);\n }\n \n Map<Integer, Long> dp_prev = new HashMap<>();\n dp_prev.put(a[0], 1L); \n for(int i = 1; i < n; i++) {\n Map<Integer, Long> dp_current = new HashMap<>();\n for(Map.Entry<Integer, Long> entry : dp_prev.entrySet()) {\n int last_or = entry.getKey();\n long count = entry.getValue();\n int new_or1 = last_or | a[i];\n dp_current.put(new_or1, (dp_current.getOrDefault(new_or1, 0L) + count) % MOD);\n if(a[i] >= last_or) {\n int new_or2 = a[i];\n dp_current.put(new_or2, (dp_current.getOrDefault(new_or2, 0L) + count) % MOD);\n }\n }\n dp_prev = dp_current;\n }\n \n long result = 0;\n for(long count : dp_prev.values()) {\n result = (result + count) % MOD;\n }\n System.out.println(result);\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 294436214,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStreamReader;\nimport java.util.*;\n \npublic class Main3 {\n static final int MOD = 998244353;\n \n public static void main(String[] args) throws IOException {\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n int n = Integer.parseInt(br.readLine());\n String[] tokens = br.readLine().split(\" \");\n int[] a = new int[n];\n for(int i = 0; i < n; i++) {\n a[i] = Integer.parseInt(tokens[i]);\n }\n Map<Integer, Long> dp_prev = new HashMap<>();\n dp_prev.put(a[0], 1L); \n for(int i = 1; i < n; i++) {\n Map<Integer, Long> dp_current = new HashMap<>();\n for(Map.Entry<Integer, Long> entry : dp_prev.entrySet()) {\n int last_or = entry.getKey();\n long count = entry.getValue();\n int new_or1 = last_or | a[i];\n dp_current.put(new_or1, (dp_current.getOrDefault(new_or1, 0L) + count) % MOD);\n if(a[i] >= last_or) {\n int new_or2 = a[i];\n dp_current.put(new_or2, (dp_current.getOrDefault(new_or2, 0L) + count) % MOD);\n }\n }\n dp_prev = dp_current;\n }\n long result = 0;\n for(long count : dp_prev.values()) {\n result = (result + count) % MOD;\n }\n System.out.println(result);\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 294429834,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#pragma GCC optimize(3)\n#pragma GCC target(\"avx\")\n#pragma GCC optimize(\"Ofast\")\n#pragma GCC optimize(\"inline\")\n#pragma GCC optimize(\"-fgcse\")\n#pragma GCC optimize(\"-fgcse-lm\")\n#pragma GCC optimize(\"-fipa-sra\")\n#pragma GCC optimize(\"-ftree-pre\")\n#pragma GCC optimize(\"-ftree-vrp\")\n#pragma GCC optimize(\"-fpeephole2\")\n#pragma GCC optimize(\"-ffast-math\")\n#pragma GCC optimize(\"-fsched-spec\")\n#pragma GCC optimize(\"unroll-loops\")\n#pragma GCC optimize(\"-falign-jumps\")\n#pragma GCC optimize(\"-falign-loops\")\n#pragma GCC optimize(\"-falign-labels\")\n#pragma GCC optimize(\"-fdevirtualize\")\n#pragma GCC optimize(\"-fcaller-saves\")\n#pragma GCC optimize(\"-fcrossjumping\")\n#pragma GCC optimize(\"-fthread-jumps\")\n#pragma GCC optimize(\"-funroll-loops\")\n#pragma GCC optimize(\"-fwhole-program\")\n#pragma GCC optimize(\"-freorder-blocks\")\n#pragma GCC optimize(\"-fschedule-insns\")\n#pragma GCC optimize(\"inline-functions\")\n#pragma GCC optimize(\"-ftree-tail-merge\")\n#pragma GCC optimize(\"-fschedule-insns2\")\n#pragma GCC optimize(\"-fstrict-aliasing\")\n#pragma GCC optimize(\"-fstrict-overflow\")\n#pragma GCC optimize(\"-falign-functions\")\n#pragma GCC optimize(\"-fcse-skip-blocks\")\n#pragma GCC optimize(\"-fcse-follow-jumps\")\n#pragma GCC optimize(\"-fsched-interblock\")\n#pragma GCC optimize(\"-fpartial-inlining\")\n#pragma GCC optimize(\"no-stack-protector\")\n#pragma GCC optimize(\"-freorder-functions\")\n#pragma GCC optimize(\"-findirect-inlining\")\n#pragma GCC optimize(\"-fhoist-adjacent-loads\")\n#pragma GCC optimize(\"-frerun-cse-after-loop\")\n#pragma GCC optimize(\"inline-small-functions\")\n#pragma GCC optimize(\"-finline-small-functions\")\n#pragma GCC optimize(\"-ftree-switch-conversion\")\n#pragma GCC optimize(\"-foptimize-sibling-calls\")\n#pragma GCC optimize(\"-fexpensive-optimizations\")\n#pragma GCC optimize(\"-funsafe-loop-optimizations\")\n#pragma GCC optimize(\"inline-functions-called-once\")\n#pragma GCC optimize(\"-fdelete-null-pointer-checks\")\ntypedef long long ll;\nbool stt;\nconst ll mod=998244353;\nstruct BIT{\n\tint tr[8000005],sz;\n\tvoid init(){\n\t\tfor(int i=1;i<=sz;i++){\n\t\t\ttr[i]=0;\n\t\t}\n\t}\n\tinline void add(int x,int val){\n\t\tif(val>=mod)val-=mod;\n\t\twhile(x<=sz){\n\t\t\ttr[x]+=val;\n if(tr[x]>=mod)tr[x]-=mod;\n\t\t\tx+=x&(-x);\n\t\t}\n\t\treturn ;\n\t}\n\tinline long long ask(int x){\n\t\tlong long ans=0;\n\t\twhile(x>=1){\n\t\t\tans+=tr[x];\n\t\t\tx-=x&(-x);\n\t\t}\n\t\treturn ans%mod;\n\t}\n}t;\nint n;\nint a[200005],st[20][200005],pw[200005];\nbool ed;\nint b[8000005],tot;\nint main(){\n\t//cout<<(&ed-&stt)/1024.0/1024.0<<endl;\n ios::sync_with_stdio(false);\n cin.tie(0);\n cout.tie(0);\n //freopen(\"gen.in\",\"r\",stdin);\n //freopen(\"tst.out\",\"w\",stdout);\n cin>>n;\n pw[1]=0;\n for(int i=1;i<=n;i++){\n cin>>a[i];\n st[0][i]=a[i];\n if(i-1)pw[i]=pw[i/2]+1;\n }\n for(int i=1;i<20;i++){\n for(int j=(1<<i);j<=n;j++){\n st[i][j]=st[i-1][j-(1<<i-1)]|st[i-1][j];\n //cout<<i<<\" \"<<j<<\" \"<<st[i][j]<<endl;\n }\n }\n vector<pair<int,int> > vec[n+1];\n vector<int> ask[n+1];\n for(int i=1;i<=n;i++){\n int nw=i,val=a[i];\n //cout<<i<<endl;\n while(1){\n int lm=nw;\n for(int j=pw[lm];j>=0;j--){\n j=min(pw[lm],j);\n if((st[j][lm]|val)==val){\n lm-=(1<<j)-1;\n }\n //cout<<\"!\"<<val<<\" \"<<lm<<endl;\n }\n lm--;\n vec[i].push_back({val,lm+1});\n b[++tot]=val;\n if(lm<=0)break;\n nw=lm;\n val=val|a[lm];\n }\n }\n //b[++tot]=0;\n sort(b+1,b+tot+1);\n tot=unique(b+1,b+tot+1)-b-1;\n t.sz=tot;\n /*if(tot>200000){\n \tcout<<tot;\n \treturn 0; \n\t}*/\n //cout<<tot<<endl;\n t.init();\n t.add(1,1);\n //cout<<endl;\n unordered_map<int,int> res[n+1];\n int sum=0;\n for(int i=1;i<=n;i++){\n int pre=i+1;\n //cout<<i<<endl;\n for(int j=0;j<vec[i].size();j++){\n \t//cout<<vec[i][j].first<<endl;\n vec[i][j].first=lower_bound(b+1,b+tot+1,vec[i][j].first)-b;\n //cout<<vec[i][j].first<<endl;\n //cout<<vec[i][j].first<<\" \"<<vec[i][j].second<<endl;\n if(vec[i][j].second>=2)ask[vec[i][j].second-2].push_back(vec[i][j].first);\n ask[pre-2].push_back(vec[i][j].first);\n pre=vec[i][j].second;\n }\n }\n if(tot>200000){\n \tcout<<tot;\n \treturn 0;\n\t}\n //if(ask[0].size()>)\n for(int i=0;i<ask[0].size();i++){\n res[0][ask[0][i]]=1;\n }\n ll ans=0;\n for(int i=1;i<=n;i++){\n int pre=i+1;\n ans=t.ask(tot);\n for(int j=0;j<vec[i].size();j++){\n int vv=vec[i][j].first,pp=vec[i][j].second;\n t.add(vv,res[pre-2][vv]-(pp>1?res[pp-2][vv]:0)+mod);\n pre=pp;\n }\n for(int j=0;j<ask[i].size();j++){\n res[i][ask[i][j]]=t.ask(ask[i][j]);\n }\n }\n cout<<(t.ask(tot)-ans+mod)%mod<<endl;\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 292293038,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define int long long\nusing namespace std;\nconst int mod=998244353;\n \nint n;\nint a[200010];\nmap<pair<int,int>,int>dp[2];\n \nsigned main(){\n\tios::sync_with_stdio(0);\n\tcin.tie(0);cout.tie(0);\n\tcin>>n;\n\tfor(int i=1;i<=n;i++)cin>>a[i];\n\tdp[1][{0,a[1]}]=1;\n\tfor(int i=2;i<=n;i++){\n\t\tdp[i&1].clear();\n\t\tfor(auto j:dp[(i-1)&1]){\n\t\t\tauto k=j.first;\n\t\t\tif(k.first<=k.second)dp[i&1][{k.second,a[i]}]=(dp[i&1][{k.second,a[i]}]+j.second)%mod;\n\t\t\tdp[i&1][{k.first,k.second|a[i]}]=(dp[i&1][{k.first,k.second|a[i]}]+j.second)%mod;\n\t\t}\n\t}\n\tint ans=0;\n\tfor(auto i:dp[n&1]){\n\t\tif(i.first.first<=i.first.second)ans=(ans+i.second)%mod;\n\t}\n\tcout<<ans;\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 294565604,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\nboolean MULTI_CASE = false;\nboolean ALWAYS_FLUSH = false;\n \nvoid go() {\n int n = nextInt();\n int[] a = new int[n];\n for (int i = 0; i < a.length; ++i) a[i] = nextInt();\n PriorityQueue<int[]> q = new PriorityQueue<>((x, y) ->\n x[1] != y[1] ? Integer.compare(x[1], y[1])\n : Integer.compare(x[0], y[0]));\n int[][] b = new int[0][3];\n int[][] b2 = new int[30][3];\n int[][][] c = new int[n][0][3];\n for (int i = 0; i < n; i++) {\n int b2n = 0;\n int cur = a[i];\n int r = i;\n for (int j = 0; j < b.length; j++) {\n if ((cur | b[j][0]) != cur) {\n b2[b2n][0] = cur;\n b2[b2n][1] = b[j][2] + 1;\n b2[b2n++][2] = r;\n cur |= b[j][0];\n r = b[j][2];\n }\n }\n b2[b2n][0] = cur;\n b2[b2n][1] = 0;\n b2[b2n++][2] = r;\n b = new int[b2n][3];\n c[i] = new int[b2n][3];\n for (int j = 0; j < b2n; j++) {\n b[j] = Arrays.copyOf(b2[j], 3);\n c[i][j] = Arrays.copyOf(b2[j], 3);\n }\n q.offer(new int[]{i, c[i][0][0], c[i][0][1], c[i][0][2]});\n// for (int j = 0; j < b.length; j++) {\n// q.offer(new int[]{i, b[j][0], b[j][1], b[j][2]});\n// }\n }\n int[] ci = new int[n];\n Arrays.fill(ci, 1);\n Fenwick fw = new Fenwick(n + 1);\n while (!q.isEmpty()) {\n int r1 = 0;\n int[] en = q.poll();\n int i = en[0];\n if (ci[i] < c[i].length) {\n q.offer(new int[]{i, c[i][ci[i]][0], c[i][ci[i]][1], c[i][ci[i]][2]});\n ci[i]++;\n }\n if (en[2] == 0) {\n r1 = 1;\n if (en[3] - 1 >= 0) {\n r1 = (r1 + fw.get(0, en[3] - 1)) % M;\n }\n } else {\n r1 = fw.get(en[2] - 1, en[3] - 1);\n }\n fw.add(en[0], r1);\n }\n sl(fw.get(n - 1, n - 1));\n}\n \nint M = 998244353;\n \nclass Fenwick {\n int size;\n int[] t;\n \n Fenwick(int size) {\n this.size = size;\n t = new int[size + 5];\n }\n \n void add(int pos, int val) {\n pos++; // pos++ to deal with index starting from 0.\n for (int p = pos; p <= size; p += p & -p) {\n t[p] = (t[p] + val) % M;\n }\n }\n \n int get(int left, int right) {\n return (internalQuery(right) - internalQuery(left - 1) + M) % M;\n }\n \n int internalQuery(int pos) {\n pos++; // pos++ to deal with index starting from 0.\n int res = 0;\n for (int p = pos; p >= 1; p -= p & -p) {\n res = (res + t[p]) % M;\n }\n return res;\n }\n}\n \n \n/**\n * Input.\n */\nInputStream inStream;\nbyte[] inBuff = new byte[1024];\nint inBuffCursor = 0, inBuffLen = 0;\n \nboolean isVisibleChar(int c) {\n return 33 <= c && c <= 126;\n}\n \nboolean hasNextByte() {\n if (inBuffCursor < inBuffLen) return true;\n inBuffCursor = 0;\n try {\n inBuffLen = inStream.read(inBuff);\n } catch (Exception e) {\n e.printStackTrace(System.out);\n }\n return inBuffLen > 0;\n}\n \nboolean hasNext() {\n while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;\n return hasNextByte();\n}\n \nint nextByte() {\n return hasNextByte() ? inBuff[inBuffCursor++] : -1;\n}\n \nString next() {\n if (!hasNext()) throw new RuntimeException(\"no next.\");\n StringBuilder sb = new StringBuilder();\n int b = nextByte();\n while (isVisibleChar(b)) {\n sb.appendCodePoint(b);\n b = nextByte();\n }\n return sb.toString();\n}\n \nlong nextLong() {\n if (!hasNext()) throw new RuntimeException(\"no next.\");\n long result = 0;\n boolean negative = false;\n int b = nextByte();\n if (b < '0') {\n if (b == '-') negative = true;\n else if (b != '+') throw new RuntimeException(\"long number must start with +/-.\");\n b = nextByte();\n }\n while (isVisibleChar(b)) {\n if (b < '0' || b > '9') throw new RuntimeException(\"wrong digit in long:\" + (char) b);\n // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).\n result = result * 10 + (b - '0');\n b = nextByte();\n }\n return negative ? -result : result;\n}\n \nint nextInt() {\n long x = nextLong();\n if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)\n throw new RuntimeException(\"int overflow:\" + x);\n return (int) x;\n}\n \ndouble nextDouble() {\n return Double.parseDouble(next());\n}\n \n/**\n * Output.\n */\nPrintWriter printOut = new PrintWriter(System.out);\n \nvoid so(Object obj, boolean newLine) {\n if (newLine) printOut.println(obj);\n else printOut.print(obj);\n if (ALWAYS_FLUSH) printOut.flush();\n}\n \nvoid so(Object obj) {\n so(obj, false);\n}\n \nvoid sl(Object obj) {\n so(obj, true);\n}\n \nvoid sl() {\n sl(\"\");\n}\n \n/**\n * Main.\n */\nvoid mainGo() {\n try {\n inStream = new FileInputStream(\"src/main.in\");\n ALWAYS_FLUSH = true;\n } catch (Exception e) {\n inStream = System.in;\n }\n while (hasNext()) {\n if (MULTI_CASE) {\n int T = nextInt();\n if (T == 0) break;\n for (int i = 0; i < T; ++i) {\n go();\n }\n } else {\n go();\n }\n }\n printOut.flush();\n}\n \npublic static void main(String[] args) throws Exception {\n new Main().mainGo();\n}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 292341793,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\nboolean MULTI_CASE = false;\nboolean ALWAYS_FLUSH = false;\n \nvoid go() {\n int n = nextInt();\n int[] a = new int[n];\n for (int i = 0; i < a.length; ++i) a[i] = nextInt();\n PriorityQueue<int[]>[] q = new PriorityQueue[1 << 20];\n for (int i = 0; i < q.length; i++) {\n q[i] = new PriorityQueue<>((x, y) ->\n x[1] != y[1] ? Integer.compare(x[1], y[1])\n : Integer.compare(x[0], y[0]));\n }\n int[][] b = new int[0][3];\n int[][] b2 = new int[30][3];\n for (int i = 0; i < n; i++) {\n int b2n = 0;\n int cur = a[i];\n int r = i;\n for (int j = 0; j < b.length; j++) {\n if ((cur | b[j][0]) != cur) {\n b2[b2n][0] = cur;\n b2[b2n][1] = b[j][2] + 1;\n b2[b2n++][2] = r;\n cur |= b[j][0];\n r = b[j][2];\n }\n }\n b2[b2n][0] = cur;\n b2[b2n][1] = 0;\n b2[b2n++][2] = r;\n b = new int[b2n][3];\n for (int j = 0; j < b2n; j++)\n b[j] = Arrays.copyOf(b2[j], 3);\n for (int j = 0; j < b.length; j++) {\n q[b[j][0] >> 10].offer(new int[]{i, b[j][0], b[j][1], b[j][2]});\n }\n }\n Fenwick fw = new Fenwick(n + 1);\n for (int i = 0; i < q.length; i++) {\n while (!q[i].isEmpty()) {\n int r1 = 0;\n int[] en = q[i].poll();\n if (en[2] == 0) {\n r1 = 1;\n if (en[3] - 1 >= 0) {\n r1 = (r1 + fw.get(0, en[3] - 1)) % M;\n }\n } else {\n r1 = fw.get(en[2] - 1, en[3] - 1);\n }\n fw.add(en[0], r1);\n }\n }\n sl(fw.get(n - 1, n - 1));\n}\n \nint M = 998244353;\n \nclass Fenwick {\n int size;\n int[] t;\n \n Fenwick(int size) {\n this.size = size;\n t = new int[size + 5];\n }\n \n void add(int pos, int val) {\n pos++; // pos++ to deal with index starting from 0.\n for (int p = pos; p <= size; p += p & -p) {\n t[p] = (t[p] + val) % M;\n }\n }\n \n int get(int left, int right) {\n return (internalQuery(right) - internalQuery(left - 1) + M) % M;\n }\n \n int internalQuery(int pos) {\n pos++; // pos++ to deal with index starting from 0.\n int res = 0;\n for (int p = pos; p >= 1; p -= p & -p) {\n res = (res + t[p]) % M;\n }\n return res;\n }\n}\n \n \n/**\n * Input.\n */\nInputStream inStream;\nbyte[] inBuff = new byte[1024];\nint inBuffCursor = 0, inBuffLen = 0;\n \nboolean isVisibleChar(int c) {\n return 33 <= c && c <= 126;\n}\n \nboolean hasNextByte() {\n if (inBuffCursor < inBuffLen) return true;\n inBuffCursor = 0;\n try {\n inBuffLen = inStream.read(inBuff);\n } catch (Exception e) {\n e.printStackTrace(System.out);\n }\n return inBuffLen > 0;\n}\n \nboolean hasNext() {\n while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;\n return hasNextByte();\n}\n \nint nextByte() {\n return hasNextByte() ? inBuff[inBuffCursor++] : -1;\n}\n \nString next() {\n if (!hasNext()) throw new RuntimeException(\"no next.\");\n StringBuilder sb = new StringBuilder();\n int b = nextByte();\n while (isVisibleChar(b)) {\n sb.appendCodePoint(b);\n b = nextByte();\n }\n return sb.toString();\n}\n \nlong nextLong() {\n if (!hasNext()) throw new RuntimeException(\"no next.\");\n long result = 0;\n boolean negative = false;\n int b = nextByte();\n if (b < '0') {\n if (b == '-') negative = true;\n else if (b != '+') throw new RuntimeException(\"long number must start with +/-.\");\n b = nextByte();\n }\n while (isVisibleChar(b)) {\n if (b < '0' || b > '9') throw new RuntimeException(\"wrong digit in long:\" + (char) b);\n // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).\n result = result * 10 + (b - '0');\n b = nextByte();\n }\n return negative ? -result : result;\n}\n \nint nextInt() {\n long x = nextLong();\n if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)\n throw new RuntimeException(\"int overflow:\" + x);\n return (int) x;\n}\n \ndouble nextDouble() {\n return Double.parseDouble(next());\n}\n \n/**\n * Output.\n */\nPrintWriter printOut = new PrintWriter(System.out);\n \nvoid so(Object obj, boolean newLine) {\n if (newLine) printOut.println(obj);\n else printOut.print(obj);\n if (ALWAYS_FLUSH) printOut.flush();\n}\n \nvoid so(Object obj) {\n so(obj, false);\n}\n \nvoid sl(Object obj) {\n so(obj, true);\n}\n \nvoid sl() {\n sl(\"\");\n}\n \n/**\n * Main.\n */\nvoid mainGo() {\n try {\n inStream = new FileInputStream(\"src/main.in\");\n ALWAYS_FLUSH = true;\n } catch (Exception e) {\n inStream = System.in;\n }\n while (hasNext()) {\n if (MULTI_CASE) {\n int T = nextInt();\n if (T == 0) break;\n for (int i = 0; i < T; ++i) {\n go();\n }\n } else {\n go();\n }\n }\n printOut.flush();\n}\n \npublic static void main(String[] args) throws Exception {\n new Main().mainGo();\n}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 292340180,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nM = 998244353\n \n \nclass fenwick(object):\n def __init__(self, n):\n self.n = n\n self.cul = [0]*n\n self.maxd = len(bin(n))-3\n \n def update(self,index,diff):\n i = index\n while i<self.n:\n self.cul[i] += diff\n self.cul[i] %= M\n i += (i+1)&(-i-1)\n \n \n def getaccu(self,index):\n output = 0\n i = index\n while i>=0:\n output += self.cul[i]\n output %= M\n i -= (i+1)&(-i-1)\n return output\n \n \n \ndef main():\n \n n = int(input())\n arr = list(map(int,input().split()))\n \n \n \n if n==200000 and arr[0]==268435456:\n print(520128814)\n return \n \n dp = [{} for _ in range(n+1)]\n \n \n \n \n \n latest = [-1]*30\n \n queries = [{} for _ in range(n+2)]\n \n \n \n addes = []\n \n \n nums = set()\n \n for i in range(n):\n temp = {}\n for d in range(30):\n if (1<<d) & arr[i]:\n latest[d] = i\n if latest[d] >= 0:\n if latest[d] not in temp:\n temp[latest[d]] = 0\n temp[latest[d]] += 1 << d\n \n keys = sorted([key for key in temp], reverse=True)\n keys.append(-1)\n accu = 0\n dp[i][0] = (i-1,keys[0]-1)\n queries[i-1][0] = -1\n queries[keys[0]-1][0] = -1\n \n nums.add(0)\n \n \n \n for j in range(len(keys)-1):\n accu += temp[keys[j]]\n dp[i][accu] = (keys[j]-1, keys[j+1]-1)\n queries[keys[j]-1][accu] = -1\n queries[keys[j+1]-1][accu] = -1 \n nums.add(accu)\n \n nums = sorted(nums)\n \n mapping = {}\n \n for num in nums:\n mapping[num] = len(mapping) \n \n m = len(mapping)\n \n fen = fenwick(m)\n \n \n fen.update(0,1)\n \n for key in queries[-2]:\n queries[-2][key] = 0\n for key in queries[-1]:\n queries[-1][key] = 1\n \n \n \n for i in range(n):\n for num in dp[i]:\n front,rear = dp[i][num][0], dp[i][num][1]\n \n dp[i][num] = (queries[front][num] - queries[rear][num])%M\n fen.update(mapping[num], dp[i][num])\n for q in queries[i]:\n queries[i][q] = fen.getaccu(mapping[q])\n \n \n ans = 0\n for key in dp[n-1]:\n ans += dp[n-1][key]\n ans %= M\n \n print(ans)\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nmain()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 293312324,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Divide OR Conquer\ntime_limit_ms: 3000\nmemory_limit_mb: 512\nDescription: You are given an array $$$[a_1, a_2, \\ldots a_n]$$$ consisting of integers between $$$0$$$ and $$$10^9$$$. You have to split this array into several segments (possibly one) in such a way that each element belongs to exactly one segment.\n\nLet the first segment be the array $$$[a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]$$$, the second segment be $$$[a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]$$$, ..., the last segment be $$$[a_{l_k}, a_{l_k+ 1}, \\ldots, a_{r_k}]$$$. Since every element should belong to exactly one array, $$$l_1 = 1$$$, $$$r_k = n$$$, and $$$r_i + 1 = l_{i+1}$$$ for each $$$i$$$ from $$$1$$$ to $$$k-1$$$. The split should meet the following condition: $$$f([a_{l_1}, a_{l_1 + 1}, \\ldots, a_{r_1}]) \\le f([a_{l_2}, a_{l_2+ 1}, \\ldots, a_{r_2}]) \\le \\dots \\le f([a_{l_k}, a_{l_k+1}, \\ldots, a_{r_k}])$$$, where $$$f(a)$$$ is the bitwise OR of all elements of the array $$$a$$$.\n\nCalculate the number of ways to split the array, and print it modulo $$$998\\,244\\,353$$$. Two ways are considered different if the sequences $$$[l_1, r_1, l_2, r_2, \\ldots, l_k, r_k]$$$ denoting the splits are different.\n\nInput: The first line contains an integer $$$n$$$ ($$$1 \\le n \\le 2 \\cdot 10^5$$$) — the length of the array $$$a$$$.\n\nThe second line contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 10 ^9$$$) — the elements of the given array.\n\nOutput: Print one integer — the number of ways to split the array, taken modulo $$$998\\,244\\,353$$$.\n\nExamples:\ninput:\n3\n1 2 3\noutput:\n4\n\ninput:\n5\n1000 1000 1000 1000 1000\noutput:\n16\n\ninput:\n3\n3 4 6\noutput:\n3\n\nNote: In the first two examples, every way to split the array is valid.\n\nIn the third example, there are three valid ways to split the array:\n\n- $$$k = 3$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 2, l_3 = 3, r_3 = 3$$$; the resulting arrays are $$$[3]$$$, $$$[4]$$$, $$$[6]$$$, and $$$3 \\le 4 \\le 6$$$;\n- $$$k = 2$$$; $$$l_1 = 1, r_1 = 1, l_2 = 2, r_2 = 3$$$; the resulting arrays are $$$[3]$$$ and $$$[4, 6]$$$, and $$$3 \\le 6$$$;\n- $$$k = 1$$$; $$$l_1 = 1, r_1 = 3$$$; there will be only one array: $$$[3, 4, 6]$$$.\n\nIf you split the array into two arrays $$$[3, 4]$$$ and $$$[6]$$$, the bitwise OR of the first array is $$$7$$$, and the bitwise OR of the second array is $$$6$$$; $$$7 > 6$$$, so this way to split the array is invalid.\n\nBuggy Code:\nM = 998244353\n \n \nclass fenwick(object):\n def __init__(self, n):\n self.n = n\n self.cul = [0]*n\n self.maxd = len(bin(n))-3\n \n def update(self,index,diff):\n i = index\n while i<self.n:\n self.cul[i] += diff\n self.cul[i] %= M\n i += (i+1)&(-i-1)\n \n \n def getaccu(self,index):\n output = 0\n i = index\n while i>=0:\n output += self.cul[i]\n output %= M\n i -= (i+1)&(-i-1)\n return output\n \n \n \ndef main():\n \n n = int(input())\n arr = list(map(int,input().split()))\n \n \n fuck = {2097152:309984555, 268435456:520128814, 2:947235282, 8:74705939}\n if n==200000 and arr[0] in fuck:\n print(fuck[arr[0]])\n return \n \n dp = [{} for _ in range(n+1)]\n \n \n \n \n \n latest = [-1]*30\n \n queries = [{} for _ in range(n+2)]\n \n \n \n addes = []\n \n \n nums = set()\n \n for i in range(n):\n temp = {}\n for d in range(30):\n if (1<<d) & arr[i]:\n latest[d] = i\n if latest[d] >= 0:\n if latest[d] not in temp:\n temp[latest[d]] = 0\n temp[latest[d]] += 1 << d\n \n keys = sorted([key for key in temp], reverse=True)\n keys.append(-1)\n accu = 0\n dp[i][0] = (i-1,keys[0]-1)\n queries[i-1][0] = -1\n queries[keys[0]-1][0] = -1\n \n nums.add(0)\n \n \n \n for j in range(len(keys)-1):\n accu += temp[keys[j]]\n dp[i][accu] = (keys[j]-1, keys[j+1]-1)\n queries[keys[j]-1][accu] = -1\n queries[keys[j+1]-1][accu] = -1 \n nums.add(accu)\n \n nums = sorted(nums)\n \n mapping = {}\n \n for num in nums:\n mapping[num] = len(mapping) \n \n m = len(mapping)\n \n fen = fenwick(m)\n \n \n fen.update(0,1)\n \n for key in queries[-2]:\n queries[-2][key] = 0\n for key in queries[-1]:\n queries[-1][key] = 1\n \n \n \n for i in range(n):\n for num in dp[i]:\n front,rear = dp[i][num][0], dp[i][num][1]\n \n dp[i][num] = (queries[front][num] - queries[rear][num])%M\n fen.update(mapping[num], dp[i][num])\n for q in queries[i]:\n queries[i][q] = fen.getaccu(mapping[q])\n \n \n ans = 0\n for key in dp[n-1]:\n ans += dp[n-1][key]\n ans %= M\n \n print(ans)\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nmain()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2038D",
"submission_id": 293313325,
"task_type": "TestCase-Eval-Task2"
} |
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