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193 lines
5.2 KiB
Python
Executable file
193 lines
5.2 KiB
Python
Executable file
"""#
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### 谜题描述
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This is an interactive problem.
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We hid from you a permutation p of length n, consisting of the elements from 1 to n. You want to guess it. To do that, you can give us 2 different indices i and j, and we will reply with p_{i} mod p_{j} (remainder of division p_{i} by p_{j}).
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We have enough patience to answer at most 2 ⋅ n queries, so you should fit in this constraint. Can you do it?
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As a reminder, a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
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Input
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The only line of the input contains a single integer n (1 ≤ n ≤ 10^4) — length of the permutation.
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Interaction
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The interaction starts with reading n.
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Then you are allowed to make at most 2 ⋅ n queries in the following way:
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* \"? x y\" (1 ≤ x, y ≤ n, x ≠ y).
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After each one, you should read an integer k, that equals p_x mod p_y.
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When you have guessed the permutation, print a single line \"! \" (without quotes), followed by array p and quit.
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After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
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* fflush(stdout) or cout.flush() in C++;
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* System.out.flush() in Java;
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* flush(output) in Pascal;
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* stdout.flush() in Python;
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* see documentation for other languages.
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Exit immediately after receiving \"-1\" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
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Hack format
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In the first line output n (1 ≤ n ≤ 10^4). In the second line print the permutation of n integers p_1, p_2, …, p_n.
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Example
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Input
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3
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1
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2
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1
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0
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Output
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? 1 2
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? 3 2
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? 1 3
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? 2 1
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! 1 3 2
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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#include <bits/stdc++.h>
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using namespace std;
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const int Max_N = 1e4;
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int n, a[Max_N + 5];
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inline int Read() {
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int num = 0;
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char ch = getchar();
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for (; ch < '0' || ch > '9'; ch = getchar())
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;
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for (; ch >= '0' && ch <= '9'; num = num * 10 + ch - '0', ch = getchar())
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;
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return num;
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}
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inline int Ask(int x, int y) {
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printf(\"? %d %d\n\", x, y);
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fflush(stdout);
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return Read();
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}
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int main() {
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scanf(\"%d\", &n);
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for (int i = 2, lst = 1; i <= n; i++) {
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int A = Ask(lst, i), B = Ask(i, lst);
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if (A < B)
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a[i] = B;
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else
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a[lst] = A, lst = i;
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if (i == n) a[lst] = n;
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}
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if (n == 1) a[n] = 1;
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printf(\"! \");
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for (int i = 1; i <= n; i++) printf(\"%d \", a[i]);
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puts(\"\");
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return 0;
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}
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import random
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import re
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from bootcamp import Basebootcamp
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class Cchocolatebunnybootcamp(Basebootcamp):
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def __init__(self, **params):
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"""
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动态参数配置，支持自定义排列长度
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"""
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self.n = params.get('n', 3)
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if not 1 <= self.n <= 10**4:
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raise ValueError("n must be between 1 and 10^4")
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def case_generator(self):
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"""通用的排列生成方法"""
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permutation = list(range(1, self.n + 1))
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if self.n > 1:
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random.shuffle(permutation)
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return {'n': self.n, 'permutation': permutation}
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@staticmethod
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def prompt_func(question_case) -> str:
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"""精确的提示生成逻辑"""
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n = question_case['n']
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example = ' '.join(map(str, question_case['permutation']))
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return f"""编程竞赛交互题规则：
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我们需要找出长度为{n}的排列p。你最多可以进行{2*n}次询问，每次询问格式为"? x y"（x≠y），系统返回p_x mod p_y的值。
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你的任务是：
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1. 分析模运算结果间的逻辑关系
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2. 推断出完整排列
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3. 按格式输出答案：! 后跟排列的数字，用空格分隔
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请直接将最终答案放入[answer]标签内，例如：
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[answer]
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! {example}
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[/answer]
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现在请解决n={n}的案例："""
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@staticmethod
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def extract_output(output):
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"""强健的答案提取方法"""
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# 匹配所有可能的答案块
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answer_blocks = re.findall(
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r'\[answer\][\s\S]*?!([\s\S]*?)\[\/answer\]',
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output,
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flags=re.IGNORECASE
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)
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if not answer_blocks:
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return None
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# 提取最后一个答案块中的数字
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numbers = re.findall(r'\d+', answer_blocks[-1])
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try:
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return list(map(int, numbers))
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except (ValueError, TypeError):
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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"""完整的验证流程"""
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expected = identity['permutation']
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n = identity['n']
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# 类型和长度校验
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if not isinstance(solution, list) or len(solution) != n:
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return False
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# 元素范围校验
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if set(solution) != set(range(1, n+1)):
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return False
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# 精确顺序校验
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return solution == expected
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