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111 lines
3.5 KiB
Python
Executable file
111 lines
3.5 KiB
Python
Executable file
"""#
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### 谜题描述
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Vasya bought the collected works of a well-known Berland poet Petya in n volumes. The volumes are numbered from 1 to n. He thinks that it does not do to arrange the book simply according to their order. Vasya wants to minimize the number of the disposition’s divisors — the positive integers i such that for at least one j (1 ≤ j ≤ n) is true both: j mod i = 0 and at the same time p(j) mod i = 0, where p(j) is the number of the tome that stands on the j-th place and mod is the operation of taking the division remainder. Naturally, one volume can occupy exactly one place and in one place can stand exactly one volume.
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Help Vasya — find the volume disposition with the minimum number of divisors.
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Input
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The first line contains number n (1 ≤ n ≤ 100000) which represents the number of volumes and free places.
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Output
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Print n numbers — the sought disposition with the minimum divisor number. The j-th number (1 ≤ j ≤ n) should be equal to p(j) — the number of tome that stands on the j-th place. If there are several solutions, print any of them.
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Examples
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Input
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2
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Output
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2 1
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Input
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3
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Output
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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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#!/usr/bin/python
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n = int(raw_input())
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r = range(2, n + 2)
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r[-1] = 1
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print ' '.join(map(str, r))
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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 math
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from bootcamp import Basebootcamp
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class Cdispositionbootcamp(Basebootcamp):
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def __init__(self, **params):
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self.n = params.get('n', random.randint(1, 100000))
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def case_generator(self):
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n = random.randint(1, 100000)
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if n == 1:
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p = [1]
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else:
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p = [i + 1 for i in range(n)]
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p[-1] = 1
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return {'n': n, 'p': p}
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@staticmethod
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def prompt_func(question_case):
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n = question_case['n']
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return (
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f"你有一个包含{n}本的书籍集，编号从1到{n}。你需要将这些书籍排列成一个序列p，使得除数的数目最少。"
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f"除数i的定义是：存在至少一个位置j（1 ≤ j ≤ {n}），使得j和p(j)都能被i整除。"
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f"你的任务是找到这样的排列p，并将其输出为一个由空格分隔的整数序列。"
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f"请将答案放在[answer]标签内，例如：[answer]2 3 1[/answer]"
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)
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@staticmethod
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def extract_output(output):
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start = output.rfind('[answer]')
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if start == -1:
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return None
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end = output.find('[/answer]', start)
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if end == -1:
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return None
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answer_str = output[start + len('[answer]'):end].strip()
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try:
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solution = list(map(int, answer_str.split()))
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except ValueError:
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return None
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return solution
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@classmethod
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def _verify_correction(cls, solution, identity):
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n = identity['n']
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if len(solution) != n:
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return False
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if sorted(solution) != list(range(1, n+1)):
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return False
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divisors = set()
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for j in range(1, n+1):
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p_j = solution[j-1]
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g = math.gcd(j, p_j)
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factors = set()
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for i in range(1, int(math.sqrt(g)) + 1):
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if g % i == 0:
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factors.add(i)
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factors.add(g // i)
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divisors.update(factors)
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return len(divisors) == 1
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