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internbootcamp/bootcamp/cdisposition/cdisposition.py

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+"""# 
+
+### 谜题描述
+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.
+
+Help Vasya — find the volume disposition with the minimum number of divisors.
+
+Input
+
+The first line contains number n (1 ≤ n ≤ 100000) which represents the number of volumes and free places.
+
+Output
+
+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.
+
+Examples
+
+Input
+
+2
+
+
+Output
+
+2 1 
+
+
+Input
+
+3
+
+
+Output
+
+1 3 2 
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+#!/usr/bin/python
+
+n = int(raw_input())
+r = range(2, n + 2)
+r[-1] = 1
+print ' '.join(map(str, r))
+```
+
+
+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。
+"""
+
+from bootcamp import Basebootcamp
+import random
+import math
+from bootcamp import Basebootcamp
+
+class Cdispositionbootcamp(Basebootcamp):
+    def __init__(self, **params):
+        self.n = params.get('n', random.randint(1, 100000))
+
+    def case_generator(self):
+        n = random.randint(1, 100000)
+        if n == 1:
+            p = [1]
+        else:
+            p = [i + 1 for i in range(n)]
+            p[-1] = 1
+        return {'n': n, 'p': p}
+    
+    @staticmethod
+    def prompt_func(question_case):
+        n = question_case['n']
+        return (
+            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。"
+            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。"
+            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。"
+            f"请将答案放在[answer]标签内，例如：[answer]2 3 1[/answer]"
+        )
+    
+    @staticmethod
+    def extract_output(output):
+        start = output.rfind('[answer]')
+        if start == -1:
+            return None
+        end = output.find('[/answer]', start)
+        if end == -1:
+            return None
+        answer_str = output[start + len('[answer]'):end].strip()
+        try:
+            solution = list(map(int, answer_str.split()))
+        except ValueError:
+            return None
+        return solution
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        n = identity['n']
+        if len(solution) != n:
+            return False
+        if sorted(solution) != list(range(1, n+1)):
+            return False
+        divisors = set()
+        for j in range(1, n+1):
+            p_j = solution[j-1]
+            g = math.gcd(j, p_j)
+            factors = set()
+            for i in range(1, int(math.sqrt(g)) + 1):
+                if g % i == 0:
+                    factors.add(i)
+                    factors.add(g // i)
+            divisors.update(factors)
+        return len(divisors) == 1