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internbootcamp/bootcamp/eprefixproductsequence/eprefixproductsequence.py

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+"""# 
+
+### 谜题描述
+Consider a sequence [a1, a2, ... , an]. Define its prefix product sequence <image>.
+
+Now given n, find a permutation of [1, 2, ..., n], such that its prefix product sequence is a permutation of [0, 1, ..., n - 1].
+
+Input
+
+The only input line contains an integer n (1 ≤ n ≤ 105).
+
+Output
+
+In the first output line, print \"YES\" if such sequence exists, or print \"NO\" if no such sequence exists.
+
+If any solution exists, you should output n more lines. i-th line contains only an integer ai. The elements of the sequence should be different positive integers no larger than n.
+
+If there are multiple solutions, you are allowed to print any of them.
+
+Examples
+
+Input
+
+7
+
+
+Output
+
+YES
+1
+4
+3
+6
+5
+2
+7
+
+
+Input
+
+6
+
+
+Output
+
+NO
+
+Note
+
+For the second sample, there are no valid sequences.
+
+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
+# coding: utf-8
+
+def isPrime(n):
+  if n == 2:
+    return True
+  if n % 2 == 0:
+    return False
+  for i in xrange(3, (int)(n ** 0.5) + 1, 2):
+    if n % i == 0:
+      return False
+  return True
+n = input()
+if n == 1:
+  print 'YES\n1'
+elif n == 4:
+  print 'YES\n1\n3\n2\n4'
+elif not isPrime(n):
+  print 'NO'
+else:
+  print 'YES\n1'
+  inv = [0, 1]
+  for i in xrange(2, n):
+    inv.append((n - n / i) * inv[n % i] % n)
+    print i * inv[i - 1] % n
+  print n
+```
+
+
+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 re
+import random
+from bootcamp import Basebootcamp
+
+class Eprefixproductsequencebootcamp(Basebootcamp):
+    def __init__(self, max_n=100000, min_n=1):
+        """
+        初始化训练场参数，确保n的取值范围合法。
+        
+        参数:
+            max_n (int): 生成的n的最大值（默认1e5，须小于等于1e5）
+            min_n (int): 生成的n的最小值（默认1，须大于等于1）
+        """
+        if min_n < 1:
+            raise ValueError("min_n must be at least 1")
+        if max_n > 10**5:
+            raise ValueError("max_n cannot exceed 100000")
+        if min_n > max_n:
+            raise ValueError("min_n must be <= max_n")
+        self.max_n = max_n
+        self.min_n = min_n
+    
+    @staticmethod
+    def is_prime(n):
+        """优化质数判断逻辑，提升大数效率"""
+        if n < 2:
+            return False
+        if n % 2 == 0:
+            return n == 2
+        for i in range(3, int(n**0.5) + 1, 2):
+            if n % i == 0:
+                return False
+        return True
+    
+    def case_generator(self):
+        """
+        生成n时确保覆盖所有边界条件：
+        - 强制包含n=1,4以确保特殊案例覆盖率
+        - 50%概率生成质数用于覆盖YES案例
+        """
+        candidates = []
+        if self.min_n <= 1 <= self.max_n:
+            candidates.append(1)
+        if self.min_n <= 4 <= self.max_n:
+            candidates.append(4)
+        # 随机生成至少一个质数案例（若范围内存在）
+        prime_candidate = self._find_prime_in_range()
+        if prime_candidate:
+            candidates.append(prime_candidate)
+        
+        # 随机选择n（优先特殊案例）
+        if candidates and random.random() < 0.5:
+            n = random.choice(candidates)
+        else:
+            n = random.randint(self.min_n, self.max_n)
+        
+        possible = n in (1, 4) or self.is_prime(n)
+        return {'n': n, 'possible': possible}
+    
+    def _find_prime_in_range(self):
+        """在允许范围内查找一个质数"""
+        attempts = 0
+        while attempts < 1000:
+            n = random.randint(self.min_n, self.max_n)
+            if self.is_prime(n):
+                return n
+            attempts += 1
+        return None  # 未找到时不强制要求
+    
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        """优化问题描述的数学表达式清晰度"""
+        n = question_case['n']
+        problem_desc = (
+            f"给定整数 n = {n}，寻找一个排列 P = [a₁, a₂, ..., aₙ] 满足：\n\n"
+            "条件定义：\n"
+            f"- 前缀积序列 B = [b₁, b₂, ..., bₙ]，其中 b_i = (a₁×a₂×...×a_i) mod {n}\n"
+            f"- 要求 B 是 [0, 1, 2, ..., {n-1}] 的一个排列\n\n"
+            "输出要求：\n"
+            f"1. 第一行输出是否存在解（YES/NO）\n"
+            f"2. 若存在解，输出{n}行具体排列（每行一个整数）\n\n"
+            "答案格式要求：\n"
+            "将完整输出包含在[answer]和[/answer]标记之间，例如：\n"
+            "[answer]\n"
+            "YES\n1\n3\n2\n4\n[/answer]"
+        )
+        return problem_desc
+    
+    @staticmethod
+    def extract_output(output):
+        """增强正则表达式鲁棒性，允许换行和空格"""
+        matches = re.findall(r'\[answer\](.*?)\[\/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        last_match = matches[-1].strip()
+        # 清理首尾空白行
+        return '\n'.join(line.strip() for line in last_match.splitlines() if line.strip())
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        """添加模运算结果的严格验证"""
+        n = identity['n']
+        possible = identity['possible']
+        lines = solution.strip().split('\n')
+        if not lines:
+            return False
+        
+        # 验证YES/NO与possible的一致性
+        first_line = lines[0].strip().upper()
+        if first_line == 'YES' and not possible:
+            return False
+        if first_line == 'NO' and possible:
+            return False
+        
+        # 处理NO情况
+        if first_line == 'NO':
+            return len(lines) == 1
+        
+        # 处理YES情况
+        if len(lines) != n + 1:
+            return False  # 行数不匹配
+        
+        try:
+            sequence = list(map(int, lines[1:n+1]))
+        except ValueError:
+            return False
+        
+        # 验证元素为1~n的排列
+        if sorted(sequence) != list(range(1, n+1)):
+            return False
+        
+        # 验证前缀积模运算结果
+        prefix_mod = []
+        current = 1
+        for num in sequence:
+            current = (current * num) % n
+            prefix_mod.append(current)
+        return sorted(prefix_mod) == list(range(n))