Symbol(clack:cancel)

solutions/0000_0029/0003.largestprime/euler_3.py

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+"""
+The prime factors of 13195 are 5, 7, 13and 29.
+What is the largest prime factor of the number 600851475143 ?
+"""
+
+import random
+from math import gcd
+from typing import List, Set
+
+
+def is_probable_prime(n: int, trials: int = 10) -> bool:
+    """Miller-Rabin素性测试（快速判断是否为质数）"""
+    if n < 2:
+        return False
+    if n in (2, 3):
+        return True
+    if n % 2 == 0:
+        return False
+
+    # 将 n-1 写成 d * 2^s 的形式
+    d = n - 1
+    s = 0
+    while d % 2 == 0:
+        d //= 2
+        s += 1
+
+    # 测试
+    for _ in range(trials):
+        a = random.randrange(2, n - 1)
+        x = pow(a, d, n)
+        if x == 1 or x == n - 1:
+            continue
+        for _ in range(s - 1):
+            x = pow(x, 2, n)
+            if x == n - 1:
+                break
+        else:
+            return False
+    return True
+
+
+def pollards_rho(n: int, max_iter: int = 100000) -> int | None:
+    """
+    Pollard's Rho 算法：返回n的一个非平凡因子
+
+    Args:
+        n: 待分解的合数
+        max_iter: 最大迭代次数防止无限循环
+
+    Returns:
+        n的一个因子（可能是质数也可能是合数）
+        若失败返回None
+    """
+
+    if n % 2 == 0:
+        return 2
+
+    # 随机生成多项式 f(x) = x^2 + c (mod n)
+    c = random.randrange(1, n - 1)
+
+    def f(x):
+        return (pow(x, 2, n) + c) % n
+
+    # Floyd 判圈算法
+    x = random.randrange(2, n - 1)
+    y = x
+    d = 1
+
+    iter_count = 0
+    while d == 1 and iter_count < max_iter:
+        x = f(x)  # 乌龟：走一步
+        y = f(f(y))  # 兔子：走两步
+        d = gcd(abs(x - y), n)
+        iter_count += 1
+
+    if d == n:
+        # 失败，尝试其他参数（递归或返回None）
+        return pollards_rho(n, max_iter) if max_iter > 1000 else None
+
+    return d
+
+
+def factorize(n: int | None) -> List[int | None]:
+    """
+    完整因数分解：递归分解所有质因数
+
+    Args:
+        n: 待分解的正整数
+
+    Returns:
+        质因数列表（可能含重复）
+    """
+    if n == 1:
+        return []
+    if n is None:
+        return [None]
+
+    # 如果是质数，直接返回
+    if is_probable_prime(n):
+        return [n]
+
+    # 获取一个因子
+    factor = pollards_rho(n)
+
+    if factor is None:
+        return [None]
+
+    # 递归分解
+    return factorize(factor) + factorize(n // factor)
+
+
+def get_prime_factors(n: int) -> Set[int | None]:
+    """获取所有不重复的质因数"""
+    return set(factorize(n))
+
+
+if __name__ == "__main__":
+    print(get_prime_factors(60))  # {2, 3, 5}