Sidney Zhang
65999c8456
📝 docs(README)：更新项目描述并添加核心数学理念说明 🔧 chore(pyproject.toml)：更新项目描述信息 ♻️ refactor(euler_3.py)：改进质因数分解函数并添加类型注解 💡 docs(readme)：添加第4题数学分析文档和算法说明 ✅ test(euler_3.py)：添加主函数测试用例验证质因数分解功能
119 lines
2.6 KiB
Python
119 lines
2.6 KiB
Python
"""
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The prime factors of 13195 are 5, 7, 13and 29.
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What is the largest prime factor of the number 600851475143 ?
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"""
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import random
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from math import gcd
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from typing import List, Set
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def is_probable_prime(n: int, trials: int = 10) -> bool:
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"""Miller-Rabin素性测试（快速判断是否为质数）"""
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if n < 2:
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return False
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if n in (2, 3):
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return True
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if n % 2 == 0:
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return False
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# 将 n-1 写成 d * 2^s 的形式
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d = n - 1
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s = 0
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while d % 2 == 0:
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d //= 2
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s += 1
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# 测试
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for _ in range(trials):
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a = random.randrange(2, n - 1)
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x = pow(a, d, n)
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if x == 1 or x == n - 1:
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continue
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for _ in range(s - 1):
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x = pow(x, 2, n)
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if x == n - 1:
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break
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else:
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return False
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return True
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def pollards_rho(n: int, max_iter: int = 100000) -> int | None:
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"""
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Pollard's Rho 算法：返回n的一个非平凡因子
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Args:
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n: 待分解的合数
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max_iter: 最大迭代次数防止无限循环
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Returns:
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n的一个因子（可能是质数也可能是合数）
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若失败返回None
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"""
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if n % 2 == 0:
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return 2
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# 随机生成多项式 f(x) = x^2 + c (mod n)
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c = random.randrange(1, n - 1)
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def f(x):
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return (pow(x, 2, n) + c) % n
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# Floyd 判圈算法
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x = random.randrange(2, n - 1)
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y = x
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d = 1
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iter_count = 0
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while d == 1 and iter_count < max_iter:
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x = f(x) # 乌龟：走一步
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y = f(f(y)) # 兔子：走两步
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d = gcd(abs(x - y), n)
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iter_count += 1
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if d == n:
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# 失败，尝试其他参数（递归或返回None）
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return pollards_rho(n, max_iter) if max_iter > 1000 else None
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return d
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def factorize(n: int | None) -> List[int | None]:
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"""
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完整因数分解：递归分解所有质因数
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Args:
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n: 待分解的正整数
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Returns:
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质因数列表（可能含重复）
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"""
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if n == 1:
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return []
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if n is None:
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return [None]
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# 如果是质数，直接返回
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if is_probable_prime(n):
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return [n]
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# 获取一个因子
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factor = pollards_rho(n)
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if factor is None:
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return [None]
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# 递归分解
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return factorize(factor) + factorize(n // factor)
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def get_prime_factors(n: int) -> Set[int | None]:
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"""获取所有不重复的质因数"""
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return set(factorize(n))
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if __name__ == "__main__":
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print(get_prime_factors(60)) # {2, 3, 5}
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