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"""
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number
would be 1+2+3+4+5+6+7=28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1 : 1
3 : 1, 3
6 : 1, 2, 3, 6
10 : 1, 2, 5, 10
15 : 1, 3, 5, 15
21 : 1, 3, 7, 21
28 : 1, 2, 4, 7, 14, 28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
NOTE
-> in Math
解法的核心是找到所有质因数及对应的最大幂, 根据组合数学的方法估算因数数量
-> in Coding
循环遍历
"""
import math
import random
import time
from collections import Counter
from functools import reduce
from math import gcd
from typing import List
def timer(func):
def wrapper(*args, **kwargs):
start_time = time.time()
result = func(*args, **kwargs)
end_time = time.time()
print(f"Time taken: {end_time - start_time} seconds")
return result
return wrapper
def get_factors(n):
if n == 0:
raise ValueError("0 没有因数集合")
n = abs(n) # 处理负数
factors = set()
for i in range(1, int(math.isqrt(n)) + 1):
if n % i == 0:
factors.add(i)
factors.add(n // i)
return sorted(factors)
def get_triangle_number(n: int) -> int:
return n * (n + 1) // 2
@timer
def main_coding() -> None:
n = 1
while True:
triangle_number = get_triangle_number(n)
factors = get_factors(triangle_number)
if len(factors) > 500:
print(triangle_number)
break
n += 1
def is_probable_prime(n: int, trials: int = 20) -> 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) -> dict[int | None, int]:
"""获取所有不重复的质因数"""
return dict(Counter(factorize(n)))
def zuheshu(tl: list[int]) -> int:
xt = [x + 1 for x in tl]
return reduce(lambda x, y: x * y, xt)
@timer
def main_math() -> None:
n = 1
while True:
tn = get_triangle_number(n)
factors = get_prime_factors(tn)
if factors == {}:
n += 1
continue
if zuheshu(list(factors.values())) > 500:
print(tn)
break
n += 1
if __name__ == "__main__":
print("暴力试算:")
main_coding()
print("质因数分解:")
main_math()