136 lines
3.2 KiB
Python
136 lines
3.2 KiB
Python
"""
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A unit fraction contains 1in the numerator.
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The decimal representation of the unit fractions with denominators 2 to 10 are given:
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1/2 = 0.5
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1/3 = 0.(3)
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1/4 = 0.25
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1/5 = 0.2
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1/6 = 0.1(6)
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1/7 = 0.(142857)
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1/8 = 0.125
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1/9 = 0.(1)
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1/10 = 0.1
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Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle.
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It can be seen that 1/7 has a 6-digit recurring cycle.
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Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
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"""
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import random
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import time
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def timer(func):
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def wrapper(*args, **kwargs):
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start_time = time.time()
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result = func(*args, **kwargs)
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end_time = time.time()
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print(f"{func.__name__} took {end_time - start_time:.6f} seconds")
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return result
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return wrapper
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def repeating_cycle_fermat(d: int) -> int:
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# 移除因子2和5
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while d % 2 == 0:
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d //= 2
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while d % 5 == 0:
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d //= 5
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if d == 1:
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return 0
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# 费马小定理：对于与10互质的d，10^(φ(d)) ≡ 1 mod d
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# 循环节长度一定是φ(d)的约数
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# 计算欧拉函数φ(d)
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phi = d
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n = d
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# 质因数分解求φ(d)
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p = 2
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while p * p <= n:
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if n % p == 0:
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phi -= phi // p
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while n % p == 0:
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n //= p
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p += 1 if p == 2 else 2 # 从2开始，然后检查奇数
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if n > 1:
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phi -= phi // n
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# 现在找φ的最小约数k，使得10^k ≡ 1 mod d
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min_cycle = phi
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for k in range(1, int(phi**0.5) + 1):
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if phi % k == 0:
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if pow(10, k, d) == 1:
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return k
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other = phi // k
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if pow(10, other, d) == 1 and other < min_cycle:
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min_cycle = other
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return min_cycle
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@timer
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def main_searchall(n: int) -> None:
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max_cycle = 0
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max_d = 0
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for d in range(1, n):
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cycle = repeating_cycle_fermat(d)
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if cycle > max_cycle:
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max_cycle = cycle
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max_d = d
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print(f"{max_d} have {max_cycle} digits in its decimal fraction part.")
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def is_probable_prime(n: int, trials: int = 20) -> 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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@timer
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def main_by_prime(n: int) -> None:
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for d in range(n, 0, -1):
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max_cycle = d - 1
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max_d = d
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cycle = repeating_cycle_fermat(d)
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if max_cycle == cycle:
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print(f"{max_d} have {max_cycle} digits in its decimal fraction part.")
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return None
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if __name__ == "__main__":
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n = 10000
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main_searchall(n)
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main_by_prime(n)
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