✨ 新增欧拉第34、35题解法及说明
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solutions/0035.CircularPrimes/euler_35.py
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206
solutions/0035.CircularPrimes/euler_35.py
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"""
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The number, 179, is called a circular prime because all rotations of the digits: 179, 791, and 917,
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are themselves prime.
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There are thirteen such primes below 100 : 2,3,5,7,11,13,17,31,37,71,73,79, and 97.
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How many circular primes are there below one million?
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"""
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import random
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import time
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from itertools import permutations
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from sqlite3.dbapi2 import Time
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from typing import Union
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# 预计算的小素数集合，用于快速排除
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_SMALL_PRIMES = frozenset((2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37))
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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__} time: {end_time - start_time:.6f} seconds")
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return result
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return wrapper
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def sieve_best(n: int) -> list[int]:
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"""返回 [2..n) 内所有素数"""
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if n <= 2:
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return []
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size = (n - 1) // 2
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comp = bytearray(size)
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limit = int(n**0.5) // 2
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for i in range(1, limit + 1):
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if not comp[i]:
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p = 2 * i + 1
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s = (p * p - 1) // 2
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comp[s::p] = b"\1" * ((size - s - 1) // p + 1)
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res = [2, *(2 * i + 1 for i, v in enumerate(comp) if not v)]
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res.sort()
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return [i for i in res if i != 1]
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def is_circular(n: int, primes: list[int]) -> bool:
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if n < 2:
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return False
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if n in [2, 3, 5, 7, 11]:
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return True
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digits = str(n)
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allnum = [digits[i:] + digits[:i] for i in range(len(digits))]
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for num in allnum:
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if int(num) not in primes:
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return False
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return True
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def is_probable_prime(n: Union[int, str], k: int = 20) -> bool:
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"""
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Miller-Rabin素性检测算法
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参数:
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n: 待检测的整数（支持int或数字字符串）
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k: 测试次数，误差率约为4^(-k)，默认20次（误差率约9e-13）
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返回:
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bool: 如果n很可能是素数返回True，否则返回False
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"""
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# 支持字符串输入
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if isinstance(n, str):
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n = int(n)
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# 基本检查
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if n < 2:
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return False
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if n in _SMALL_PRIMES:
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return True
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# 检查小素数的倍数
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for p in _SMALL_PRIMES:
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if n % p == 0:
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return False
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# 特殊情况：偶数（已排除2）
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if n % 2 == 0:
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return False
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# 将 n-1 写成 2^s * d 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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# 边界：如果 n-1 可以直接作为基数，避免随机数生成问题
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if n < 4:
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return n in (2, 3)
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# Miller-Rabin测试
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for _ in range(k):
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# 生成随机基数，范围 [2, n-2]
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a = random.randrange(2, n - 1)
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# 计算 x = a^d mod n
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x = pow(a, d, n)
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# 如果 x == 1 或 x == n-1，通过本轮测试
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if x == 1 or x == n - 1:
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continue
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# 重复平方检查
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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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# 所有平方后都没有得到 n-1，n是合数
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return False
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return True
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# 优化的确定素数版本（适用于小整数）
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def is_prime_small(n: int) -> bool:
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"""确定性素数检测，适用于n < 2^64"""
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if n < 2:
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return False
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if n % 2 == 0:
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return n == 2
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if n % 3 == 0:
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return n == 3
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# 检查小素数
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for p in _SMALL_PRIMES:
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if n % p == 0:
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return n == p
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# 6k±1 优化检查
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i = 5
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w = 2
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while i * i <= n:
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if n % i == 0:
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return False
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i += w
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w = 6 - w # 在2和4之间切换：5,7,11,13,17,19...
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return True
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# 智能选择函数
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def is_prime(n: Union[int, str], k: int = 20) -> bool:
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"""
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智能素数检测：对小整数使用确定性算法，对大整数使用Miller-Rabin
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参数:
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n: 待检测的整数
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k: Miller-Rabin测试次数（仅对大整数有效）
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返回:
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bool: 如果是素数返回True
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"""
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if isinstance(n, str):
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n = int(n)
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# 小整数使用确定性算法
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if n < 2**64:
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return is_prime_small(n)
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# 大整数使用Miller-Rabin
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return is_probable_prime(n, k)
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@timer
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def main_quick(max: int) -> None:
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res = set()
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for i in range(2, max):
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if is_prime(i):
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if i in res:
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continue
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else:
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s = str(i)
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allnums = [s[i:] + s[:i] for i in range(len(s))]
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flag = 0
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for j in allnums:
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num = int("".join(j))
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if is_prime(num):
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flag += 1
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if flag == len(allnums):
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res.update(allnums)
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print(len(res))
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@timer
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def main(n: int) -> None:
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primes = sieve_best(n)
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circular_primes = [p for p in primes if is_circular(p, primes)]
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print(len(circular_primes))
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if __name__ == "__main__":
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num = 1000000
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main(num)
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main_quick(num)
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80
solutions/0035.CircularPrimes/euler_35_better.py
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80
solutions/0035.CircularPrimes/euler_35_better.py
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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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# ---------- 1. PRIMES ----------
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MAX = 1_000_000
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is_p = bytearray(b"\1") * MAX # 1 表示素数，0 表示合数（与原来相反，方便理解）
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is_p[0] = is_p[1] = 0 # 0和1不是素数
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# 优化：使用标准埃拉托斯特尼筛法
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for p in range(2, int(MAX**0.5) + 1):
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if is_p[p]:
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is_p[p * p : MAX : p] = b"\0" * ((MAX - p * p - 1) // p + 1)
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# ---------- 2. SCROLLING ----------
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def rotations(n: int) -> list[int]:
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"""返回 n 的所有旋转（数学移位版，无字符串开销）"""
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if n < 10:
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return [n]
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s = str(n)
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k = len(s)
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res = []
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for i in range(k):
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# 使用切片旋转，避免重复计算
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rotated = int(s[i:] + s[:i])
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res.append(rotated)
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return res
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# ---------- 3. MAIN ----------
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@timer
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def main(limit: int = MAX) -> int:
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# 预过滤：只要含 0,2,4,5,6,8 就可以跳过（除了2和5本身）
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bad_digits = set("024568")
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total = 0
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counted = set() # 用于去重，避免重复计数
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for p in range(2, limit):
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if is_p[p]: # p是素数
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# 快速数字过滤（2和5是特例）
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if p != 2 and p != 5:
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s = str(p)
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if not bad_digits.isdisjoint(s):
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continue
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# 如果已经统计过，跳过
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if p in counted:
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continue
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# 检查所有旋转
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rs = rotations(p)
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# 检查所有旋转数是否都是素数
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valid = True
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for r in rs:
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if r >= limit or not is_p[r]:
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valid = False
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break
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if valid:
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# 如果是循环素数，统计并标记已计数
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total += len(set(rs)) # 使用set避免重复旋转
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counted.update(rs)
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return total
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if __name__ == "__main__":
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print(f"循环素数数量: {main()}")
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6
solutions/0035.CircularPrimes/readme.md
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6
solutions/0035.CircularPrimes/readme.md
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# 判断素数
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这个问题也很有和判断素数，以及如何更好的记录判断结果。
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当然，如何更快计算出滚动的数字也很重要。
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个人感觉这是个算法问题更多的问题，数学占比并不多。
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