81 lines
2.2 KiB
Python
81 lines
2.2 KiB
Python
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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