135 lines
3.3 KiB
Python
135 lines
3.3 KiB
Python
import math
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import time
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from functools import lru_cache
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from typing import Callable
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def timer(func: Callable) -> Callable:
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def wrapper(*args, **kwargs):
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start_time = time.perf_counter()
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result = func(*args, **kwargs)
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end_time = time.perf_counter()
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elapsed = end_time - start_time
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print(f"Function {func.__name__} execution time: {elapsed:.4f} seconds")
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return result
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return wrapper
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def maxpower(a: int, n: int) -> int:
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"""计算最大的整数c，使得a^c ≤ n"""
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res = int(math.log(n) / math.log(a))
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# 处理边界情况
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if pow(a, res + 1) <= n:
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res += 1
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if pow(a, res) > n:
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res -= 1
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return res
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@lru_cache(maxsize=None)
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def lcm(a: int, b: int) -> int:
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"""计算最小公倍数（使用缓存优化）"""
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gcd_val = math.gcd(a, b)
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return a // gcd_val * b
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def recurse(
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lc: int, index: int, sign: int, left: int, right: int, thelist: list[int]
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) -> int:
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"""容斥原理的递归实现"""
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if lc > right:
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return 0
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res = sign * (right // lc - (left - 1) // lc)
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# 递归处理剩余元素
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for i in range(index + 1, len(thelist)):
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res += recurse(lcm(lc, thelist[i]), i, -sign, left, right, thelist)
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return res
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def dd(left: int, right: int, a: int, b: int, check: list[bool]) -> int:
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"""双层容斥计算"""
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res = right // b - (left - 1) // b
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thelist = [i for i in range(a, b) if check[i]]
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for i in range(len(thelist)):
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res -= recurse(lcm(b, thelist[i]), i, 1, left, right, thelist)
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return res
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def compute_counts(n: int) -> tuple[list[int], int, int]:
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"""计算前缀和数组"""
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sqn = int(math.isqrt(n)) # 使用isqrt替代sqrt，返回整数
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maxc = maxpower(2, n)
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# 初始化数组
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counts = [0] * (maxc + 1)
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counts[1] = n - 1
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# 主计算循环
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for c in range(2, maxc + 1):
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check = [True] * (maxc + 1)
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umin = (c - 1) * n + 1
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umax = c * n
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# 优化筛法：使用步长跳过非倍数
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for i in range(c, maxc // 2 + 1):
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check[i * 2 : maxc + 1 : i] = [False] * ((maxc - i * 2) // i + 1)
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# 只处理质数（check[f]为True）
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for f in range(c, maxc + 1):
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if check[f]:
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counts[f] += dd(umin, umax, c, f, check)
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# 计算前缀和
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for c in range(2, maxc + 1):
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counts[c] += counts[c - 1]
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return counts, sqn, maxc
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def compute_final_answer(n: int) -> int:
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"""计算最终答案"""
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counts, sqn, _ = compute_counts(n)
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ans = 0
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coll = 0
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used = [False] * (sqn + 1)
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# 统计答案
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for i in range(2, sqn + 1):
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if not used[i]:
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c = maxpower(i, n)
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ans += counts[c]
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u = i
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for j in range(2, c + 1):
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u *= i
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if u <= sqn:
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used[u] = True
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else:
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coll += c - j + 1
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break
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# 最终调整
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ans += (n - sqn) * (n - 1)
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ans -= coll * (n - 1)
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return ans
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@timer
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def main(n: int = 10**6) -> None:
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answer = compute_final_answer(n)
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print(f"n = {n}, Answer = {answer}")
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if __name__ == "__main__":
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main(10**7)
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