✨ feat(0029.DistinctPowers)：添加欧拉项目第29题解决方案
📝 docs(0029.DistinctPowers)：添加问题描述和解题思路文档 ✨ feat(euler_29.py)：实现基础解决方案和高效算法 ✨ feat(euler_29_best.py)：实现基于容斥原理的最优解决方案
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solutions/0029.DistinctPowers/README.md
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solutions/0029.DistinctPowers/README.md
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# Distinct Powers
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只需要比较所有可能重复的底和幂，找到有多少这样的a^b就能知道有多少重复。
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这个逻辑最为简单，我自己的实现免费处理较大额的底和幂的情况，这点我暂时没想到好方法。
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当看到 [WP(Page 5)](https://projecteuler.net/post_id=92910) 的方法，我才明白自己的问题在哪。
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这类问题真的是，单纯解出来不算什么，如何使用数学更简单更快捷的解出来，才是难的。
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核心关键点是组合数学的[容斥原理](https://zh.wikipedia.org/wiki/%E6%8E%92%E5%AE%B9%E5%8E%9F%E7%90%86)。
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因为幂的数学特点，可能需要多次应用容斥原理，以确保不重复计算。这也是我自己方法和WP方法的差距所在。
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solutions/0029.DistinctPowers/euler_29.py
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solutions/0029.DistinctPowers/euler_29.py
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"""
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Consider all integer combinations of a^b for 2<=a<=5 and 2<=b<=5 :
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2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32
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3^2 = 9 3^3 = 27 3^4 = 81 3^5 = 243
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4^2 = 16 4^3 = 64 4^4 = 256 4^5 = 1024
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5^2 = 25 5^3 = 125 5^4 = 625 5^5 = 3125
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If they are then placed in numerical order, with any repeats removed,
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we get the following sequence of 15 distinct terms:
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4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
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How many distinct terms are in the sequence generated by a^b for 2<=a<=100 and 2<=b<=100 ?
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"""
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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__} runtime: {end_time - start_time} seconds")
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return result
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return wrapper
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def count_distinct_terms_efficient(limit_a, limit_b):
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# 预计算每个 a 的最小底数表示
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base_map = {}
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power_map = {}
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for a in range(2, limit_a + 1):
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base_map[a] = a
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power_map[a] = 1
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# 找出所有可以表示为幂的数
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for power in range(2, 7): # 2^7=128>100，所以最多到6次幂
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base = 2
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while base**power <= limit_a:
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value = base**power
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# 如果这个值还没有被更小的底数表示
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if base_map[value] == value:
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base_map[value] = base
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power_map[value] = power
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base += 1
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# 使用集合存储 (base, exponent) 对
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seen = set()
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for a in range(2, limit_a + 1):
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base = base_map[a]
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power = power_map[a]
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# 如果 a 是某个数的幂，我们需要检查 base 是否还能分解
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# 例如：16 = 4^2，但 4 = 2^2，所以 16 = 2^4
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# 我们需要找到最终的 base
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while base_map[base] != base:
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power = power * power_map[base]
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base = base_map[base]
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for b in range(2, limit_b + 1):
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seen.add((base, power * b))
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return len(seen)
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def do_compute(a_top: int, b_top: int, a_down: int = 2, b_down: int = 2) -> int:
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tmp = set()
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for a in range(a_down, a_top + 1):
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for b in range(b_down, b_top + 1):
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tmp.add(a**b)
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return len(tmp)
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@timer
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def main():
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"""not bad"""
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print(count_distinct_terms_efficient(100, 100))
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@timer
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def main2():
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print(do_compute(100, 100))
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if __name__ == "__main__":
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main()
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main2()
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solutions/0029.DistinctPowers/euler_29_best.py
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solutions/0029.DistinctPowers/euler_29_best.py
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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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