Sidney Zhang
266544ac47
📝 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
93 lines
2.5 KiB
Python
93 lines
2.5 KiB
Python
"""
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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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