97 lines
2.6 KiB
Python
97 lines
2.6 KiB
Python
"""
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Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
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If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.
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For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, and 110; therefore d(220) = 284.
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The proper divisors of 284 are 1, 2, 4, 71, and 142; so d(284) = 220.
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Evaluate the sum of all the amicable numbers under 10000.
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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__} took {end_time - start_time:.8f} seconds")
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return result
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return wrapper
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def sieve_primes(limit):
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"""埃拉托斯特尼筛法生成质数列表"""
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is_prime = [True] * (limit + 1)
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is_prime[0:2] = [False, False]
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primes = []
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for i in range(2, limit + 1):
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if is_prime[i]:
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primes.append(i)
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for j in range(i * i, limit + 1, i):
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is_prime[j] = False
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return primes
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def prime_factors_with_sieve(n, primes):
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"""使用预计算的质数列表进行分解"""
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factors = {}
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temp = n
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for p in primes:
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if p * p > temp:
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break
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while temp % p == 0:
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factors[p] = factors.get(p, 0) + 1
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temp //= p
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if temp > 1:
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factors[temp] = factors.get(temp, 0) + 1
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return factors
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def get_divisors_optimized(n):
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"""优化版本：预计算质数加速分解"""
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if n < 2:
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return [1] if n == 1 else []
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# 计算需要的质数范围
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limit = int(n**0.5) + 1
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primes = sieve_primes(limit)
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# 质因数分解
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factors = prime_factors_with_sieve(n, primes)
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# 生成除数
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divisors = [1]
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for p, exp in factors.items():
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powers = [p**e for e in range(exp + 1)]
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divisors = [a * b for a in divisors for b in powers]
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return sorted(divisors)
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def sum_of_divisors(n):
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"""计算一个数的所有真因子之和"""
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return sum(get_divisors_optimized(n)) - n
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def find_amicable_numbers(limit):
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"""查找小于给定上限的全部亲和数对"""
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amicable_pairs = set()
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for a in range(2, limit):
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b = sum_of_divisors(a)
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if a != b and sum_of_divisors(b) == a:
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amicable_pairs.add((min(a, b), max(a, b)))
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return amicable_pairs
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@timer
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def sum_of_amicable_numbers(limit):
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"""计算小于给定上限的全部亲和数之和"""
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return sum(a + b for a, b in find_amicable_numbers(limit))
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if __name__ == "__main__":
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print(sum_of_amicable_numbers(10000))
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