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  • Problem 21

    Problem 21

    https://projecteuler.net/problem=21

    Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).

    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.

    如果a的因子之和等于b,b的因子之和等于a,并且a不等于b,那么a和b称为亲和数。

    For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

    Evaluate the sum of all the amicable numbers under 10000.

    计算10000以下的所有亲和数之和。

    import time
    
    
    def is_amicable_number(num1, num2, divisors_num1, divisors_num2):
        if sum(divisors_num1) != num2:
            return False
        if sum(divisors_num2) != num1:
            return False
        return True
    
    
    def find_divisors(num):
        divisors = []
        for i in range(1, num//2+1):
            if num % i == 0:
                divisors.append(i)
        return divisors
    
    
    start_time = time.ctime()
    amicable_numbers = []
    for x in range(1, 10001):
        x_divisors = find_divisors(x)
        for y in range(x//2, x):
            y_divisors = find_divisors(y)
            print('Searching for amicable numbers({x}?{y})...'.format(x=x, y=y))
            if is_amicable_number(x, y, x_divisors, y_divisors):
                print('Amicable numbers:', x, ':', y)
                amicable_numbers.append([x, y])
    end_time = time.ctime()
    print(start_time)
    print(end_time)
    print(amicable_numbers)
    tot = 0
    for i in amicable_numbers:
        tot += sum(i)
    print(tot)

    结果:

    10000以下的亲和数:[[284, 220], [1210, 1184], [2924, 2620], [5564, 5020], [6368, 6232]]
    31626

    Resistance is Futile!
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  • 原文地址:https://www.cnblogs.com/noonjuan/p/11031349.html
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