【笔记】欧拉计划第一题

Q:If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

A:(python)

x, y, max = map(int, input('what is your arguments?\n').split())
#input().split()   map(int, input().split())   input_list = map(int, input().split())
#import sys   lines = sys.stdin.;read().splitlines()
def sum_of_multiples(x, y, max):
    sum = 0
    if x <= 0 or y <= 0 or max < min(x, y):
        print('input error!')
        return 0
    if max > x * y:
        for i in range(1, x*y+1):
            if i % x == 0 or i % y == 0:
                sum  = sum + i
        n = int(max / (x * y))
        sum = n * (n - 1) / 2 * x * y + n * sum
        for i in range(n*x*y+1, max+1):
            sum = sum + i
        return int(sum)
    else:
        for i in range(1, max+1):
            if i % x == 0 or i % y == 0:
                sum  = sum + i
        return sum
print(sum_of_multiples(x, y, max))

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转载自blog.csdn.net/xxxxfengheheda/article/details/78152751