Project Euler Problem 1 - Multiples of 3 and 5

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.

对给定的max和k，小于max的k的倍数的个数为

　　c = (max - 1) / k

值分别为

　　k, 2k, ... , ck,

则总和为

　　k(1 + 2 + ... + c) = k(1+c)c/2

Python代码：

max = 1000 - 1
a = 3; b = 5
c = a * b

def f(k):
    return (1 + k) * k / 2

res = f(max / a) * a + f(max / b) * b - f(max / c) * c

print res

时间复杂度O(1)

空间复杂度O(1)