01 ：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
注意：由于能被15整除的數(shù)都能被3和5整除冲秽，所以應(yīng)該分別求出能被3和5整除的數(shù)的和再減去能被15整除的和赌髓。
clear; clc
n =[3, 5];
m1 = fix(999/n(1));
m2 = fix(999/n(2));
m3 = fix(999/(n(1)*n(2)));
sum1 = 0; sum2 = 0; sum3 = 0;
for i = 1:1:m1
? ? sum1 = sum1 + n(1)*i;
end
for j = 1:1:m2
? ? sum2 = sum2 + n(2)*j;
end
for k = 1:1:m3
? ? sum3 = sum3 + (n(1)*n(2))*k;
end
sum = sum1 + sum2 - sum3
