作者:mobiledu2502886633 | 来源:互联网 | 2023-10-13 03:18
Ifwelistallthenaturalnumbersbelow10thataremultiplesof3or5,weget3,5,6and9.Thesumofthesemu
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.
题目大意:
10以下的自然数中,属于3和5的倍数的有3,5,6和9,它们之和是23.
找出1000以下的自然数中,属于3和5的倍数的数字之和。
#include
#include <string.h>
#include
void solve()
{
int sum,i;
sum=0;
for(i=3; i<1000; i++)
{
if(i%3==0 || i%5==0)
{
sum+=i;
}
}
printf("%d\n",sum);
}
int main()
{
solve();
return 0;
}