题解 | A50141.求和

#include <bits/stdc++.h>
using namespace std;
long long n , sum , a[100010];
int main(){
	cin >> n;
    for(int i = 1;i <= n;i++){
        cin >> a[i];
    }
    for(int i = 1;i <= n;i++){
        if(a[i]%5 == 0){ //判断是否为5的倍数
            sum += a[i];
        }
    }
    cout << sum;
    return 0;
}