题解 | A92265.数列求和

#include <bits/stdc++.h>
using namespace std;
long long n , a[200010] , sum; //sum为后面求和做准备
int main(){
    cin >> n;
    for(int i = 1;i <= n;i++){
        cin >> a[i];
    }
    for(int i = 1;i <= n;i++){
        sum += a[i]; //求和 
    }
    cout << sum;
    return 0;
}