题解 | #【模板】前缀和#

解题思路：

• 直接存储前n项的和。对于查询$q_{[ltor]}=su{m}_{[0tor]}-su{m}_{[0tol-1]}q\_\{[l\backslash \; to\backslash \; r]\}\; =\; sum\_\{[0\backslash \; to\backslash \; r]\}\; -\; sum\_\{[0\backslash \; to\backslash \; l-1]\}$,时间复杂度O(n+q)，空间复杂度O(n)

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n, q;
    cin>>n>>q;
    vector<long long>v(n+1);
    long long  sum = 0;
    for(int i = 1; i <= n; ++i){
        scanf("%lld", &v[i]);
        sum += v[i];
        v[i] = sum;
    }
    for(int i = 0; i < q; ++i){
        int start, end;
        cin>>start>>end;
        cout<<v[end] - (start == 1? 0: v[start-1])<<endl;
    }
    return 0;
}