第二条题解

#include<stdio.h>
signed main(int argc,char** argv){
    int n;
    scanf("%d\n",&n);
    int a[100010]{};
    for(int i=1/*i建议为1*/;i<=n;i++)scanf("%d ",&a[i]);
    long long pre[100010]{};
    for(int i=1/*i建议为1*/;i<=n;i++)pre[i]=pre[i-1]+a[i]; //前缀和：pre[i]=pre[i-1]+a[i]
    int m;
    scanf("\n%d\n",&m);
    while(m--){
        int l,r;
        scanf("%d %d\n",&l,&r);
        printf("%lld\n",pre[r]-pre[l-1]);
    }
}