题面见COGS 分析：这道题，又双叒叕是区间操作，果断线段树啊。刚刚跟某人@dch探(si)讨(bi)关于线段树大法好还是二叉索引树大法好的问题，肯定线段树大法好啦！！！ 话说二叉索引树是什么鬼啦，太难了，%dch，%jjq。 线段树大法好！！！

#include 
#include 
#define LL long long
using namespace std;
const int MX = 10005;
int n, m;
int num[MX];
struct node{
    int le, ri;
    LL sum;
}tree[MX*4];
inline int leftt(int a)
{
    return a<<1;
}
inline int rightt(int a)
{
    return (a<>1;
}
void biu(int cur,int l,int r)
{//建树
    if(l==r)
    {
        tree[cur].sum=num[l];
        tree[cur].le=tree[cur].ri=l;
        return;
    }
    else
    {
        biu(leftt(cur),l,mid(l,r));
        biu(rightt(cur),mid(l,r)+1,r);
        tree[cur].sum=tree[leftt(cur)].sum+tree[rightt(cur)].sum;
        tree[cur].le=l;
        tree[cur].ri=r;
    }
    return;
}
inline LL ask(int cur, int l, int r)
{//查询
    if(l==tree[cur].le&&r==tree[cur].ri)
    {
        return tree[cur].sum;
    }
    else
    {
        if(r=tree[rightt(cur)].le)
            return ask(rightt(cur),l,r);
        else
            return ask(leftt(cur),l,tree[leftt(cur)].ri)+ask(rightt(cur),tree[rightt(cur)].le,r);
    }
}
inline int get_num()
{
    int ans=0;
    char tmp;
    tmp=getchar();
    while(tmp'9')   tmp=getchar();
    while(tmp='0')
    {
        ans=ans*10+tmp-48;
        tmp=getchar();
    }
    return ans;
}
int main()
{
    //freopen("sum.in","r",stdin);
    //freopen("sum.out","w",stdout);
    int x, y;
    n=get_num();
    LL tmp=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&num[i]);
    }
    biu(1,1,n);
    m=get_num();
    for(int i=1;i<=m;i++)
    {
        x=get_num();
        y=get_num();
        tmp=ask(1,x,y);
        cout<<tmp<<"\n";
    }
    return 0;
}