这个成段的编写复杂度很低，不需要加大空间复杂度，便于处理成段加，询问每个位置的值的操作：
#include <bits/stdc++.h>
using namespace std ;
typedef long long ll ;
typedef long double ld ;
typedef unsigned long long ull ;
#ifdef _WIN32
#define LLD "%I64d"
#else
#define LLD "%lld"
#endif
#define pi (acos(-1.0))
#define F first
#define S second
#define lson (o<<1),l,mid
#define rson (o<<1|1),mid+1,r
#define MP make_pair
const double eps = 1e-9 ;
const int inf = 0x3f3f3f3f ;
const ll INF = (ll)4e18 ;const int M = (int)1e5+10 ;
struct BIT {ll u[M] ;int n ;void clr(int _n) { n = _n ;memset (u,0,sizeof(int)*(n+1)) ;}void add (int x , ll v) {for (; x<=n ; x += x&-x) u[x] += v ;}void add (int l , int r , ll v) {add(l,v) ; add(r+1,-v) ;}ll get (int x , ll ret=0) {for (; x>0 ; x -= x&-x) ret += u[x] ;return ret ;}
}bit ;int main () {int a[20] = {0,3,7,4,8,9,11,5,4,9,1} ;for (int i=1 ; i<=10 ; i++) printf ("%-4d" , a[i]) ; puts ("");bit.clr (10) ;for (int i=10 ; i>0 ; i--) {a[i] = a[i]-a[i-1] ;bit.add(i,i,a[i]) ;}for (int i=1 ; i<=10 ; i++) printf ("%-4d",a[i]) ; puts ("");for (int i=1 ; i<=10 ; i++) printf ("%-4d" , bit.u[i]) ; puts ("") ;int Q ;scanf ("%d" , &Q) ;while (Q --) {int x ;scanf ("%d" , &x) ;printf ("%d\n" , bit.get(x) ) ;}return 0 ;
}
