ZOJ 3635 Cinema in Akiba（線段樹）

ZOJ 3635 Cinema in Akiba（線段樹）

Cinema in Akiba (CIA) is a small but very popular cinema in Akihabara. Every night the cinema is full of people. The layout of CIA is very interesting, as there is only one row so that every audience can enjoy the wonderful movies without any annoyance by other audiences sitting in front of him/her.

The ticket for CIA is strange, too. There are n seats in CIA and they are numbered from 1 to n in order. Apparently, n tickets will be sold everyday. When buying a ticket, if there are k tickets left, your ticket number will be an integer i (1 ≤ i ≤ k), and you should choose the ith empty seat (not occupied by others) and sit down for the film.

On November, 11th, n geeks go to CIA to celebrate their anual festival. The ticket number of the ith geek is a_i. Can you help them find out their seat numbers?

Input

The input contains multiple test cases. Process to end of file.
The first line of each case is an integer n (1 ≤ n ≤ 50000), the number of geeks as well as the number of seats in CIA. Then follows a line containing n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n - i + 1), as described above. The third line is an integer m (1 ≤ m ≤ 3000), the number of queries, and the next line is m integers, q₁, q₂, ..., q_m (1 ≤ q_i ≤ n), each represents the geek's number and you should help him find his seat.

Output

For each test case, print m integers in a line, seperated by one space. The ith integer is the seat number of the q_ith geek.

Sample Input

3
1 1 1
3
1 2 3
5
2 3 3 2 1
5
2 3 4 5 1

Sample Output

1 2 3
4 5 3 1 2

題意：就是一排人看電影，票上安排給某人的位置是第幾個空位。讓你找出某些人的對應座位號。

線段樹保存空位個數進行處理。

#include
#include
#include
#include
#include
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=50000+100;
int sum[maxn<<2];
int ans[maxn];
int n,m;
void pushup(int rs)
{
    sum[rs]=sum[rs<<1]+sum[rs<<1|1];
}
void build(int rs,int l,int r)
{
    sum[rs]=r-l+1;
    if(l==r)  return ;
    int mid=(l+r)>>1;
    build(rs<<1,l,mid);
    build(rs<<1|1,mid+1,r);
    pushup(rs);
}
int update(int rs,int l,int r,int x)
{
 //    cout<<"2333 "<>1;
     if(x<=sum[rs<<1])  ret=update(rs<<1,l,mid,x);
     else            ret=update(rs<<1|1,mid+1,r,x-sum[rs<<1]);
     pushup(rs);
     return ret;
}
int main()
{
    int x;
    while(~scanf("%d",&n))
    {
        CLEAR(sum,0);
        CLEAR(ans,0);
        build(1,1,n);
        REPF(i,1,n)
        {
            scanf("%d",&x);
            ans[i]=update(1,1,n,x);
        }
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d",&x);
            printf(m==0?"%d\n":"%d ",ans[x]);
        }
    }
    return 0;
}