ZOJ 3635 Cinema in Akiba【树状数组+二分查找】

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

题目大意：如果有k剩票,您的机票号码将整数我(1≤我≤k),你应该选择第i空的座位(不被他人)和坐下来为电影。给出m， m次询问，每次p顾客最后的位置；

思路：树状数组+二分查找； 感觉树状数组做的题目还是太少了，或者没有找到树状数组的真谛，加强学习；
开始的时候都是空位，那么add(i, 1）;统计i位置前面的空座位；对于每次i空位置被占一个，那么i位置之后的每个位置的空位少了一个，那么add(i, -1);
对于找和sum的位置的时候用到二分查找, 超找出和为i的位置，L=1, R=n, 每次计算SUM(mid)进行更新左右点；

代码如下：

#include<string.h>
#include<stdio.h>
#include<stdio.h>
#include<string.h>
#define N 50005 
int c[N], n, place[N]; 
int Lowbit(int t)  
{
    return t&(t^(t-1));  
}
int Sum(int end)   
{
    int sum = 0;
    while(end > 0)
    {
        sum += c[end];
        end -= Lowbit(end);
    }
    return sum;
}
void add(int li, int val) 
{
    while(li<=n)
    {
        c[li] += val;
        li += Lowbit(li);
    }
} 
int main()
{
    int m, i, j, q, a;
    while(scanf("%d", &n)!=EOF)
    {
        memset(place, 0, sizeof(place));
        memset(c, 0, sizeof(c)); 
        for(i=1; i<=n; i++)
            add(i, 1);
        for(i=1; i<=n; i++)
        {
            scanf("%d", &a);
            int l=1, r=n;
            while(l<r)
            {
                int mid=(l+r)/2;
                if(Sum(mid)>=a)
                    r=mid;
                else
                    l=mid+1;
            }
            place[i]=l;
            add(l, -1);
        }
        scanf("%d", &m);
        for(i=1; i<=m; i++)
        {
            scanf("%d", &q); 
            if(i!=m)
                printf("%d ", place[q]);
            else
                printf("%d\n", place[q]);
        }
    }
}