ZOJ 3635 Cinema in Akiba（线段树）

最新推荐文章于 2018-05-09 16:13:00 发布

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最新推荐文章于 2018-05-09 16:13:00 发布

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原文链接：http://www.cnblogs.com/yxwkf/p/5092113.html

本文介绍了一个小而受欢迎的电影院Cinema in Akiba (CIA) 的独特布局和购票方式。每个座位都有编号，并且每天售出n张票。当购买一张票时，票上会标记出当前剩余座位的第k个空位。本文详细解释了如何通过线段树数据结构解决此问题，并提供了代码实现。具体地，文章展示了如何通过输入参数计算出特定座位号，以帮助观众快速找到自己的座位。

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

Sample Output

1 2 3
4 5 3 1 2

题意：就是一排人看电影，票上安排给某人的位置是第几个空位。让你找出某些人的相应座位号。

线段树保存空位个数进行处理。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
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 "<<l<<" "<<r<<endl;
     if(l==r)
     {
         sum[rs]=0;
         return l;
     }
     int ret;
     int mid=(l+r)>>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;
}

转载于:https://www.cnblogs.com/yxwkf/p/5092113.html