Black Box《优先队列》

Description

Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box; GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1

N Transaction i Black Box contents after transaction Answer 
       (elements are arranged by non-descending)   
 1 ADD(3)      0 3   
 2 GET         1 3                                    3 
 3 ADD(1)      1 1, 3   
 4 GET         2 1, 3                                 3 
 5 ADD(-4)     2 -4, 1, 3   
 6 ADD(2)      2 -4, 1, 2, 3   
 7 ADD(8)      2 -4, 1, 2, 3, 8   
 8 ADD(-1000)  2 -1000, -4, 1, 2, 3, 8   
 9 GET         3 -1000, -4, 1, 2, 3, 8                1 
 10 GET        4 -1000, -4, 1, 2, 3, 8                2 
 11 ADD(2)     4 -1000, -4, 1, 2, 2, 3, 8   

It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.

Input

Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.

Output

Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.

Sample Input

7 4
3 1 -4 2 8 -1000 2
1 2 6 6

Sample Output

3
3
1
2

这题的题意：7 是7个数；4是指分4次输入
；第一次·输入1个，并取出第一小的，
第二次输入2个（总共输入两个，要加上第一次输入的不分），取第二小的
第三次输入6个 取第3小的&…………

要是按照输入输出那样一点点的执行，会超时；
用另个优先队列，一个小的在前，一个大的在前，下面我上代码，自己模拟一下看看；

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
int main()
{
    priority_queue<int,vector<int>,greater<int> >qb;
    priority_queue<int,vector<int>,less<int> >qa;
    int m,n,a[30005],b,i,j;
    scanf("%d %d",&m,&n);
    for(i=0; i<m; i++)
        scanf("%d",&a[i]);
    j=0;
    for(i=0; i<n; i++)
    {
        scanf("%d",&b);
        while(j<b)
            qa.push(a[j++]);//压入大的在前的队列
        while(qa.size()>i)
        {
            qb.push(qa.top());
            qa.pop();
        }
        printf("%d\n",qb.top());
        qa.push(qb.top());//经过下面两部操作，大的在前的队列中的数不会影响接下来的取值
        qb.pop();
    }
    return 0;
}