Black Box

Black Box

Time Limit: 1000MS
Memory Limit: 10000KTotal Submissions: 7301
Accepted: 2976

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 43 1 -4 2 8 -1000 21 2 6 6

Sample Output

3312

题意为找第几个所给数中的第几小的数

例如：第一个输入的6就是指找3,1,-4,2，8，-1000这六个数中的第三小，即1

<span style="font-size:18px;">#include <iostream>#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>#include<queue>#include<set>#include<string>using namespace std;int main(){    priority_queue<int,vector<int>,greater<int> >p1;    priority_queue<int>p2;    int n,m,i,j,x,y,t;    int a[30001];    scanf("%d%d",&n,&m);    for(i=0;i<n;i++)     scanf("%d",&a[i]);     y=0;     for(i=0;i<m;i++)     {         scanf("%d",&x);         while(y<x)         {            p1.push(a[y]);            if(!p2.empty()&&p1.top()<p2.top())            {                t=p1.top();                p1.pop();                p1.push(p2.top());                p2.pop();                p2.push(t);            }            y++;         }         printf("%d\n",p1.top());         p2.push(p1.top());         p1.pop();     }}</span>