poj1442 Black Box

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10378		Accepted: 4276

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

这个题的大意就是给你两个数列我们用a,b表示，第一个数列表示你要加入黑盒的值，然后第二个数列表示在插入b[i]个数时，第i小的数是几，比如说示例上：

第一个数是1也就是插入一个数（3）时，第i（1）小的数是3.

第二个数是2也就是插入两个数（3,1）时第i（2）小的数是3.

第三个数是6也就是插入六个数（3,1,-4,2,8,-1000)第i（3）小的数是1.

然后就是想一想怎么做。单纯输出第几大可能不难，但这是实时更新的数列，所以我们用优先队列来更新我们的动态数组，但光用一个肯定是不行的，因为优先队列输出头顶元素好用，每次重复压入，出列，肯定是吃不消的。这里我们可以考虑用两个优先队列，来把这个动态的数组分割开，用一个优先队列来存储前i小的数，它不是要求每次输出第i个小的数么，我们暂且称之为in队列，表示前i小的数，这样我们把一个q队列分割成两个，in表示前i小的数，out表示后面大的数，那么，我们每次只需要输出out的顶元素就可以了，输出之后把out的顶元素压入in队列中。在压入之前我们要考虑确保in队列中的每个元素都是比out队列中元素要小的，不然肯定是错的。

利用in队列的最大值比out队列中的最小值小，便可以确保in队列中的每个值都比out队列中的全部的值要小了。

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