I - The kth great number

Description

Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling giddy. Now, try to help Xiao Bao.

 

Input

There are several test cases. For each test case, the first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to output the kth great number.

 

Output

The output consists of one integer representing the largest number of islands that all lie on one line.

 

Sample Input

    
    

     
     8 3
I 1
I 2
I 3
Q
I 5
Q
I 4
Q 
    
    

 

Sample Output

    
    

     
     1
2
3 
    
    

Hint

Xiao Ming won't ask Xiao Bao the kth great number when the number of the written number is smaller than k. (1=<k<=n<=1000000). 
         

#include <iostream>
#include <queue>
#include <cstdio>
#include <vector>
using namespace std;

int main()
{
    priority_queue<int,vector<int>,greater<int> >num;
    int n1,n2,number;
    char s[5];
    while((scanf("%d%d",&n1,&n2))!=EOF)
    {
       while (!num.empty())
        num.pop();
        for(int i = 0;i < n1; i++)
        {
            scanf("%s",&s);
            if(s[0]=='I')
            {
                scanf("%d",&number);
                num.push(number);
                if(num.size()>n2)
                    num.pop();
            }
            else
            {

               cout<<num.top()<<endl;
            }
        }
    }
    return 0;
}

一看就是用优先队列 然后求第K大的数字 这个思路比较奇妙 可以一个一个加进来 因为优先队列都是由小到大排好序的 所以只要加进来数值超过K  那么就把最小的弹出去  一直让它保留K个数  然后最小的  就是我们所求的第K大的数值