The kth great number

The kth great number

Time Limit: 1000 MS Memory Limit: 65536 K Total Submit: 90(27 users) Total Accepted: 55(27 users) Rating: Special Judge: No

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

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

本题考查对priority_queue的使用

priority_queue 对于基本类型的使用方法相对简单。他的模板声明带有三个参数:
priority_queue<Type, Container, Functional>
其中Type 为数据类型， Container 为保存数据的容器，Functional 为元素比较方式。
Container 必须是用数组实现的容器，比如 vector, deque 但不能用 list.
STL里面默认用的是 vector. 比较方式默认用 operator< , 所以如果你把后面俩个参数缺省的话，
优先队列就是大顶堆，队头元素最大。如果要用到小顶堆，则一般要把模板的三个参数都带进去。
STL里面定义了一个仿函数 greater<>，对于基本类型可以用这个仿函数声明小顶堆

以上这段话来自大神博客：http://www.cnblogs.com/flyoung2008/articles/2136485.html

代码：

#include<cstdio>
#include<cmath>
#include<queue>
#include<iostream>
using namespace std;
priority_queue<int> a,b;
int main()
{
    int h,n,k,x;
    char m;
    while(~scanf("%d%d",&n,&k))
    {
        while(!a.empty())
        {
            a.pop();
        }
        while(n--)
        {
            getchar();
            scanf("%c",&m);
            if(m==‘I‘)
            {
                scanf("%d",&x);
                a.push(x);
            }
            else if(m==‘Q‘)
            {
                h=k-1;
                b=a;
                while(h--)
                {
                    b.pop();
                }
                printf("%d\n",b.top());
            //    getchar();
            //    printf("%d\n",a.top());
            }
        }
    }
    return 0;
}