杭电4006（multiset的应用） 之 The kth great number

Problem 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=

题目大意：

I表示记下后面的那个数，Q则表示输出当前记下的第k大的数

分析：因为元素允许相同，所以用multiset存储，注意n最大为1e6,直接输出倒数第三个数会WA，正确的做法是删除多余的数

错误代码：

#include 
#include 
#include 
#include 
using namespace std;

multiset  s;

int main()
{
    int n,k,num;
    char ch;
    while(scanf("%d %d",&n,&k)==2)
    {
        while(!s.empty()) s.clear();
        getchar();
        for(int i=0;i>num;
                getchar();
                s.insert(num);
            }
            else if(ch=='Q')
            {
                multiset:: iterator p;
                p=s.end();
                p--;
                p--;
                p--;
                cout<<*p<

AC代码如下：

#include 
#include 
#include 
#include 
using namespace std;

multiset  s;

int main()
{
    int n,k,num;
    char ch;
    while(scanf("%d %d",&n,&k)==2)
    {
        while(!s.empty()) s.clear();
        getchar();
        for(int i=0;i>num;
                getchar();
                s.insert(num);
                if(s.size()>k) s.erase(s.begin());
            }
            else if(ch=='Q')
            {
                multiset:: iterator p;
                p=s.begin();
                cout<<*p<