Problem Description

A simple problem

Problem Description You have a multiple set,and now there are three kinds of operations: 1 x : add number x to set 2 : delete the minimum number (if the set is empty now,then ignore it) 3 : query the maximum number (if the set is empty now,the answer is 0)

Input

The first line contains a number N (N≤106),representing the number of operations.

Next N line ,each line contains one or two numbers,describe one operation. The number in this set is not greater than 109.

Output

For each operation 3,output a line representing the answer.

Sample Input

6

1 2 1 3 3 1 3 1 4 3

Sample Output

3

4

题目大意：给你一个数，表示有几行，然后给你一个a和x，如果a等于1的话，就向里面增加一个数，如果a等于2的话 ，就删除集合中最小的一个数，如果等于a==3 的话，就输出最大的数；

解题思路：STL,集合；

具体见代码：

#include 
     
      #include 
      
       #include 
       
        using namespace std;set
        
         ::iterator it;set
         
           s;int main(){    long long n, a, x;    scanf("%lld",&n);    s.clear();    while(n--)    {        scanf("%lld",&a);        if(a == 1)        {            scanf("%lld",&x);            s.insert(x);        }        else if(a == 2)        {            if(!s.empty())               s.erase(s.begin());        }        else if(a == 3)        {            if(s.empty())                puts("0");            else            {                it=s.end();                it--;                printf("%lld\n",*it);            }        }    }    return 0;}