G - MZL's simple problem

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)

InputThe first line contains a number NN (N≤106N≤106),representing the number of operations. 
Next NN line ,each line contains one or two numbers,describe one operation. 
The number in this set is not greater than 109109.OutputFor 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

这题TLE了很多次，一直不明白为什么，到后来才发现是因为输出用了cout，这题还要注意，因为这是集合，所以里面的元素是不能重复的

#include <cstdio>
#include <iostream>
#include <set>
using namespace std;
typedef long long ll;
const int inf = 1e6 +10;
set<int>s;
int main()
{
    int num;
    scanf("%d", &num);
    while(num --){
        int way;
        scanf("%d", &way);
        if(way == 1){
            int val;
            scanf("%d", &val);
            s.insert(val);
        }else if(way == 2){
            if(!s.empty()){
               s.erase(s.begin());
            }
        }else {
            if(!s.empty()){
                set<int>::iterator it = s.end();
                it --;
                printf("%d\n", *it);
            }else {
                printf("0\n");
            }
        }
    }
    return 0;
}