2015 多校 #5 1007 MZL's simple problem

MZL's simple problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2541    Accepted Submission(s): 694

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

 

水题。

模拟就好。

话说puts() 不是挺快的。。。

结果还tle了一次。

然后改成 printf 就A了。。。

/*************************************************************************
    > File Name: code/multi/#5/1007.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月04日 星期二 12时39分42秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=1E6+7;
multiset<int>se;
multiset<int>::iterator it;
int main()
{
    int n;
    int cmd;
    int x;
    scanf("%d",&n);
    for ( int i = 0 ; i < n;  i++)
    {
    scanf("%d",&cmd);
    if (cmd==1)
    {
        scanf("%d",&x);
        se.insert(x);
    }
    if (cmd==2&&!se.empty())
    {
        se.erase(se.begin());
    }
    if (cmd==3)
    {
        if (se.empty())
        {
        printf("0
");
        }
        else
        {
        it = se.end();
        it--;
        printf("%d
",*it);
        }
    }
    }
    return 0;
}