其他-HDOJ-5349-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): 2000 Accepted Submission(s): 781

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

不要用数组存储，及时处理就行，水题。

//
// main.cpp
// 150804-1007
//
// Created by 袁子涵 on 15/8/4.
// Copyright (c) 2015年 袁子涵. All rights reserved.
//

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>

using namespace std;

long long int N;

long long int a,b,c;

unsigned long int sum=1;


long long int maxs=0;

long long int mins=0;

int main(int argc, const char * argv[]) {
    scanf("%lld",&N);
    long long int T=N;
    while (T--) {
         scanf("%lld",&a);
        if (a==1) {
            scanf("%lld",&b);
            sum++;
            if (b>maxs) {
                maxs=b;
            }
            if (sum==2) {
                maxs=b;
                mins=b;
            }
            continue;
        }
        else
        {
            if (a==2) {
                if (sum==1) {
                    continue;
                }
                if (sum==2) {
                    sum=1;
                    maxs=0;
                    continue;
                }
                sum--;
                continue;
            }
            if (a==3) {
                if (sum==1) {
                    printf("0\n");
                    continue;
                }
                printf("%lld\n",maxs);
                continue;
            }
        }
    }
    return 0;
}