2015 Multi-University Training Contest 5 hdu 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): 740    Accepted Submission(s): 357

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

 

Source

2015 Multi-University Training Contest 5

 

解题：由于我们只需要查询最大值，所以最大值总是最后一个被删的，如果当前集合只有一个元素，那么肯定是删最大的元素了，否则，随便删就是了，不影响最大值

 

 #include <bits/stdc++.h>
 using namespace std;
 int main() {
     int n,op,val;
     while(~scanf("%d",&n)) {
         int sz = ,maxval = INT_MIN;
         while(n--) {
             scanf("%d",&op);
             switch(op) {
             case :
                 scanf("%d",&val);
                 maxval = max(val,maxval);
                 sz++;
                 break;
             case :
                 sz = max(,sz-);
                 if(!sz) maxval = INT_MIN;
                 break;
             case :
                 printf("%d\n",sz?maxval:);
                 break;
             default:
                 ;
             }
         }
     }
     return ;
 }