HDOJ2275(Kiki & Little Kiki 1)

Problem Description
Kiki is considered as a smart girl in HDU, many boys fall in love with her! Now, kiki will finish her education, and leave school, what a pity! One day, zjt meets a girl, who is like kiki very much, in the campus, and calls her little kiki. Now, little kiki want to design a container, which has two kinds of operation, push operation, and pop operation.  
Push M:  
  Push the integer M into the container.
Pop M:
  Find the maximal integer, which is not bigger than M, in the container. And pop it from the container. Specially, for all pop operations, M[i] is no bigger than M[i+1].
Although she is as smart as kiki, she still can't solve this problem! zjt is so busy, he let you to help little kiki to solve the problem. Can you solve the problem?

 

 

Input
The input contains one or more data sets. At first line of each input data set is an integer N (1<= N <= 100000) indicate the number of operations. Then N lines follows, each line contains a word (“Push” or “Pop”) and an integer M. The word “Push” means a push operation, while “Pop” means a pop operation. You may assume all the numbers in the input file will be in the range of 32-bit integer.
 

 

Output
For each pop operation, you should print the integer satisfy the condition. If there is no integer to pop, please print “No Element!”. Please print a blank line after each data set.
 

 

Sample Input
9

Push 10

Push 20

Pop 2

Pop 10

Push 5

Push 2

Pop 10

Pop 11

Pop 19

3

Push 2

Push 5

Pop 2
 

 

Sample Output
No Element!

10

5

2

No Element!



2
复制代码
#include <iostream>
#include 
<cstring>
#include 
<set>
using namespace std;

typedef 
char CMD[5];
multiset
<int> S;
multiset
<int>::iterator p, q;

int main()
{
    
int n, num;
    CMD cmd;
    
while(scanf("%d"&n) != EOF)
    
{
        S.clear();
        
while(n--)
        
{
            getchar();
            scanf(
"%s%d", cmd, &num);
            
if(strcmp(cmd, "Push"== 0)
            
{
                S.insert(num);
            }

            
else
            
{
                
if(num < *(S.begin()))
                
{
                    printf(
"No Element!\n");
                    
continue;
                }

                p 
= S.find(num);
                
if(p != S.end())
                
{
                    printf(
"%d\n",num);
                    S.erase(p);
                }

                
else
                
{
                    S.insert(num);
                    q 
= p = S.find(num);
                    q
--;
                    printf(
"%d\n"*q);
                    S.erase(p);
                    S.erase(q);
                }

            }

        }

        printf(
"\n");
    }

    
return 0;
}

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