find your present (2)

Problem Description

In the new year party, everybody will get a "special present".Now it's your turn to get your special present, a lot of presents now putting on the desk, and only one of them will be yours.Each present has a card number on it, and your present's card number will be the one that different from all the others, and you can assume that only one number appear odd times.For example, there are 5 present, and their card numbers are 1, 2, 3, 2, 1.so your present will be the one with the card number of 3, because 3 is the number that different from all the others.

 

Input

The input file will consist of several cases. 
Each case will be presented by an integer n (1<=n<1000000, and n is odd) at first. Following that, n positive integers will be given in a line, all integers will smaller than 2^31. These numbers indicate the card numbers of the presents.n = 0 ends the input.

 

Output

For each case, output an integer in a line, which is the card number of your present.

 

Sample Input

5

1 1 3 2 2

3

1 2 1

0

 

Sample Output

3

2

 

Hint

use scanf to avoid Time Limit Exceeded

 

题目内存限制：1024K，所以不能简单地用数组存然后再处理。

为了节约内存，可以用STL里面的set,map等容器。

当容器里没有这个元素的时候，就插入这个元素，否则，删除这个元素。

最后，容器中肯定只剩下一个元素，便是我们所要的结果。

#include <set>
#include <stdio.h>
using namespace std;

int main(){
    int n;
    int i;
    int temp;
    set<int> S;
    
    while(1){
        scanf("%d",&n);
        
        if(n==0)
            break;
            
        for(i=0;i<n;i++){
            scanf("%d",&temp);
            
            if(S.find(temp)==S.end())
                S.insert(temp);
                
            else
                S.erase(temp);
        }
        
        printf("%d
",*S.begin());
        S.clear();
    }
    
    return 0;
}