水题不解释，思路是用map记录一下出现次数，数据也很水

#include <cstdio>
#include <map>
using namespace std;

int main()
{
    int n;
    map<int,int> present;
    while( scanf("%d", &n) && n)
    {
        present.clear();
        int m;
        while(n--)
        {
            scanf("%d", &m);
            if(!present.count(m))
              present[m] = 1;
            else
              present[m]++;
        }
        for( map<int, int>::iterator i = present.begin(); i != present.end(); ++i) 
            if(i->second % 2)
            {
                printf("%d\n",i->first);
                break;
            }
    }
    return 0;
}