好吧，這題我開始先排序超時了，後來用了異或，再次見識到位運算的強大。。。

我們知道 A^A=0 A^0=A 所以只要把輸入的數字全部異或 那麼留下來的就是奇數個的數字。

find your present (2)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/1024 K (Java/Others)
Total Submission(s): 14445 Accepted Submission(s): 5464


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

HintHint 
use scanf to avoid Time Limit Exceeded

Author 8600
Source HDU 2007-Spring Programming Contest - Warm Up （1）

#include
int main()
{
    int T;
    while(scanf("%d",&T)&&T)
    {
        int ans=0;
        int a;
        for(int i=1;i<=T;i++)
        {
            scanf("%d",&a);
            ans^=a;
        }
        printf("%d\n",ans);
    }
    return 0;
}