零起点学算法82——find your present

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include<algorithm>
using namespace std;
int main()
{
int n;
while (scanf("%d",&n)&&n)
{
int a,b= 0;//0与任何非零数n相异或均得到n
while (n--)
{
scanf("%d",&a);
b ^= a;
}
printf("%d\n",b);
}
return 0;
}

注：异或运算满足交换律，并且两个相同的数 ^ 后会变0，所以所有偶数次的数运算后都会变0，奇数次n^后会变成n。

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