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.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<=200, and n is odd) at first. Following that, n positive integers will be given in a line. 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

题意：找出一行数中独立的数！

按位异或的3个特点:

(1) 0^0=0,0^1=1 0异或任何数＝任何数 (2) 1^0=1,1^1=0 1异或任何数－任何数取反 (3) 任何数异或自己＝把自己置0

先说一下异或运算的运算法则：

1. a ^ b = b ^ a 2. a ^ b ^ c = a ^ (b ^ c) = (a ^ b) ^ c 3. d = a ^ b ^ c 可以推出 a = d ^ b ^ c 4. a ^ b ^ a = b

对于性质1，显而易见。

对于性质2和4，就是可以查找出一组数列中具有奇数个数的数。比如： 题目：有2n+1个数，其中有n个数出现过两次，只有一个数字出现过一次。要求是找出这个数字。

import java.util.Scanner;public class Main{    public static void main(String[] args) {        Scanner sc = new Scanner(System.in);        while(sc.hasNext()){            int n =sc.nextInt();            if(n==0){                return ;            }            int m =0;            int s;            while(n-->0){                s = sc.nextInt();                m = m^s;            }            System.out.println(m);        }    }}