G - find your present (2)

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题意：

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
2use scanf to avoid Time Limit Exceeded

Hint

Hint

题意：

在这些数据中，只有个数会出现奇数次，别的数都是出现偶数次，让你找出出现奇数次的那个数；

思路：

因为题目的特殊性，用亦或是最简单的并且不会超时；

这道题中只有一个数字出现奇数次，凡是出现偶数次的都会在亦或中变成0；

同时亦或的三个特点是：

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

代码如下：
 

#include<stdio.h>
int n,m,s;
int main()
{while(~scanf("%d",&n)&&n){int i,s=0;for(i=0; i<n; i++){scanf("%d",&m);s=s^m;}printf("%d\n",s);}return 0;
}

 

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