HDU 2095 find your present (2)

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

Hint

Hint

 

use scanf to avoid Time Limit Exceeded

思路解析：

木有想到用异或，这个方法这的很巧妙啊！！

 

异或运算法则　　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

 

代码：

#include<stdio.h>
int n;
int a;
int b;
int main()
{
    while(~scanf("%d",&n),n)
    {
        scanf("%d",&b);
        for(int c=1;c<n;c++)
        {
            scanf("%d",&a);
            b^=a;
        }
        printf("%d\n",b);
    }
}