find your present (2) v【 异或运算的巧用】

find your present (2)

Time Limit: 2000/1000 MS (Java/Others)    Memory
Limit: 32768/32768 K (Java/Others)

Total Submission(s): 17684    Accepted Submission(s): 6772


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
For each case, output an integer in a line, which is the card number of your present.
 

[align=left]Sample Input[/align]

5
1 1 3 2 2
3
1 2 1
0

 

[align=left]Sample Output[/align]

3
2

HintHint
use scanf to avoid Time Limit Exceeded思路： x^x=0; 0^x=x; y^0=y;有因为 除了要求的数字，其他的数字都是有偶数个，他们之间的^一定都为0，但是 0^x=x; 所以x是不会变的；代码

#include<stdio.h>
int main()
{
int n,a,sp;
while(scanf("%d",&n)&&(n!=0))
{
sp=0;
while(n--)
{
scanf("%d",&a);
sp^=a;
}
printf("%d\n",sp);
}
return 0;
}