hdu-oj 1563 Find your present!

Find your present!

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

Total Submission(s): 2690    Accepted Submission(s): 1785


[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.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<=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.
 

[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 方法很多：(1）查找，数组设为499999超时了！ac不了!
（2）you can assume that only one number appear odd times.根据这句话可以想想更高效的算法，思考中……
（3）用异或运算符
异或运算符的用法：
整数的异或是先把它们化成二进制，再按位异或。比如3^5,     3=011,5=101,两数按位异或后为
110，即6。
      几个数异或满足交换律。2^3^2=2^2^3=0^3=3.
      两个相同的数异或为0，普通数都出现了偶数次，所以它们异或后都是0，而0与那个特别数异或后还是那个特殊数。#include<stdio.h>
//异或运算的运用
int main()
{
int t,a,sum;
while(scanf("%d",&t)&&t!=0)
{
scanf("%d",&sum);
t--;
while(t--)
{
scanf("%d",&a);
sum^=a;
}
printf("%d\n",sum);
}
return 0;
}