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• # 杭电2095--find your present (2) （异或）

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18235    Accepted Submission(s): 7021

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`

Author
8600

Source

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//这题巧妙的运用了位运算的方法，同一个数位运算两次，化为0；多个数的话就留下了出现次数是奇数的那个数；
``` 1 #include <stdio.h>
2 int main()
3 {
4     int n ;
5     while(scanf("%d", &n) , n)
6     {
7         int i, num, temp = 0 ;
8         for(i=0; i<n; i++)
9         {
10             scanf("%d", &num) ;
11             temp = temp ^ num ;
12         }
13         printf("%d
", temp) ;
14     }
15     return 0 ;
16  } ```
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• 原文地址：https://www.cnblogs.com/soTired/p/4641717.html