1175: 零起点学算法82——find your present

1175: 零起点学算法82——find your present

Time Limit: 1 Sec  Memory Limit: 2 MB   64bit IO Format: %lld
Submitted: 1804  Accepted: 418
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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

Source

零起点学算法

 

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

很巧妙的异或位运算方法！！

and you can assume that only one number appear odd times 这句话很重要。英语很重要啊。

odd 奇数

even 偶数