HDU.2095（异或运算）

find your present (2)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21862 Accepted Submission(s): 8634

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

use scanf to avoid Time Limit Exceeded

题意分析

找出所输入n个数中出现次数为奇数的数字，并将其输出。注意此题的数字范围比较大，最大2^31，如果此时开数组的话，必然超空间，并且查找也麻烦。用异或运算较为简便。

代码总览

/*
    Title:HDU.2095
    Author:pengwill
    Date:2016-11-12
*/
#include <stdio.h>
int main()
{
    int n,temp,a;
    while(scanf("%d",&n)!= EOF && n){
        temp = 0;
        while(n--){
            scanf("%d",&a);
            temp = temp ^ a;
        }
        printf("%d
",temp);
    }
    return 0;
}