HDU2095 find your present (2)

find your present (2)

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

                                                                          Total Submission(s): 23256    Accepted Submission(s): 9242

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

51 1 3 2 231 2 10

 

Sample Output

32
Hint
Hint
 use scanf to avoid Time Limit Exceeded

 

Author

8600

 

Source

HDU 2007-Spring Programming Contest - Warm Up （1）

题意就是在新年聚会上，每个人都会得到一份特殊的礼物，现在轮到你去得到自己特殊的礼物。桌子上有许多礼物，每个礼物上面有一个数字卡片，而你的礼物上的数字卡片和其他的都有所不同，并且我们可以假设只有一个数字出现奇数次，现在请你找出这个特殊的礼物？

用异或算法非常简单，普通的话数据比较大，容易超时。航电上还有一道类似的题1563，测试数据都一样我还以为是一道题，但那个题是找出哪个数只出现过一次。。我还一度以为我第一次翻译错了。

#include<stdio.h>int main(){    int n;    while(scanf("%d",&n),n!=0)    {        int ans=0,x;        while(n--)        {            scanf("%d",&x);            ans^=x;//异或用法，满足交换律        }        printf("%d\n",ans);    }    return 0;}