HDU 2095 find your present (2)

find your present (2)

Time Limit: 1000/2000 MS (Java/Others)Memory Limit: 32768/1024 K (Java/Others)
Total Submission(s): 6852Accepted Submission(s): 2509

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

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

 

Recommend

8600

呢个题目，我用左我地一个师兄叫linda既牛人既一种异或运算既妙用。

其实就系一个数n， n^n = 0 即系如果有累异或变量tsu 对n 进行偶数次异或，甘样 tsu = 0。

另一方面如果进行奇数次异或，tsu= n。

尼条题目就系要搵到出现奇数次既数字，所以直接不断异或就ok。

贴代码：

4313813

2011-08-02 16:08:28

Accepted

2095

218MS

188K

227 B

C++

10SGetEternal{（。）（。）}!

#include <iostream>
using namespace std;

int main()
{
    int n, a, pre;

    while (scanf("%d", &n), n)
    {
        pre = 0;
        while (n--)
        {
            scanf("%d", &a);
            pre ^= a;
        }
        printf("%d\n", pre);
    }

    return 0;
}

代码系够短，不过^运算其实系好费时间噶。