hdu2095 find your present (2)

find your present (2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/1024 K (Java/Others)
Total Submission(s): 15975    Accepted Submission(s): 6079

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

 

初看这道题目可以想到用平衡树等数据结构来判重，事实上我们确实可以用c++的set或map容器来解。

但是这题其实还有一个更巧妙地方法，即使用位运算来解题。

注意到有

0 xor n=n

n xor n=0

所以，a xor b xor a=b

#include
#include
using namespace std;

int main()
{
    int n,x,a;
    while(scanf("%d",&n)!=EOF && n!=0)
    {
        x=0;
        while(n--)
        {
            scanf("%d",&a);
            x=x^a;
        }
        printf("%d\n",x);
    }
    return 0;
}

#include
#include
using namespace std;

int main()
{
    int n,x,a;
    while(scanf("%d",&n)!=EOF && n!=0)
    {
        x=0;
        while(n--)
        {
            scanf("%d",&a);
            x=x^a;
        }
        printf("%d\n",x);
    }
    return 0;
}