HDOJ 2095

find your present (2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/1024 K (Java/Others) Total Submission(s): 13802    Accepted Submission(s): 5194

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

 

异或运算^

目给的数据量特别大 如果之前试过用hash算的话，（既开一个1000000的数组来保存每一个数字出现得次数,然后扫描值为1的） 很可能会Memory Limit Exceeded 。 
 所以才想到了用异或。 异或：就是按位比较相同则为0不同则是1；   比如   3^5,   3=011,5=101,两数按位异或后为110，即6。 
偶数次出现的异或结果为0 基数次结果为1  
然后  异或满足交换率。 再举一个例子。
 如 数据  1 2 3 2 1       先让result=0     那么可以看成是  result^1^2^3^2^1     
 交换律     result^1^1^2^2^3      
 很明显      1^1 和  2^2 都为 0      
所以最后得    result^3 =0^3 =3(二进制 101）

9867992

2013-12-20 22:29:40

Accepted

2095

218MS

256K

210 B

C++

泽泽

#include<stdio.h>
int main()
{
    int n,x,i,t,m;
    while(scanf("%d",&t)!=EOF&&t)
    {
        scanf("%d",&n);
        t--;
        while(t--)
        {
            scanf("%d",&m);
            n=m^n;
        }
        printf("%d\n",n);
    }
    return 0;
    
}