【菜鸟做水题】：杭电 find your present（用异或的解题思想）

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

 

 

#include<iostream>
using namespace std;

int main()
{
    int num,n,result,i;
    while(scanf("%d",&n),n){
        result=0;            //如果重复出现，异或运算结果为0或者1，与0再做一次运算确保最终结果就为单独出现的元素
        for(i=0;i<n;++i){
            scanf("%d",&num);
            result^=num;// 异或运算，相同则为1，不同则为0
        }
        printf("%d\n",result);
    }
    return 0;
}