hdu-1563 Find your present!

关于寻找数组中的特别数字的小技巧

Find your present!

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2335    Accepted Submission(s): 1524

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.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<=200, and n is odd) at first. Following that, n positive integers will be given in a line. 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
   
   
   
   

 

题目大意：找出一串数字中比较特别的数字。

解题思路：这个题目中可以用到一个小技巧，就是用异或运算，这样能快速找出来那个特别的数字。因为异或运算是“同0异1”。a^a=0   a^0=a  而且异或运算具有交换性和结合性

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
int a[220];
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        int ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            ans^=a[i];
        }
        printf("%d\n",ans);
    }
    return 0;
}