HDU 2095 find your present (2)

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原文链接：http://www.cnblogs.com/GODLIKEING/p/3431305.html

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#java

本文介绍了一个通过编程解决问题的方法：在一串整数中找到唯一出现奇数次的数字。使用异或运算来实现，此方法能够有效地处理大数据量，并避免了时间限制超出的问题。

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find your present (2)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/1024 K (Java/Others) Total Submission(s): 13413 Accepted Submission(s): 5049

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）

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思路：异或运算“^”，异或运算是（二进制）按位相同取0，按位不同取1，例如5 — 101，3 — 011,5 ^ 3 = 110 = 6，我、而且满足交换律，result ^ 2 ^ 1 ^ 1 ^ 2 = result ^ 2 ^ 2 ^ 1 ^ 1，而2 ^ 2 = 0，所以最后的结果是所有偶数次数的都为零，最后的结果就是所求

代码：

#include<iostream>
using namespace std;
int main()
{
    int num,n,result,i;
    while(scanf("%d",&n),n){
        result=0;
        for(i=0;i<n;++i){
            scanf("%d",&num);
            result^=num;
        }
        printf("%d\n",result);
    }
    return 0;
}