HDU - 2095 - find your present (2)-优快云博客

本文介绍了一个使用异或运算解决寻找数组中唯一出现奇数次元素的问题。通过利用异或运算的特性，即相同数异或为0，不同数异或保持不变的特点，有效地解决了问题，避免了使用大量内存或复杂的数据结构。

先上题目：

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）

Recommend

8600

　　题目的意思就是给你最多1000000-1个数，找出里面出现奇数次的那个数。

　　一开始题目没有看清就开始做，开了个大数组，结果就内存超了= =。（题目还非常好心地提示里用scanf不要用cin = =）

　　当然看到人家的题解有人用map来做，可能这个会和用什么输入有关。我是看了别人的题解发现可以用异或运算来做，于是就用了异或运算。

　　这里用到了异或运算的性质： A^0=A A^A=0。

　　出现偶数的数都会变成0，而奇数次出现的数会因为异或0保留下来。

　　题目已经说明会出现奇数次的数只有一个，因此可以放心用异或。

上代码:

 1 #include <stdio.h>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     int a,n,t;
 7     while(scanf("%d",&n),n)
 8     {
 9         t=0;
10         while(n--)
11         {
12             scanf("%d",&a);
13             t=t^a;
14         }
15         printf("%d\n",t);
16     }
17     return 0;
18 }