find your present (2) （位异或）

find your present (2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21397    Accepted Submission(s): 8383

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

51 1 3 2 231 2 10

 

Sample Output

32
Hint
Hint
 use scanf to avoid Time Limit Exceeded

 

题意：

给你n个数字，已知只有一个数字出现了奇数次，其他数字都出现了偶数次，要求你找出这个特别的数字。

解题思路：

题目内存限制：1024K，所以不能简单地用数组存然后再处理。

为了节约内存，可以用STL里面的set,map等容器。

当容器里没有这个元素的时候，就插入这个元素，否则，删除这个元素。

最后，容器中肯定只剩下一个元素，便是我们所要的结果。

[cpp] view plain

    #include <set>      #include <stdio.h>      using namespace std;      int main()      {          int n,x;          set <int> S;          while(scanf("%d",&n),n)          {              while(n--)              {                  scanf("%d",&x);                  if(S.find(x) == S.end())    //没找到，插入                      S.insert(x);                  else                        //找到了，删除                      S.erase(x);              }              printf("%d\n",*S.begin());              S.clear();          }          return 0;      }  

其实，这题还有个更好的方法————位异或。

我们先了解一下位异或的运算法则吧：

1、a^b = b^a。

2、(a^b)^c = a^(b^c)。

3、a^b^a = b。

对于一个任意一个数n，它有几个特殊的性质：

1、0^n = n。

2、n^n = 0。

所以可以通过每次异或运算，最后剩下的值就是出现奇数次的那个数字。

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

PS：话说通过位运算的这些性质，可以通过不借用中间变量实现a,b的值的交换。

    void swap(int &a,int &b)      {          a ^= b;          b ^= a;          a ^= b;      }