HDOJ2095 find your present (2)

find your present (2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25477    Accepted Submission(s): 10064

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

Hint

Hint

use scanf to avoid Time Limit Exceeded

题目大意：在输入的数列中 找唯一出现奇数次的一个数。

正如你所看到的这一句：

use scanf to avoid Time Limit Exceeded

 我用java换了几种方法，异或应该是最为简单的了，但还是超时了，估计是Scanner接收数据太慢了

有人用java过了的，但是不多，我也找不到方法了（已解决）

C，C++还是比较容易过的

这道题的难点或者说是技巧就在与异或的运用

java过不了的问题已解决（2017-12-1）

java：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;

public class Main{
  public static void main(String[] args) throws IOException {
    StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    while (in.nextToken() != StreamTokenizer.TT_EOF) {
      int n = (int) in.nval;
      if (n == 0) {
        break;
      }
      int m = 0;
      int s;
      while (n-- > 0) {
        in.nextToken();
        s = (int) in.nval;
        m = m ^ s;
      }
      out.println(m);
      out.flush();
    }
  }
}

C++代码：

#include<cstdio>  
    int main()  
    {  
        int n;  
        while(scanf("%d",&n),n)  
        {  
           int ans=0,m;  
            while(n--)  
            {  
                scanf("%d",&m);  
                ans=ans^m;  
            }  
            printf("%d\n",ans);  
        }  
    }