OJ-HDU 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): 28605    Accepted Submission(s): 11214
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

思路1.

使用Map或者是Set解决,如果有就把原来的删除,若没有就添加,最后肯定只有一个元素,输出就行

思路2.

看了别人的发现的好牛P的方法...........使用异或运算

1)0^n=n

2)n^n=0

由于我使用的Java所以即使优化到O(n)了也还是TLE,但是拿c一下就过了,贴一下Java代码:

package hdu经典100题;

import java.util.Scanner;

public class P2095 {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		while(in.hasNext()) {
			int n = in.nextInt();
			if(n==0) {
				break;
			}
			int result = 0;
			for (int i = 0; i < n; i++) {
				result^=in.nextInt();
			}
			System.out.println(result);
		}
	}
}