杭电2095 find your present (2)

在新年派对上,每个人都会收到一份独特的礼物。通过解析一系列带有不同编号的礼物,找出那个唯一出现一次的编号作为您的礼物。本文详细介绍了如何使用扫描算法解决此类问题,提供了一个简洁高效的解决方案。

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
 
#include<stdio.h>

int main()
{
	int n,a,i;
	while(~scanf("%d",&n),n)
	{
		int s=0;
		for(i=1;i<=n;++i)
		{
			scanf("%d",&a);
			s^=a;//在一排数中找到独一无二的一个数
		}
		printf("%d\n",s);
	}
	
return 0;
}

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