*寒假水71——find your present (2) 【位运算~异或(^)】-优快云博客

本文链接：https://blog.youkuaiyun.com/chengchencheng/article/details/79342332

在一个新年派对上，每位参与者都将获得一份特殊礼物。礼物通过编号卡片标识，每张卡片上的编号除了一份礼物外都是成对出现的。参与者需要找出那份唯一编号的特殊礼物。

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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.

InputThe 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.OutputFor each case, output an integer in a line, which is the card number of your present.Sample Input

Sample Output

3
2


        
  
use scanf to avoid Time Limit Exceeded

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

题解：相同为1，不同为0.

n^n^a=a。