J - find your present (2)

find your present(2)

题目描述

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.

Iutput

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

use scanf to avoid Time Limit Exceeded

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思路：

题意：找出一组数中出现次数为奇数的那个数； 在网上学到了一种很巧妙地方法：新输入的数不断的关键数数做比较，如果不同，则关键数为当前输入的数，否则关键数为0，用异或实现。 原理：除了我们要找的那个数为奇数个外，其余的数都是偶数个，所以每一次的异或操作其实就是让相同的数抵消，保留下与前一个数不同的数，偶数个数的数最终全部被抵消完，只剩下奇数个的那个数成为最后一个关键数。

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代码

#include <cstdio> using namespace std; int main() { int n; while(scanf("%d",&n),n){ int i; int ans,temp; ans=0; for(i=0;i<n;i++){ scanf("%d",&temp); ans=ans^temp; } printf("%d\n",ans); } return 0; }