标签:位运算

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.

输入

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.

输出

For each case, output an integer in a line, which is the card number of your present.

样例输入

5
1 1 3 2 2
3
1 2 1
0

样例输出

3
2

提示

Use scanf to avoid Time Limit Exceeded

简单的异或计算

AC代码

/* 异或计算 */
/* 异或计算偶数项会被抵消 */
#include <bits/stdc++.h>using namespace std;int main()
{int n , a , x = 0; // 初始化while(~scanf("%d",&n) , n) // 用scanf读入,防止时间超限{x = 0; // 每次记得清零while(n --){scanf("%d",&a); // 用scanf!!!x = x ^ a; // 异或计算}cout << x << endl; // 输出记得换行}return 0;
}