异或运算的基础有点忘记了
先介绍一下。。2个数异或 就是对于每一个二进制位进行位运算
具有2个特殊的性质
1、一个数异或本身恒等于0，如5^5恒等于0；
2、一个数异或0恒等于本身，如5^0恒等于5。
3 满足交换律
1.交换数字
这个性质能利用与交换数字先考虑加减改变法。a=a+bb=a-b;a=a-b;;
实际上以第一个 A（与a区别）=a+b 作为临时的参数A（实际a+b） 完成b=A-b=a+b-b=a;a=A-a=(经过上面的运算 a==b了)=A-b;
用异或也是同理A=a^bb=a^b^b=aa=A^a(经过上面的运算 a==b了)=A^b第一个应用讨论完毕2.hdu2095
find your present (2) |
| Problem DescriptionIn 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
5
|
Sample Output
3
|
and you can assume that only one number appear odd times.注意这句话。。。只有ans会出现odd次
根据这几个性质知
1、一个数异或本身恒等于0，如5^5恒等于0；
2、一个数异或0恒等于本身，如5^0恒等于5。
3 满足交换律
8^2^2^8^3^3^8^8^5=5; 用来寻找一串数字中唯一odd次出现的数
所以显然用异或运算是很完美的解法
代码就不贴了
