判断二进制数中的1有奇数个还是偶数个 - 计算机的艺术 - 博客频道 - CSDN.NET


判断二进制数中的1有奇数个还是偶数个 - 计算机的艺术 - 博客频道 - CSDN.NET
http://www.matrix67.com/blog/archives/266
判断(32位)整数的二进制表示中的1有奇数个还是偶数个

第一次异或的结果,第i个位置上是1代表原数中第i位和第i+1位有一个为1,也就是奇数个1。也就是说,某个位为1,代表从它开始向左连续两位中1的个数是奇数(异或的本质)。
第二次异或,注意是两位两位的比较了,所有x是和x>>2做异或运算,比如,若结果第6位为1,那么代表上一次的结果的第6位和第8位有奇数个1,也就是原数中第6,7,8,9位中有奇数个1。
同样,到第五次运算结束后,过第0位为1代表0到31位共奇数个1。
整个过程就是不断的压缩信息,先用1位代表从本位开始向左的2位中1的奇偶,然后向左的4位,向左的8位。。。。

二进制运算是很强大的,平时应多注意思考和总结,恰当的使用二进制运算可以大大提高程序的效率。

最直接的思路就是求二进制数中1的个数,然后确定是偶数还是奇数。
  1. bool OddOnes(int x)   
  2. {  
  3.     int cnt = 0;  
  4.     while(x)  
  5.     {  
  6.         cnt++;  
  7.         x &= x-1;  
  8.     }  
  9.     return cnt & 1;  
  10. }  
先求x二进制中1的个数,方法就是通过x&= x-1去掉最右端的1,直到x变为0,去掉了几次1就是原x二进制中含有多少个1;然后确定个数为奇数还是偶数。

为了说明上面这段代码的原理,我们还是拿1314520出来说事。1314520的二进制为101000000111011011000,第一次异或操作的结果如下:

    00000000000101000000111011011000
X   0000000000010100000011101101100
—————————————
    00000000000111100000100110110100


    得到的结果是一个新的二进制数,其中右起第i位上的数表示原数中第i和i+1位上有奇数个1还是偶数个1。比如,最右边那个0表示原数末两位有偶数个1,右起第3位上的1就表示原数的这个位置和前一个位置中有奇数个1。对这个数进行第二次异或的结果如下:

     00000000000111100000100110110100
X      000000000001111000001001101101
—————————————
    00000000000110011000101111011001


    结果里的每个1表示原数的该位置及其前面三个位置中共有奇数个1,每个0就表示原数对应的四个位置上共偶数个1。一直做到第五次异或结束后,得到的二进制数的最末位就表示整个32位数里有多少个1,这就是我们最终想要的答案。
http://blog.csdn.net/pinghegood/article/details/8194681
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