二进制形式特殊的数
a&(a-1):去掉a的二进制表示的最后一个1之后得到的数,如果是2的n次幂则得到的会是0
0xaaaaaaaa = 10101010101010101010101010101010 (偶数位为1,奇数位为0)
0x55555555 = 1010101010101010101010101010101 (偶数位为0,奇数位为1)
0x33333333 = 110011001100110011001100110011 (1和0每隔两位交替出现)
0xcccccccc = 11001100110011001100110011001100 (0和1每隔两位交替出现)
0x0f0f0f0f = 00001111000011110000111100001111 (1和0每隔四位交替出现)
0xf0f0f0f0 = 11110000111100001111000011110000 (0和1每隔四位交替出现)