位操作

x+y = (x&y) + (x|y)   //在第k位: x_k=1,y_k=1,则(x&y)_k=1,(x|y)_k=1; x_k=0,y_k=1,则(x&y)_k=0,(x|y)_k=1; 所以做加法不影响结果

(x|y) = (x&y) + (x^y)

x+y = 2*(x&y) + (x^y)

ë(x+y)/2û = (x&y) + (x^y)>>1 //为了防止x+y溢出

 

Ref

[1] http://aggregate.ee.engr.uky.edu/MAGIC/