[第二轮] 3/18-3/24 CareerCup 5.4

Explain what the following code does:((n & (n-1)) == 0).

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http://www.1point3acres.com/bbs/thread-48094-1-1.html
http://www.1point3acres.com/bbs/thread-32423-1-1.html

check n is the kth power of 2 or not.

check if n is the 2^x...

intuitively, it checks whether only one bit of n (in binary representation) is set to be 1, hence this means whether n is some power of 2, i.e. whether n=2^x ( where x is positive integer)

2^n
假设n的二进制为 1abcd...... (a,b,c,d= 0/1)
当abcd..... = 0000... 则 n - 1 = 0111...1 则 n&(n-1) = 0
当abcd..... != 0000...., n -1 = 1a'b'c'd'..... 则 n&(n-1) = 1a''b''c''d'' != 0

https://github.com/1094401996/Ca ... rc/fivedot4/5.4.txt

check whether  any bit is '1' in n.

power of 2

Check if the n is the power of Two

it it all 0 IFF n and n-1 has no bits to be 1 at the same time.
1:  n is all 0
2. n is 10000...0
so it test if n is the power of 2 or not.

can u make a testament like this to test 3^n, or mod 2?