<> Xiao Ming asked me a question ：
<> How to use it python Judge whether a number is 2 Of n To the power of ?
<> My code is as follows ：
n = int(input()) if n & (n - 1) == 0: print('True') else: print('False')
<> My thoughts
Seeing this problem, the first thing I think of is to do it in a circle , Should be able to do it , However, if the number is large, it is estimated that the memory will overflow .
switching sites , Change a few rules
for i in range(0, 10): print(bin(2 ** i)) result ： 0b1 0b10 0b100 0b1000 0b10000
0b100000 0b1000000 0b10000000 0b100000000 0b1000000000
It's obvious , If a number is 2 Of n Power , Then convert the number to binary and then to string, and there is only one of them 1. So I found this way :
n = str(bin(n)) if n.count('1') == 1: print(' What's the number 2 Of n Power ')
however , If it's just like this, the robustness is too poor from the perspective of programming
We can have an in-depth observation ：
for i in range(0, 10): print(bin(2 ** i - 1)) result ： 0b0 0b1 0b11 0b111 0b1111
0b11111 0b111111 0b1111111 0b11111111 0b111111111
Get the law ： If it's a number i by 2 Of n Power , be i&(i-1)=0.
notes ：& ： Bitwise and operators ： Two values involved in the operation , If both corresponding bits are 1, The result of this bit is 1, Otherwise 0.
Of course, we need to do some judgment and exception handling , But here's how , You don't have to care about the details .
<> Learning experience
An interesting example of a combination of operators and judgments , A review of binary computing . The joy of solving problems feels good , keep trying , There will always be growth, won't there ?
<> come on. !!!
<> You can do that ! You always believe in yourself like this !