Who's the boss--a puzzle like Joseph's ring

an intellectual question similar to the Joseph Ring

This is a question to be asked in the interview, in fact, it is very easy to think about it. But at that time only gave two minutes, failed to give a direct solution. Problem Description

There were 156 pirates on an island, they decided to choose a person to do the boss, because so many people, choose who when the boss has not been elected, so they come up with a way to put these 156 people from 1 to 156 numbers, and then press 1, 2, 1, 2, ... Count off, Report 1 of the exit election, the rest of the people continue to be numbered, and then the ' 1 ' people kicked out, the last remaining is the boss. So, excuse me, who is the boss? ideas for solving problems

This question is a bit like the classic Joseph Ring, but it's actually easier than the Joseph Ring. We can use x 2 = 0来 as a condition of judgment, because each time it is the person who reported ' 1 '. So the program can be written in the following way:
(python)

def calc_out_2 (num):
    nlist = range (1, num)
    cnt = 1 while
    len (nlist)!= 1:
        nlist = [x/2 to x in Nlist if X 2 = 0]
        cnt *= 2 return
    cnt

calc_out_2 (156)

Load B, use functional programming, you can write this:

def calc_out_3 (num):
    nlist = range (1, num)
    cnt = 1
    while Len (nlist)!= 1:
        nlist = map (lambda x:x/2, fi Lter (lambda x:x%2 = = 0, nlist)
        CNT *= 2 return
    cnt
calc_out_1 (156)

Again careful analysis, because every time is ' 1 ' out, so the definition of f (n) for n Pirates Select the boss of the function.

F (2) = 2
F (3) = 2
F (4) = 4
F (5) = 4
F (6) = 4
F (7) = 4
F (8) = 8
F (9) = 8
...

Let's see the rules:

F (n) = max (2^x) if 2^x < n

Rewrite the Python code as follows:

Import math as M
log = M.log

def calc_out_1 (num): Return
    POW (2, int (log (num)/log (2))

calc_out_1 (156)