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| -rw-r--r-- | 3.1.py | 92 |
1 files changed, 92 insertions, 0 deletions
@@ -0,0 +1,92 @@ +#!/usr/bin/python2 + +# Fuel Injection Perfection +# ========================= + +# Commander Lambda has asked for your help to refine the automatic quantum +# antimatter fuel injection system for her LAMBCHOP doomsday device. It's a +# great chance for you to get a closer look at the LAMBCHOP - and maybe sneak +# in a bit of sabotage while you're at it - so you took the job gladly. + +# Quantum antimatter fuel comes in small pellets, which is convenient since the +# many moving parts of the LAMBCHOP each need to be fed fuel one pellet at a +# time. However, minions dump pellets in bulk into the fuel intake. You need to +# figure out the most efficient way to sort and shift the pellets down to a +# single pellet at a time. + +# The fuel control mechanisms have three operations: + +# 1) Add one fuel pellet + +# 2) Remove one fuel pellet + +# 3) Divide the entire group of fuel pellets by 2 (due to the destructive +# energy released when a quantum antimatter pellet is cut in half, the safety +# controls will only allow this to happen if there is an even number of +# pellets) + +# Write a function called solution(n) which takes a positive integer as a +# string and returns the minimum number of operations needed to transform the +# number of pellets to 1. The fuel intake control panel can only display a +# number up to 309 digits long, so there won't ever be more pellets than you +# can express in that many digits. + +# For example: +# solution(4) returns 2: 4 -> 2 -> 1 +# solution(15) returns 5: 15 -> 16 -> 8 -> 4 -> 2 -> 1 + +def change(b, n): + a = int("".join([str(x) for x in b]), 2) + n + return [int(i) for i in list("{0:b}".format(a))] + +def solution(n): + if n == "1": return 0 + if n == "2": return 1 + + b = [int(i) for i in list("{0:b}".format(int(n, 10)))] + + c = 0 + + flag = True + while flag == True: + if (len(b) <= 2): + flag = False + continue + if b[-1] == 0: + c += 1 + b.pop() + else: + if b[-2] == 1: + c += 1 + b = change(b, 1) + else: + c += 1 + b = change(b, -1) + + if b[1] == 1: + c += 2 + else: + c += 1 + + return c + +def funny_add1(b): + b.reverse() + + carry = 0 + for i in range(0, len(b)): + if b[i] == 0: + b[i] = 1 + break + else: #if b[i] == 1: + b[i] = 0 + carry = 1 + if i == len(b) - 1 and carry == 1: + b.append(1) + + b.reverse() + return b + +print(solution("19")) +print(solution("4")) # => 2 +print(solution("15")) # => 5 |