I'm trying to figure out how many beers I can buy with 10 RMB after recycling every bottle I get. It's obvious to me that I'm doing something wrong, procedurally, but it's not occurring to me what that is. I'm currently reading "How To Think Like a Computer Scientist: Think Python" on chapter 9. I feel like this should be an easy program for me, but I'm not sure how to loop in the recycling portion of the app. What would be the most concise way to rinse and repeat beer purchases?
Basically, one beer costs 2 RMB. 2 bins gets 1 RMB. 4 caps gets 1 RMB. I'm starting out with 10 RMB. How many beers can I buy (recycling all the bins and caps)?
#5 bottles 5 caps
#= 3 rmb + 1 caps 1 bottles
#6th bottle bought
#= 2rmb + 2 caps
#7th bottle bought
#= 0rmb + 3 caps 1 bottles.
import math
def countbeers(rmb):
beers = 0;
caps = 0;
bins = 0;
bcost = 2;
for i in range (0,rmb):
beers += 1/2
for i in range (0,math.floor(beers)):
caps += 1
bins += 1
rmb = rmb - bcost
for i in range (0,caps):
rmb += 1/4
for i in range (0,bins):
rmb += 1/2
# if rmb > 2 what goes here, trying to loop back through
return beers
print(countbeers(10))
#5 bottles 5 caps
#= 3 wallet + 1 caps 1 bottles
#6th bottle bought
#= 2wallet + 2 caps
#7th bottle bought
#= 0wallet + 3 caps 1 bottles.
import math
global beers
global caps
global bins
global bcost
beers = 0
caps = 0
bins = 0
bcost = 2
def buybeers(wallet):
beers = 0
for i in range (0,wallet):
beers += 1/2
wallet -= 2
return beers
def drinkbeers(beers):
for i in range (0,math.floor(beers)):
caps += 1
bins += 1
wallet = wallet - bcost
return wallet, caps, bins
def recycle(caps, bins):
for i in range (0,caps):
wallet += 1/4
for i in range (0,bins):
wallet += 1/2
return wallet
def maxbeers(wallet):
if wallet > 2:
buybeers(wallet)
if math.floor(beers) > 1:
drinkbeers(beers)
if caps > 4 | bins > 2:
recycle(caps, bins)
return wallet
wallet = int(input("How many wallet do you have?"))
maxbeers(wallet)
if wallet >= 2:
maxbeers(wallet)
elif wallet < 2:
print(beers)
Your main problem is that you are not looping. Every beer you bought from rmb
gives you one more bottle, and one more cap. This new bottle and cap might be enough to earn you another rmb
, which might be enough for another beer. Your implementation handles this to a limited extent, since you call maxbeers multiple times, but it will not give the correct answer if you give it a truckload of beers, i.e. 25656 bottles.
If you know the number of rmb
you have, you can do the calculation by hand on paper and write this:
def maxbeers(rmb):
return 7 # totally correct, I promise. Checked this by hand.
but that's no fun. What if rmb
is 25656
?
Assuming we can exchange:
2 bottles -> 1 rmb
4 caps -> 1 rmb
2 rmb -> 1 beer + 1 cap + 1 bottle
we calculate it like this, through simulation:
def q(rmb):
beers = 0
caps = 0
bottles = 0
while rmb > 0:
# buy a beer with rmb
rmb -= 2
beers += 1
caps += 1
bottles += 1
# exchange all caps for rmb
while caps >= 4:
rmb += 1
caps -= 4
# exchange all bottles for rmb
while bottles >= 2:
rmb += 1
bottles -= 2
return beers
for a in range(20):
print("rmb:", a, "beers:", q(a))
Then we can buy 20525 beers.