mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-24 05:21:09 +00:00
54 lines
1.1 KiB
Python
54 lines
1.1 KiB
Python
# -*- coding: utf-8 -*-
|
||
from __future__ import print_function
|
||
try:
|
||
raw_input # Python 2
|
||
except NameError:
|
||
raw_input = input # Python 3
|
||
'''
|
||
Coin sums
|
||
Problem 31
|
||
In England the currency is made up of pound, £, and pence, p, and there are
|
||
eight coins in general circulation:
|
||
|
||
1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
|
||
It is possible to make £2 in the following way:
|
||
|
||
1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
|
||
How many different ways can £2 be made using any number of coins?
|
||
'''
|
||
|
||
|
||
def one_pence():
|
||
return 1
|
||
|
||
|
||
def two_pence(x):
|
||
return 0 if x < 0 else two_pence(x - 2) + one_pence()
|
||
|
||
|
||
def five_pence(x):
|
||
return 0 if x < 0 else five_pence(x - 5) + two_pence(x)
|
||
|
||
|
||
def ten_pence(x):
|
||
return 0 if x < 0 else ten_pence(x - 10) + five_pence(x)
|
||
|
||
|
||
def twenty_pence(x):
|
||
return 0 if x < 0 else twenty_pence(x - 20) + ten_pence(x)
|
||
|
||
|
||
def fifty_pence(x):
|
||
return 0 if x < 0 else fifty_pence(x - 50) + twenty_pence(x)
|
||
|
||
|
||
def one_pound(x):
|
||
return 0 if x < 0 else one_pound(x - 100) + fifty_pence(x)
|
||
|
||
|
||
def two_pound(x):
|
||
return 0 if x < 0 else two_pound(x - 200) + one_pound(x)
|
||
|
||
|
||
print(two_pound(200))
|