47 lines
1020 B
Python
47 lines
1020 B
Python
#!/usr/bin/env python
|
|
"""
|
|
Created on 24 Feb 2021
|
|
|
|
@author: David Doblas Jiménez
|
|
@email: daviddoji@pm.me
|
|
|
|
Solution for problem 031 of Project Euler
|
|
https://projecteuler.net/problem=31
|
|
"""
|
|
|
|
from itertools import product
|
|
|
|
from utils import timeit
|
|
|
|
|
|
@timeit("Problem 031")
|
|
def compute():
|
|
"""
|
|
In the United Kingdom the currency is made up of pound (£) and pence (p).
|
|
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:
|
|
|
|
1x£1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p
|
|
|
|
How many different ways can £2 be made using any number of coins?
|
|
|
|
"""
|
|
|
|
ans = 0
|
|
coins = [2, 5, 10, 20, 50, 100]
|
|
bunch_of_coins = product(*[range(0, 201, i) for i in coins])
|
|
|
|
for money in bunch_of_coins:
|
|
if sum(money) <= 200:
|
|
ans += 1
|
|
|
|
# consider also the case for 200 coins of 1p
|
|
return ans + 1
|
|
|
|
|
|
if __name__ == "__main__":
|
|
print(f"Result for Problem 031: {compute()}")
|