forked from TheAlgorithms/Python
-
Notifications
You must be signed in to change notification settings - Fork 0
/
sol1.py
65 lines (41 loc) · 1.43 KB
/
sol1.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
"""
Coin sums
Problem 31: https://projecteuler.net/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() -> int:
return 1
def two_pence(x: int) -> int:
return 0 if x < 0 else two_pence(x - 2) + one_pence()
def five_pence(x: int) -> int:
return 0 if x < 0 else five_pence(x - 5) + two_pence(x)
def ten_pence(x: int) -> int:
return 0 if x < 0 else ten_pence(x - 10) + five_pence(x)
def twenty_pence(x: int) -> int:
return 0 if x < 0 else twenty_pence(x - 20) + ten_pence(x)
def fifty_pence(x: int) -> int:
return 0 if x < 0 else fifty_pence(x - 50) + twenty_pence(x)
def one_pound(x: int) -> int:
return 0 if x < 0 else one_pound(x - 100) + fifty_pence(x)
def two_pound(x: int) -> int:
return 0 if x < 0 else two_pound(x - 200) + one_pound(x)
def solution(n: int = 200) -> int:
"""Returns the number of different ways can n pence be made using any number of
coins?
>>> solution(500)
6295434
>>> solution(200)
73682
>>> solution(50)
451
>>> solution(10)
11
"""
return two_pound(n)
if __name__ == "__main__":
print(solution(int(input().strip())))