COMING SOON!

COMING SOON!

```
"""Coin sums
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?
Hint:
> There are 100 pence in a pound (£1 = 100p)
> There are coins(in pence) are available: 1, 2, 5, 10, 20, 50, 100 and 200.
> how many different ways you can combine these values to create 200 pence.
Example:
to make 6p there are 5 ways
1,1,1,1,1,1
1,1,1,1,2
1,1,2,2
2,2,2
1,5
to make 5p there are 4 ways
1,1,1,1,1
1,1,1,2
1,2,2
5
"""
def solution(pence: int) -> int:
"""Returns the number of different ways to make X pence using any number of coins.
The solution is based on dynamic programming paradigm in a bottom-up fashion.
>>> solution(500)
6295434
>>> solution(200)
73682
>>> solution(50)
451
>>> solution(10)
11
"""
coins = [1, 2, 5, 10, 20, 50, 100, 200]
number_of_ways = [0] * (pence + 1)
number_of_ways[0] = 1 # base case: 1 way to make 0 pence
for coin in coins:
for i in range(coin, pence + 1, 1):
number_of_ways[i] += number_of_ways[i - coin]
return number_of_ways[pence]
if __name__ == "__main__":
assert solution(200) == 73682
```