此代码将一笔钱作为Int
并返回一个元组列表,表示所需金额或金币所需金额最少。
purse :: Int -> [(String, Int)]
purse x
| x == 0 = [("$0", 0)]
| div x 200 >= 1 = before x 200 : after x 200
| div x 100 >= 1 = before x 100 : after x 100
| div x 50 >= 1 = before x 50 : after x 50
| div x 20 >= 1 = before x 20 : after x 20
| div x 10 >= 1 = before x 10 : after x 10
| div x 5 >= 1 = before x 5 : after x 5
| div x 2 >= 1 = before x 2 : after x 2
| div x 1 >= 1 = before x 1 : after x 1
where before x y = ("$" ++ show x, div x y)
after x y = if mod x y > 0
then purse (mod x y)
else []
例如,purse 18
将返回:
[("$10", 1), ("$5", 1), ("$2", 1), ("$1", 1)]
我的问题是:这一系列警卫是否可以被移除/分解,而且该功能只是根据账单/硬币列表工作,如where bills = [200, 100, 50, 20...]
?
在Python中我会做类似的事情(不是确切的工作解决方案,但你明白了):
purse(x):
for bill in [200, 100, 50, 20, 10, 5, 2, 1]:
if x / bill >= 1:
return [x // bill] + purse(x % bill)
答案 0 :(得分:5)
purse = go [200,100,50,20,10,5,2,1] where
go (c:cs) x | x < c = go cs x
go (c:cs) x = ("$" ++ show c, div x c) : go cs (mod x c)
go [] _ = []
purse x = filter ((>0) . snd) $ snd $ mapAccumL foo x [200,100,50,20,10,5,2,1]
where foo x c = (mod x c, ("$" ++ show c, div x c))