在哈斯克爾中分解守衛

Question

這段代碼將一筆錢作爲Int並返回一個元組列表，表示所需的最小數量的必要賬單或硬幣，以達到該總額。

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將返回：

[("$10", 1), ("$5", 1), ("$2", 1), ("$1", 1)] 

我的問題是：可能這個系列守衛被移除/工廠化函數只是基於票據/硬幣的列表上工作，像where bills = [200, 100, 50, 20...] ？

在Python我會做這樣的事情（沒有確切的工作解決方案，但你的想法）：

purse(x): 
    for bill in [200, 100, 50, 20, 10, 5, 2, 1]: 
     if x/bill >= 1: 
      return [x // bill] + purse(x % bill) 

Answer 1

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))