Haskell - 更有效的方法来完成算法？

Question

我创建了一个算法,该算法采用非负的Int值,表示总分钟数,并返回给出(天,小时,分钟)对应的三元组.

这是我的代码:

calcdays :: Int -> Int
calcdays x = x `div` (24*60)

calchours :: Int -> Int
calchours x = (x - ((calcdays x)*24*60)) `div` 60

calcmins :: Int -> Int
calcmins x = (x - ((calcdays x)*24*60) - ((calchours x)*60))

dayshoursmins :: Int -> (Int,Int,Int)
dayshoursmins x = (calcdays x, calchours x, calcmins x)

仅使用基本的Haskell操作(守卫,div,mod等),是否有更简单的方法来编程此功能？

编辑:

我使用下面的建议使这段代码更简单,虽然不像qoutRem解决方案那么简单,我想我可能会发布它:

calcdays :: Int -> Int
calcdays x = x `div` (24*60)

calchours :: Int -> Int
calchours x = (x `mod` (24*60)) `div` 60

calcmins :: Int -> Int
calcmins x = (x `mod` (24*60)) `mod` 60

dayshoursmins :: Int -> (Int,Int,Int)
dayshoursmins x = (calcdays x, calchours x, calcmins x)

Answer 1

我想有点想这样

dayshoursmins x = (d,hr,mr) where
 (h,mr) = quotRem x 60
 (d,hr) = quotRem h 24