haskell - Can mapEvery be implemented with foldr -

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For a function that maps the function to each NTH element in a list: 

 < Code> mapEvery :: int - & gt; (A -> A) - & gt; [A] - & gt; [A] mapEvery nf = zipWith ($) (drop 1 cycle. N $ f le: repeat id)   
 Is it possible to implement it with  foldr  Like normal  map ?  
 Edit: In the title, 'folder' has been changed to 'folder' Autocorrect ...   
 
 
 Here's a solution  
  mapEvery :: Int - & gt; (A -> A) - & gt; [A] - & gt; [A] The map in the form of every F = go plateau (const []) as 1 as an M go m == n = f a: as 1. Otherwise, as a = A: (M + 1)   
 it uses " foldl  as  foldr " as On the right side of the list as you fold Basically, if we type the  foldr  to  (a -> gt - r -> gt) - & gt; R - & gt; [A] - & gt; R  then we do  r  as  int -> gt; [A]  Where the integer we pass is the current number of elements which we have passed without calling the function.

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