Another Haskell solution:
dont_give_me_five :: [Int] -> Int dont_give_me_five = foldl (\x y -> if notElem '5' $ show y then x + y else x) 0
Or a little bit shorter in the same language:
dont_give_me_five :: [Int] -> Int dont_give_me_five = foldl1 (\x y -> if notElem '5' $ show y then x + y else x)
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Another Haskell solution:
Or a little bit shorter in the same language: