Сократил мою прогу для поиска циклов в десятичных дробях

исходная задача с Эйлера https://projecteuler.net/problem=26

import Data.List

findCycle xs = (take firstOccurenceInd tmp, drop firstOccurenceInd tmp)
  where repeatingElement = xs !! (length tmp)
        tmp = last $ takeWhile (check . reverse) (inits xs)
        firstOccurenceInd = length $ takeWhile (\x -> x /= repeatingElement) xs
        check [] = True
        check [x] = True
        check (x:xs) = if x `elem` xs then False else True

process n = (n, "0." ++ (tail prefixPart) ++ cyclicPart)
  where 
    prefixPart = concat $ map (\x -> show $ fst x) (fst $ cycleInfo) 
    cyclicPart = "(" ++ (concat $ map (\x -> show $ fst x) (snd $ cycleInfo)) ++ ")"
    cycleInfo = findCycle $ iterate divisionStep (0,1)
    divisionStep (_,x) = ( 10*x `quot` n, 10*x `mod` n)

main = do 
  print $ map process [2..10]
  print $ map process [100..120]

Выдает

[(2,"0.5(0)"),
(3,"0.(3)"),
(4,"0.25(0)"),
(5,"0.2(0)"),
(6,"0.1(6)"),
(7,"0.(142857)"),
(8,"0.125(0)"),
(9,"0.(1)"),
(10,"0.1(0)")]
....

[(100,"0.01(0)"),
(101,"0.(0099)"),
(102,"0.0(0980392156862745)"),
(103,"0.(0097087378640776699029126213592233)"),
(104,"0.009(615384)"),
...
(113,"0.(0088495575221238938053097345132743362831858407079646017699115044247787610619469026548672566371681415929203539823)"),
...

Но народ на Эйлере - монстры, они решили там задачу 26 еще радикально быстрее, даже не вычисляя сами по себе циклы, а использую математическое знание:

http://mathworld.wolfram.com/FullReptendPrime.html
the period of a number n's reciprocal is the smallest k such that n divides 10^k - 1.

так что у монстров решение такое

import Data.List

nums = [ n | n <- [3,5..], n `mod` 5 /= 0 ]

period n =
    head $ [ p | p <- [1..], (10^p - 1) `mod` n == 0 ]

answer =
    fst $
    maximumBy (\(_,a) (_,b) -> compare a b) $
    map (\n -> (n,period n)) $
    takeWhile (<1000) nums

main = print answer

На то они и монстры!