content: {- 
   - Solution to Project Euler problem 71
   - Copyright (c) Project Nayuki. All rights reserved.
   - 
   - https://www.nayuki.io/page/project-euler-solutions
   - https://github.com/nayuki/Project-Euler-solutions
   -}
  
  import Data.Ratio ((%), numerator)
  import qualified EulerLib
  
  
  {- 
   - We consider each (integer) denominator d from 1 to 1000000 by brute force.
   - For a given d, what is the largest integer n such that n/d < 3/7?
   - 
   - * If d is a multiple of 7, then the integer n' = (d / 7) * 3 satisfies n'/d = 3/7.
   -   Hence we choose n = n' - 1 = (d / 7) * 3 - 1, so that n/d < 3/7.
   -   Since (d / 7) * 3 is already an integer, it is equal to floor(d * 3 / 7),
   -   which will unifie with the next case. Thus n = floor(d * 3 / 7) - 1.
   - * Otherwise d is not a multiple of 7, so choosing n = floor(d * 3 / 7)
   -   will automatically satisfy n/d < 3/7, and be the largest possible n
   -   due to the definition of the floor function.
   - 
   - When we choose n in this manner, it might not be coprime with d. In other words,
   - the simplified form of the fraction n/d might have a denominator smaller than d.
   - 
   - Let's process denominators in ascending order. Each denominator generates a pair
   - of integers (n, d) that conceptually represents a fraction, without simplification.
   - Whenever the current value of n/d is strictly larger than the previously saved value,
   - we save this current value of (n, d).
   - 
   - If we handle denominators in this way - starting from 1, counting up consecutively -
   - then it is guaranteed that our final saved pair (n, d) is in lowest terms. This is
   - because if (n, d) is not in lowest terms, then its reduced form (n', d') would have
   - been saved when the smaller denominator d' was processed, and because n/d is
   - not larger than n'/d' (they are equal), the saved value would not be overwritten.
   - Hence in this entire computation we can avoid explicitly simplifying any fraction at all.
   -}
  main = putStrLn (show ans)
  ans = numerator $ foldl1 max [((div (d * 3) 7) - (EulerLib.boolToInt ((mod d 7) == 0))) % d | d <- [1 .. 10^6]]
  