「HaskellでProject Euler(Problem 4～6)」ブログ用

euler4.hs

import Data.List
 
euler4 = maxPalindrome 100 999
 
maxPalindrome :: (Enum a, Num a, Ord a, Show a) => a -> a -> a
maxPalindrome n m =
  let list = sortBy (\x y -> compare y x) [x * y | x <- [n..m], y <- [n..m]] in
  head $ filter isPalindrome list
    where
      isPalindrome x = let s = show x in s == reverse s

euler5-1.hs

euler5 = smallestMultiple 20

smallestMultiple :: Integral a => a -> a
smallestMultiple n = let list = [1..(n - 1)] in
  smallestMultiple' n $ filter (\x -> all (\y -> y `mod` x /= 0 || x == y) list) list
  where
    smallestMultiple' n' list
      | all (\x -> n' `mod` x == 0) list = n'
      | otherwise                        = smallestMultiple' (n + n') list

euler5-2.hs

euler5 = smallestMultiple 20

smallestMultiple :: Integral a => a -> a
smallestMultiple n = head $ filter (\c -> (all (\d -> c `mod` d == 0) divisors)) candidates
  where
    divisors =
      let list = [1..(n - 1)] in
      reverse $ (filter (\x -> all (\y -> y `mod` x /= 0 || x == y) list) list)
    candidates = [n, (n * 2)..(foldl1 (*) divisors)]

euler5-3.hs

-- ghc -O euler5-3.hs と「-O」最適化オプションをつけてコンパイルすることで、高速になる

main = print euler5

euler5 = smallestMultiple 20

smallestMultiple :: Integral a => a -> a
smallestMultiple n = head $ filter (\c -> (all (\d -> c `mod` d == 0) divisors)) candidates
  where
    divisors = take (fromIntegral n `div` 2) [(n - 1), (n -2)..1]
    candidates = [n, (n * 2)..]

euler5-4.hs

-- ghc -O euler5-4.hs と「-O」最適化オプションをつけてコンパイルすることで、高速になる

main = print euler5
 
euler5 = smallestMultiple 20
 
smallestMultiple :: Integral a => a -> a
smallestMultiple n = head [c| c <- [n, (n * 2)..], and [c `mod` d == 0| d <- divisors]]
  where
    divisors = take (fromIntegral n `div` 2) [(n - 1), (n -2)..1]

euler6.hs

euler6 = (sumSquare 100) - (squareSum 100)

sumSquare :: (Enum a, Num a) => a -> a
sumSquare n = (foldl1 (+) [1..n]) ^ 2

squareSum :: (Enum a, Num a) => a -> a
squareSum n = foldl1 (+) [x ^ 2 |x <- [1..n]]