import Data.Set
euler :: Int
euler = sum [ x | x <- nums ]
where
nums = Data.Set.toList (Data.Set.union (Data.Set.fromList [3,6..999])
(Data.Set.fromList [5,10..999]))
I am learning Haskell and hope you don't mind me asking this. Is there a nicer way to get a list holding all natural numbers below one thousand that are multiples of 3 or 5? (Eg with zip or map?)
Edit:
import Data.List
euler :: Int
euler = sum (union [3,6..999] [5,10..999])
Thanks for your help, guys.
使用列表理解:
sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]
您也可以使用硬编码版本:
sum $ [3, 6 .. 999] ++ [5, 10 .. 999] ++ [-15, -30 .. -999]
这将为您提供您要求的列表:
filter (\x -> (x `mod` 3 == 0) || (x `mod` 5 == 0)) [1..999]
Here's one.
mults35 = union [3,6..999] [5,10..999]
where
union (x:xs) (y:ys) = case (compare x y) of
LT -> x : union xs (y:ys)
EQ -> x : union xs ys
GT -> y : union (x:xs) ys
union xs [] = xs
union [] ys = ys
Here's another, less efficient way:
import Data.List
nub . sort $ ([3,6..999] ++ [5,10..999])
(we don't have to use fully qualified names if we have the import statement).
Also interesting is to find the multiples of only 3 and 5:
m35 = 1 : (map (3*) m35 `union` map (5*) m35)
sum [x | x <- [1..999], let m k = (x`mod`k==0), m 3 || m 5]
A more general solution for a list of numbers instead of just 3 and 5:
addMultiples :: [Int] -> Int -> Int
addMultiples multiplesOf upTo = sum[n | n <- [1..upTo-1], or (map ((0==) . mod n) multiplesOf)]
Here's one that is super-fast. Try it with values of over a billion.
eu x = sum[div (n*(p*(p+1))) 2 | n<-[3,5,-15], let p = div (x-1) n]
I imagine it can be shortened further.
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