Project Euler #1 using Haskell

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]

---

你也可以使用硬编码版本：

sum $ [3, 6 .. 999] ++ [5, 10 .. 999] ++ [-15, -30 .. -999]

---

这会给你你要求的清单：

filter (x -> (x `mod` 3 == 0) || (x `mod` 5 == 0)) [1..999]

链接地址: http://www.djcxy.com/p/80422.html

上一篇: C vs Haskell Collat​​z猜想速度比较

下一篇: 项目欧拉＃1使用Haskell