Euler, ex. 1 F#.fs

#light

(* If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9.
   The sum of these multiples is 23.
   Find the sum of all the multiples of 3 or 5 below 1000. *)

(* Easiest way: Use a set *)
let multiples = new Set<int>([0..3..1000] @ [0..5..1000])
let answer = Set.fold (+) 0 multiples

(* More compact *)
let answer = Set.fold (+) 0 (new Set<int>([0..3..1000] @ [0..5..1000]))

(* More efficient *)
let multiplesOf3ButNot5 = [for x in [0..3..1000] do
                           if x % 5 <> 0 then yield x]
let sumOfList x = List.fold (+) 0 x
let answer = sumOfList(multiplesOf3ButNot5) + sumOfList([0..5..995]) (* What, no exclusive range? *)

(* Lazy sequence *)
let multiples = seq { for x in  0
                  |> Seq.unfold( fun(x) -> Some(x, x+1)) do
                  if x % 3 = 0 || x % 5 = 0 then yield x }
let answer =  Seq.takeWhile (fun(x) -> x < 1000) multiples |> Seq.fold (+) 0