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type cout_carac = int * int * int
type t = {
value : int
; couts : cout_carac
; bonus : int
}
let create : ?bonus:int -> int -> cout_carac -> t =
fun ?(bonus = 0) value couts -> { value; couts; bonus }
let incr ?(step = 1) t = { t with bonus = t.bonus + step }
(*
Evaluate the cost for the successives upgrades.
I’m pretty sure this can be transformed into a linear function, but I do not see how…
c0 * t.bonus
+ max 0 ((((t.bonus - 1) * 2) - 1) * c1)
+ ?
*)
let cout : t -> int =
fun t ->
let c0, c1, c2 = t.couts in
let rec c acc t =
match t with
| 0 -> acc
| 1 -> acc + c0
| 2 -> c (acc + c0 + c1) (t - 1)
| 3 -> c (acc + c0 + (c1 * 2)) (t - 1)
| n -> c (acc + c0 + (c1 * 2) + ((n - 3) * c2)) (t - 1)
in
c 0 t.bonus
let value t = t.value + t.bonus
let repr : Format.formatter -> t -> unit =
fun out t -> Format.fprintf out "%d (+%d)" (value t) t.bonus
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