Here's a solution with F#:
#light
type Appetizer = { name : string; cost : int }
let menu = [
{name="fruit"; cost=215}
{name="fries"; cost=275}
{name="salad"; cost=335}
{name="wings"; cost=355}
{name="moz sticks"; cost=420}
{name="sampler"; cost=580}
]
// Choose: list -> list -> int -> list>
let rec Choose allowedMenu pickedSoFar remainingMoney =
if remainingMoney = 0 then
// solved it, return this solution
[ pickedSoFar ]
else
// there's more to spend
[match allowedMenu with
| [] -> yield! [] // no more items to choose, no solutions this branch
| item :: rest ->
if item.cost <= remainingMoney then
// if first allowed is within budget, pick it and recurse
yield! Choose allowedMenu (item :: pickedSoFar) (remainingMoney - item.cost)
// regardless, also skip ever picking more of that first item and recurse
yield! Choose rest pickedSoFar remainingMoney]
let solutions = Choose menu [] 1505
printfn "%d solutions:" solutions.Length
solutions |> List.iter (fun solution ->
solution |> List.iter (fun item -> printf "%s, " item.name)
printfn ""
)
(*
2 solutions:
fruit, fruit, fruit, fruit, fruit, fruit, fruit,
sampler, wings, wings, fruit,
*)
