% 2017 Round 1A, Problem C, in Picat, by Afa Zhou
% https://code.google.com/codejam/contest/5304486/dashboard#s=p2&a=1
% Play the Dragon

import util,planner.

main =>
    T = read_line().to_int(),
    foreach (TC in 1..T)
        L = [to_int(Token) : Token in read_line().split()],
        L = [Hd|_],
        State = to_array([Hd|L]),
        (best_plan_unbounded(State,Plan) ->
            printf("Case #%w: %w\n", TC, len(Plan))
        ;
            printf("Case #%w: IMPOSSIBLE\n", TC)
        )
    end.

final({_,Hd,Ad,Hk,Ak,B,D}), Hk == 0 => true.

action({Hd0,Hd,Ad,Hk,Ak,B,D}, NS,A,Cost) ?=>
    A = a, Cost = 1,
    Hk1 = Hk-Ad,
    (Hk1 > 0 -> Hd1 = Hd-Ak, Hd1 > 0; Hd1 = Hd),
    NS = {Hd0,Hd1,Ad,max(Hk1,0),Ak,B,D}.
action({Hd0,Hd,Ad,Hk,Ak,B,D}, NS,A,Cost) ?=>
    A = b, Cost = 1,
    Ad+B =< Hk,
    Hd1 = Hd-Ak, Hd1 > 0,
    NS = {Hd0,Hd1,Ad+B,Hk,Ak,B,D}.
action({Hd0,_Hd,Ad,Hk,Ak,B,D}, NS,A,Cost) ?=>
    A = c, Cost = 1,
    Hd1 = Hd0-Ak, Hd1 > 0,
    NS = {Hd0,Hd1,Ad,Hk,Ak,B,D}.
action({Hd0,Hd,Ad,Hk,Ak,B,D}, NS,A,Cost) =>
    A = d, Cost = 1,
    Ak1 = max(Ak-D,0),
    Hd1 = Hd-Ak1, Hd1 > 0,
    NS = {Hd0,Hd1,Ad,Hk,Ak1,B,D}.

picat-lang.org