/*

  Euler Problem 12 in SICStus Prolog

  Problem 12
  """
  The sequence of triangle numbers is generated by adding the natural numbers. 
  So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. 
  The first ten terms would be:

  1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

  Let us list the factors of the first seven triangle numbers:

       1: 1
       3: 1,3
       6: 1,2,3,6
      10: 1,2,5,10
      15: 1,3,5,15
      21: 1,3,7,21
      28: 1,2,4,7,14,28

  We can see that the 7th triangle number, 28, is the first triangle number 
  to have over five divisors.

  Which is the first triangle number to have over five-hundred divisors?")
  """


  Model created by Hakan Kjellerstrand, hakank@gmail.com
  See also my SICStus Prolog page: http://www.hakank.org/sicstus_prolog/

*/

:- ensure_loaded(hakank_utils).

go :-
        L = [
            euler12a % ,
            % euler12b
            ],
        run_problems(L).


%%
%% 0.145s
%%
euler12a :- 
        e12a(0,0,Num,0,_Len),
        writeln(Num).

e12a(_N,Num,Num,Len,Len) :- Len > 500.
e12a(N,Num0,Num,Len0,Len) :-
        Len0 < 500,
        Num1 is Num0+N+1, % the N'th triangle number
        num_divisors(Num1,Len1),
        N1 is N+1,
        e12a(N1,Num1,Num,Len1,Len).

%%
%% 0.165s
%%
euler12b :-
    Limit = 500,
    Found = _,
    ( for(I,1, 15000),
      param(Limit,Found) do
      p12b(I,T,Len),
      (var(Found), Len >= Limit) ->
          writeln(T),
          Found = true 
      ;
          true
      ).

p12b(N, T, Len) :-
        triangle(N,T),
        num_divisors(T,Len).