Schoolar farewell task #2

program first_half_twelve_factorial_terms_for_a217626;

label
  The_end;

(*

Devised and released for documentation purposes under the terms of use of
"The On-Line Encyclopedia Of Integer Sequences" (OEIS.org).

For details please visit: http://www.oeis.org


Dear user/reader:
-----------------

When it is used properly, this code should help you to find the first ( (12!/2)-1 ) terms
of A217626, a task for which you need to have stored at least the first ( (12!/2)+1 ) terms
of A215940.

The idea behind this processing was taken from a test session under PARI/GP like:

x=[6,0,1,2,3,4,5,7,8,9,10,11];
z=sum(y=1,12,x[y]*12^(12-y));
print(z);

This program is not expensive in lines of code and extra details such as tech checkings...
It is assumed here that you know what you are doing when messing with it, for this
subtle reason previously mentioned, this code goes without any warranty whatsoever.

*)

const
  base=12;
  denominator=base-1; (*Unconditionally should be the quantity (base-1) in decimal*)
  power12: array [1..base] of int64=
  (base*base*base*base*base*base*base*base*base*base*base,
   base*base*base*base*base*base*base*base*base*base,
   base*base*base*base*base*base*base*base*base,
   base*base*base*base*base*base*base*base,
   base*base*base*base*base*base*base,
   base*base*base*base*base*base,
   base*base*base*base*base,
   base*base*base*base,
   base*base*base,
   base*base,
   base,
   1);

var
input_f:text;
input_s:string;
delta_A_operand,
delta_B_operand,
delta_quotient,
prev_quotient: int64;
register: array [1..100] of int64;

function f(x:char):byte;
var
  ans:byte;
begin
  if (x in ['a'..'z']) then begin
    ans:=10+ord(x)-ord('a');
  end else if (x in ['0'..'9']) then begin
    ans:=ord(x)-ord('0');
  end else begin
    ans:= 255;
  end;
  f:=ans;
end;

function g(y:string):int64;
var
  ans:int64;
  k:byte;
begin
  ans:=0;
  if (length(y) = base) then for k:= 1 to base do begin
    ans:= ans + f(y[k])*power12[k];
  end;
  g:=ans;
end;

function p(q:longint):longint;
var
  ans,
  u:longint;
begin
  ans:=1;
  if (q >= 1) then begin
     for u:= 1 to q do begin
    ans:= ans*10;
     end;
  end;
  p:=ans;
end;

function h(b:longint; z:int64):int64;
var
  ans,
  cz:int64;
  k,
  w:longint;
begin
  ans:=0;
  cz:=z;
  k:=1;
  while (cz >= b) do begin
    register[k]:= cz mod b;
    cz:= cz div b;
    k:=k+1;
  end;
  register[k]:=cz;
  for w:= k downto 1 do begin
    ans:= ans+register[w]*p(w-1);
  end;
  h:=ans;
end;

(*
    The filename of the data intended to be processed here should be
    passed as the first and unique parameter in the command line
    invocation to this program. The output is made directly to the
    console, so normally you should redirect it to a file in order
    to save the answer (Just a term of A217626 in each line, like
    in a b-file but without including the offsets at left)
*)

begin
  if (paramcount() = 1) then begin
    assign(input_f, paramstr(1));
    (*$i-*)reset(input_f);(*$i+*)
    if (IOResult = 0) then begin
      readln(input_f, input_s);
      delta_A_operand:= g(input_s);
      (*
      writeln('0');
      *)
      prev_quotient:= 0;
      while (not eof(input_f)) do begin
    readln(input_f,input_s);
    delta_B_operand:= g(input_s);
    delta_quotient:=h(base,(delta_B_operand - delta_A_operand) div denominator);
    writeln(delta_quotient-prev_quotient);
    (* Toggle the following comment switch in order to obtain either A215940 (OFF) or A217626 (ON). Default is OFF; *)
    (**)
    prev_quotient:= delta_quotient;
    (**)
      end;
    end else begin
      writeln('Unexpected error when processing: ',paramstr(1),'; **ABORTING**');
      goto The_end;
    end;
    close(input_f);
  end else begin
    writeln('You must specify one and only one filename as invocation parameter. Please try again.');
  end;
  The_end:
end.