I know that my algorithm exactly correct but only mathematically but not in program

var n1:longword;
function max(n:longword):word;
begin
  max:=max+(n mod 2);
  if (n<>1) then
  if n mod 2=0 then
    max(n div 2)
  else
  begin
    max(n div 2);
    max((n div 2)+1);
  end;
end;

begin
  readln(n1);
  writeln(max(n1));
  readln;
end.

Maybe I do not fully understand how do recursive functions operate.

Edited by author 02.04.2016 15:07

If you change word to longint, it returns even more~
Even if you leave just
function max(n:longword):word;
begin
 max:=max+(n mod 2);
end;
, it will return an enormous number. Because, you see, in this very first row, you try to assign the function its own value (plus something). But it might be initialized in its own mysterious ways. So it's not how you do it.

So uh... i solved this one long ago without any recursion, and i'm not quite sure what you're trying to do, but i modified it a bit, so at least it's trying to make some sense (even though it returns the wrong result still; but it's not that bad anymore, and maybe you'll be able to fix it into a proper version). So there you go.

function max(n:longint):longint;
var funcRes: longint;
begin
    funcRes:=(n mod 2);
    if (n <> 1) then
        if n mod 2 = 0 then inc(funcRes, max(n div 2))
    else begin
        inc(funcRes, max(n div 2));
        inc(funcRes, max((n div 2)+1));
    end;
    max:=funcRes;
end;

Thank you very much. I understood

No problem. If you want to use recursion for calculations, maybe i'd suggest something like this:

var i, n1, r1, maximum: longint;
function x(p: longint): longint;
begin
    if p <= 1 then x:=p else
    if p mod 2 = 0 then x:=x(p div 2)
                   else x:=x(p div 2) + x(p div 2 + 1);
end;
begin
    readln(n1);
    maximum:=0;
    for i:=0 to n1 do begin
        r1:=x(i);
        if maximum < r1 then maximum:=r1;
    end;
    writeln(maximum);
    readln;
end.

But i think it's kinda slower than if you'd just precalculated the array, and then precalculated another array for maximums.
It was quite likely harder back in the days, with 64kb turbo pascal limitations and all, so maybe you'd have to use smarter ways, rather than those blunt ones nowadays~

I meant I understood that  I've used wrong way to get maximum. By the way I accidentally started to write a program that prints out the number from this sequence. And namely this way I think would be good to solve it. I was interested to write such a program and that's what I got:
var n:longword; r:word;

procedure max(n1,n2:word; var res:word);
begin
  if n1 and (n1-1)=0 then inc(res) else
  if n1=3 then inc(res,2) else
  if n1 mod 2=0 then max(n1 div 2,0,res) else max(n1 div 2, (n1 div 2)+1,res);
  if n2<>0 then
  if n2 and (n2-1)=0 then inc(res) else
  if n2=3 then inc(res,2) else
  if n2 mod 2=0 then max(n2 div 2,0,res) else max(n2 div 2, (n2 div 2)+1,res);

end;



begin
  readln(n);
  if n=1 then writeln(1) else
  if n and (n-1)=0 then writeln(1) else
  if n mod 2=0 then begin max(n div 2,0,r); writeln(r); end else
  begin
    max(n div 2, (n div 2)+1,r);
    writeln(r);
  end;
  readln;
end.

Also div 2 can be shl 1

Edited by author 02.04.2016 20:42

Edited by author 02.04.2016 20:44