I do not understand why my solution gets wr6
like written correct do not understand why wr6(
this text my programm
------
var s:string;
    trol,tram,p:integer;
begin
  tram:=0;trol:=0;
  while not eof() do begin
    readln(s);
    s:=' '+s+' ';
    p:=pos('tram',s);
    if p>0 then begin
      if s[p-1]<>' ' then p:=0;
      if (p<>0)and (s[p+4]<>' ')then p:=0;
    end;
    while p>0 do begin
      inc(tram);
      delete(s,p,4);
      p:=pos('tram',s);
      if p>0 then begin
        if s[p-1]<>' ' then p:=0;
        if (p<>0)and (s[p+4]<>' ')then p:=0;
      end;
    end;

    p:=pos('trolleybus',s);
    if p>0 then begin
      if s[p-1]<>' ' then p:=0;
      if (p<>0)and (s[p+10]<>' ')then p:=0;
    end;
    while p>0 do begin
      inc(trol);
      delete(s,p,10);
      p:=pos('trolleybus',s);
      if p>0 then begin
        if s[p-1]<>' ' then p:=0;
        if (p<>0)and (s[p+10]<>' ')then p:=0;
      end;
    end;
  end;

  if trol=tram then writeln('Bus driver')
  else
  if tram>trol then writeln('Tram driver')
  else writeln('Trolleybus driver');
end.

Possibly, the shortest test on which this prog fails is:

trama tram

Yes, you are right, after he changed me again wr6
I do not understand that why
this code:

var s,cur:string;
    trol,tram,i,fr:integer;
begin
  tram:=0;trol:=0;
  while not eof() do begin
    readln(s);
    s:=' '+s+' ';
    cur:='';fr:=0;
    for i:=1 to length(s) do begin
      if s[i]=' ' then begin
        inc(fr);
        if fr=2 then begin
         fr:=1;
         if cur='tram' then inc(tram);
         if cur='trolleybus' then inc(trol);
         cur:='';
        end;
      end
      else cur:=cur+s[i];
    end;
  end;

  if trol=tram then writeln('Bus driver')
  else
  if tram>trol then writeln('Tram driver')
  else writeln('Trolleybus driver');
end.

Edited by author 03.04.2008 03:34

This is even simpler to fail:

tram!

on test tram! my programm prints 'Bus driver' because No there tram.

Thanks I AC