Total length of all strings is no more then 10^7 characters.

Simple programm reads such file 4-6 seconds!!!
Time limit 2 seconds!

>
> Simple programm reads such file 4-6 seconds!!!
> Time limit 2 seconds!
>

Judge system doesn't count time of reading or writing.

> Total length of all strings is no more then 10^7
characters.
>
> Simple programm reads such file 4-6 seconds!!!
> Time limit 2 seconds!
>

You are absolutely wrong, modern hard disks read 30-40 megs
per second!

Just use the following schema, like I did:

#define BUFSZ 16384
int pos=BUFSZ;
char buf[BUFSZ];

void getnextchar(char&c)
{
    if(pos==BUFSZ) {
        pos=0;
        for(int i=0;i<BUFSZ;++i) buf[i]='\n';
        fread(buf,BUFSZ,1,stdin);
    }
    c=buf[pos++];
}

> Judge system doesn't count time of reading or writing.

Really?
Hm.. :)
I wonder how i can find out the time my program works, if
i'm reading and processing data in series?
I also got Time Limit, though algorithm is surely a
primitive o(n)! Even when i make a buffer of 255 bytes.

I think the bad thing is: this program is made for being
solved on C, not on Pascal. Because you can make the buffer
of any length on C and work with stdin as with any binary
file, that means read the amount of bytes you want.
But in Pascal and OP, Input is a TEXT FILE!!! And you can't
read more, than 255 bytes at a time.
How could some people wrote it on Pascal and have it
accepted, i wonder?

Hey, guys! How did you read the input????

And another thing i do not like here is: time limits are
counted in non-standart manner. Common ACM rules say all
problems are judged on 386-586, so people can freely
evaluate the estimate work time, even if they write
programs on different machines (it is so, usually!).

So, it would be good, if somebody would show us some
information about the judge system. I mean the work-time of
empty cycle from 1 to 10^6, work-time of reading 10^6
bytes, etc.
Because there were some problems, that i hadn't submitted,
because they hadn't met the desired time on my machine. And
i lost too much time invain, trying to improve the
algorithm, but finally got "accepted" without any hope to!

Var  ...
    k,i:integer;
    Buf:string[255];
    Buflen:integer;

BEGIN
 Readln(n);
  ...
 for i:=1 to n do
 begin
  ...
  k:=0;
  read(buf);
  buflen:=length(buf);

  if buf='' then
  begin Writeln('YES'); continue; end;

  repeat

   inc(k);
   ....
    {here i process one char buf[k] in CONSTANT time!!!}
   ....
   if k=buflen then
    begin read(buf); buflen:=length(buf); k:=0; end;

  until (buf='');
  ...
  while buf<>'' do read(buf);
  readln;
 end
END.


So, i process data in series of 255 bytes - and still, it
works too slow... :(
The reading is too slow!

var
  BUF : array [1..buf_size] of char;
  buft, bufp : longint;
  tmp : word;
  fin : file;

procedure getchar(var c : char);
begin
  if bufp > buft then
  begin
    bufp := 1;
{$IFDEF BP}
    blockread(fin, BUF, buf_size, tmp);
    buft := tmp;
{$ELSE}
    blockread(fin, BUF, buf_size, buft);
{$ENDIF}
    if buft = 0 then
    begin
      c := #10;
      exit;
    end;
  end;
  c := BUF[bufp];
  inc(bufp);
  if c = #13 then getchar(c);
end;

It seems that blockread is fast enough. Btw I think the
bigest speedup comes when you skip the chars till the end
of the line when you are sure the dialog is incorrect. Also
if you are using blockread you would have to read n char by
char (parse it), and note that delphi insists that the
variable where it returns the chars read is longint, while
BP7.0 wants it to be word(i got some compile errors that
way).

Yeah, thanks for help,
but i knew quite well, that it would be fast with
blockread, but the problem is: we do not read from the
file, we read from INPUT.
"BlockRead(f,.." doesn't work with INPUT, because the type
of INPUT is TEXT.
Perhaps i just do not know something, but what do you do to
assign your FIN:FILE to INPUT?

I would appreciate that part of help

Don't you know that the file name '' refers to standard
input or output?
So just do this :
var
  fin : file;
begin
  assign(fin, '');
  reset(fin, 1) {1 is important for blockread}
  .............
  .............
  close(fin);
end.

You may use the same technique to write to stdout with
blockwrite (if you reset '' means stdin if you rewrite it
means stdout).

WWWWHHHHAAAATTTT????????
I THOUGHT AS MUCH!!
Thanks, now everything's clear - i'll do it in an hour...
Thanks thanks and thakns again.
I do think now, that every problem has simple, fast and
right solution... :)))))

> > Judge system doesn't count time of reading or writing.
>
> Really?

Yes. The judge counts only so called "User Time" on NT
system. This make it less dependent on harddisk speed.

> Hm.. :)
> I wonder how i can find out the time my program works, if
> i'm reading and processing data in series?
> I also got Time Limit, though algorithm is surely a
> primitive o(n)! Even when i make a buffer of 255 bytes.

Let's count. Input size is 10^7. It contains 1000 lines.
So, there is line with at least 10^7/1000 = 10000
characters. May be longer. Why are you using any buffers?

> I think the bad thing is: this program is made for being
> solved on C, not on Pascal.

Author of this problem understands difference between C and
Pascal. All problems are solvable on any language.

> And another thing i do not like here is: time limits are
> counted in non-standart manner. Common ACM rules say all
> problems are judged on 386-586, so people can freely
> evaluate the estimate work time, even if they write
> programs on different machines (it is so, usually!).
>
> So, it would be good, if somebody would show us some
> information about the judge system. I mean the work-time
of
> empty cycle from 1 to 10^6, work-time of reading 10^6
> bytes, etc.

Thank you. You say wise things. We'll work on it.

> Because there were some problems, that i hadn't
submitted,
> because they hadn't met the desired time on my machine.

It's hard enough to predict worktime on our judge, because
of my first answer. You should try.