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Discussion of Problem 1195. Ouths and Crosses

whi i got WA. it is simple
Posted by I am david. Tabo. 24 Oct 2002 23:10
const    fi     = '1195.inp';

         max    = 3;

         so     = 8;
         line   :array[1..so,1..3,1..2] of byte
                =( ( (1,1), (1,2), (1,3) ),
                   ( (2,1), (2,2), (2,3) ),
                   ( (3,1), (3,2), (3,3) ),
                   ( (1,1), (2,1), (3,1) ),
                   ( (1,2), (2,2), (3,2) ),
                   ( (1,3), (2,3), (3,3) ),
                   ( (1,1), (2,2), (3,3) ),
                   ( (1,3), (2,2), (3,1) ) );

var      a,b        :array[1..3,1..3] of char;
         kq         :byte;


procedure input;
var   f         :text;
      i,j       :byte;
      st        :string;
begin
  {assign(f, fi); reset(f);}
  for i := 1 to 3 do
  begin
    readln({f}, st);
    while st[1] = ' ' do delete(st,1,1);
    while st[ length(st) ] = ' ' do delete(st, length(st), 1);

    for j := 1 to 3 do a[i,j] := st[j];
  end;
  {close(f);}
end;



procedure out;
begin
  case kq of
  0 : writeln('Draw');
  1 : writeln('Crosses win');
  2 : writeln('Ouths win');
  end;
end;



function thang :boolean;
var     i,j    :byte;
        c      :char;
        ok     :boolean;
begin
  thang := true;

  for i := 1 to so do
  begin
   ok := true;

   c  := b[ line[i,1,1], line[i,1,2] ];
   if c = '#' then ok := false;

   for j := 1 to 3 do
     if b[ line[i,j,1], line[i,j,2] ] <> c
     then ok := false;

   if ok then exit;
  end;

  thang := false;
end;




function cross :boolean;
var     i,j             :byte;
begin
  cross := true;

  for i := 1 to 3 do
    for j := 1 to 3 do
     if a[i,j] = '#' then
     begin
       b := a;
       b[i,j] := 'X';
       if thang then
         begin  kq := 1;
                exit;
         end;
     end;

  cross := false;
end;



procedure outh;
var     i,j,u,v         :byte;
        ok              :boolean;
begin
  for i := 1 to 3 do
    for j := 1 to 3 do
     if a[i,j] = '#' then
     begin
       b := a; b[i,j] := 'X';

       ok := false;
       for u := 1 to 3 do
         for v := 1 to 3 do
           if b[u,v] = '#' then
           begin
             b[u,v] := 'O';
             if thang then ok := true;
             b[u,v] := '#';
           end;

       if not ok then exit;
     end;

  kq := 2;
end;



procedure solve;
begin
  kq := 0;
  if cross then exit;
  outh;
end;


begin
  input;
  solve;
  out;
end.