Объяснение работы программы по задаче

program Provider;
  const MAX_N_CH = $40; INFTY = $7FFFFFFF;
  var heads : text; 
  fn: string;   
  chns : array[1..MAX_N_CH] of text;
  allfree, allbusy : text; 
  curr_val: array[1..MAX_N_CH] of longint;
  is_odd : array[1..MAX_N_CH] of boolean;
  NCh, i, minidx, used : byte;
  mintime : longint; 
Begin
  assign(heads, 'C:\..\provider.dat'); 
  reset(heads);
  readln(heads, NCh); 
  used := 0;
  for i := 1 to NCh do 
    is_odd[i] := true; 
  for i := 1 to NCh do 
  begin
    readln(heads, fn); 
    assign(chns[i], fn); 
    reset(chns[i]);
    if eof(chns[i]) then 
      curr_val[i] := INFTY
    else readln(chns[i], curr_val[i]);
  end;
  assign(allfree, 'C:\..\allfree.sol'); 
  rewrite(allfree);
  assign(allbusy, 'C:\..\allbusy.sol'); 
  rewrite(allbusy);
  write (allfree, '0');
  repeat
    mintime := curr_val[1]; 
    minidx := 1;
    for i := 2 to NCh do
      if (mintime > curr_val [i]) or (mintime = curr_val [i]) and (is_odd[i]) then 
      begin
        mintime := curr_val[i]; 
        minidx := i;
      end;
    if mintime = INFTY then 
      break;  
    if eof(chns[minidx]) then 
      curr_val[minidx] := INFTY
    else
    readln(chns[minidx], curr_val[minidx]);
    if is_odd[minidx] then 
    begin 
      used := used+1;
      if used = NCh then 
        write(allbusy, mintime);
      if used = 1 then  
        writeln(allfree,' ', mintime)
    end
    else 
    begin   
      used := used-1;
      if used = 0 then 
        write(allfree, mintime);
      if used = NCh-1 then 
        writeln(allbusy,' ', mintime)
    end;
    is_odd [minidx] := not is_odd [minidx] ; 
  until mintime = INFTY;
  if used = 0 then 
    writeln(allfree, ' now'); 
  if used = NCh then 
    writeln(allbusy,' now'); 
  close(allfree); 
  close(allbusy);
  for i := 1 to NCh do 
    close(chns[i])
end.