谁能在一秒内把13皇后的解得个数求出？我用了对称，可还是不行
var
  s:array[1..14]of integer;
  a,b,c:array[-14..28]of boolean;
  able:array[1..13,1..13]of boolean;
  i,j,k,l,m,n,sum:longint;
procedure serch(m:integer);
var
  i:integer;
begin
  if sum=3 then exit;
  if m=n+1 then begin inc(sum);
                      if sum<=3 then begin write(s[1]);for i:=2 to n do write(' ',s[i]);
                      writeln;end;
                      exit;
                end;
  for i:=1 to n do
    if (a[i])and(b[m-i])and(c[m+i]) then begin
                                           a[i]:=false;b[m-i]:=false;c[m+i]:=false;
                                           s[m]:=i;
                                           serch(m+1);
                                           a[i]:=true;b[m-i]:=true;c[m+i]:=true;
                                         end;
end;
procedure greatserch(m:integer);
var
  i,k:integer;
begin
  if m=n+1 then begin inc(sum);
                      exit;
                end;
  k:=n;
  if m=1 then begin k:=k div 2; end;
  for i:=1 to k do
    if (a[i])and(b[m-i])and(c[m+i]) then begin
                                           a[i]:=false;b[m-i]:=false;c[m+i]:=false;
                                           greatserch(m+1);
                                           a[i]:=true;b[m-i]:=true;c[m+i]:=true;
                                         end;
end;
procedure superserch(m:integer);
var
  i,a1,b1:integer;
begin
  if m=n+1 then begin inc(sum,2);
                      exit;
                end;
  b1:=n;a1:=1;
  if m=1 then begin a1:=n div 2+1; b1:=n div 2+1; end;
  if m=2 then b1:=b1 div 2;
  for i:=a1 to b1 do
    if (a[i])and(b[m-i])and(c[m+i]) then begin
                                           a[i]:=false;b[m-i]:=false;c[m+i]:=false;
                                           superserch(m+1);
                                           a[i]:=true;b[m-i]:=true;c[m+i]:=true;
                                         end;
end;
begin
  assign(input,'checker.in');reset(input);
  assign(output,'checker.out');rewrite(output);
  readln(n);sum:=0;
  for i:=-14 to 28 do begin a[i]:=true;b[i]:=true;c[i]:=true; end;
  fillchar(able,sizeof(able),true);
  serch(1);
  sum:=0;
  for i:=-14 to 28 do begin a[i]:=true;b[i]:=true;c[i]:=true; end;
  greatserch(1);
  sum:=sum*2;
  if n mod 2=1 then superserch(1);
  writeln(sum);
  close(input);close(output);
end.
是usaco的c1.4 checker
我已经等了很久了，谁帮帮我！