回溯法第6题—0/1字符串问题

2

4

var a:array[1..40]of byte;//0/1序列
    tot:longint;
    L:integer;//位指针
    n:integer;
function judge(L:integer):boolean;
var x,y,z:integer;r,q,p:integer;
begin
  x:=L;y:=L-1;z:=L-2;//指针初始化
  while z>0 do
   begin
    p:=x;q:=y;r:=z;
    while (r<y)and(a[p]=a[q])and(a[q]=a[r]) do begin inc(p);inc(q);inc(r);end;//遍历子串
    if r=y then exit(false);//如果三个子串完全一样就返回false
    dec(x,1);dec(y,2);dec(z,3);
   end;
  exit(true);
end;
procedure search(L:integer);
begin
   if L>n then begin inc(tot,2); exit; end;
   a[L]:=0;
   if judge(L) then search(L+1);
   a[L]:=1;
   if judge(L) then search(L+1);
 end;
begin
  readln(n);
  a[1]:=0;
  L:=1;
  tot:=0;
  search(2);
  writeln(tot);
end.