CATALAN – SPOJ

Đề bài: http://vn.spoj.com/problems/CATALAN/

Thuật toán:

• Bài này sử dụng phương pháp đệ quy có nhớ. Bạn có thể đọc code để hiểu thêm

Code:

const
  fi='';
  fo='';
  maxn=21;
var
  f : array[1..2*maxn,0..2*maxn] of int64;
  i,j,n : longint;
  x,kq2 : array[1..2*maxn] of longint;
  res,kq1,k : int64;
procedure enter;
  begin
    assign(input,fi);reset(input);
    readln(n);
    for i:=1 to 2*n+1 do read(x[i]);
    readln(k);
    close(input);
  end;
function tinh(i,tr : longint) : int64;
  var sl : int64;
  begin
    if f[i,tr]>-1 then exit(f[i,tr]);
    if i>2*n+1 then
      if tr=0 then exit(1) else exit(0);
    sl := 0;
    sl := sl + tinh(i+1,tr+1);
    if tr>0 then
    sl := sl + tinh(i+1,tr-1);
    f[i,tr] := sl;
    exit(sl);
  end;
procedure truyvan1;
  begin
    kq1 := 0;
    for i:=2 to 2*n do
      if x[i]>x[i-1] then
        if x[i]-2>=0 then
          kq1 := kq1 + tinh(i+1,x[i]-2);
    inc(kq1);
  end;
procedure lankq(i : longint);
  begin
    if i>2*n then exit;
    if kq2[i-1]-1>=0 then
      begin
        if k>tinh(i+1,kq2[i-1]-1) then
          begin
            k:=k-tinh(i+1,kq2[i-1]-1);
            kq2[i] := kq2[i-1]+1;
            lankq(i+1);
            exit;
          end;
      end
      else
        begin
          kq2[i] := kq2[i-1]+1;
          lankq(i+1);
          exit;
        end;
    kq2[i] := kq2[i-1]-1;
    lankq(i+1);
  end;
procedure truyvan2;
  begin
    kq2[1] := 0;
    lankq(2);
  end;
procedure process;
  begin
    fillchar(f,sizeof(f),255);
    res := tinh(2,0);
    f[1,0] := res;
    truyvan1;
    truyvan2;
  end;
procedure print;
  begin
    assign(output,fo);rewrite(output);
    //writeln(res);
    writeln(kq1);
    for i:=1 to 2*n+1 do write(kq2[i],' ');
    close(output);
  end;
begin
  enter;
  process;
  print;
end.