递推第2题—凸多边形的三角形剖分

var c:array[2..22]of longint;//当Catalan数列中n=22时就超出长整型，需要用到高精度
      i,n,k:integer;
begin
 fillchar(c,sizeof(c),0); c[2]:=1;
 readln(n);
 for i:=3 to n do
   for k:=2 to i-1 do
    inc(c[i],[k]*c[i-k+1]);
 writeln(c[n]);
end.

program p2_2(input,output);
const max=21;
var c:array[2..max] of longint;
    n,i,k:integer;
    total:longint;
begin
  write(‘input n=‘);readln(n);
  c[2]:=1;
  for i:=3 to n do
    begin
      c[i]:=0;
      for k:=2 to i-1 do c[i]:=c[i]+c[k]*c[i-k+1];
    end;
  writeln(‘catalan=‘,c[n]);
end.

program p2_2a(input,output);
const maxn=1000;
type arraytype=array[0..maxn] of longint;
var i,j,n:longint;
    h:arraytype;
procedure mul(var h:arraytype;k:longint);
  var i:longint;
  begin
    for i:=0 to maxn do h[i]:=h[i]*k;
    for i:=0 to maxn-1 do
      begin
        h[i+1]:=h[i+1]+h[i] div 10;
        h[i]:=h[i] mod 10
      end
  end;
procedure devide(var h:arraytype;k:longint);
  var d,i,r:longint;
  begin
    r:=0;
    for i:=maxn downto 0 do
      begin
        d:=10*r+h[i];
        h[i]:=d div k;
        r:=d mod k
      end
  end;
begin {Main program}
  write(‘Input n:‘);
  readln(n);
  for i:=1 to maxn do h[i]:=0;
  h[0]:=1;
  for i:=3 to n-1 do
    begin
      mul(h,4*i-6);
      devide(h,i)
    end;
  i:=maxn;
  while (i>0) and (h[i]=0) do i:=i-1;
  for j:=i downto 0 do write(h[j]);
  writeln
end.