NO OF BINARY SEARCH TREE POSSIBLE ON NODES IN C

/* following prog. takes no of elements as input and return no of possible BST on that number */

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

 long int bst(long int a);
 long int BST[100];
 //array to store solutions of subproblems.assume that n should less the 100 here.
 //you can increase this value.
 long int n;

 int main()
{
int i;
long int total;
printf("\n enter no. of elements = \n");//take the no for which u want to calculate no of BST.
scanf("%ld",&n);
//initial set all subproblems sol. to 0.
 for (i=2;i<=n-1;i++)
 BST[i]=0;

BST[0]= 1;//set no of BST for nodes 1,0, equal 1.
BST[1] = 1;

//pass the value to function bst.
total = bst(n);
//total holds the final result

printf("\n the possible no of binary search tree on %ld elements is %ld",n,total);
return 0;
}

long int bst(long int a)
{
    int i;long int count;
//if no_of elements 0 or 1 return 1.
    if(a == 0 || a == 1 )
    return 1;
//if solution of subproblem is there return the value.
   else if(BST[a] != 0)
   return BST[a];

  else
 {
      count=0;
  for (i=1;i<=a;i++)
 {
     //choose 1 to n roots at one time then call for both subtrees left and right recursivley.
  count= count+ bst(i-1)*bst(a-i);
  printf(" \n the count is  on nodes is = %ld \n",count);
 }
 BST[a]=count;
 return count;


  }
}