binary expansion in C# - converting BigInteger to binary form

This source code will do the job.
Add appropriate controls and fire the number: 174050332293622031404857552280219410364023488927386650641
and you'll get:
01110001100100101011100101011111111111001000110110100111100001100011000100000001000111101101011010110010010011001101110101010111001111111001011101111010000100011110011110010100100000010001
with a bunch of more noughts in front.
What a happy day. :)

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Numerics;

namespace binaryexpansion
{
    public partial class Form1 : Form
    {   
 
        public Form1()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {
            BigInteger t = BigInteger.Parse(inputnr.Text); 
            selResult.Text = calcBinary(t);

            //the line below would work if t was an integer - so nice and easy, but no dice with BigIntegers
            //string t2 = Convert.ToString(t, 2);
             
        }

      
        private string calcBinary(BigInteger tee) {

            BigInteger t = tee;
            //4 times the length because it takes about 3 times the number of digits to write a number i binary form
            int len = 4*(inputnr.Text).Length;
            
            string fullnum= " ";
            for (int i = len; i >= 0; i--) 
            {
                int inum;
                
                BigInteger num =  (t >> i)&1;
                if(num == 1){
                //ugly but easy
                inum = 1;}
                else if(num == 0){
                inum = 0;}
                else{
                inum=666;
                }
                fullnum =  fullnum  + Convert.ToString(inum);
                
            }
            return fullnum;

            }
        }
    }

Double and Add algorithm

Very much like the fast powering algorithm - binary expansion can be used to calculate Q= nP fast.
ex n=947 = 2^9+2^8+2^7+2^5+2^4+2^1+2^0

which means that 947P = P+2P+16P+32P+128P+256P+512P. Thats 9 doublings and 6 additions, much better than 946 additions....
if we allow subtraction as well (ternary expansion) its even better.

want to calculate nP?

algorithm:

1. input point P from E(Fp) and n>=1
2. set Q=P and R=Ø (point at infinity)
3. loop while n>0{
     if n = 1 (mod2) set R= R+Q
     set Q=2Q and n=n/2 (floor)}
4.return R (which equals nP)

I've now posted working sourcecode available at http://christelbach.com
(source code contained in master thesis)

for instance this:
//double and add algorithm

private BigInteger[] calculatenP(BigInteger x, BigInteger y, BigInteger n, BigInteger A, BigInteger modp)
{
BigInteger[] Q = { x, y };
BigInteger[] R = { 0, 0 };
while (n > 0)
{
   BigInteger t = BigInteger.ModPow(n, 1, 2); 
   if ((t) == 1)
   {
   R = addPoints(Q[0], Q[1], R[0], R[1], modp);
   } 
   Q = doublePoints(Q[0], Q[1], A, modp);
   n = n / 2;
}
BigInteger[] resPoint = { R[0], R[1] };
return resPoint;
}

Look in the pdf file for the methods addPoints() and doublePoints()