You could simply have done this to convert from decimal to binary:

-1→D

While D≤0

Input D

End

0→A

For(Z,1,1+iPart(log(D)/log(2

10^(Z-1)*2fPart(D/2)+A→A

iPart(D/2→D

End

Disp A

DelVar ADelVar DDelVar Z

D is the number in decimal, A is the answer in binary. 1+iPart(log(D)/log(2 is used to calcule how many time you need to divide by 2 for the loop while 10^(Z-1)*2fPart(D/2)+A→A store 1 or 0 for this step in A . iPart(D/2→D divides the number in Decimal to calculate the next step..

I know its not fully optimized cause i'm not an expert in optimization but your program was way too big for a binary converter.

With this routine, you can also substitute the number 2 for the number you want or even a variable to have a converter in the base you want like in octodecimal.