Binary code-swiss cheese

Question

I have made a program that converts a number to binary and if the number has more zeros than ones then I display "Too light." If it has more ones than zeros it displays too heavy. If they are equal it says "just right."
The program works on some numbers and not on others. I don't know what's wrong with the code. Any help is appreciated. Thanks in advance
import java.lang.*;
import java.lang.String.*;
import java.io.*;
public class Goldilocks
{//begins class
     public static void main(String[] args)throws IOException
     {//begin main
          String userInA;
          int binnum;
          String bin;
          int i = 0;
          int o = 0;
          InputStreamReader inStream =
                       new InputStreamReader( System.in ) ;
          BufferedReader stdin =
               new BufferedReader( inStream );
               do{//begin loop
          System.out.print("Please enter a number(A to quit): ");
          userInA = stdin.readLine();
          if(userInA.equals("A")){
               System.out.print("Thank you");
               System.exit(0);
          binnum = Integer.parseInt(userInA);
          bin = Integer.toBinaryString(binnum);
          System.out.println(bin);
          for(int n = 0; n<=bin.length()-1; n++){
               if(bin.charAt(n) == '1'){
                    i++;}
               else if(bin.charAt(n) == '0'){
                    o++;}
          if(i > o){
               System.out.println("Too heavy.");
          else if(o>i){
               System.out.println("Too light.");
          else{
                               System.out.println("Just right.");
               }while(userInA != "A");//end loop
     }//end main
}//ends class

Answer 1

You're welcome :-)
You should also add some verification for bad user inputs:
try {
    binnum = Integer.parseInt(userInA);
} catch (NumberFormatException e) {
    System.out.println("Bad number format");
    continue;