Count number of ones till N in binary

Problem

Given a number n find total number of ones in the binary representation of the numbers 1, 2, 3, 4, ..., n

For example, if n=5

1=1 (number of ones=1)
2=10 (number of ones=1)
3=11 (number of ones=2)
4=100 (number of ones=1)
5=101 (number of ones=2)

output
7 (1+1+2+1+2)

Solution

Initialize a count to 1 for odd numbers and 0 for even numbers. If the bth bit is set then add to a count 2^(b-1)*b+n%2^b+1;

Code

public class NumberOfOnes
{

 /**
  * given n find the total number of ones in the binary representation of
  * numbers 1,2,3,...,n
  * 
  * @param args
  */
 public static void main(String[] args)
 {
  System.out.println(getNumberOfOnes(30));
 }

 public static int getNumberOfOnes(int num)
 {
  int p = 1, cnt = 0;

  if ((num & 1) != 0)
   cnt++;

  while (1 << (p - 1) < num)
  {
   if ((num & (1 << p)) != 0)
    cnt += (1 << (p - 1)) * (p) + num % (1 << p) + 1;
   p++;
  }
  return cnt;
 }

}