Forming the nth binary sequence

Initial approach (not efficient):
 
Starting from the sequence 0, this algorithm iterates k times. 

To get the next sequence, in each iteration we do the following.
 check the last index of 0 in the number

• If it is found change that to 1 and all the digits after that to 0
• If it is not found, the next sequence contains all 0 which is one  digit longer than previous

Efficient approach:

Observe the following. we can have
‘2’ single digit numbers
‘4’ two digit numbers
‘8’ three digit numbers
…
‘2^i’ i digit numbers

We can find the number of digits in k^th sequence by calculating the power of 2 which is less than k .

For example if k is 5, the smallest power (s) of 2 less thank k is 4. So k^th sequence has 2 digits.

To form the the sequence we need to find the binary representation of 
k – ( 2^1 + 2^2 + …+2^s)
 
This SPOJ problem is based on the above algorithm.
 
Here is the C++ program which implements this.