human
Count the numbers < N which have equal number of divisors as K | Function to return the count of the divisors of a number.
Count the number of 2 s that divide n.
n must be odd at this point . So we can skip one element.
While i divides n.
This condition is to handle the case when n is a prime number > 2.
Count the total elements that have divisors exactly equal to as that of k 's.
Exclude k from the result if it is smaller than n ..
Driver code"
Solution in Python:
 ... Give me a code to count the numbers < N which have equal number of divisors as K and the other steps required?
assistant
#Here is code to count the numbers < N which have equal number of divisors as K and the other steps required.
 def countDivisors ( n ) :
	x , ans = 0 , 1
	while ( n % 2 == 0 ) :
		x += 1
		n = n / 2
	ans = ans * ( x + 1 )
	for i in range ( 3 , int ( n ** 1 / 2 ) + 1 , 2 ) :
		x = 0
		while ( n % i == 0 ) :
			x += 1
			n = n / i
		ans = ans * ( x + 1 )
	if ( n > 2 ) :
		ans = ans * 2
	return ans
def getTotalCount ( n , k ) :
	k_count = countDivisors ( k )
	count = 0
	for i in range ( 1 , n ) :
		if ( k_count == countDivisors ( i ) ) :
			count += 1
	if ( k < n ) :
		count = count - 1
	return count
if __name__ == ' _ _ main _ _ ' :
	n , k = 500 , 6
	print ( getTotalCount ( n , k ) )