human
Check if a number has an odd count of odd divisors and even count of even divisors | Function to find the count of even and odd factors of N.
Loop runs till square root.
Condition to check if the even factors of the number N is is even and count of odd factors is odd.
Driver Code"
How can the above be solved in Python?
 Can you provide a code to check if a number has an odd count of odd divisors and even count of even divisors and the other steps required?
assistant
#Here is code to check if a number has an odd count of odd divisors and even count of even divisors and the other steps required.
 def checkFactors ( N ) :
	ev_count = 0 ; od_count = 0 ;
	for i in range ( 1 , int ( pow ( N , 1 / 2 ) ) + 1 ) :
		if ( N % i == 0 ) :
			if ( i == N / i ) :
				if ( i % 2 == 0 ) :
					ev_count += 1 ;
				else :
					od_count += 1 ;
			else :
				if ( i % 2 == 0 ) :
					ev_count += 1 ;
				else :
					od_count += 1 ;
				if ( ( N / i ) % 2 == 0 ) :
					ev_count += 1 ;
				else :
					od_count += 1 ;
	if ( ev_count % 2 == 0 and od_count % 2 == 1 ) :
		print ( " Yes " + " " ) ;
	else :
		print ( " No " + " " ) ;
if __name__ == ' _ _ main _ _ ' :
	N = 36 ;
	checkFactors ( N ) ;