human
Count number of 1 s in the array after N moves | Function to count number of 1 's in the array after performing N moves.
If index is multiple of move number.
arr [ j - 1 ] = 1 Convert 0 to 1.
arr [ j - 1 ] = 0 Convert 1 to 0.
Count number of 1 's.
count += 1 count number of 1 's.
Driver Code.
Initialize all elements to 0"
How can the above be solved in Python?

Give me a code to count number of 1 s in the array after N moves and the other steps required?
assistant
#Here is code to count number of 1 s in the array after N moves and the other steps required.
 def countOnes ( arr , N ) :
	for i in range ( 1 , N + 1 , 1 ) :
		for j in range ( i , N + 1 , 1 ) :
			if ( j % i == 0 ) :
				if ( arr [ j - 1 ] == 0 ) :
				else :
	count = 0
	for i in range ( N ) :
		if ( arr [ i ] == 1 ) :
	return count
if __name__ == ' _ _ main _ _ ' :
	arr = [ 0 for i in range ( 10 ) ]
	ans = countOnes ( arr , N )
	print ( ans )