human
Can you provide a code to count of N and the other steps required given the following:
Count of N | Function to calculate the total count of N - digit numbers such that the sum of digits at even positions and odd positions are divisible by A and B respectively.
For single digit numbers.
Largest possible number.
Count of possible odd digits.
Count of possible even digits.
Calculate total count of sequences of length even_count with sum divisible by A where first digit can be zero.
Calculate total count of sequences of length odd_count with sum divisible by B where cannot be zero.
Return their product as answer.
Driver Code"
How can the above be solved in Python?
assistant
#Here is code to count of N and the other steps required.
 def count ( N , A , B ) :
	if ( N == 1 ) :
		return 9 // B + 1
	max_sum = 9 * N
	odd_count = N // 2 + N % 2
	even_count = N - odd_count
	dp = [ [ 0 for x in range ( max_sum + 1 ) ] for y in range ( even_count ) ]
	for i in range ( 10 ) :
		dp [ 0 ] [ i % A ] += 1
	for i in range ( 1 , even_count ) :
		for j in range ( 10 ) :
			for k in range ( max_sum + 1 ) :
				if ( dp [ i - 1 ] [ k ] > 0 ) :
					dp [ i ] [ ( j + k ) % A ] += dp [ i - 1 ] [ k ]
	dp1 = [ [ 0 for x in range ( max_sum ) ] for y in range ( odd_count ) ]
	for i in range ( 1 , 10 ) :
		dp1 [ 0 ] [ i % B ] += 1
	for i in range ( 1 , odd_count ) :
		for j in range ( 10 ) :
			for k in range ( max_sum + 1 ) :
				if ( dp1 [ i - 1 ] [ k ] > 0 ) :
					dp1 [ i ] [ ( j + k ) % B ] += dp1 [ i - 1 ] [ k ]
	return dp [ even_count - 1 ] [ 0 ] * dp1 [ odd_count - 1 ] [ 0 ]
if __name__ == " _ _ main _ _ " :
	N = 2
	A = 2
	B = 5
	print ( count ( N , A , B ) )