human
Count of N | Function to find count of N - digit numbers with single digit XOR.
dp [ i ] [ j ] stores the number of i - digit numbers with XOR equal to j.
For 1 - 9 store the value.
Iterate till N.
Calculate XOR.
Store in DP table.
Initialize count.
Print answer.
Driver Code.
Given number N.
Function Call"
Solution in Python:
assistant
#Here is code to count of N and the other steps required.
 def countNums ( N ) :
	dp = [ [ 0 for i in range ( 16 ) ] for j in range ( N ) ] ;
	for i in range ( 1 , 10 ) :
		dp [ 0 ] [ i ] = 1 ;
	for i in range ( 1 , N ) :
		for j in range ( 0 , 10 ) :
			for k in range ( 0 , 16 ) :
				xor = j ^ k ;
				dp [ i ] [ xor ] += dp [ i - 1 ] [ k ] ;
	count = 0 ;
	for i in range ( 0 , 10 ) :
		count += dp [ N - 1 ] [ i ] ;
	print ( count ) ;
if __name__ == ' _ _ main _ _ ' :
	N = 1 ;
	countNums ( N ) ;