human
Count of distinct N | Function to find the count of distinct arrays of size n having elements in range [ 1 , k ] and all adjacent elements ( P , Q ) follows ( P <= Q ) or ( P % Q > 0 ).
Stores the divisors of all integers in the range [ 1 , k ].
Calculate the divisors of all integers using the Sieve.
Stores the dp states such that dp [ i ] [ j ] with i elements having j as the last element of array.
Initialize the dp array.
Calculate the dp states using the derived relation.
Calculate the sum for len - 1.
Subtract dp [ len - 1 ] [ j ] for each factor of j from [ 1 , K ].
Calculate the final result.
Return the resultant sum.
Driver Code"
Solution in Python:
assistant
#Here is code to count of distinct N and the other steps required.
 def countArrays ( n , k ) :
	divisors = [ [ ] for i in range ( k + 1 ) ]
	for i in range ( 1 , k + 1 , 1 ) :
		for j in range ( 2 * i , k + 1 , i ) :
			divisors [ j ] . append ( i )
	dp = [ [ 0 for i in range ( k + 1 ) ] for j in range ( n + 1 ) ]
	for j in range ( 1 , k + 1 , 1 ) :
		dp [ 1 ] [ j ] = 1
	for x in range ( 2 , n + 1 , 1 ) :
		sum = 0
		for j in range ( 1 , k + 1 , 1 ) :
			sum += dp [ x - 1 ] [ j ]
		for y in range ( 1 , k + 1 , 1 ) :
			dp [ x ] [ y ] = sum
			for d in divisors [ y ] :
				dp [ x ] [ y ] = ( dp [ x ] [ y ] - dp [ x - 1 ] [ d ] )
	sum = 0
	for j in range ( 1 , k + 1 , 1 ) :
		sum += dp [ n ] [ j ]
	return sum
if __name__ == ' _ _ main _ _ ' :
	N = 2
	K = 3
	print ( countArrays ( N , K ) )