human
Count of triplets having sum of product of any two numbers with the third number equal to N | Function to find the SPF [ i ] using the Sieve Of Erastothenes.
Stores whether i is prime or not.
Initializing smallest factor as 2 for all even numbers.
Iterate for all odd numbers < N.
SPF of i for a prime is the number itself.
Iterate for all the multiples of the current prime number.
The value i is smallest prime factor for i * j.
Function to generate prime factors and its power.
Current prime factor of N.
Stores the powers of the current prime factor.
Find all the prime factors and their powers.
Return the total count of factors.
Function to count the number of triplets satisfying the given criteria.
Stores the count of resultant triplets.
Add the count all factors of N - z to the variable CountTriplet.
Return total count of triplets.
Driver Code.
S [ i ] stores the smallest prime factor for each element i.
Find the SPF [ i ].
Function Call"
Solution in Python:
 ... What is a code to count of triplets having sum of product of any two numbers with the third number equal to N and the other steps required?
assistant
#Here is code to count of triplets having sum of product of any two numbers with the third number equal to N and the other steps required.
 def sieveOfEratosthenes ( N , s ) :
	prime = [ False ] * ( N + 1 )
	for i in range ( 2 , N + 1 , 2 ) :
		s [ i ] = 2
	for i in range ( 3 , N + 1 , 2 ) :
		if ( prime [ i ] == False ) :
			s [ i ] = i
			for j in range ( i , int ( N / i ) + 1 , 2 ) :
				if ( prime [ i * j ] == False ) :
					prime [ i * j ] = True
					s [ i * j ] = i
def generatePrimeFactors ( N ) :
	curr = s [ N ]
	cnt = { s [ N ] : 1 }
	while ( N > 1 ) :
		N //= s [ N ]
		if N and s [ N ] :
			if cnt . get ( s [ N ] , 0 ) == 0 :
				cnt [ s [ N ] ] = 1
			else :
				cnt [ s [ N ] ] += 1
	if 0 in cnt :
		cnt . pop ( 0 )
	totfactor = 1
	for i in cnt . values ( ) :
		totfactor *= i + 1
	return totfactor
def countTriplets ( N ) :
	CountTriplet = 0
	for z in range ( 1 , N + 1 ) :
		p = generatePrimeFactors ( N - z )
		if p > 1 :
			CountTriplet += p
	return CountTriplet + 1
N = 10
s = [ 0 ] * ( N + 1 )
sieveOfEratosthenes ( N , s )
print ( countTriplets ( N ) )