human
Find and Count total factors of co | Function to return the count of numbers which are divisible by both A and B in the range [ 1 , N ] in constant time.
Compute the count of numbers divisible by A in the range [ 1 , N ].
Compute the count of numbers divisible by B in the range [ 1 , N ].
Adding the counts which are divisible by A and B.
The above value might contain repeated values which are divisible by both A and B . Therefore , the count of numbers which are divisible by both A and B are found.
The count computed above is subtracted to compute the final count.
Function to return the sum of numbers which are divisible by both A and B in the range [ 1 , N ].
Set to store the numbers so that the numbers are not repeated.
For loop to find the numbers which are divisible by A and insert them into the set.
For loop to find the numbers which are divisible by A and insert them into the set.
For loop to iterate through the set and find the sum.
Driver code"
How can the above be solved in Python?

Can you provide a code to find and Count total factors of co and the other steps required?
assistant
#Here is code to find and Count total factors of co and the other steps required.
 def countOfNum ( n , a , b ) :
	cnt_of_a , cnt_of_b , cnt_of_ab , sum = 0 , 0 , 0 , 0
	cnt_of_a = n // a
	cnt_of_b = n // b
	sum = cnt_of_b + cnt_of_a
	cnt_of_ab = n // ( a * b )
	sum = sum - cnt_of_ab
	return sum
def sumOfNum ( n , a , b ) :
	i = 0
	sum = 0
	ans = dict ( )
	for i in range ( a , n + 1 , a ) :
		ans [ i ] = 1
	for i in range ( b , n + 1 , b ) :
		ans [ i ] = 1
	for it in ans :
		sum = sum + it
	return sum
if __name__ == ' _ _ main _ _ ' :
	N = 88
	A = 11
	B = 8
	count = countOfNum ( N , A , B )
	sumofnum = sumOfNum ( N , A , B )
	print ( sumofnum % count )