human
Count of subsequences whose product is a difference of square of two integers | Function to count all the contiguous subsequences whose product is expressed as the square of the difference of two integers.
Creating vectors to store the remainders and the subsequences.
Iterating through the array.
Finding the remainder when the element is divided by 4.
Bringing all the elements in the range [ 0 , 3 ].
If the remainder is 2 , store the index of the.
If the remainder is 2 , store the index of the.
Finding the total number of subsequences.
If there are no numbers which yield the remainder 2.
Iterating through the vector.
If the element is 2 , find the nearest 2 or 0 and find the number of elements between them.
Returning the count.
Driver Code"
How can the above be solved in Python?
assistant
#Here is code to count of subsequences whose product is a difference of square of two integers and the other steps required.
 def CntcontSubs ( a , n ) :
	prod = 1
	vect = [ ]
	vect . append ( ( 0 , 2 ) )
	two , zero = [ ] , [ ]
	for i in range ( n ) :
		a [ i ] = a [ i ] % 4
		if ( a [ i ] < 0 ) :
			a [ i ] = a [ i ] + 4
		if ( a [ i ] == 2 ) :
			two . append ( i + 1 )
		if ( a [ i ] == 0 ) :
			zero . append ( i + 1 )
		if ( a [ i ] == 0 or a [ i ] == 2 ) :
			vect . append ( ( i + 1 , a [ i ] ) )
	vect . append ( ( n + 1 , 2 ) )
	total = ( n * ( n + 1 ) ) // 2
	if ( len ( two ) == 0 ) :
		return total
	else :
		Sum = 0
		pos1 , pos2 , pos3 = - 1 , - 1 , - 1
		sz = len ( vect )
		for i in range ( 1 , sz - 1 ) :
			if ( vect [ i ] [ 1 ] == 2 ) :
				Sum += ( ( vect [ i ] [ 0 ] - vect [ i - 1 ] [ 0 ] ) * ( vect [ i + 1 ] [ 0 ] - vect [ i ] [ 0 ] ) - 1 )
		return ( total - Sum - len ( two ) )
a = [ 5 , 4 , 2 , 9 , 8 ]
n = len ( a )
print ( CntcontSubs ( a , n ) )