human
Count Arithmetic Progressions having sum N and common difference equal to 1 | Function to count all possible AP series with common difference 1 and sum of elements equal to N.
Stores the count of AP series.
Traverse through all factors of 2 * N.
Check for the given conditions.
Increment count.
Prcount - 1.
Given value of N.
Function call to count required number of AP series"
How can the above be solved in Python?
 Can you provide a code to count Arithmetic Progressions having sum N and common difference equal to 1 and the other steps required?
assistant
#Here is code to count Arithmetic Progressions having sum N and common difference equal to 1 and the other steps required.
 def countAPs ( N ) :
	count = 0
	i = 1
	while ( i * i <= 2 * N ) :
		res = 2 * N
		if ( res % i == 0 ) :
			op = res / i - i + 1
			if ( op % 2 == 0 ) :
				count += 1
			if ( i * i != res and ( i - res / i + 1 ) % 2 == 0 ) :
				count += 1
		i += 1
	print ( count - 1 )
N = 963761198400
countAPs ( N )