Count Arithmetic Progressions having sum N and common difference equal to 1

Given an integer N, the task is to count all arithmetic progression with common difference equal to 1 and the sum of all its elements equal to N.

Examples:

Input: N = 12
Output: 3
Explanation: 
Following three APs satisfy the required conditions:

• {3, 4, 5}
• {−2, −1, 0, 1, 2, 3, 4, 5}
• {−11, −10, −9, …, 10, 11, 12]}

Input: N = 963761198400
Output: 1919

Approach: The given problem can be solved using the following properties of AP:

• The sum of an AP series having N terms with first term as A and common difference 1 is given by the formula S = (N / 2 )* ( 2 *A + N − 1).
• Solving the above equation:

S = N / 2 [2 *A + N − 1].

Rearranging this and multiplying both sides by 2
=> 2 * S = N * (N + 2 * A − 1)

Rearranging further
=> A = ((2 * N / i ) − i + 1) / 2.

Now, iterate over all the factors of 2 * N, and check for each factor i, whether (2 * N/i) − i + 1 is even or not.

Follow the steps below to solve the problem:

• Initialize a variable, say count, to store the number of AP series possible having given conditions.
• Iterate through all the factors of 2 * N and for every i^th factor, check if (2 * N / i) − i + 1 is even.
• If found to be true, increment the count.
• Print count - 1 as the required answer, as the sequence consisting only of N needs to be discarded.

Below is the implementation of the above approach:

1919

Time Complexity: O(√N)
Auxiliary Space: O(1)