Find number of factors of N when location of its two factors whose product is N is given

Given a and b which represent the location of two factors of N whose product is equal to the number N when the factors are arranged in ascending order. The task is to find the total number of factors of N.
Examples:

Input : a = 2, b = 3 
Output : 4 
N = 6 
factors are {1, 2, 3, 6} 
No of factors are 4

Input : a = 13, b = 36 
Output : 48  

Approach: The solution is based on observation. 
Assume N = 50. Then N has 6 factors 1, 2, 5, 10, 25 and 50. On multiplying 1 and 50 will always give the value 50( N ). Also, by multiplying 2 and 25 gives us the value N, similarly by multiplying 5 and 10 we get the value N. So, here we can see that by multiplying two factors when arranged in an increasing order gives the value N. The multiplication must be done in a manner: 1st factor and the last factor on multiplication gives N, 2nd factor and 2nd last factor on multiplication gives N and so on.
With this pattern, we can find a way to calculate the number of factors. Say, the 1st and 4th factor on multiplication gives N. This means that there are 4 factors ( 1st, 2nd, 3rd, and 4th ). If the product of 2nd and 3rd factor gives N then we can say there must be a factor in the 1st position and on the 4th position. 
Therefore, the number of factors will be equal to a + b – 1.

Below is the implementation of the above approach: 

C++

// C++ program to implement
// the above problem
#include <bits/stdc++.h>
using namespace std;

// Function to find the number of factors
void findFactors(int a, int b)
{
    int c;
    c = a + b - 1;

    // print the number of factors
    cout << c;
}

// Driver code
int main()
{
    // initialize the factors position
    int a, b;
    a = 13;
    b = 36;

    findFactors(a, b);

    return 0;
}

Java

// Java program to implement
// the above problem
class GFG
{

// Function to find the number of factors
static void findFactors(int a, int b)
{
    int c;
    c = a + b - 1;

    // print the number of factors
    System.out.print(c);
}

// Driver code
public static void main(String[] args)
{
    // initialize the factors position
    int a, b;
    a = 13;
    b = 36;

    findFactors(a, b);
}
}

// This code is contributed by Princi Singh

Python3

# Python 3 program to implement
# the above problem

# Function to find the number of factors
def findFactors(a, b):
    c = a + b - 1

    # print the number of factors
    print(c)

# Driver code
if __name__ == '__main__':

    # initialize the factors position
    a = 13
    b = 36
    findFactors(a, b)

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to implement
// the above problem
using System;

class GFG
{

// Function to find the number of factors
static void findFactors(int a, int b)
{
    int c;
    c = a + b - 1;

    // print the number of factors
    Console.Write(c);
}

// Driver code
public static void Main(String[] args)
{
    // initialize the factors position
    int a, b;
    a = 13;
    b = 36;

    findFactors(a, b);
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript program to implement
// the above problem

// Function to find the number of factors
function findFactors(a, b)
{
    let c;
    c = a + b - 1;

    // print the number of factors
    document.write(c);
}

// Driver code

// initialize the factors position
let a, b;
a = 13;
b = 36;

findFactors(a, b);

</script>

Output:

48

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