Given two integers **S** and **D**, the task is to count the number of Arithmetic Progressions possible with sum **S** and common difference **D**.

**Examples:**

Input:S = 12, D = 1Output:4Explanation:Following 4 arithmetic progressions with sum 12 and common difference 1 are possible:

- {1, 2}
- {3, 4, 5}
- {-2, -1, 0, 1, 2, 3, 4, 5}
- {-11, -10, -9, …, 10, 11, 12}

Input:S = 1, D = 1Output:2

**Approach:** The given problem can be solved based on the following observations:

- The sum of the AP series is given by:

where,

Sis the sum of the AP series,ais the first term of the series,Nis the number of terms in the series,dis a common difference

- After rearranging the above expressions:

=> 2*S = N*(2*a + (N – 1)*d) … (1)

=> …(2)

- From the above two expressions:
- The idea is to consider all the factors of
**2*S**and check if there exists any factor**F**such that the product of**F**and**(2*a + (F – 1)*d)**is equal to**2 * S**. If found to be true, then count that factor for one of the possible**AP**having the given sum**S**. - If there exists any factor
**F**, such that**(D * F – (2 * S / F) + D)**is divisible by**2**, then count that factor for one of the possible**AP**having the given sum**S**.

- The idea is to consider all the factors of

Follow the steps below to solve the problem:

- Initialize a variable, say
**answer**, to store the count of**AP**s with sum**S**and common difference**D**. - Iterate over the range
**[1, √2*S]**and check if**2 * S**is divisible by**i**, then perform the following steps:- If the value of
**((2 * S / i) + 1 – i * D)**is divisible by**2**, then increment**answer**by**1**. - If the value of
**(i * D – S / i + 1)**is divisible by**2**, then increment**answer**by**1**.

- If the value of
- After completing the above steps, print the value of
**answer**as the resultant count of**AP**s.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to count the number of APs` `// with sum S and common difference D` `int` `countAPs(` `int` `S, ` `int` `D)` `{` ` ` `// Multiply S by 2` ` ` `S = S * 2;` ` ` `// Stores the count of APs` ` ` `int` `answer = 0;` ` ` `// Iterate over the factors of 2*S` ` ` `for` `(` `int` `i = 1; i <= ` `sqrt` `(S); i++) {` ` ` `// Check if i is the factor` ` ` `// or not` ` ` `if` `(S % i == 0) {` ` ` `// Conditions to check if AP` ` ` `// can be formed using factor F` ` ` `if` `(((S / i) - D * i + D) % 2 == 0)` ` ` `answer++;` ` ` `if` `((D * i - (S / i) + D) % 2 == 0)` ` ` `answer++;` ` ` `}` ` ` `}` ` ` `// Return the total count of APs` ` ` `return` `answer;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `S = 12, D = 1;` ` ` `cout << countAPs(S, D);` ` ` `return` `0;` `}` |

## Java

`// Java program for above approach` `/*package whatever //do not write package name here */` `import` `java.io.*;` `class` `GFG` `{` ` ` `// Function to count the number of APs` `// with sum S and common difference D` `static` `int` `countAPs(` `int` `S, ` `int` `D)` `{` ` ` ` ` `// Multiply S by 2` ` ` `S = S * ` `2` `;` ` ` `// Stores the count of APs` ` ` `int` `answer = ` `0` `;` ` ` `// Iterate over the factors of 2*S` ` ` `for` `(` `int` `i = ` `1` `; i <= Math.sqrt(S); i++) {` ` ` `// Check if i is the factor` ` ` `// or not` ` ` `if` `(S % i == ` `0` `) {` ` ` `// Conditions to check if AP` ` ` `// can be formed using factor F` ` ` `if` `(((S / i) - D * i + D) % ` `2` `== ` `0` `)` ` ` `answer++;` ` ` `if` `((D * i - (S / i) + D) % ` `2` `== ` `0` `)` ` ` `answer++;` ` ` `}` ` ` `}` ` ` `// Return the total count of APs` ` ` `return` `answer;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `S = ` `12` `, D = ` `1` `;` ` ` `System.out.println(countAPs(S, D));` `}` `}` `// This code is contributed by susmitakundugoaldanga.` |

## Python3

`# Python3 program for the above approach` `# Function to count the number of APs` `# with sum S and common difference D` `def` `countAPs(S, D):` ` ` ` ` `# Multiply S by 2` ` ` `S ` `=` `S ` `*` `2` ` ` `# Stores the count of APs` ` ` `answer ` `=` `0` ` ` `# Iterate over the factors of 2*S` ` ` `for` `i ` `in` `range` `(` `1` `, S):` ` ` `if` `i ` `*` `i > S:` ` ` `break` ` ` ` ` `# Check if i is the factor` ` ` `# or not` ` ` `if` `(S ` `%` `i ` `=` `=` `0` `):` ` ` `# Conditions to check if AP` ` ` `# can be formed using factor F` ` ` `if` `(((S ` `/` `/` `i) ` `-` `D ` `*` `i ` `+` `D) ` `%` `2` `=` `=` `0` `):` ` ` `answer ` `+` `=` `1` ` ` `if` `((D ` `*` `i ` `-` `(S ` `/` `/` `i) ` `+` `D) ` `%` `2` `=` `=` `0` `):` ` ` `answer ` `+` `=` `1` ` ` `# Return the total count of APs` ` ` `return` `answer` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `S, D ` `=` `12` `, ` `1` ` ` `print` `(countAPs(S, D));` ` ` `# This code is contributed by mohit kumar 29.` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to count the number of APs` ` ` `// with sum S and common difference D` ` ` `static` `int` `countAPs(` `int` `S, ` `int` `D)` ` ` `{` ` ` `// Multiply S by 2` ` ` `S = S * 2;` ` ` `// Stores the count of APs` ` ` `int` `answer = 0;` ` ` `// Iterate over the factors of 2*S` ` ` `for` `(` `int` `i = 1; i <= Math.Sqrt(S); i++) {` ` ` `// Check if i is the factor` ` ` `// or not` ` ` `if` `(S % i == 0) {` ` ` `// Conditions to check if AP` ` ` `// can be formed using factor F` ` ` `if` `(((S / i) - D * i + D) % 2 == 0)` ` ` `answer++;` ` ` `if` `((D * i - (S / i) + D) % 2 == 0)` ` ` `answer++;` ` ` `}` ` ` `}` ` ` `// Return the total count of APs` ` ` `return` `answer;` ` ` `}` ` ` `// Driver code` ` ` `static` `void` `Main()` ` ` `{` ` ` `int` `S = 12, D = 1;` ` ` `Console.Write(countAPs(S, D));` ` ` `}` `}` `// This code is contributed by sanjoy_62.` |

**Output:**

4

**Time Complexity:** O(√S) **Auxiliary Space:** O(1)

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