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# Count Arithmetic Progressions having sum S and common difference equal to D

• Last Updated : 20 Jul, 2022

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

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

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

Approach: The given problem can be solved based on the following observations: where,
S is the sum of the AP series,
a is the first term of the series,
N is the number of terms in the series,
d is 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.

Follow the steps below to solve the problem:

• Initialize a variable, say answer, to store the count of APs 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.
• After completing the above steps, print the value of answer as the resultant count of APs.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach   #include  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.

## Javascript

 

Output:

4

Time Complexity: O(√S)
Auxiliary Space: O(1),  since no extra space has been taken.

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