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Count Arithmetic Progressions having sum S and common difference equal to D
  • Last Updated : 10 Mar, 2021

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:



S = \frac{N}{2}*(2*a + (N - 1)*d)

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)

=>a = \frac{\frac{2*S}{N} - (N - 1)*d}{2}           …(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 <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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