# Find N fractions that sum upto a given fraction N/D

Given a fraction N/D, the task is to split this fraction into N parts such that their sum is equal to the fraction N/D, i.e., Note: Represents the terms in terms of fractions, instead of floating point numbers.

Input: N = 4, D = 2
Output: 4/5, 1/5, 1/3, 4/6
Explanation: Therefore, it is a valid set of fractions such that their sum is Input: N = 3, D = 4
Output: 1/2, 1/10, 3/20
Explanation: Therefore, it is a valid set of fractions such that their sum is ## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The key observation in the problem is that the first fraction numerator can be and then further denominators can be using the below recurrence relation. Below is the implementation of the above approach:

## C++

 // C++ implementation to split the  // fraction into N parts  #include  using namespace std;     // Function to split the fraction  // into the N parts  void splitFraction(int n, int d)  {      int ar[n];      int first = d + n - 1;      ar = first;         // Loop to find the N - 1      // fraction      for(int i = 1; i < n; i++)      {         int temp = --first;         first++;            ar[i] = first * temp;         --first;      }         // Loop to print the Fractions      for(int i = 0; i < n; i++)      {         if (ar[i] % n == 0)          {             cout << "1/" << ar[i] / n << ", ";         }         else         {             cout << n << "/" << ar[i] << ", ";         }      }  }     // Driver Code  int main()  {      int N = 4;      int D = 2;         // Function Call      splitFraction(N, D);  }     // This code is contributed by Bhupendra_Singh

## Java

 // Java implementation to split the  // fraction into N parts     import java.util.Scanner;     class Solution {         // Function to split the fraction      // into the N parts      public static void     splitFraction(int n, int d)      {             long ar[] = new long[n];          long first = d + n - 1;          ar = first;             // Loop to find the N - 1          // fraction          for (int i = 1; i < n; i++) {              ar[i] = first * (--first);          }             // Loop to print the Fractions          for (int i = 0; i < n; i++) {              if (ar[i] % n == 0) {                  System.out.print(                      "1/" + ar[i] / n                      + ", ");              }              else {                  System.out.print(                      n + "/" + ar[i]                      + ", ");              }          }      }         // Driver Code      public static void main(          String[] args) throws Exception      {          int N = 4;          int D = 2;             // Function Call          splitFraction(N, D);      }  }

## C#

 // C# implementation to split the  // fraction into N parts  using System;     class GFG{     // Function to split the fraction  // into the N parts  public static void splitFraction(int n, int d)  {      long []ar = new long[n];      long first = d + n - 1;      ar = first;         // Loop to find the N - 1      // fraction      for(int i = 1; i < n; i++)      {         ar[i] = first * (--first);      }         // Loop to print the Fractions      for(int i = 0; i < n; i++)       {         if (ar[i] % n == 0)         {             Console.Write("1/" + ar[i] / n + ", ");         }         else        {             Console.Write(n + "/" + ar[i] + ", ");         }      }  }     // Driver Code  public static void Main(String[] args)   {      int N = 4;      int D = 2;         // Function Call      splitFraction(N, D);  }  }     // This code is contributed by SoumikMondal

Output:

4/5, 1/5, 1/3, 4/6,


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Improved By : bgangwar59, SoumikMondal