# Find sum of inverse of the divisors when sum of divisors and the number is given

Given an integer N and the sum of its divisors. The task is to find the sum of the inverse of the divisors of N.

Examples:

Input: N = 6, Sum = 12
Output: 2.00
Divisors of N are {1, 2, 3, 6}
Sum of inverse of divisors is equal to (1/1 + 1/2 + 1/3 + 1/6) = 2.0

Input: N = 9, Sum = 13
Output: 1.44

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: Calculate all the divisors of the given integer N. Then calculate the sum of the inverse of the calculated divisors. This approach would give TLE when the value of N is large.
Time Complexity: O(sqrt(N))

Efficient Approach: Let the number N has K divisors say d1, d2, …, dK. It is given that d1 + d2 + … + dK = Sum
The task is to calculate (1 / d1) + (1 / d2) + … + (1 / dK).
Multiply and divide the above equation by N. The equation becomes [(N / d1) + (N / d2) + … + (N / dK)] / N

Now it is easy to see that N / di would represent some another divisor of N for all 1 ≤ i ≤ K. The numerator is equal to the sum of the divisors. Hence sum of inverse of the divisors is equal to Sum / N.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the ` `// sum of inverse of divisors ` `double` `SumofInverseDivisors(``int` `N, ``int` `Sum) ` `{ ` ` `  `    ``// Calculating the answer ` `    ``double` `ans = (``double``)(Sum)*1.0 / (``double``)(N); ` ` `  `    ``// Return the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 9; ` ` `  `    ``int` `Sum = 13; ` ` `  `    ``// Function call ` `    ``cout << setprecision(2) << fixed ` `         ``<< SumofInverseDivisors(N, Sum); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` `import` `java.math.*; ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to return the ` `// sum of inverse of divisors ` `static` `double` `SumofInverseDivisors(``int` `N, ``int` `Sum) ` `{ ` ` `  `    ``// Calculating the answer ` `    ``double` `ans = (``double``)(Sum)*``1.0` `/ (``double``)(N); ` ` `  `    ``// Return the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` ` `  `    ``int` `N = ``9``; ` `    ``int` `Sum = ``13``; ` ` `  `    ``// Function call ` `    ``System.out.println (SumofInverseDivisors(N, Sum)); ` `} ` `} ` ` `  `// This code is contributed by jit_t. `

## Python

 `# Python implementation of above approach ` ` `  `# Function to return the ` `# sum of inverse of divisors ` `def` `SumofInverseDivisors( N, ``Sum``): ` ` `  `    ``# Calculating the answer ` `    ``ans ``=` `float``(``Sum``)``*``1.0` `/``float``(N); ` ` `  `    ``# Return the answer ` `    ``return` `round``(ans,``2``); ` ` `  ` `  `# Driver code ` `N ``=` `9``; ` `Sum` `=` `13``; ` `print` `SumofInverseDivisors(N, ``Sum``); ` ` `  `# This code is contributed by CrazyPro `

## C#

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `         `  `// Function to return the ` `// sum of inverse of divisors ` `static` `double` `SumofInverseDivisors(``int` `N, ``int` `Sum) ` `{ ` ` `  `    ``// Calculating the answer ` `    ``double` `ans = (``double``)(Sum)*1.0 / (``double``)(N); ` ` `  `    ``// Return the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main () ` `{ ` `     `  `    ``int` `N = 9; ` `    ``int` `Sum = 13; ` ` `  `    ``// Function call ` `    ``Console.Write(SumofInverseDivisors(N, Sum)); ` `} ` `} ` ` `  `// This code is contributed by ajit `

Output:

```1.44
```

Time Complexity: O(1)

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Improved By : jit_t, CrazyPro

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