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

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 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` `if` `__name__ ``=``=` `"__main__"` `:` `  ``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`

Javascript

 ``

Output:

`1.44`

Time Complexity: O(1)

Auxiliary Space: O(1)

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