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 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++

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// C++ implementation of above approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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.

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Python

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# 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

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C#

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// 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

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Output:

1.44

Time Complexity: O(1)



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Improved By : jit_t, CrazyPro