当给出除数之和和个数时，求除数的反数之和

给定整数N及其除数之和。任务是找到N的除数的逆之和。

例子：

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

天真的方法：计算给定整数N的所有除数。然后计算所计算的除数的逆之和。当N的值较大时，此方法将给TLE。
时间复杂度： O(sqrt(N))

高效的方法：让N个数为K的除数为d ₁ ，d ₂ ，…，d _K。假设d ₁ + d ₂ +…+ d _K =总和
任务是计算(1 / d ₁ )+(1 / d ₂ )+…+(1 / d _K ) 。
将上述方程式乘以N并除以。等式变为[(N / d ₁ )+(N / d ₂ )+…+(N / d _K )] / N

现在很容易看出，对于所有1≤i≤K ， N / d _i代表N的另一个约数。分子等于除数之和。因此，除数的逆之和等于Sum / N。

下面是上述方法的实现：

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

输出：

1.44

时间复杂度： O(1)