给定除数和数时求除数的倒数之和

原文:https://www . geesforgeks . org/find-除数的倒数之和-当除数之和和被给定时/

给定一个整数 N 及其除数之和。任务是求 N 的除数的倒数之和。 例:

输入: N = 6，和= 12 输出:2.00 N 的除数为{1，2，3，6} 除数的倒数之和等于(1/1 + 1/2 + 1/3 + 1/6) = 2.0 输入: N = 9，和= 13 输出: 1.44

天真法:计算给定整数的所有除数 N 。然后计算所计算除数的倒数之和。当 N 的值较大时，这种方法将给出 TLE。 时间复杂度: O(sqrt(N)) 高效途径:让数 N 有 K 除数说 d 1 ，d 2 ，…，d K 。给出d1+d2+…+dK= Sum 任务是计算(1/d1)+(1/d2)+…+(1/dK)。 用 N 乘除上述方程。等式变成[(N/d1)+(N/d2)+…+(N/dK)]/NT52】现在很容易看出N/dIT56】将代表所有 1 ≤ i ≤ K 的 N 的另一个除数。分子等于除数之和。因此，除数的倒数之和等于和/ N 。 以下是上述方法的实施:

C++

// 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;
}

Java 语言(一种计算机语言，尤用于创建网站)

// 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.

计算机编程语言

# 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

java 描述语言

<script>

// JavaScript implementation of above approach

// Function to return the
// sum of inverse of divisors
function SumofInverseDivisors(N, Sum)
{

    // Calculating the answer
    let ans = (Sum)*1.0 / (N);

    // Return the answer
    return ans;
}

// Driver code

    let N = 9;

    let Sum = 13;

    // Function call
    document.write(SumofInverseDivisors(N, Sum).toFixed(2));

// This code is contributed by Surbhi Tyagi.

</script>

Output: 

1.44

时间复杂度: O(1)

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