为 N 的每个除数获得的欧拉总函数和

给定一个正整数N ，任务是找到给定数N的所有除数的欧拉总函数之和。

例子：

Input: N = 3
Output: 3
Explanation:
Divisors of 3 are {1, 3}. The Euler totient function for the values 1 and 3 are 1 and 2 respectively.
Therefore, the required sum is 1 + 2 = 3.

Input: N = 6
Output: 6

朴素方法：给定问题可以通过找到N的所有除数来解决，然后打印每个除数的欧拉总函数值的总和作为结果。

时间复杂度： O(N * sqrt(N))
辅助空间： O(1)

有效方法：上述方法也可以通过使用欧拉总函数的性质进行优化，该属性表明所有除数的欧拉总函数的所有值的总和为N 。

因此， N的欧拉函数的所有值之和就是这个数本身。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the sum of Euler
// Totient Function of divisors of N
int sumOfDivisors(int N)
{
    // Return the value of N
    return N;
}
 
// Driver Code
int main()
{
    int N = 5;
    cout << sumOfDivisors(N);
 
    return 0;
}

Java

// Java program for the above approach
public class GFG {
 
    // Function to find the sum of Euler
    // Totient Function of divisors of N
    static int sumOfDivisors(int N)
    {
        // Return the value of N
        return N;
    }
    // Driver code
    public static void main(String[] args)
    {
        int N = 5;
        System.out.println(sumOfDivisors(N));
    }
}
// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach
 
# Function to find the sum of Euler
# Totient Function of divisors of N
def sumOfDivisors(N):
     
    # Return the value of N
    return N
 
# Driver Code
if __name__ == '__main__':
     
    N = 5
     
    print (sumOfDivisors(N))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
     
  // Function to find the sum of Euler
    // Totient Function of divisors of N
    static int sumOfDivisors(int N)
    {
        // Return the value of N
        return N;
    }
 
// Driver code
static void Main()
{
     int N = 5;
     Console.Write(sumOfDivisors(N));
     
}
}
 
// This code is contributed by sanjoy_62.

Javascript

  
// Js program for the above approach
  
    // Function to find the sum of Euler
    // Totient Function of divisors of N
    function sumOfDivisors(N){
 
        // Return the value of N
        return N;
    }
 
    // Driver Code
    let N = 5;
    document.write(sumOfDivisors(N));
  

输出：

5

时间复杂度： O(1)
辅助空间： O(1)