从1到N的所有适当除数的总和

给定正整数N ，任务是找到的值$\sum_{x=1}^{x=N} F(x)$其中函数F(x)可以定义为’ x ‘的所有适当除数的和。
例子：

Input: N = 4 
Output: 5 
Explanation: 
Sum of all proper divisors of numbers: 
F(1) = 0 
F(2) = 1 
F(3) = 1 
F(4) = 1 + 2 = 3 
Total Sum = F(1) + F(2) + F(3) + F(4) = 0 + 1 + 1 + 3 = 5
Input: N = 5 
Output: 6 
Explanation: 
Sum of all proper divisors of numbers: 
F(1) = 0 
F(2) = 1 
F(3) = 1 
F(4) = 1 + 2 = 3 
F(5) = 1 
Total Sum = F(1) + F(2) + F(3) + F(4) + F(5) = 0 + 1 + 1 + 3 + 1 = 6 
 

幼稚的方法：想法是分别找到[1，N]范围内每个数字的适当除数的总和，然后将它们相加以找到所需的总和。
下面是上述方法的实现：

C++

// C++ implemenation to find sum of all
// proper divisor of number up to N
#include 
using namespace std;
 
// Utility function to find sum of
// all proper divisor of number up to N
int properDivisorSum(int n)
{
    int sum = 0;
 
    // Loop to iterate over all the
    // numbers from 1 to N
    for (int i = 1; i <= n; ++i) {
 
        // Find all divisors of
        // i and add them
        for (int j = 1; j * j <= i; ++j) {
            if (i % j == 0) {
                if (i / j == j)
                    sum += j;
                else
                    sum += j + i / j;
            }
        }
 
        // Subtracting 'i' so that the
        // number itself is not included
        sum = sum - i;
    }
    return sum;
}
 
// Driver Code
int main()
{
    int n = 4;
    cout << properDivisorSum(n) << endl;
 
    n = 5;
    cout << properDivisorSum(n) << endl;
 
    return 0;
}

Java

// Java implemenation to find sum of all
// proper divisor of number up to N
class GFG {
     
    // Utility function to find sum of
    // all proper divisor of number up to N
    static int properDivisorSum(int n)
    {
        int sum = 0;
     
        // Loop to iterate over all the
        // numbers from 1 to N
        for (int i = 1; i <= n; ++i) {
     
            // Find all divisors of
            // i and add them
            for (int j = 1; j * j <= i; ++j) {
                if (i % j == 0) {
                    if (i / j == j)
                        sum += j;
                    else
                        sum += j + i / j;
                }
            }
     
            // Subtracting 'i' so that the
            // number itself is not included
            sum = sum - i;
        }
        return sum;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int n = 4;
        System.out.println(properDivisorSum(n));
     
        n = 5;
        System.out.println(properDivisorSum(n)) ;
     
    }
}
 
// This code is contributed by Yash_R

Python3

# Python3 implemenation to find sum of all
# proper divisor of number up to N
 
# Utility function to find sum of
# all proper divisor of number up to N
def properDivisorSum(n):
 
    sum = 0
 
    # Loop to iterate over all the
    # numbers from 1 to N
    for i in range(n+1):
 
        # Find all divisors of
        # i and add them
        for j in range(1, i + 1):
            if j * j > i:
                break
            if (i % j == 0):
                if (i // j == j):
                    sum += j
                else:
                    sum += j + i // j
 
        # Subtracting 'i' so that the
        # number itself is not included
        sum = sum - i
 
    return sum
 
# Driver Code
if __name__ == '__main__':
 
    n = 4
    print(properDivisorSum(n))
 
    n = 5
    print(properDivisorSum(n))
 
# This code is contributed by mohit kumar 29

C#

// C# implemenation to find sum of all
// proper divisor of number up to N
using System;
 
class GFG {
     
    // Utility function to find sum of
    // all proper divisor of number up to N
    static int properDivisorSum(int n)
    {
        int sum = 0;
     
        // Loop to iterate over all the
        // numbers from 1 to N
        for (int i = 1; i <= n; ++i) {
     
            // Find all divisors of
            // i and add them
            for (int j = 1; j * j <= i; ++j) {
                if (i % j == 0) {
                    if (i / j == j)
                        sum += j;
                    else
                        sum += j + i / j;
                }
            }
     
            // Subtracting 'i' so that the
            // number itself is not included
            sum = sum - i;
        }
        return sum;
    }
     
    // Driver Code
    public static void Main (string[] args)
    {
        int n = 4;
        Console.WriteLine(properDivisorSum(n));
     
        n = 5;
        Console.WriteLine(properDivisorSum(n)) ;   
    }
}
 
// This code is contributed by Yash_R

Javascript

 
//Javascript implemenation to find sum of all
// proper divisor of number up to N 
 
// Utility function to find sum of
// all proper divisor of number up to N
function properDivisorSum(n)
{
    let sum = 0;
 
    // Loop to iterate over all the
    // numbers from 1 to N
    for (let i = 1; i <= n; ++i) {
 
        // Find all divisors of
        // i and add them
        for (let j = 1; j * j <= i; ++j) {
            if (i % j == 0) {
                if (i / j == j)
                    sum += j;
                else
                    sum += j + i / j;
            }
        }
 
        // Subtracting 'i' so that the
        // number itself is not included
        sum = sum - i;
    }
    return sum;
}
 
// Driver Code
  
    let n = 4;
    document.write(properDivisorSum(n) + "<br>");
 
    n = 5;
    document.write(properDivisorSum(n) + "<br>");
 
     
// This code is contributed by Mayank Tyagi
     

输出：

5
6

时间复杂度： O(N *√N)
辅助空间： O(1)
高效的方法：观察函数的模式后，可以看到“对于给定的数字N，在[1，N]范围内的每个数字’x’发生(N / x)次” 。
例如：

Let N = 6 => G(N) = F(1) + F(2) + F(3) + F(4) + F(5) + F(6) 
x = 1 => 1 will occurs 6 times (in F(1), F(2), F(3), F(4), F(5) and F(6)) 
x = 2 => 2 will occurs 3 times (in F(2), F(4) and F(6)) 
x = 3 => 3 will occur 2 times (in F(3) and F(6)) 
x = 4 => 4 will occur 1 times (in F(4)) 
x = 5 => 5 will occur 1 times (in F(5)) 
x = 6 => 6 will occur 1 times (in F(6)) 
 

从上面的观察中，可以容易地观察到，数字x仅以其小于或等于N的倍数出现。因此，对于[1，N]中x的每个值，我们只需要查找此类倍数的计数，然后将其乘以x即可。然后将此值添加到最终总和中。
下面是上述方法的实现：

C++

// C++ implementation to find sum of all
// proper divisor of numbers up to N
 
#include 
using namespace std;
 
// Utility function to find sum of
// all proper divisor of number up to N
int properDivisorSum(int n)
{
    int sum = 0;
 
    // Loop to find the proper
    // divisor of every number
    // from 1 to N
    for (int i = 1; i <= n; ++i)
        sum += (n / i) * i;
 
    return sum - n * (n + 1) / 2;
}
 
// Driver Code
int main()
{
    int n = 4;
    cout << properDivisorSum(n) << endl;
 
    n = 5;
    cout << properDivisorSum(n) << endl;
    return 0;
}

Java

// Java implementation to find sum of all
// proper divisor of numbers up to N
  
// Utility function to find sum of
// all proper divisor of number up to N
 
class GFG
{
    static int properDivisorSum(int n)
    {
        int sum = 0;
        int i;
        // Loop to find the proper
        // divisor of every number
        // from 1 to N
        for (i = 1; i <= n; ++i)
            sum += (n / i) * i;
      
        return sum - n * (n + 1) / 2;
    }
      
    // Driver Code
    public static void main(String []args)
    {
        int n = 4;
        System.out.println(properDivisorSum(n));
      
        n = 5;
        System.out.println(properDivisorSum(n));
         
    }
}

Python3

# Python3 implementation to find sum of all
# proper divisor of numbers up to N
 
# Utility function to find sum of
# all proper divisor of number up to N
def properDivisorSum(n):
     
    sum = 0
     
    # Loop to find the proper
    # divisor of every number
    # from 1 to N
    for i in range(1, n + 1):
        sum += (n // i) * i
         
    return sum - n * (n + 1) // 2
 
 
# Driver Code
n = 4
print(properDivisorSum(n))
 
n = 5
print(properDivisorSum(n))
 
# This code is contributed by shubhamsingh10

C＃

// C# implementation to find sum of all
// proper divisor of numbers up to N
   
// Utility function to find sum of
// all proper divisor of number up to N
using System;
 
class GFG
{
    static int properDivisorSum(int n)
    {
        int sum = 0;
        int i;
        // Loop to find the proper
        // divisor of every number
        // from 1 to N
        for (i = 1; i <= n; ++i)
            sum += (n / i) * i;
       
        return sum - n * (n + 1) / 2;
    }
       
    // Driver Code
    public static void Main(String []args)
    {
        int n = 4;
        Console.WriteLine(properDivisorSum(n));
       
        n = 5;
        Console.WriteLine(properDivisorSum(n));
          
    }
}
 
// This code is contributed by 29AjayKumar

Java脚本

 
// Javascript implementation to find sum of all
// proper divisor of numbers up to N
 
// Utility function to find sum of
// all proper divisor of number up to N
function properDivisorSum(n)
{
    var sum = 0;
 
    // Loop to find the proper
    // divisor of every number
    // from 1 to N
    for (var i = 1; i <= n; ++i)
        sum += parseInt(n / i) * i;
 
    return sum - n * ((n + 1) / 2);
}
 
// Driver Code
var n = 4;
document.write(properDivisorSum(n)+"<br>");
n = 5;
document.write(properDivisorSum(n)+"<br>");
 
// This code is contributed by rutvik_56.

输出：

5
6

时间复杂度： O(N)
辅助空间： O(1)