从1到N的所有除数之和|套装2

给定正整数N ，任务是找到前N个自然数的除数之和。

例子：

Input: N = 4 
Output: 15 
Explanation: 
Sum of divisors of 1 = (1) 
Sum of divisors of 2 = (1+2) 
Sum of divisors of 3 = (1+3) 
Sum of divisors of 4 = (1+2+4) 
Hence, total sum = 1 + (1+2) + (1+3) + (1+2+4) = 15 

Input: N = 5 
Output: 21 
Explanation: 
Sum of divisors of 1 = (1) 
Sum of divisors of 2 = (1+2) 
Sum of divisors of 3 = (1+3) 
Sum of divisors of 4 = (1+2+4) 
Sum of divisors of 5 = (1+5) 
Hence, total sum = (1) + (1+2) + (1+3) + (1+2+4) + (1+5) = 21 

对于线性时间方法，请参阅从1到N的所有除数之和

方法：
为了优化上述文章中的方法，我们需要寻找具有对数复杂度的解决方案。在最终答案中，数字D被多次添加。让我们尝试观察重复加法的模式。
考虑N = 12 ：

D	Number of times added
1	12
2	6
3	4
5, 6	2
7, 8, 9, 10, 11, 12	1

从上面的模式中，观察到每个数字D被添加( N / D )次。另外，有多个具有相同(N / D)的D。因此，我们可以得出结论，对于给定的N和特定的i ，将从( N /( i + 1 ))+ 1到( N / i )的数字相加i次。

Illustration: 
 

1. N = 12, i = 1 
   (N/(i+1))+1 = 6+1 = 7 and (N/i) = 12 
   All numbers will be 7, 8, 9, 10, 11, 12 and will be added 1 time only.
2. N = 12, i = 2 
   (N/(i+1))+1 = 4+1 = 5 and (N/i) = 6 
   All numbers will be 5, 6 and will be added 2 times.

现在，假设A =(N /(i + 1))， B =(N / i)
从A到1的数字总和=从1到B的数字总和–从1到A的数字总和
另外，不仅仅是每次将i递增1，而是查找下一个像这样的i，i = N /(N /(i + 1))

下面是上述方法的实现：

C++

// C++ program for
// the above approach
#include
using namespace std;
 
int mod = 1000000007;
  
// Functions returns sum
// of numbers from 1 to n
int linearSum(int n)
{
    return (n * (n + 1) / 2) % mod;
}
  
// Functions returns sum
// of numbers from a+1 to b
int rangeSum(int b, int a)
{
    return (linearSum(b) -
            linearSum(a)) % mod;
}
 
// Function returns total
// sum of divisors
int totalSum(int n)
{
     
    // Stores total sum
    int result = 0;
    int i = 1;
  
    // Finding numbers and
    //its occurence
    while(true)
    {
          
        // Sum of product of each
        // number and its occurence
        result += rangeSum(n / i, n / (i + 1)) *
                          (i % mod) % mod;
          
        result %= mod;
         
        if (i == n)
            break;
             
        i = n / (n / (i + 1));
    }
    return result;
}       
  
// Driver code
int main()
{
    int N = 4;
    cout << totalSum(N) << endl;
  
    N = 12;
    cout << totalSum(N) << endl;
    return 0;
}
 
// This code is contributed by rutvik_56

Java

// Java program for
// the above approach
class GFG{
     
static final int mod = 1000000007;
 
// Functions returns sum
// of numbers from 1 to n
public static int linearSum(int n)
{
    return (n * (n + 1) / 2) % mod;
}
   
// Functions returns sum
// of numbers from a+1 to b
public static int rangeSum(int b, int a)
{
    return (linearSum(b) -
            linearSum(a)) % mod;
}
  
// Function returns total
// sum of divisors
public static int totalSum(int n)
{
      
    // Stores total sum
    int result = 0;
    int i = 1;
   
    // Finding numbers and
    //its occurence
    while(true)
    {
           
        // Sum of product of each
        // number and its occurence
        result += rangeSum(n / i,
                           n / (i + 1)) *
                          (i % mod) % mod;
           
        result %= mod;
          
        if (i == n)
            break;
              
        i = n / (n / (i + 1));
    }
    return result;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 4;
    System.out.println(totalSum(N));
   
    N = 12;
    System.out.println(totalSum(N));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for
# the above approach
 
mod = 1000000007
 
# Functions returns sum
# of numbers from 1 to n
def linearSum(n):
    return n*(n + 1)//2 % mod
 
# Functions returns sum
# of numbers from a+1 to b
def rangeSum(b, a):
    return (linearSum(b) - (
          linearSum(a))) % mod
 
# Function returns total
# sum of divisors
def totalSum(n):
 
    # Stores total sum
    result = 0
    i = 1
 
    # Finding numbers and
    # its occurence
    while True:
         
        # Sum of product of each
        # number and its occurence
        result += rangeSum(
            n//i, n//(i + 1)) * (
                  i % mod) % mod;
         
        result %= mod;
        if i == n:
            break
        i = n//(n//(i + 1))
         
    return result       
 
# Driver code
 
N= 4
print(totalSum(N))
 
N= 12
print(totalSum(N))

C#

// C# program for
// the above approach
using System;
 
class GFG{
     
static readonly int mod = 1000000007;
 
// Functions returns sum
// of numbers from 1 to n
public static int linearSum(int n)
{
    return (n * (n + 1) / 2) % mod;
}
   
// Functions returns sum
// of numbers from a+1 to b
public static int rangeSum(int b, int a)
{
    return (linearSum(b) -
            linearSum(a)) % mod;
}
  
// Function returns total
// sum of divisors
public static int totalSum(int n)
{
     
    // Stores total sum
    int result = 0;
    int i = 1;
   
    // Finding numbers and
    //its occurence
    while(true)
    {
           
        // Sum of product of each
        // number and its occurence
        result += rangeSum(n / i,
                           n / (i + 1)) *
                          (i % mod) % mod;
           
        result %= mod;
          
        if (i == n)
            break;
              
        i = n / (n / (i + 1));
    }
    return result;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 4;
    Console.WriteLine(totalSum(N));
   
    N = 12;
    Console.WriteLine(totalSum(N));
}
}
 
// This code is contributed by Amit Katiyar

输出：

15
127

时间复杂度： O(log N)