从1到N的所有除数之和|套装3

给定正整数N ，任务是找到从1到N的所有数字的除数之和。
例子：

Input: N = 5 
Output: 21 
Explanation: 
Sum of divisors of all numbers from 1 to 5 = 21. 
Divisors of 1 -> 1 
Divisors of 2 -> 1, 2 
Divisors of 3 -> 1, 3 
Divisors of 4 -> 1, 2, 4 
Divisors of 5 -> 1, 5, hence Sum = 21

Input: N = 6 
Output: 33 
Explanation: 
Sum of divisors of all numbers from 1 to 6 = 33. 
Divisors of 1 -> 1 
Divisors of 2 -> 1, 2 
Divisors of 3 -> 1, 3 
Divisors of 4 -> 1, 2, 4 
Divisors of 5 -> 1, 5 
Divisors of 6 -> 1, 2, 3, 6, hence sum = 33 

朴素和线性方法：有关朴素和线性方法，请参阅从1到n的所有除数之和。
对数方法：指从1到N的所有除数的和。设置2为对数时间法。

高效方法：
请按照以下步骤解决问题：

• 我们可以观察到，从1数每x到N，发生在总结它的最高倍数是≤ñ。
• 因此，通过公式x * floor(N / x)计算每个x的贡献，
• 可以看出，对于一系列连续数l ₁ ，l ₂ ，l ₃ ….l _r ， floor(N / i)相同。因此，代替为每个i计算l _i * floor(N / i) ，而是计算(l ₁ + l ₂ + l ₃ +…。+ l _r )* floor(N / l ₁ ) ，从而降低了计算复杂度。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
#define int long long int
#define m 1000000007
 
// Function to find the sum
// of all divisors of all
// numbers from 1 to N
void solve(long long n)
{
 
    // Stores the sum
    long long s = 0;
 
    for (int l = 1; l <= n;) {
 
        // Marks the last point of
        // occurence with same count
        int r = n / floor(n / l);
 
        int x = (((r % m) * ((r + 1)
                             % m))
                 / 2)
                % m;
        int y = (((l % m) * ((l - 1)
                             % m))
                 / 2)
                % m;
        int p = ((n / l) % m);
 
        // Calculate the sum
        s = (s + (((x - y) % m) * p) % m
             + m)
            % m;
 
        s %= m;
        l = r + 1;
    }
 
    // Return the result
    cout << (s + m) % m;
}
 
// Driver Code
signed main()
{
    long long n = 12;
    solve(n);
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{
 
static final int m = 1000000007;
 
// Function to find the sum
// of all divisors of all
// numbers from 1 to N
static void solve(long n)
{
  // Stores the sum
  long s = 0;
 
  for (int l = 1; l <= n;)
  {
    // Marks the last point of
    // occurence with same count
    int r = (int)(n /
             Math.floor(n / l));
 
    int x = (((r % m) *
             ((r + 1) %
               m)) / 2) % m;
    int y = (((l % m) *
             ((l - 1) %
               m)) / 2) % m;
    int p = (int)((n / l) % m);
 
    // Calculate the sum
    s = (s + (((x - y) %
                m) * p) %
                m + m) % m;
 
    s %= m;
    l = r + 1;
  }
 
  // Return the result
  System.out.print((s + m) % m);
}
 
// Driver Code
public static void main(String[] args)
{
  long n = 12;
  solve(n);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 Program to implement
# the above approach
import math
m = 1000000007
 
# Function to find the sum
# of all divisors of all
# numbers from 1 to N
def solve(n):
   
    # Stores the sum
    s = 0;
    l = 1;
    while(l < n + 1):
       
        # Marks the last poof
        # occurence with same count
        r = (int)(n /
             math.floor(n / l));
 
        x = ((((r % m) *
              ((r + 1) % m)) / 2) % m);
        y = ((((l % m) *
              ((l - 1) % m)) / 2) % m);
        p = (int)((n / l) % m);
 
        # Calculate the sum
        s = ((s + (((x - y) % m) *
                     p) % m + m) % m);
 
        s %= m;
        l = r + 1;
 
    # Return the result
    print (int((s + m) % m));
 
# Driver Code
if __name__ == '__main__':
   
    n = 12;
    solve(n);
 
# This code is contributed by Rajput-Ji

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
static readonly int m = 1000000007;
 
// Function to find the sum
// of all divisors of all
// numbers from 1 to N
static void solve(long n)
{
   
  // Stores the sum
  long s = 0;
 
  for(int l = 1; l <= n;)
  {
     
    // Marks the last point of
    // occurence with same count
    int r = (int)(n /(Math.Floor((double)n/l)));
 
    int x = (((r % m) *
             ((r + 1) %
             m)) / 2) % m;
    int y = (((l % m) *
             ((l - 1) %
             m)) / 2) % m;
    int p = (int)((n / l) % m);
 
    // Calculate the sum
    s = (s + (((x - y) %
               m) * p) %
                m + m) % m;
 
    s %= m;
    l = r + 1;
  }
   
  // Return the result
  Console.Write((s + m) % m);
}
 
// Driver Code
public static void Main(String[] args)
{
  long n = 12;
   
  solve(n);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// Javascript program implementation
// of the approach
 
let m = 1000000007;
  
// Function to find the sum
// of all divisors of all
// numbers from 1 to N
function solve(n)
{
  // Stores the sum
  let s = 0;
  
  for (let l = 1; l <= n;)
  {
    // Marks the last polet of
    // occurence with same count
    let r = (n /
             Math.floor(n / l));
  
    let x = Math.floor(((r % m) *
             ((r + 1) %
               m)) / 2) % m;
    let y = Math.floor(((l % m) *
             ((l - 1) %
               m)) / 2) % m;
    let p = (Math.floor(n / l) % m);
  
    // Calculate the sum
    s = (s + (((x - y) %
                m) * p) %
                m + m) % m;
  
    s %= m;
    l = r + 1;
  }
  
  // Return the result
  document.write((s + m) % m);
}
  
// Driver Code
     
    let n = 12;
  solve(n);
  
 // This code is contributed by splevel62.

输出

127

时间复杂度： O(√N)
辅助空间： O(1)