📜  从1到N的所有除数之和|套装3

📅  最后修改于: 2021-06-26 22:36:13             🧑  作者: Mango

给定正整数N ,任务是找到从1N的所有数字的除数之和。
例子:

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xN,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 rfloor(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


输出
127

时间复杂度: O(√N)
辅助空间: O(1)