从1到n的所有除数之和

给定正整数n。找到的价值

$\sum_{i=1}^{i=n} F(i)$

其中将编号i的函数F(i)定义为“ i ”的所有除数之和。

例子 ：

Input: 4
Output: 15
Explanation
F(1) = 1
F(2) = 1 + 2 = 3
F(3) = 1 + 3 = 4
F(4) = 1 + 2 + 4 = 7
ans = F(1) + F(2) + F(3) + F(4)
    = 1 + 3 + 4 + 7
    = 15
Input: 5
Output: 21

天真的方法是遍历每个数字(1到n)，找到所有除数，并不断用该除数更新和。请参阅此内容以了解更多信息。

C++

// C++ program to find sum of all
// divisor of number up to 'n'
#include
using namespace std;
 
// Utility function to find sum of
// all divisor of number up to 'n'
int divisorSum(int n)
{
    int sum = 0;
 
    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;
            }
        }
    }
    return sum;
}
 
// Driver code
int main()
{
    int n = 4;
    cout << " " << divisorSum(n) << endl;
     
    n = 5;
    cout << " " << divisorSum(n);
     
    return 0;
}

Java

// JAVA program to find sum of all
// divisor of number up to 'n'
import java.io.*;
 
class GFG {
 
    // Utility function to find sum of
    // all divisor of number up to 'n'
    static int divisorSum(int n)
    {
        int sum = 0;
 
        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;
                }
            }
        }
        return sum;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 4;
        System.out.println(divisorSum(n));
        n = 5;
        System.out.println(divisorSum(n));
    }
}
 
/*This code is contributed by Nikita tiwari.*/

Python3

# Python3 code to find sum of all
# divisor of number up to 'n'
 
# Utility function to find sum of
# all divisor of number up to 'n'
def divisorSum( n ):
    sum = 0
     
    for i in range(1, n + 1):
         
        # Find all divisors of i
        # and add them
        j = 1
        while j * j <= i:
            if i % j == 0:
                if i / j == j:
                    sum += j
                else:
                    sum += j + i / j
            j = j + 1
    return int(sum)
 
# Driver code
n = 4
print( divisorSum(n))
n = 5
print( divisorSum(n))
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to find sum of all
// divisor of number up to 'n'
using System;
 
class GFG {
 
    // Utility function to find sum of
    // all divisor of number up to 'n'
    static int divisorSum(int n)
    {
        int sum = 0;
 
        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;
                }
            }
        }
        return sum;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 4;
        Console.WriteLine(divisorSum(n));
        n = 5;
        Console.WriteLine(divisorSum(n));
    }
}
 
/*This code is contributed by vt_m.*/

Javascript

// Javascript program to find sum of all
// divisor of number up to 'n'
 
// Utility function to find sum of
// all divisor of number up to 'n'
 
function divisorSum(n)
{
    let sum = 0;
 
    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;
            }
        }
    }
    return sum;
}
 
// Driver code
let n = 4;
document.write(divisorSum(n) + "<br>");
n = 5;
document.write(divisorSum(n) + "<br>");
 
// This code is contributed by _saurabh_jaiswal

---

输出 ：

15
21

时间复杂度： O(n√(n)))
辅助空间： O(1)

高效的方法是观察函数并关联模式。对于给定的数字n，从1到n的每个数字都会以小于n的最高倍数表示其存在。例如，

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

从上面的观察中，可以很容易地观察到数字i仅以小于或等于n的倍数出现。因此，我们只需要找到倍数，然后将其乘以i就可以得到最终总和。取(n / i)的底数，然后将其与i相乘即可轻松地在O(1)时间内完成。

C++

// C++ program to find sum of all
// divisor of number up to 'n'
#include
using namespace std;
 
// Utility function to find sum of
// all divisor of number up to 'n'
int divisorSum(int n)
{
    int sum = 0;
    for (int i = 1; i <= n; ++i)
        sum += (n / i) * i;
    return sum;
}
 
// Driver code
int main()
{
    int n = 4;
    cout <<" "<< divisorSum(n)<

C

// C program to find sum of all
// divisor of number up to 'n'
#include 
 
// Utility function to find sum of
// all divisor of number up to 'n'
int divisorSum(int n)
{
    int sum = 0;
    for (int i = 1; i <= n; ++i)
        sum += (n / i) * i;
    return sum;
}
 
// Driver code
int main()
{
    int n = 4;
    printf("%d\n", divisorSum(n));
    n = 5;
    printf("%d", divisorSum(n));
    return 0;
}

Java

// Java program to find sum of all
// divisor of number up to 'n'
import java.io.*;
 
class GFG {
 
    // Utility function to find sum of
    // all divisor of number up to 'n'
    static int divisorSum(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; ++i)
            sum += (n / i) * i;
        return sum;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 4;
        System.out.println(divisorSum(n));
        n = 5;
        System.out.println(divisorSum(n));
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

Python3

# Python3 code to find sum of all
# divisor of number up to 'n'
 
# Utility function to find sum of
# all divisor of number up to 'n'
def divisorSum( n ):
    sum = 0
    for i in range(1, n + 1):
        sum += int(n / i) * i
    return int(sum)
     
# Driver code
n = 4
print( divisorSum(n))
n = 5
print( divisorSum(n))
 
# This code is contributed by "Sharad_Bhardwaj".

C＃

// C# program to find sum of all
// divisor of number up to 'n'
using System;
 
class GFG {
 
    // Utility function to find sum of
    // all divisor of number up to 'n'
    static int divisorSum(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; ++i)
            sum += (n / i) * i;
        return sum;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 4;
        Console.WriteLine(divisorSum(n));
        n = 5;
        Console.WriteLine(divisorSum(n));
    }
}
 
/*This code is contributed by vt_m.*/

Java脚本

// Javascript program to find sum of all
// divisor of number up to 'n'
 
// Utility function to find sum of
// all divisor of number up to 'n'
function divisorSum(n)
{
    let sum = 0;
    for (let i = 1; i <= n; ++i)
        sum += Math.floor(n / i) * i;
    return sum;
}
 
// Driver code
let n = 4;
document.write(divisorSum(n) + "
");
n = 5;
document.write(divisorSum(n) + "
");
 
// This code is contributed by _saurabh_jaiswal.

输出 ：

15
21

时间复杂度： O(n)
辅助空间： O(1)

更有效的解决方案：

我们需要计算

$\sum_{i=1}^{i=N} (i*\lfloor N/i \rfloor )$

为了评估O(sqrt(N))中的上述表达式，我们使用了Harmonic Lemma。

考虑整数除法上的谐波序列：{N / 1，N / 2，N / 3，…..，N / N}

引理指出，以上序列不增加，并且最多存在2 * sqrt(N)个不同的元素。

考虑楼板(N / i)= k。因此，k≤N/i≤k+ 1。从中我们得到最大=地板(N / k)。因此，我们可以找到floor(N / i)恒定的i值范围。并且使用谐波引理我们知道最多为2 * sqrt(N)项，因此我们可以通过编程以O(sqrt(N))复杂度进行计算。请考虑以下示例，以更好地进行说明。

C++

// C++ program to calculate sum of divisors
// of numbers from 1 to N in O(sqrt(N)) complexity
#include 
using namespace std;
 
#define ll long long
#define mod 1000000007
 
/*
Function to calculate x^y using
Modular exponentiation
Refer to https://www.geeksforgeeks.org/
modular-exponentiation-power-in-modular-arithmetic/
*/
ll power(ll x, ll y, ll p)
{
     
    // re x^y if p not specified
    // else (x^y)%p
    ll res = 1;
    x = x % p;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return (res + p) % p;
}
 
// Function to find modular
// inverse of a under modulo m
// Assumption: m is prime
ll modinv(ll x)
{
    return power(x, mod - 2, mod);
}
 
// Function to calculate sum from 1 to n
ll sum(ll n)
{
    // sum 1 to n = (n*(n+1))/2
    ll retval = ((((n % mod) * ((n + 1) %
        mod)) % mod) * modinv(2)) % mod;
    return retval;
}
 
ll divisorSum(ll n)
{
    ll l = 1;
    ll ans = 0;
 
    while (l <= n)
    {
        ll k = n / l;
        ll r = n / k;
        k %= mod;
         
        // For i=l to i=r, floor(n/i) will be k
        ans += ((sum(r) - sum(l - 1) %
                        mod) * k) % mod;
         
        // Since values can be very large
        // we need to take mod at every step
        ans %= mod;
        l = r + 1;
    }
    ans = ans % mod;
      // ans can be negative
      // for example n = 831367 ans would be -534577982
    if (ans < 0){
        return ans+mod;
    }else{
        return ans;
    }
}
 
/* Driver program to test above function */
int main()
{
    int n = 5;
    cout << "The sum of divisors of all \
                numbers from 1 to " << n << " is: " \
                            << divisorSum(n) << '\n';
 
    n = 14;
    cout << "The sum of divisors of all \
                numbers from 1 to " << n << " is: " \
                            << divisorSum(n) << '\n';
}

Java

// Java program to calculate
// sum of divisors of numbers
// from 1 to N in O(sqrt(N))
// complexity
import java.util.*;
class Main{
     
static int mod = 1000000007;
     
/*
Function to calculate x^y using 
Modular exponentiation
Refer to https://www.geeksforgeeks.org/
modular-exponentiation-power-in-
modular-arithmetic/
*/
public static long power(long x,
                         long y,
                         long p)
{
  // re x^y if p not specified 
  // else (x^y)%p
  long res = 1;
  x = x % p;
   
  while (y > 0)
  {
    if ((y & 1) != 0)
      res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return (res + p) % p;
}
 
// Function to find modular 
// inverse of a under modulo m
// Assumption: m is prime
public static long modinv(long x)
{
  return power(x, mod - 2, mod);
}
 
// Function to calculate sum
// from 1 to n
public static long sum(long n)
{
  // sum 1 to n = (n*(n+1))/2
  long retval = ((((n % mod) * ((n + 1) %
                    mod)) % mod) * modinv(2)) %
                    mod;
  return retval;
}
       
public static long divisorSum(long n)
{
  long l = 1;
  long ans = 0;
 
  while (l <= n)
  {
    long k = n / l;
    long r = n / k;
    k %= mod;
 
    // For i=l to i=r,
    // floor(n/i) will be k
    ans += ((sum(r) - sum(l - 1) %
             mod) * k) % mod;
 
    // Since values can be very
    // large we need to take mod
    // at every step
    ans %= mod;
    l = r + 1;
  }
  ans = ans % mod;
  return ans;
}
 
// Driver code   
public static void main(String[] args)
{
  int n = 5;
  System.out.println("The sum of divisors of" +
                     " all numbers from 1 to " +
                     n + " is: " + divisorSum(n));
 
  n = 14;
  System.out.println("The sum of divisors of all" +
                     " numbers from 1 to " + n +
                     " is: " + divisorSum(n));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python program to calculate
# sum of divisors of numbers
# from 1 to N in O(sqrt(N))
# complexity
mod = 1000000007;
 
# Function to calculate x^y using Modular exponentiation Refer to
# https:#www.geeksforgeeks.org/ modular-exponentiation-power-in-
# modular-arithmetic/
def power(x, y, p):
     
    # re x^y if p not specified
    # else (x^y)%p
    res = 1;
    x = x % p;
 
    while (y > 0):
        if ((y & 1) != 0):
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
     
    return (res + p) % p;
 
# Function to find modular
# inverse of a under modulo m
# Assumption: m is prime
def modinv(x):
    return power(x, mod - 2, mod);
 
# Function to calculate sum
# from 1 to n
def sum(n):
   
    # sum 1 to n = (n*(n+1))/2
    retval = ((((n % mod) * ((n + 1) % mod)) % mod) * modinv(2)) % mod;
    return retval;
 
def divisorSum(n):
    l = 1;
    ans = 0;
 
    while (l <= n):
        k = n // l;
        r = n // k;
        k %= mod;
 
        # For i=l to i=r,
        # floor(n/i) will be k
        ans += ((sum(r) - sum(l - 1) % mod) * k) % mod;
 
        # Since values can be very
        # large we need to take mod
        # at every step
        ans %= mod;
        l = r + 1;
     
    ans = ans % mod;
    return ans;
 
# Driver code
if __name__ == '__main__':
    n = 5;
    print("The sum of divisors of all numbers from 1 to " , n , " is: " ,int( divisorSum(n)));
 
    n = 14;
    print("The sum of divisors of all numbers from 1 to ", n ," is: " , int(divisorSum(n)));
 
# This code contributed by aashish1995 Write

C＃

// C# program to calculate
// sum of divisors of numbers
// from 1 to N in O(sqrt(N))
// complexity
using System;
 
class GFG{
     
static int mod = 1000000007;
 
/*
Function to calculate x^y using 
Modular exponentiation
Refer to https://www.geeksforgeeks.org/
modular-exponentiation-power-in-
modular-arithmetic/
*/
static long power(long x, long y, long p)
{
     
    // re x^y if p not specified 
    // else (x^y)%p
    long res = 1;
    x = x % p;
     
    while (y > 0)
    {
        if ((y & 1) != 0)
            res = (res * x) % p;
             
        y = y >> 1;
        x = (x * x) % p;
    }
    return (res + p) % p;
}
 
// Function to find modular 
// inverse of a under modulo m
// Assumption: m is prime
static long modinv(long x)
{
    return power(x, mod - 2, mod);
}
 
// Function to calculate sum
// from 1 to n
static long sum(long n)
{
     
    // sum 1 to n = (n*(n+1))/2
    long retval = ((((n % mod) * ((n + 1) %
                  mod)) % mod) * modinv(2)) %
                  mod;
    return retval;
}
    
static long divisorSum(long n)
{
    long l = 1;
    long ans = 0;
     
    while (l <= n)
    {
        long k = n / l;
        long r = n / k;
        k %= mod;
         
        // For i=l to i=r,
        // floor(n/i) will be k
        ans += ((sum(r) - sum(l - 1) %
                   mod) * k) % mod;
         
        // Since values can be very
        // large we need to take mod
        // at every step
        ans %= mod;
        l = r + 1;
    }
    ans = ans % mod;
    return ans;
}
 
// Driver code
static void Main()
{
    int n = 5;
    Console.WriteLine("The sum of divisors of" +
                      " all numbers from 1 to " +
                      n + " is: " + divisorSum(n));
     
    n = 14;
    Console.WriteLine("The sum of divisors of all" +
                      " numbers from 1 to " + n +
                      " is: " + divisorSum(n));
}
}
 
// This code is contributed by divyesh072019

输出：

The sum of divisors of all numbers from 1 to 5 is: 21
The sum of divisors of all numbers from 1 to 14 is: 165

时间复杂度： O(sqrt(N))

辅助空间： O(1)