Suma de todos los divisores de 1 a n

Given a positive integer n. Find the value of 

donde la función F(i) para el número i se define como la suma de todos los divisores de ‘ i ‘. 

Ejemplos: 

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

 El enfoque ingenuo es recorrer para cada número (1 a n), encontrar todos los divisores y seguir actualizando la suma con ese divisor. Mira esto para entender más. 

// C++ program to find sum of all
// divisor of number up to 'n'
#include<bits/stdc++.h>
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 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 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# 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.*/

<?php
// PHP 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)
{
    $sum = 0;
 
    for ($i = 1; $i <= $n; ++$i)
    {
 
        // Find all divisors of i
        // and add them
        for ($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
$n = 4;
echo "\n", divisorSum($n), "\n";
$n = 5;
echo divisorSum($n), "\n";
 
// This code is contributed by aj_36
?>

<script>
// 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
</script>

Producción :

15
21

Complejidad temporal: O(n√(n))) 
Espacio auxiliar: O(1)

El enfoque eficiente es observar la función y correlacionar el patrón. Para un número n dado, todo número del 1 al n contribuye con su presencia hasta el mayor múltiplo menor que n. Por ejemplo, 

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)

De la observación anterior, se puede observar fácilmente que el número i ocurre solo en sus múltiplos menores o iguales que n . Por lo tanto, solo necesitamos encontrar el conteo de múltiplos y luego multiplicarlo con i para obtener la contribución total en la suma final. Se puede hacer fácilmente en tiempo O(1) tomando el piso de (n / i) y luego multiplicándolo con i para la suma. 

// C++ program to find sum of all
// divisor of number up to 'n'
#include<bits/stdc++.h>
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)<<endl;
    n = 5;
    cout <<" "<< divisorSum(n)<< endl;
    return 0;
}
 
// This code is contributed by shivanisinghss2110

// C program to find sum of all
// divisor of number up to 'n'
#include <stdio.h>
 
// 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 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 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# 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.*/

<?php
// PHP 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)
{
    $sum = 0;
    for ( $i = 1; $i <= $n; ++$i)
        $sum += floor($n / $i) * $i;
    return $sum;
}
 
// Driver code
$n = 4;
echo divisorSum($n),"\n";
$n = 5;
echo divisorSum($n),"\n";
 
// This code is contributed by anuj_67.
?>

// 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) + "<br>");
n = 5;
document.write(divisorSum(n) + "<br>");
 
// This code is contributed by _saurabh_jaiswal.

Producción :

15
21

Complejidad temporal: O(n) 
Espacio auxiliar: O(1)

Solución más eficiente:

tenemos que calcular 

Para evaluar la expresión anterior en O(sqrt(N)) hacemos uso del lema armónico. 

Considere la secuencia armónica en la división de enteros: {N/1, N/2, N/3, ….. ,N/N}

El lema establece que la sucesión anterior no es creciente y que hay como máximo 2*sqrt(N) elementos diferentes.

Considere piso(N/i) = k. Así, k <= N/i < k+1. De esto obtenemos mayor = piso (N/k). Por lo tanto, podemos encontrar un rango de valores de i para el cual piso(N/i) es constante. Y usando The Harmonic Lemma sabemos que serán como máximo 2*sqrt(N) términos, por lo que podemos calcularlo programáticamente en O(sqrt(N)) complejidad. Considere el siguiente ejemplo para una mejor aclaración. 

// C++ program to calculate sum of divisors
// of numbers from 1 to N in O(sqrt(N)) complexity
#include <iostream>
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 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

# 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# 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

<script>
 
// Javascript program to calculate
// sum of divisors of numbers
// from 1 to N in O(sqrt(N))
// complexity
 
var mod = 10007;
     
/*
Function to calculate x^y using 
Modular exponentiation
Refer to https://www.geeksforgeeks.org/
modular-exponentiation-power-in-
modular-arithmetic/
*/
function power(x, y, p)
{
  // re x^y if p not specified 
  // else (x^y)%p
  var 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
function modinv(x)
{
  return power(x, mod - 2, mod);
}
 
// Function to calculate sum
// from 1 to n
function sum(n)
{
  // sum 1 to n = (n*(n+1))/2
  var retval = Math.floor((((n % mod) * ((n + 1) %
                    mod)) % mod) * modinv(2)) %
                    mod;
  return retval;
}
       
function divisorSum(n)
{
  var l = 1;
  var ans = 0;
 
  while (l <= n)
  {
    var k = n / l;
    var r = n / k;
    k %= mod;
 
    // For i=l to i=r,
    // floor(n/i) will be k
    ans += Math.floor((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   
  var n = 5;
  document.write("The sum of divisors of" +
                     " all numbers from 1 to " +
                     n + " is: " + divisorSum(n) +"<br>");
 
  n = 14;
  document.write("The sum of divisors of all" +
                     " numbers from 1 to " + n +
                     " is: " + divisorSum(n));
 
// This code is contributed by shivanisinghss2110
</script>

Producción:

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

Complejidad del tiempo: O(sqrt(N))

Espacio auxiliar: O(1)

Otro enfoque sqrt(n):

Dondequiera que se use división en el siguiente artículo, significa división de enteros.

Comencemos con un ejemplo, supongamos que n = 20, ahora veamos cómo cada número del 1 al 20 aparece como el factor de algún otro número.

 1 : 1 * 1, 1 * 2, 1 * 3, 1 * 4..., 1 * (20 / 1)
 2 : 2 * 1, 2 * 2, 2 * 3, 2 * 4,..., 2 * (20 / 2)
 3 : 3 * 1, 3 * 2, 3 * 3, 3 * 4...., 3 * (20 / 3)

nuestro objetivo es sumar cada número cada vez que aparece como el factor de algún otro número. Por ejemplo 3 aparece como el factor de (3 * 1), (3 * 2), (3 * 3)…, (3 * (20 / 3)). Ahora empecemos desde 1 y sumamos 1 a nuestra suma cada vez que aparezca y también sumaremos todos los números que salieron con 1, haremos lo mismo con 2 y cuando lleguemos a 3 ya habremos agregado 3 a nuestra suma cuando aparecía con 1 y 2 ahora solo sumaremos 3 cuando aparezca con números mayores a 2 es decir 3, 4, 5, 6 también sumaremos los números que aparecían con 3 así que sumaremos 4, 5 y 6 también (observe que aquí no sumaremos 3 dos veces debido a 3 * 3). Del mismo modo, cuando llegamos a 4, ya hemos agregado 4 cuando apareció con 1, 2 y 3, por lo que lo agregaremos solo cuando aparezca con números >= y sumaremos los números que aparecen con 4.

Finalmente, podemos decir que cuando estamos en un número i, ya hemos procesado los números del 1 al i – 1 y, por lo tanto, hemos agregado i cada vez que aparece con los números 1 a i – 1, por lo que esta vez solo necesitamos agregar i cada vez que aparece con números >= i también hay que sumar todos los números que aparecen junto con i y son > i.

Por lo tanto, para cada número i queremos agregar los siguientes términos a nuestra suma

t1 : (add i each time it appears with numbers >= itself) -> i * (num / i - (i - 1)) 

(recall i will appear with numbers 1 to num / i 
and we have already added i each time it appeared with a numbers less than itself)

t2 : (add numbers that appear with i) -> (i + 1) + (i + 2) ... + (num / i) 
(numbers 1 to num / i will appear with i but 
we have already processed numbers 1 to i - 1 and added them 
when they appeared with i so now we only have to add the numbers 
that appear with i and are greater than i, 
here we will not add i itself because when i appears with itself 
it should be added only once and we have added it once in t1)

we need to calculate t2 in O(1) time, here's how to do that
t2 = (i + 1) + (i + 2) + (i + 3) + ... + (num / i)
add and subtract 1 + 2 + 3 ... + i
=> t2 = 1 + 2 + 3 + ... + i + (i + 1) + (i + 2) + ... + (num / i) - (1 + 2 + 3 + ... + i)
=> t2 = (1 + 2 + 3 + .. + (num / i)) - (1 + 2 + 3 .. + i)
=> t2 = ((num / i) * (num / i + 1)) / 2 - (i * (i + 1)) / 2 

Finalmente, veamos los números que son mayores que sqrt(num). Estos números solo aparecerán con números menores que sqrt(num). Digamos que x es un número mayor que sqrt(num)

we have,
x > sqrt(num)
multiply sqrt(num) on both sides
=> x * sqrt(num) > sqrt(num) * sqrt(num)
=> x * sqrt(num) > num

queremos agregar x cada vez que aparece, de la prueba anterior vemos que x multiplicado por la raíz de num en sí mismo es mayor que num, por lo tanto, x solo aparecerá con números menores que la raíz de num, por lo que si procesamos todos los números de 1 a sqrt (num) sumaremos cada vez que aparezca x. Por ejemplo, tome n = 100, ahora considere 11, 11 * 10 > 100, por lo que 11 aparece solo con 1 a 9, es decir, como un factor de 11, 22, 33,…, 99. Lo mismo es cierto para el resto de los números que son mayores que 10. solo aparecerán con números menores a 10 y por lo tanto solo necesitamos procesar los números del 1 al 10 para sumar los números mayores a 10 para n = 100.

Finalmente, nuestra solución es esta 

for each i in 1 to sqrt(num) //no need to visit numbers greater than the root
    add t1 and t2 to the sum

a continuación se muestra el código c ++

#include <bits/stdc++.h>
using namespace std;
long long sum_all_divisors(long long num)
{
    long long sum = 0;
    for (long long i = 1; i <= sqrt(num); i++) {
        long long t1 = i * (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself
        long long t2 = (((num / i) * (num / i + 1)) / 2) - ((i * (i + 1)) / 2); // adding numbers that appear with i and are greater than i
        sum += t1 + t2;
    }
    return sum;
}
int main()
{
    int n;
    long long sum = sum_all_divisors(n);
    cout << sum << '\n';
    return 0;
}

import java.io.*;
 
class GFG {
 
public static int sum_all_divisors(int num)
{
    int sum = 0;
    for (int i = 1; i <= Math.sqrt(num); i++) {
        int t1 = i * (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself
        int t2 = (((num / i) * (num / i + 1)) / 2) - ((i * (i + 1)) / 2); // adding numbers that appear with i and are greater than i
        sum += t1 + t2;
    }
    return sum;
}
   
// Driver code 
public static void main (String[] args)
{
    int n = 1;
    int sum = sum_all_divisors(n);
    System.out.println(sum);
}
}
 
// This code is contributed by shivanisinghss2110

import math
def sum_all_divisors(num):
 
    sum = 0;
    for i in range(1,math.floor(math.sqrt(num))+1):
        t1 = i * (num / i - i + 1) # adding i every time it appears with numbers greater than or equal to itself
        t2 = (((num / i) * (num / i + 1)) / 2) - ((i * (i + 1)) / 2) # adding numbers that appear with i and are greater than i
        sum += t1 + t2;
     
    return sum;
 
n = 1
sum = sum_all_divisors(n)
print(sum)
 
# This code is contributed by shivanisinghss2110

using System;
 
class GFG {
 
public static int sum_all_divisors(int num)
{
    int sum = 0;
    for (int i = 1; i <= Math.Sqrt(num); i++) {
        int t1 = i * (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself
        int t2 = (((num / i) * (num / i + 1)) / 2) - ((i * (i + 1)) / 2); // adding numbers that appear with i and are greater than i
        sum += t1 + t2;
    }
    return sum;
}
   
// Driver code 
public static void Main (String[] args)
{
    int n = 1;
    int sum = sum_all_divisors(n);
    Console.Write(sum);
}
}
 
// This code is contributed by shivanisinghss2110

<script>
function sum_all_divisors(num)
{
    var sum = 0;
    for (var i = 1; i <= Math.sqrt(num); i++) {
        var t1 = i * (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself
        var t2 = (((num / i) * (num / i + 1)) / 2) - ((i * (i + 1)) / 2); // adding numbers that appear with i and are greater than i
        sum += t1 + t2;
    }
    return sum;
}
 
    var n;
    var sum = sum_all_divisors(n);
    document.write( sum );
     
// This code is contributed by shivanisinghss2110
</script>

Complejidad del tiempo: O(sqrt(N))

Espacio auxiliar: O(1)