Encuentra el número de divisores de todos los números en el rango [1, n]

Dado un número entero N . La tarea es encontrar el número de divisores de todos los números en el rango [1, N] . 

Ejemplos: 

Entrada: N = 5 
Salida: 1 2 2 3 2 
divisores(1) = 1 
divisores(2) = 1 y 2 
divisores(3) = 1 y 3 
divisores(4) = 1, 2 y 4 
divisores(5) = 1 y 5

Entrada: N = 10 
Salida: 1 2 2 3 2 4 2 4 3 4 
 

Enfoque: Cree una array arr[] del tamaño (N + 1) donde arr[i] almacena la cantidad de divisores de i . Ahora, para cada j del rango [1, N] , incremente todos los elementos que son divisibles por j . 
Por ejemplo, si j = 3 , actualice arr[3], arr[6], arr[9], …

A continuación se muestra la implementación del enfoque anterior:  

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the number of divisors
// of all numbers in the range [1, n]
void findDivisors(int n)
{
 
    // Array to store the count
    // of divisors
    int div[n + 1];
    memset(div, 0, sizeof div);
 
    // For every number from 1 to n
    for (int i = 1; i <= n; i++) {
 
        // Increase divisors count for
        // every number divisible by i
        for (int j = 1; j * i <= n; j++)
            div[i * j]++;
    }
 
    // Print the divisors
    for (int i = 1; i <= n; i++)
        cout << div[i] << " ";
}
 
// Driver code
int main()
{
    int n = 10;
    findDivisors(n);
 
    return 0;
}

// Java implementation of the approach
 
class GFG
{
     
    // Function to find the number of divisors
    // of all numbers in the range [1, n]
    static void findDivisors(int n)
    {
     
        // Array to store the count
        // of divisors
        int[] div = new int[n + 1];
     
        // For every number from 1 to n
        for (int i = 1; i <= n; i++)
        {
     
            // Increase divisors count for
            // every number divisible by i
            for (int j = 1; j * i <= n; j++)
                div[i * j]++;
        }
     
        // Print the divisors
        for (int i = 1; i <= n; i++)
            System.out.print(div[i]+" ");
    }
     
    // Driver code
    public static void main(String args[])
    {
        int n = 10;
        findDivisors(n);
    }
}
 
// This code is contributed by Ryuga

# Python3 implementation of the approach
# Function to find the number of divisors
# of all numbers in the range [1,n]
def findDivisors(n):
     
    # List to store the count
    # of divisors
    div = [0 for i in range(n + 1)]
     
    # For every number from 1 to n
    for i in range(1, n + 1):
         
        # Increase divisors count for
        # every number divisible by i
        for j in range(1, n + 1):
            if j * i <= n:
                div[i * j] += 1
 
    # Print the divisors
    for i in range(1, n + 1):
        print(div[i], end = " ")
 
# Driver Code
if __name__ == "__main__":
    n = 10
    findDivisors(n)
 
# This code is contributed by
# Vivek Kumar Singh

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to find the number of divisors
// of all numbers in the range [1, n]
static void findDivisors(int n)
{
 
    // Array to store the count
    // of divisors
    int[] div = new int[n + 1];
 
    // For every number from 1 to n
    for (int i = 1; i <= n; i++)
    {
 
        // Increase divisors count for
        // every number divisible by i
        for (int j = 1; j * i <= n; j++)
            div[i * j]++;
    }
 
    // Print the divisors
    for (int i = 1; i <= n; i++)
        Console.Write(div[i]+" ");
}
 
// Driver code
static void Main()
{
    int n = 10;
    findDivisors(n);
}
}
 
// This code is contributed by mits

<?php
// PHP implementation of the approach
 
// Function to find the number of divisors
// of all numbers in the range [1, n]
function findDivisors($n)
{
 
    // Array to store the count
    // of divisors
    $div = array_fill(0, $n + 2, 0);
     
    // For every number from 1 to n
    for ($i = 1; $i <= $n; $i++)
    {
 
        // Increase divisors count for
        // every number divisible by i
        for ($j = 1; $j * $i <= $n; $j++)
            $div[$i * $j]++;
    }
 
    // Print the divisors
    for ($i = 1; $i <= $n; $i++)
        echo $div[$i], " ";
}
 
// Driver code
$n = 10;
findDivisors($n);
 
// This code is contributed
// by Arnab Kundu
?>

<script>
 
// Javascript implementation of the approach
 
// Function to find the number of divisors
// of all numbers in the range [1, n]
function findDivisors(n)
{
 
    // Array to store the count
    // of divisors
    let div = new Array(n + 1).fill(0);
 
    // For every number from 1 to n
    for(let i = 1; i <= n; i++)
    {
         
        // Increase divisors count for
        // every number divisible by i
        for(let j = 1; j * i <= n; j++)
            div[i * j]++;
    }
 
    // Print the divisors
    for(let i = 1; i <= n; i++)
        document.write(div[i] + " ");
}
 
// Driver code
let n = 10;
findDivisors(n);
 
// This code is contributed by souravmahato348
 
</script>

1 2 2 3 2 4 2 4 3 4

Complejidad de Tiempo: O(n ^3/2 )
Espacio Auxiliar: O(n)