Cuente números en el rango LR que son divisibles por todos sus dígitos distintos de cero

Dado un rango l – r (inclusive), cuente los números que son divisibles por todos sus dígitos distintos de cero.
Ejemplos: 
 

Input : 1 9 
Output : 9
Explanation: 
all the numbers are divisible by 
their digits in the range 1-9.

Input : 10 20 
Output : 5
Explanation: 
10, 11, 12, 15, 20 

Enfoque: 
1. Ejecute un ciclo para generar cada número de l y r. 
2. Compruebe si cada dígito distinto de cero de ese número divide el número o no. 
3. Lleva la cuenta de todos los números que son completamente divisibles por sus dígitos. 
4. Imprime el conteo de números.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ program to
// Count numbers in
// range L-R that are
// divisible by
// all of its non-zero
// digits
#include <bits/stdc++.h>
using namespace std;
 
// check if the number is
// divisible by the digits.
bool check(int n)
{
    int m = n;
    while (n) {
        int r = n % 10;
        if (r > 0)
            if ((m % r) != 0)
                return false;       
        n /= 10;
    }
 
    return true;
}
 
// function to calculate the
// number of numbers
int count(int l, int r)
{
    int ans = 0;
    for (int i = l; i <= r; i++)
        if (check(i))
            ans += 1;   
    return ans;
}
 
// Driver function
int main()
{
    int l = 10, r = 20;
    cout << count(l, r);
    return 0;
}

// Java program to Count
// numbers in range L-R
// that are divisible by
// all of its non-zero
// digits
import java.io.*;
 
class GFG {
 
    // check if the number
    // is divisible by the
    // digits.
    static boolean check(int n)
    {
        int m = n;
     
        while (n != 0)
        {
            int r = n % 10;
         
            if (r > 0)
                if ((m % r) != 0)
                    return false;    
         
            n /= 10;
        }
     
        return true;
    }
     
    // function to calculate
    // the number of numbers
    static int count(int l, int r)
    {
        int ans = 0;
         
        for (int i = l; i <= r; i++)
            if (check(i))
                ans += 1;
        return ans;
    }
     
    // Driver function
    public static void main(String args[])
    {
        int l = 10, r = 20;
         
        System.out.println(count(10, 20));
    }
}
 
// This code is contributed by Nikita Tiwari.

# Python 3 program
# to Count numbers in
# range L-R that are
# divisible by all of
# its non-zero digits
 
 
# check if the number is
# divisible by the digits.
def check(n) :
    m = n
    while (n != 0) :
        r = n % 10
        if (r > 0) :
            if ((m % r) != 0) :
                return False   
        n = n // 10
     
    return True
     
 
# function to calculate the
# number of numbers
def count(l, r) :
    ans = 0
    for i in range(l, r+1) :
        if (check(i)) :
            ans = ans + 1
    return ans
 
# Driver function
l = 10
r = 20
print(count(l, r))
 
# This code is contributed by Nikita Tiwari.

// Java program to Count
// numbers in range L-R
// that are divisible by
// all of its non-zero
// digits
using System;
 
class GFG {
 
    // check if the number
    // is divisible by the
    // digits.
    static bool check(int n)
    {
        int m = n;
     
        while (n != 0)
        {
            int r = n % 10;
         
            if (r > 0)
                if ((m % r) != 0)
                    return false;
         
            n /= 10;
        }
     
        return true;
    }
     
    // function to calculate
    // the number of numbers
    static int count(int l, int r)
    {
        int ans = 0;
         
        for (int i = l; i <= r; i++)
            if (check(i))
                ans += 1;
        return ans;
    }
     
    // Driver function
    public static void Main()
    {
        int l = 10, r = 20;
         
        Console.WriteLine(count(l, r));
    }
}
 
// This code is contributed by Vt_m.

<?php
// PHP program to Count numbers
// in range L-R that are
// divisible by all of its
// non-zero digits
 
// check if the number is
// divisible by the digits.
function check($n)
{
    $m = $n;
    while ($n) {
        $r = $n % 10;
        if ($r > 0)
            if (($m % $r) != 0)
                return false;    
        $n /= 10;
    }
 
    return true;
}
 
// function to calculate the
// number of numbers
function countIn($l, $r)
{
    $ans = 0;
    for ($i = $l; $i <= $r; $i++)
        if (check($i))
            $ans += 1;
             
    return $ans;
 
}
 
// Driver function
$l = 10; $r = 20;
echo countIn($l, $r);
 
// This code is contributed ajit
?>

<script>
 
// Javascript program to Count numbers
// in range L-R that are
// divisible by all of its
// non-zero digits
 
// check if the number is
// divisible by the digits.
function check(n)
{
     let m = n;
    while (n) {
        let r = n % 10;
        if (r > 0)
            if ((n % r) != 0)
                return false;   
        n /= 10;
    }
 
    return true;
}
 
// function to calculate the
// number of numbers
function countIn(l, r)
{
    let ans = 0;
    for (let i = l; i <= r; i++)
        if (check(i))
            ans += 1;
             
    return ans;
 
}
 
// Driver function
let l = 10;
let r = 20;
document.write(countIn(l, r));
         
// This code is contributed by sravan kumar
 
</script>

Producción: 

5

Complejidad de tiempo: O((rl) * log ₁₀ r), donde r representa el límite superior del rango y l denota el límite inferior del rango dado
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.