Find count of digits in a number that divide the number

Given a positive integer n. The task is to find count of digits of number which evenly divides the number n.
Examples: 

Input : n = 12
Output : 2
1 and 2 divide 12.

Input : n = 1012
Output : 3
1, 1 and 2 divide 1012.

The idea is to find each digit of the number n by modulus 10 and then check whether it divides n or not. Accordingly, increment the counter. Notice that the digit can be 0, so take care of that case.
Below is implementation of this approach: 

C++

// C++ program to count number of digits
// that divides the number.
#include <bits/stdc++.h>
using namespace std;

// Return the number of digits that divides
// the number.
int countDigit(int n)
{
    int temp = n, count = 0;
    while (temp != 0) {
        // Fetching each digit of the number
        int d = temp % 10;
        temp /= 10;

        // Checking if digit is greater than 0
        // and can divides n.
        if (d > 0 && n % d == 0)
            count++;
    }

    return count;
}

// Driven Program
int main()
{
    int n = 1012;

    cout << countDigit(n) << endl;
    return 0;
}

Java

// Java program to count number of digits
// that divides the number.

class Test {
    // Return the number of digits that divides
    // the number.
    static int countDigit(int n)
    {
        int temp = n, count = 0;
        while (temp != 0) {
            // Fetching each digit of the number
            int d = temp % 10;
            temp /= 10;

            // Checking if digit is greater than 0
            // and can divides n.
            if (d > 0 && n % d == 0)
                count++;
        }

        return count;
    }

    // Driver method
    public static void main(String args[])
    {
        int n = 1012;
        System.out.println(countDigit(n));
    }
}

Python3

# Python3 code to count number of
# digits that divides the number.

# Return the number of digits
# that divides the number.
def countDigit (n):
    temp = n
    count = 0
    while temp != 0:

        # Fetching each digit
        # of the number
        d = temp % 10
        temp //= 10

        # Checking if digit is greater
        # than 0 and can divides n.
        if d > 0 and n % d == 0:
            count += 1
    return count

# Driven Code
n = 1012
print(countDigit(n))

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to count number of digits
// that divides the number.
using System;

class GFG {

    // Return the number of digits that
    // divides the number.
    static int countDigit(int n)
    {
        int temp = n, count = 0;
        while (temp != 0) {

            // Fetching each digit of
            // the number
            int d = temp % 10;
            temp /= 10;

            // Checking if digit is
            // greater than 0 and can
            // divides n.
            if (d > 0 && n % d == 0)
                count++;
        }

        return count;
    }

    // Driver method
    public static void Main()
    {
        int n = 1012;

        Console.Write(countDigit(n));
    }
}

// This code is contributed by parashar.

PHP

<?php
// PHP program to count
// number of digits
// that divides the number.

// Return the number of
// digits that divides
// the number.
function countDigit($n)
{
    $temp = $n;
    $count = 0;

    while ($temp != 0)
    {

        // Fetching each digit
        // of the number
        $d = $temp % 10;
        $temp /= 10;

        // Checking if digit
        // is greater than 0
        // and can divides n.
        if ($d > 0 && $n % $d == 0)
        $count++;
    }

    return $count;
}

    // Driver Code
    $n = 1012;
    echo countDigit($n), "\n";

// This code is contributed by ajit
?>

Javascript

<script>
// javascript program to count number of digits
// that divides the number.
// Return the number of digits that divides
    // the number.

function countDigit(n)
{
    var temp = n, count = 0;
    while (temp != 0)
    {

        // Fetching each digit of the number
        var d = temp % 10;
        temp /= 10;

        // Checking if digit is greater than 0
        // and can divides n.
        if (d > 0 && n % d == 0)
            count++;
    }

    return count;
}

// Driver method
var n = 1012;
document.write(countDigit(n));

// This code is contributed by Amit Katiyar
</script>

Output: 

3

Time Complexity: O(d) where d is the number of digits in a number.
Auxiliary Space: O(1)

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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