Count integers in a range which are divisible by their euler totient value

Given 2 integers L and R, the task is to find out the number of integers in the range [L, R] such that they are completely divisible by their Euler totient value.
Examples: 
 

Input: L = 2, R = 3 
Output: 1 
 (2) = 2 => 2 %  (2) = 0 
 (3) = 2 => 3 %  (3) = 1 
Hence 2 satisfies the condition.
Input: L = 12, R = 21 
Output: 3 
Only 12, 16 and 18 satisfy the condition. 
 

Approach: We know that the euler totient function of a number is given as follows: 
 

\phi(n) = n * (1 - \frac{1}{p_1}) * (1 - \frac{1}{p_2}) * ... * (1 - \frac{1}{p_k})

Rearranging the terms, we get: 
 

\frac{n}{\phi(n)} = \frac{p_1 * p_2 * ... * p_k}{(p_1 - 1) * (p_2 -1) * ... * (p_k -1)}

If we take a close look at the RHS, we observe that only 2 and 3 are the primes that satisfy n %  = 0. This is because for primes p₁ = 2 and p₂ = 3, p₁ - 1 = 1 and p₂ - 1 = 2. Hence, only numbers of the form 2^p3^q where p >= 1 and q >= 0 need to be counted while lying in the range [L, R].
Below is the implementation of the above approach:
 

// C++ implementation of the above approach.
#include <bits/stdc++.h>

#define ll long long
using namespace std;

// Function to return a^n
ll power(ll a, ll n)
{
    if (n == 0)
        return 1;

    ll p = power(a, n / 2);
    p = p * p;

    if (n & 1)
        p = p * a;

    return p;
}

// Function to return count of integers
// that satisfy n % phi(n) = 0
int countIntegers(ll l, ll r)
{

    ll ans = 0, i = 1;
    ll v = power(2, i);

    while (v <= r) {

        while (v <= r) {

            if (v >= l)
                ans++;
            v = v * 3;
        }

        i++;
        v = power(2, i);
    }

    if (l == 1)
        ans++;

    return ans;
}

// Driver Code
int main()
{
    ll l = 12, r = 21;
    cout << countIntegers(l, r);

    return 0;
}

// Java implementation of the above approach.
class GFG 
{

// Function to return a^n
static long power(long a, long n)
{
    if (n == 0)
        return 1;

    long p = power(a, n / 2);
    p = p * p;

    if (n%2== 1)
        p = p * a;

    return p;
}

// Function to return count of integers
// that satisfy n % phi(n) = 0
static int countIntegers(long l, long r)
{

    long ans = 0, i = 1;
    long v = power(2, i);

    while (v <= r) 
    {
        while (v <= r) 
        {

            if (v >= l)
                ans++;
            v = v * 3;
        }

        i++;
        v = power(2, i);
    }

    if (l == 1)
        ans++;

    return (int) ans;
}

// Driver Code
public static void main(String[] args)
{
    long l = 12, r = 21;
    System.out.println(countIntegers(l, r));
}
}

// This code contributed by Rajput-Ji

# Python3 implementation of the approach 

# Function to return a^n 
def power(a, n): 

    if n == 0: 
        return 1

    p = power(a, n // 2) 
    p = p * p 

    if n & 1: 
        p = p * a 

    return p 

# Function to return count of integers 
# that satisfy n % phi(n) = 0 
def countIntegers(l, r): 

    ans, i = 0, 1
    v = power(2, i) 

    while v <= r: 

        while v <= r: 

            if v >= l: 
                ans += 1
            
            v = v * 3

        i += 1
        v = power(2, i) 
    
    if l == 1: 
        ans += 1

    return ans 

# Driver Code 
if __name__ == "__main__":

    l, r = 12, 21
    print(countIntegers(l, r)) 
    
# This code is contributed 
# by Rituraj Jain

// C# implementation of the above approach.
using System;

class GFG 
{

// Function to return a^n
static long power(long a, long n)
{
    if (n == 0)
        return 1;

    long p = power(a, n / 2);
    p = p * p;

    if (n % 2 == 1)
        p = p * a;

    return p;
}

// Function to return count of integers
// that satisfy n % phi(n) = 0
static int countIntegers(long l, long r)
{

    long ans = 0, i = 1;
    long v = power(2, i);

    while (v <= r) 
    {
        while (v <= r) 
        {

            if (v >= l)
                ans++;
            v = v * 3;
        }

        i++;
        v = power(2, i);
    }

    if (l == 1)
        ans++;

    return (int) ans;
}

// Driver Code
public static void Main()
{
    long l = 12, r = 21;
    Console.WriteLine(countIntegers(l, r));
}
}

/* This code contributed by PrinciRaj1992 */

<?php
// PHP implementation of the above approach

// Function to return a^n 
function power($a, $n) 
{ 
    if ($n == 0) 
        return 1; 

    $p = power($a, $n / 2); 
    $p = $p * $p; 

    if ($n & 1) 
        $p = $p * $a; 

    return $p; 
} 

// Function to return count of integers 
// that satisfy n % phi(n) = 0 
function countIntegers($l, $r) 
{ 
    $ans = 0 ;
    $i = 1; 
    $v = power(2, $i); 

    while ($v <= $r)
    { 
        while ($v <= $r) 
        { 
            if ($v >= $l) 
                $ans++; 
            $v = $v * 3; 
        } 

        $i++; 
        $v = power(2, $i); 
    } 

    if ($l == 1) 
        $ans++; 

    return $ans; 
} 

// Driver Code 
$l = 12;
$r = 21; 

echo countIntegers($l, $r); 

// This code is contributed by Ryuga
?>

<script>

// JavaScript implementation of the above approach.    

// Function to return a^n
    function power(a , n) {
        if (n == 0)
            return 1;

        var p = power(a, parseInt(n / 2));
        p = p * p;

        if (n % 2 == 1)
            p = p * a;

        return p;
    }

    // Function to return count of integers
    // that satisfy n % phi(n) = 0
    function countIntegers(l , r) {

        var ans = 0, i = 1;
        var v = power(2, i);

        while (v <= r) {
            while (v <= r) {

                if (v >= l)
                    ans++;
                v = v * 3;
            }

            i++;
            v = power(2, i);
        }

        if (l == 1)
            ans++;

        return parseInt( ans);
    }

    // Driver Code
    
        var l = 12, r = 21;
        document.write(countIntegers(l, r));

// This code contributed by Rajput-Ji

</script>

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