Minimum number of power terms with sum equal to n

Given two positive integer n and x. The task is to express n as sum of powers of x (x^a1 + x^a2 +.....+ x^a3) such that the number of powers of x (x^a1, x^a2, ....., x^a3) should be minimum. Print the minimum number of power of x used to make sum equal to n.
Examples: 
 

Input : n = 5, x = 3
Output : 3
5 = 30 + 30 + 31.
We use only 3 power terms of x { 30, 30, 31} 

Input : n = 13, x = 4
Output : 4
13 = 40 + 41 + 41 + 41.
We use only four power terms of x.

Input : n = 6, x = 1
Output : 6

If x = 1, then answer will be n only (n = 1 + 1 +.... n times). 
The idea is to use Horner's method. Any number n can be expressed as, n = x * a + b where 0 <= b <= x-1. Now since b is between 0 to x - 1, then b should be expressed as sum of x⁰ b times. 
Further a can be decomposed in similar manner and so on.
Algorithm to solve this problem: 

1. Initialize a variable ans to 0.
2. While n > 0
   a) ans = ans + n % x
   b) n = n/x
3. Return ans. 

Below is the implementation of above idea : 
 

// C++ program to calculate minimum number
// of powers of x to make sum equal to n.
#include <bits/stdc++.h>
using namespace std;

// Return minimum power terms of x required
int minPower(int n, int x)
{
    // if x is 1, return n since any power
    // of 1 is 1 only.
    if (x==1)
        return n;

    // Consider n = a * x  + b where a = n/x
    // and b = n % x.
    int ans = 0;
    while (n > 0)
    {
        // Update count of powers for 1's added
        ans += (n%x);

        // Repeat the process for reduced n
        n /= x;
    }

    return ans;
}

// Driven Program
int main()
{
    int n = 5, x = 3;
    cout << minPower(n, x) << endl;
    return 0;
}

// Java program to calculate
// minimum numberof powers of 
// x to make sum equal to n.

class GFG
{
    // Return minimum power 
    // terms of x required
    static int minPower(int n, int x)
    {
        // if x is 1, return n since 
        // any power of 1 is 1 only.
        if (x==1)
            return n;
    
        // Consider n = a * x + b where 
        // a = n/x and b = n % x.
        int ans = 0;
        while (n > 0)
        {
            // Update count of powers 
            // for 1's added
            ans += (n % x);
    
            // Repeat the process for reduced n
            n /= x;
        }
    
        return ans;
    }
    
    // Driver code
    public static void main (String[] args)
    {
        int n = 5, x = 3;
        System.out.println(minPower(n, x));
    }
}

// This code is contributed by Anant Agarwal.

# Python program to
# calculate minimum number
# of powers of x to make
# sum equal to n.

# Return minimum power
# terms of x required
def minPower(n,x):

    # if x is 1, return
    # n since any power
    # of 1 is 1 only.
    if (x==1):
        return n
 
    # Consider n = a * x  + b where a = n/x
    # and b = n % x.
    ans = 0
    while (n > 0):

        # Update count of powers for 1's added
        ans += (n%x)
 
        # Repeat the process for reduced n
        n //= x

 
    return ans
    
# Driver code
n = 5 
x = 3
print(minPower(n, x))

# This code is contributed
# by Anant Agarwal.

// C# program to calculate
// minimum numberof powers 
// of x to make sum equal 
// to n.
using System;

class GFG
{
    
    // Return minimum power 
    // terms of x required
    static int minPower(int n, int x)
    {
        // if x is 1, return n since 
        // any power of 1 is 1 only.
        if (x == 1)
            return n;
    
        // Consider n = a * x + b where 
        // a = n / x and b = n % x.
        int ans = 0;
        while (n > 0)
        {
            // Update count of 
            // powers for 1's 
            // added
            ans += (n % x);
    
            // Repeat the process 
            // for reduced n
            n /= x;
        }
    
        return ans;
    }
    
    // Driver code
    public static void Main ()
    {
        int n = 5, x = 3;
        Console.WriteLine(minPower(n, x));
    }
}

// This code is contributed by vt_m.

<?php
// PHP program to calculate minimum number
// of powers of x to make sum equal to n.

// Return minimum power 
// terms of x required
function minPower($n, $x)
{
    
    // if x is 1, return n since 
    // any power of 1 is 1 only.
    if ($x == 1)
        return $n;

    // Consider n = a * x + b 
    // where a = n/x and b = n % x.
    $ans = 0;
    while ($n > 0)
    {
        // Update count of powers
        // for 1's added
        $ans += ($n % $x);

        // Repeat the process
        // for reduced n
        $n /= $x;
    }

    return $ans;
}

// Driver Code
$n = 5; $x = 3;
echo(minPower($n, $x));

// This code is contributed by Ajit.
?>

<script>

// JavaScript program to calculate minimum number
// of powers of x to make sum equal to n.

// Return minimum power terms of x required
function minPower(n, x)
{
    // if x is 1, return n since any power
    // of 1 is 1 only.
    if (x==1)
        return n;

    // Consider n = a * x + b where a = n/x
    // and b = n % x.
    let ans = 0;
    while (n > 0)
    {
        // Update count of powers for 1's added
        ans += (n%x);

        // Repeat the process for reduced n
        n = Math.floor(n / x);
    }

    return ans;
}

// Driven Program

    let n = 5, x = 3;
    document.write(minPower(n, x) + "<br>");

// This code is contributed by Surbhi Tyagi.

</script>

Output: 

3

Time complexity: O(log_xn)

Auxiliary space:O(1)