Representation of a number in powers of other

Given two numbers w and m, we need to determine whether it is possible to represent m in terms of powers of w. The powers of number w can be added or subtracted to obtain m and each powers of w can be used only once .
Examples:

Input : 3 7
Output : Yes
As 7 = 9 - 3 + 1 (3^2 - 3^1 + 3^0 )
so it is possible .

Input : 100 50
Output : No
As 50 is less than 100 so we can never
represent it in the powers of 100 .

Here we have to represent m in terms of powers of w used only once so it can be shown through the following equation . 
c0 + c1*w^1 + c2*w^2 + ... = m ------ (Equation 1)
Where each c0, c1, c2 ... are either -1 (for subtracting that power of w ), 0 (not using that power of w ), 1 (for adding that power of w ) .
=> c1*w^1 + c2*w^2 + ... = m - c0 
=> w(c1 + c2*w^1 + c3*w^2 + ... ) = m - c0 
=> c1 + c2*w^1 + c3*w^2 + ... = (m - c0)/w ------ (Equation 2)
Now, notice equation 1 and equation 2 -- we are trying to solve the same problem all over again. So we have to recurse till m > 0 . For such a solution to exist (m — ci) must be a multiple of w, where ci is the coefficient of the equation . The ci can be -1, 0, 1 . So we have to check for all three possibilities ( ( m - 1 ) % w == 0), ( ( m + 1 ) % w == 0) and ( m % w == 0) . If it is not, then there will not be any solution. 

C++

// CPP program to check if m can be represented
// as powers of w.
#include <bits/stdc++.h>
using namespace std;

bool asPowerSum(int w, int m)
{
    while (m) {
        if ((m - 1) % w == 0) 
            m = (m - 1) / w;
       else if ((m + 1) % w == 0) 
            m = (m + 1) / w;
        
        else if (m % w == 0) 
            m = m / w;
        
        else
            break; // None of 3 worked.
    }

    // If m is not zero means, it can't be 
    // represented in terms of powers of w.
    return (m == 0);
}

// Driver code
int main()
{
    int w = 3, m = 7;
    if (asPowerSum(w, m))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
   return 0;
}

Java

// Java program to check if m can 
// be represented as powers of w.

class GFG
{
    static boolean asPowerSum(int w, int m)
    {
        while (m > 0) 
        {
            if ((m - 1) % w == 0) 
                m = (m - 1) / w;
            
            else if ((m + 1) % w == 0) 
                m = (m + 1) / w;
            
            else if (m % w == 0) 
                m = m / w;
            
            else
                break; // None of 3 worked.
        }
    
        // If m is not zero means, it can't be 
        // represented in terms of powers of w.
        return (m == 0);
    } 
    
    // Driver function
    public static void main (String[] args)
    {
        int w = 3, m = 7;
        if (asPowerSum(w, m))
            System.out.println("Yes");
        else
            System.out.println("No"); 
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 program to check if m can 
# be represented as powers of w.
def asPowerSum(w, m):
    while (m > 0):
        if ((m - 1) % w == 0):
            m = (m - 1) / w;
        
        elif ((m + 1) % w == 0):
            m = (m + 1) / w;
        
        elif (m % w == 0):
            m = m / w;
        
        else:
            break; # None of 3 worked.
    
    # If m is not zero means, it can't be
    # represented in terms of powers of w.
    return (m == 0);

# Driver code
w = 3; 
m = 7;
if (asPowerSum(w, m)):
    print("Yes");
else:
    print("No");

# This code is contributed by mits

C#

// C# program to check if 
// m can be represented
// as powers of w.
using System;

class GFG
{
    static bool asPowerSum(int w, 
                           int m)
    {
        while (m > 0) 
        {
            if ((m - 1) % w == 0) 
                m = (m - 1) / w;
            
            else if ((m + 1) % w == 0) 
                m = (m + 1) / w;
            
            else if (m % w == 0) 
                m = m / w;
            
            else
                break; // None of 3 worked.
        }
    
        // If m is not zero means, 
        // it can't be represented
        // in terms of powers of w.
        return (m == 0);
    } 
    
    // Driver Code
    static public void Main ()
    {
        int w = 3, m = 7;
        if (asPowerSum(w, m))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No"); 
    }
}

// This code is contributed 
// by akt_mit.

PHP

<?php
// PHP program to check if m can 
// be represented as powers of w.

function asPowerSum($w, $m)
{
    while ($m) 
    {
        if (($m - 1) % $w == 0) 
            $m = ($m - 1) / $w;
    else if (($m + 1) % $w == 0) 
            $m = ($m + 1) / $w;
        
        else if ($m % $w == 0) 
            $m = $m / $w;
        
        else
            break; // None of 3 worked.
    }

    // If m is not zero means, it can't be 
    // represented in terms of powers of w.
    return ($m == 0);
}

// Driver code
$w = 3; 
$m = 7;
if (asPowerSum($w, $m))
    echo "Yes\n";
else
    echo "No\n";

// This code is contributed by mits
?>

JavaScript

<script>

// Javascript program to check if m can 
// be represented as powers of w.

    function asPowerSum(w, m)
    {
        while (m > 0) 
        {
            if ((m - 1) % w == 0) 
                m = (m - 1) / w;
              
            else if ((m + 1) % w == 0) 
                m = (m + 1) / w;
              
            else if (m % w == 0) 
                m = m / w;
              
            else
                break; // None of 3 worked.
        }
      
        // If m is not zero means, it can't be 
        // represented in terms of powers of w.
        return (m == 0);
    } 
    
// Driver code

        let w = 3, m = 7;
        if (asPowerSum(w, m))
            document.write("Yes");
        else
            document.write("No");
      
      // This code is contributed by sanjoy_62.
</script>

Output:

Yes

Time complexity: O(m)
Auxiliary space: O(1)

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