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 (xa1 + xa2 +…..+ xa3) such that the number of powers of x (xa1, xa2, ….., xa3) 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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?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.
?>

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Output:

3

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

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Improved By : vt_m, jit_t



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