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Check if a number N can be expressed as the sum of powers of X or not

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Given two positive numbers N and X, the task is to check if the given number N can be expressed as the sum of distinct powers of X. If found to be true, then print “Yes”, Otherwise, print “No”.

Examples:

Input: N = 10, X = 3
Output: Yes
Explanation:
The given value of N(= 10) can be written as (1 + 9) = 30 + 32. Since all the power of X(= 3) are distinct. Therefore, print Yes.

Input: N= 12, X = 4
Output: No

 

Approach: The given problem can be solved by checking if the number N can be written in base X or not. Follow the steps below to solve the problem:

  • Iterate a loop until the value of N is at least 0 and perform the following steps:
    • Calculate the value of remainder rem when N is divided by X.
    • If the value of rem is at least 2, then print “No” and return.
    • Otherwise, update the value of N as N / X.
  • After completing the above steps, if there doesn’t exist any termination, then print “Yes” as the result at N can be expressed in the distinct power of X.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the number N
// can be expressed as the sum of
// different powers of X or not
bool ToCheckPowerofX(int n, int x)
{
    // While n is a positive number
    while (n > 0) {
 
        // Find the remainder
        int rem = n % x;
 
        // If rem is at least 2, then
        // representation is impossible
        if (rem >= 2) {
            return false;
        }
 
        // Divide the value of N by x
        n = n / x;
    }
 
    return true;
}
 
// Driver Code
int main()
{
    int N = 10, X = 3;
    if (ToCheckPowerofX(N, X)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
 
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to check if the number N
// can be expressed as the sum of
// different powers of X or not
static boolean ToCheckPowerofX(int n, int x)
{
     
    // While n is a positive number
    while (n > 0)
    {
         
        // Find the remainder
        int rem = n % x;
 
        // If rem is at least 2, then
        // representation is impossible
        if (rem >= 2)
        {
            return false;
        }
 
        // Divide the value of N by x
        n = n / x;
    }
    return true;
}
 
// Driver Code
public static void main (String[] args)
{
    int N = 10, X = 3;
    if (ToCheckPowerofX(N, X))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by sanjoy_62


Python3




# Python3 program for the above approach
 
# Function to check if the number N
# can be expressed as the sum of
# different powers of X or not
def ToCheckPowerofX(n, x):
     
    # While n is a positive number
    while (n > 0):
         
        # Find the remainder
        rem = n % x
 
        # If rem is at least 2, then
        # representation is impossible
        if (rem >= 2):
            return False
 
        # Divide the value of N by x
        n = n // x
 
    return True
 
# Driver Code
if __name__ == '__main__':
     
    N = 10
    X = 3
     
    if (ToCheckPowerofX(N, X)):
        print("Yes")
    else:
        print("No")
         
# This code is contributed by bgangwar59


C#




// C# program for the above approach
using System;
class GFG{
     
 
// Function to check if the number N
// can be expressed as the sum of
// different powers of X or not
static bool ToCheckPowerofX(int n, int x)
{
     
    // While n is a positive number
    while (n > 0)
    {
         
        // Find the remainder
        int rem = n % x;
 
        // If rem is at least 2, then
        // representation is impossible
        if (rem >= 2)
        {
            return false;
        }
 
        // Divide the value of N by x
        n = n / x;
    }
    return true;
}
 
// Driver code
public static void Main(String []args)
{
     int N = 10, X = 3;
    if (ToCheckPowerofX(N, X))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
// This code is contributed by code_hunt,


Javascript




<script>
 
// JavaScripts program for the above approach
 
// Function to check if the number N
// can be expressed as the sum of
// different powers of X or not
function ToCheckPowerofX(n, x)
{
    // While n is a positive number
    while (n > 0) {
 
        // Find the remainder
        var rem = n % x;
 
        // If rem is at least 2, then
        // representation is impossible
        if (rem >= 2) {
            return false;
        }
 
        // Divide the value of N by x
        n = n / x;
    }
 
    return true;
}
 
// Driver Code
    var N = 10, X = 3;
    if (ToCheckPowerofX(N, X)) {
        document.write("Yes");
    }
    else {
        document.write("No");
    }
 
</script>


Output: 

Yes

 

Time Complexity: O(log N)
Auxiliary Space: O(1)



Last Updated : 02 Jun, 2021
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