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++ 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 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 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# 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, |
<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> |
Yes
Time Complexity: O(log N)
Auxiliary Space: O(1)