Cuente el número de 0 en base K Representación de un número

Dado un número N, la tarea es encontrar el número de ceros en la representación base K del número dado, donde K > 1 .

Ejemplos:

Entrada: N = 10, K = 3
Salida: 1
Explicación: La representación de base 3 de 10 es 101. 
Por lo tanto, el número de 0 en 101 es 1.

Entrada: N = 8, K = 2
Salida: 3
Explicación: La representación de base 2 de 8 es Por lo tanto, el número de 0 es 3. 

Enfoque: este problema se puede resolver convirtiendo el número a la forma base K. 
El pseudocódigo para convertir un número N de base 10 a base K :

// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of zeros
// in K base representation of N
int countZero(int N, int K)
{
    // Making a variable to store count
    int count = 0;
 
    // Looping till n becomes 0
    while (N > 0) {
 
        // Check if the current digit
        // is 0 or not.
        if (N % K == 0)
            count++;
        N /= K;
    }
    return count;
}
 
// Driver code
int main()
{
    int N = 8;
    int K = 2;
 
    // Function call
    cout << countZero(N, K) << endl;
    return 0;
}
 
// This code is contributed by Ashish Kumar

// C code to implement the approach
#include<stdio.h>
 
// Function to count the number of zeros
// in K base representation of N
int countZero(int N, int K)
{
   
  // Making a variable to store count
  int count = 0;
 
  // Looping till n becomes 0
  while (N > 0) {
 
    // Check if the current digit
    // is 0 or not.
    if (N % K == 0)
      count++;
    N /= K;
  }
  return count;
}
 
// Driver code
int main()
{
  int N = 8;
  int K = 2;
 
  // Function call
  int ans = countZero(N,K);
  printf("%d",ans);
  return 0;
}
 
// This code is contributed by ashishsingh13122000.

// Java code to implement the approach
import java.io.*;
 
class GFG
{
 
  // Function to count the number of zeros
  // in K base representation of N
  public static int countZero(int N, int K)
  {
 
    // Making a variable to store count
    int count = 0;
 
    // Looping till n becomes 0
    while (N > 0) {
 
      // Check if the current digit
      // is 0 or not.
      if (N % K == 0)
        count++;
      N /= K;
    }
    return count;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 8;
    int K = 2;
 
    // Function call
    System.out.println(countZero(N, K));
  }
}
 
// This code is contributed by Rohit Pradhan

# Python code to implement the approach
 
# Function to count the number of zeros
# in K base representation of N
def countZero(N, K):
   
    # Making a variable to store count
    count = 0
 
    # Looping till n becomes 0
    while (N > 0):
 
        # Check if the current digit
        # is 0 or not.
        if (N % K == 0):
            count += 1
        N //= K
 
    return count
 
# Driver code
N = 8
K = 2
 
# Function call
print(countZero(N, K))
 
# This code is contributed by shinjanpatra

// C# code to implement the approach
using System;
 
class GFG {
 
  // Function to count the number of zeros
  // in K base representation of N
  static int countZero(int N, int K)
  {
 
    // Making a variable to store count
    int count = 0;
 
    // Looping till n becomes 0
    while (N > 0) {
 
      // Check if the current digit
      // is 0 or not.
      if (N % K == 0)
        count++;
      N /= K;
    }
    return count;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 8;
    int K = 2;
 
    // Function call
    Console.Write(countZero(N, K));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

<script>
      // JavaScript code for the above approach
 
      // Function to count the number of zeros
      // in K base representation of N
      function countZero(N, K)
      {
       
          // Making a variable to store count
          let count = 0;
 
          // Looping till n becomes 0
          while (N > 0) {
 
              // Check if the current digit
              // is 0 or not.
              if (N % K == 0)
                  count++;
              N = Math.floor(N / K);
          }
          return count;
      }
 
      // Driver code
      let N = 8;
      let K = 2;
 
      // Function call
      document.write(countZero(N, K) + '<br>');
 
  // This code is contributed by Potta Lokesh
  </script>

3

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