Cuente el número de ceros finales en la representación binaria de un número usando Bitset

Dado un número. La tarea es contar el número de ceros finales en la representación binaria de un número usando un conjunto de bits.
Ejemplos: 
 

Input : N = 16
Output : 4
Binary representation of N is 10000. Therefore,
number of zeroes at the end is 4.

Input : N = 8
Output : 3

Enfoque: simplemente establecemos el número en el conjunto de bits y luego iteramos desde 0 índices del conjunto de bits, tan pronto como obtengamos 1, romperemos el ciclo porque no hay un cero final después de eso.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ program to count number of trailing zeros
// in Binary representation of a number
// using Bitset
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count number of trailing zeros in
// Binary representation of a number
// using Bitset
int CountTrailingZeros(int n)
{
    // declare bitset of 64 bits
    bitset<64> bit;
 
    // set bitset with the value
    bit |= n;
 
    int zero = 0;
 
    for (int i = 0; i < 64; i++) {
        if (bit[i] == 0)
            zero++;
        // if '1' comes then break
        else
            break;
    }
 
    return zero;
}
 
// Driver Code
int main()
{
    int n = 4;
 
    int ans = CountTrailingZeros(n);
 
    cout << ans << "\n";
 
    return 0;
}

// Java program to count number of trailing zeros
// in Binary representation of a number
// using Bitset
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
      
    // Function to count number of trailing zeros in
    // Binary representation of a number
    // using Bitset
    static int CountTrailingZeros(int n)
    {
         
        String bit = Integer.toBinaryString(n);
        StringBuilder bit1 = new StringBuilder();
        bit1.append(bit);
        bit1=bit1.reverse();
        int zero = 0;
      
        for (int i = 0; i < 64; i++) {
            if (bit1.charAt(i) == '0')
                zero++;
            // if '1' comes then break
            else
                break;
        }
      
        return zero;
    }
      
    // Driver Code
    public static void main(String []args)
    {
        int n = 4;
      
        int ans = CountTrailingZeros(n);
      
        System.out.println(ans);
    }
}
 
// This code is contributed by chitranayal

# Python3 program to count
# number of trailing zeros in
# Binary representation of a number
 
# Function to count number of
# trailing zeros in Binary
# representation of a number
def CountTrailingZeros(n):
     
    # declare bitset of 64 bits
    bit = bin(n)[2:]
    bit = bit[::-1]
 
    zero = 0;
 
    for i in range(len(bit)):
        if (bit[i] == '0'):
            zero += 1
             
        # if '1' comes then break
        else:
            break
 
    return zero
 
# Driver Code
n = 4
 
ans = CountTrailingZeros(n)
 
print(ans)
 
# This code is contributed
# by Mohit Kumar

// C# program to count number of trailing zeros
// in Binary representation of a number
// using Bitset
using System;
class GFG
{
 
  // Function to count number of trailing zeros in
  // Binary representation of a number
  // using Bitset
  static int CountTrailingZeros(int n)
  {
    string bit=Convert.ToString(n, 2);
    char[] charArray = bit.ToCharArray();
    Array.Reverse( charArray );
    string bit1 = new string( charArray );
    int zero = 0;
    for (int i = 0; i < 64; i++)
    {
      if (bit1[i] == '0')
      {
        zero++;
      }
 
      // if '1' comes then break
      else
      {
        break;
      }
    }
    return zero;
  }
 
  // Driver Code
  static public void Main ()
  {
    int n = 4;
    int ans = CountTrailingZeros(n);
    Console.WriteLine(ans);
  }
}
 
// This code is contributed by avanitrachhadiya2155

<script>
 
// JavaScript program to count number of trailing zeros
// in Binary representation of a number
// using Bitset
     
    // Function to count number of trailing zeros in
    // Binary representation of a number
    // using Bitset
    function CountTrailingZeros(n)
    {
        let bit = n.toString(2);
        let bit1=bit.split("");
        bit1=bit1.reverse();
        let zero = 0;
       
        for (let i = 0; i < 64; i++) {
            if (bit1[i] == '0')
                zero++;
            // if '1' comes then break
            else
                break;
        }
       
        return zero;
    }
     
    // Driver Code
    let n = 4;
    let ans = CountTrailingZeros(n);
    document.write(ans);
 
     
 
// This code is contributed by rag2127
 
</script>

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