打印数字的’K’最低有效位

给出了数字N。我们需要打印其“ K”个最低有效位。

例子 ：

Input : num = 10, k = 4 
Output : 1
Explanation : Binary Representation 
of 10 is 1010. 4th LSB is 1.

Input : num = 16, k = 3
Output : 0
Explanation : Binary Representation 
of 16 is 10000. 3rd LSB is 0.

我们可以通过以下步骤轻松解决此问题：

1. 向左移动数字“ 1”(K-1)次。
2. 这将产生一个数字，除了第K个比特外，所有其他未设置的比特。现在，我们将对给定数字对移位后的数字进行逻辑“与”运算。
3. 除第“ K”位以外的所有位都将产生0，而“ K”位将取决于数字。这是因为1 AND 1为1。0AND 1为0。

C++

// CPP code to print 'K'th LSB
#include 
using namespace std;
  
//Function returns 1 if set, 0 if not
bool LSB(int num, int K)
{
    return (num & (1 << (K-1)));
}
  
//Driver code
int main() 
{
    int num = 10, K = 4;
      
    //Function call
    cout << LSB(num, K);
      
    return 0;
}

java

// java code to print 'K'th LSB
import java .io.*;
  
class GFG {
      
    // Function returns 1 if set, 0 if not
    static boolean LSB(int num, int K)
    {
        boolean x = (num & (1 << (K-1))) != 0;
        return (x);
    }
      
    // Driver code
     public static void main(String[] args) 
    {
        int num = 10, K = 4;
          
        //Function call
        if(LSB(num, K)) 
            System.out.println("1") ;
          
        else
            System.out.println("0");
    } 
}
  
// This code is contributed by Anuj_67

Python

# Python code to print 'K'th LSB
  
# Function returns 1 if set, 0 if not
def LSB(num, K):
    return bool(num & (1 << (K - 1) ))
  
# Driver code
num, k = 10, 4
  
res = LSB(num, k)
if res :
    print 1
else:
    print 0
  
#This code is contributed by Sachin Bisht

C#

// C# code to print 'K'th LSB
using System;
  
class GFG {
      
    // Function returns 1 if set, 0 if not
    static bool LSB(int num, int K)
    {
        bool x = (num & (1 << (K-1))) != 0;
        return (x);
    }
      
    // Driver code
    static void Main() 
    {
        int num = 10, K = 4;
          
        //Function call
        if(LSB(num, K)) 
            Console.Write("1") ;
          
        else
            Console.Write("0");
    } 
}
  
// This code is contributed by Anuj_67

输出 ：

1