在二进制表示中 LSB 为 0 的范围 [L, R] 中的数字计数

给定两个整数L和R 。任务是找出范围[L, R]中所有二进制表示的最低有效位为0 的数字的计数。

例子：

Input: L = 10, R = 20  
Output: 6  

Input: L = 7, R = 11  
Output: 2   

天真的方法：解决此问题的最简单方法是检查范围[L, R] 中的每个数字，如果二进制表示中的最低有效位为0。

下面是上述方法的实现：

C++

// C++ implementation of the approach
 
#include 
using namespace std;
 
// Function to return the count
// of required numbers
int countNumbers(int l, int r)
{
    int count = 0;
 
    for (int i = l; i <= r; i++) {
        // If rightmost bit is 0
        if ((i & 1) == 0) {
            count++;
        }
    }
    // Return the required count
    return count;
}
 
// Driver code
int main()
{
    int l = 10, r = 20;
 
    // Call function countNumbers
    cout << countNumbers(l, r);
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG{
     
// Function to return the count
// of required numbers
static int countNumbers(int l, int r)
{
    int count = 0;
    for(int i = l; i <= r; i++)
    {
         
        // If rightmost bit is 0
        if ((i & 1) == 0)
            count += 1;
    }
     
    // Return the required count
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int l = 10, r = 20;
     
    // Call function countNumbers
    System.out.println(countNumbers(l, r));
}
}
 
// This code is contributed by MuskanKalra1

Python3

# Python3 implementation of the approach
 
# Function to return the count
# of required numbers
def countNumbers(l, r):
     
    count = 0
     
    for i in range(l, r + 1):
         
        # If rightmost bit is 0
        if ((i & 1) == 0):
            count += 1
             
    # Return the required count
    return count
 
# Driver code
l = 10
r = 20
 
# Call function countNumbers
print(countNumbers(l, r))
 
# This code is contributed by amreshkumar3

C#

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to return the count
    // of required numbers
    static int countNumbers(int l, int r)
    {
        int count = 0;
        for (int i = l; i <= r; i++) {
 
            // If rightmost bit is 0
            if ((i & 1) == 0)
                count += 1;
        }
 
        // Return the required count
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int l = 10, r = 20;
 
        // Call function countNumbers
        Console.WriteLine(countNumbers(l, r));
    }
}
 
// This code is contributed by subham348.

Javascript

 
// JavaScript implementation of the approach
 
// Function to return the count
// of required numbers
function countNumbers(l, r)
{
    let count = 0;
 
    for (let i = l; i <= r; i++) {
        // If rightmost bit is 0
        if ((i & 1) == 0) {
            count++;
        }
    }
    // Return the required count
    return count;
}
 
// Driver code
    let l = 10, r = 20;
 
    // Call function countNumbers
    document.write(countNumbers(l, r));
 

输出

6

时间复杂度： O(r – l)
辅助空间： O(1)

有效的方法：这个问题可以通过使用比特的属性来解决。只有偶数的最右边位为0 。可以使用此公式((R / 2) – (L – 1) / 2)在O(1)时间内找到计数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count
// of required numbers
int countNumbers(int l, int r)
{
    // Count of numbers in range
    // which are divisible by 2
    return ((r / 2) - (l - 1) / 2);
}
 
// Driver code
int main()
{
    int l = 10, r = 20;
    cout << countNumbers(l, r);
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG{
     
// Function to return the count
// of required numbers
static int countNumbers(int l, int r)
{
     
    // Count of numbers in range
    //  which are divisible by 2
    return ((r / 2) - (l - 1) / 2);
}
 
// Driver Code
public static void main(String[] args)
{
    int l = 10;
    int r = 20;
     
    System.out.println(countNumbers(l, r));
}
}
 
// This code is contributed by MuskanKalra1

Python3

# Python3 implementation of the approach
 
# Function to return the count
# of required numbers
def countNumbers(l, r):
 
    # Count of numbers in range
    #  which are divisible by 2
    return ((r // 2) - (l - 1) // 2)
 
# Driver code
l = 10
r = 20
 
print(countNumbers(l, r))
 
# This code is contributed by amreshkumar3

C＃

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to return the count
// of required numbers
static int countNumbers(int l, int r)
{
   
    // Count of numbers in range
    // which are divisible by 2
    return ((r / 2) - (l - 1) / 2);
}
 
// Driver code
public static void Main()
{
    int l = 10, r = 20;
    Console.Write(countNumbers(l, r));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

 
// Javascript implementation of the approach
 
// Function to return the count
// of required numbers
function countNumbers(l, r)
{
     
    // Count of numbers in range
    // which are divisible by 2
    return(parseInt(r / 2) -
          parseInt((l - 1) / 2));
}
 
// Driver code
let l = 10, r = 20;
 
document.write(countNumbers(l, r));
 
// This code is contributed by subhammahato348
 

输出

6

时间复杂度： O(1)
辅助空间： O(1)