计算最右置位为K的N之前的数字

给定两个正整数N和K ，任务是对范围[1，N]中的数字进行计数，其范围从右起的^第K位，即LSB，是最右边的设置位。

例子：

Input: N = 15, K = 2
Output: 4
Explanation:
(2)₁₀ = (010)₂, (6)₁₀ = (110)₂, (10)₁₀ = (1010)₂, (14)₁₀ = (1110)₂ have the 2^{nd bit from the right is set.}

Input: N = 10 K = 3
Output: 3

天真的方法：想法是在[1，N]范围内进行迭代，并对该范围内的每个数字检查最右边的设置位的位置是否为K并打印这些数字的计数。
时间复杂度： O(N LogN)
辅助空间： O(1)

高效的方法：想法是在每一步中找到位置最靠右的位置i处的数字。请按照以下步骤解决问题：

• 使用变量i遍历范围[1，K] 。
• 找出最右边的设置位是i ^th的数字。
• 从N减去该数字。
• 对i的所有值重复上述步骤。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the numbers in the
// range [1, N] whose rightmost set bit is K
int countNumberHavingKthBitSet(int N, int K)
{
    // Stores the number whose
    // rightmost set bit is K
    int numbers_rightmost_setbit_K;
 
    for (int i = 1; i <= K; i++) {
 
        // Numbers whose rightmost set bit is i
        int numbers_rightmost_bit_i = (N + 1) / 2;
 
        // Subtracting the number whose
        // rightmost set bit is i, from N
        N -= numbers_rightmost_bit_i;
 
        // Since i = k, then the number whose
        // rigthmost set bit is K is stored
        if (i == K) {
            numbers_rightmost_setbit_K
                = numbers_rightmost_bit_i;
        }
    }
 
    cout << numbers_rightmost_setbit_K;
}
 
// Driver Code
int main()
{
    int N = 15;
    int K = 2;
    countNumberHavingKthBitSet(N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.Arrays;
class GFG
{
      
// Function to count the numbers in the
// range [1, N] whose rightmost set bit is K
static void countNumberHavingKthBitSet(int N, int K)
{
    // Stores the number whose
    // rightmost set bit is K
    int numbers_rightmost_setbit_K = 0;
    for (int i = 1; i <= K; i++)
    {
  
        // Numbers whose rightmost set bit is i
        int numbers_rightmost_bit_i = (N + 1) / 2;
  
        // Subtracting the number whose
        // rightmost set bit is i, from N
        N -= numbers_rightmost_bit_i;
  
        // Since i = k, then the number whose
        // rigthmost set bit is K is stored
        if (i == K)
        {
            numbers_rightmost_setbit_K
                = numbers_rightmost_bit_i;
        }
    }
    System.out.println(numbers_rightmost_setbit_K);
}
  
// Driver Code
static public void main(String args[])
{
    int N = 15;
    int K = 2;
    countNumberHavingKthBitSet(N, K);
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# Function to count the numbers in the
# range [1, N] whose rightmost set bit is K
def countNumberHavingKthBitSet(N, K):
     
    # Stores the number whose
    # rightmost set bit is K
    numbers_rightmost_setbit_K = 0
 
    for i in range(1, K + 1):
 
        # Numbers whose rightmost set bit is i
        numbers_rightmost_bit_i = (N + 1) // 2
 
        # Subtracting the number whose
        # rightmost set bit is i, from N
        N -= numbers_rightmost_bit_i
 
        # Since i = k, then the number whose
        # rigthmost set bit is K is stored
        if (i == K):
            numbers_rightmost_setbit_K = numbers_rightmost_bit_i
 
    print (numbers_rightmost_setbit_K)
 
# Driver Code
if __name__ == '__main__':
    N = 15
    K = 2
    countNumberHavingKthBitSet(N, K)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG
{
 
  // Function to count the numbers in the
  // range [1, N] whose rightmost set bit is K
  static void countNumberHavingKthBitSet(int N, int K)
  {
    // Stores the number whose
    // rightmost set bit is K
    int numbers_rightmost_setbit_K = 0;
    for (int i = 1; i <= K; i++)
    {
 
      // Numbers whose rightmost set bit is i
      int numbers_rightmost_bit_i = (N + 1) / 2;
 
      // Subtracting the number whose
      // rightmost set bit is i, from N
      N -= numbers_rightmost_bit_i;
 
      // Since i = k, then the number whose
      // rigthmost set bit is K is stored
      if (i == K)
      {
        numbers_rightmost_setbit_K
          = numbers_rightmost_bit_i;
      }
    }
    Console.WriteLine(numbers_rightmost_setbit_K);
  }
 
  // Driver Code
  static public void Main(String []args)
  {
    int N = 15;
    int K = 2;
    countNumberHavingKthBitSet(N, K);
  }
}
 
 
// This code is contributed by 29AjayKumar

输出：

4

时间复杂度： O(K)
辅助空间： O(1)