kth number

Here we are going to use to clear K^th bit in the binary representation of a given number.

在这里，我们将使用 清除给定数字的二进制表示形式中的^第 K位 。

Problem Statement: To write a C++ program to clear K^th bit of a number.

问题陈述：编写一个C ++程序来清除数字的^第 K位。

Constraints: 1<=n<=100

约束：1 <= n <= 100

Example:

例：

Input:    Enter number: 11    Enter k: 4    Output:    original number before clearing: 11    new number after clearing: 3

Problem Explanation:

问题说明：

Suppose the given number is 11 and the bit to be cleared is 4^th bit that is 3 left to the LSB.

假定给定数字为11 ，要清除的位为^第 4位，即LSB剩下3 。

Now the binary representation of 11 is:

现在11的二进制表示为：

n = 00001011    mask = 00001000 (left shift 1, three times (4-1) as k=4)    Bitwise complement of mask= ~mask = 11110111    n &(~ mask) = 00000011    Hence decimal representation of new number is 3.

This can be explained as follows:

可以解释如下：

Performing & with the ~mask clears Kth bit only as Kth bit in (~mask) is 0.

用〜掩模进行＆清除第K位只作为第K在(〜掩模 )为0位。

If X represents bits of n then, 0 & X is always 0. For remaining bits of original number, & with 1 gives the bit itself.

如果X表示n的位，则0和X始终为0。对于原始数字的其余位， ＆加上1表示该位本身。

Therefore performing bitwise AND of original number with (~mask) clears Kth bit in a number.

因此与执行逐位原来数目的AND(〜mask)将清除第K位的数字。

Algorithm:

算法：

1. Input the number and K^th bit to be cleared.

   输入要清除的数字和^第 K位。

2. Left shift 1 - (K-1) times to create a mask where only K^th bit is set.

   左移1- (K-1)次以创建仅设置^第 K位的掩码。

3. Take bitwise complement of the mask.

   取掩码的按位补码。

4. Perform bitwise AND of original number with this mask to clear the K^th bit.

   这种面膜进行逐位原来的号码，并清除^第 K ^个位。

5. Output the result after bitwise AND in decimal form.

   以十进制形式按位与后输出结果。

C++ Implementation:

C ++实现：

#include 
    
     using namespace std;int clearKthBit(int n,int k){
   	// left shift 1 , (k-1) times to get a mask 	// in which only kth bit is set	int m=1<<(k-1); 	// bitwise complement of mask	m=~(m);	// new number after clearing kth bit 	return (n&m);   }//driver program to check the codeint main(){
   	int num,k;	cout<<"Enter number: ";	cin>>num;	cout<<"Enter k: ";	cin>>k;	cout<<"original number before clearing: "<
     
      <
     
    

Output

输出量

Enter number: 11Enter k: 4original number before clearing: 11new number after clearing: 3

翻译自:

kth number