给定范围内的未设置位

给定一个非负数n和两个值l和r 。问题是要取消设置n的二进制表示形式中范围在l到r之间的位，即要取消设置从最右边的第l位到最右边的第r位的位。

约束： 1 <= l <= r <= n的二进制表示形式的位数。

例子：

Input : n = 42, l = 2, r = 5
Output : 32
(42)10 = (101010)2
(32)10 = (100000)2
The bits in the range 2 to 5 in the binary
representation of 42 have been unset.

Input : n = 63, l = 1, r = 4
Output : 48

方法：以下是步骤：

1. 计算num =(1 <<((sizeof(int)* 8 – 1))–1。这将产生最高的正整数num 。 num中的所有位都将被设置。
2. 在num中切换范围从l到r的位。请参阅这篇文章。
3. 现在，执行n = n＆num 。这将取消设置n中从l到r的范围中的位。
4. 返回n 。

注意： sizeof(int)已用作int数据类型的输入。对于大型输入，可以使用long int或long long int数据类型代替int 。

C/C++

// C++ implementation to unset bits in the given range
#include 
  
using namespace std;
  
// function to toggle bits in the given range
unsigned int toggleBitsFromLToR(unsigned int n,
                                unsigned int l,
                                unsigned int r)
{
    // calculating a number 'num' having 'r' number of bits
    // and bits in the range l to r are the only set bits
    int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
  
    // toggle the bits in the range l to r in 'n'
    // and return the number
    return (n ^ num);
}
  
// function to unset bits in the given range
unsigned int unsetBitsInGivenRange(unsigned int n,
                                   unsigned int l, unsigned int r)
{
    // 'num' is the highest positive integer number
    // all the bits of 'num' are set
    unsigned int num = (1 << (sizeof(int) * 8 - 1)) - 1;
  
    // toggle the bits in the range l to r in 'num'
    num = toggleBitsFromLToR(num, l, r);
  
    // unset the bits in the range l to r in 'n'
    // and return the number
    return (n & num);
}
  
// Driver program to test above
int main()
{
    unsigned int n = 42;
    unsigned int l = 2, r = 5;
    cout << unsetBitsInGivenRange(n, l, r);
    return 0;
}

Java

// Java implementation to unset bits in the given range
import java.io.*;
  
class GFG 
{
    // Function to toggle bits in the given range
    static int toggleBitsFromLToR(int n, int l, int r)
    {
        // calculating a number 'num' having 'r' number of bits
        // and bits in the range l to r are the only set bits
        int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
   
        // toggle the bits in the range l to r in 'n'
        // and return the number
        return (n ^ num);
    }
      
    // Function to unset bits in the given range
    static int unsetBitsInGivenRange(int n, int l, int r)
    {
        // 'num' is the highest positive integer number
        // all the bits of 'num' are set
        int num = (1 << (4 * 8 - 1)) - 1;
   
        // toggle the bits in the range l to r in 'num'
        num = toggleBitsFromLToR(num, l, r);
   
        // unset the bits in the range l to r in 'n'
        // and return the number
        return (n & num);
    }
      
    // driver program
    public static void main (String[] args) 
    {
        int n = 42;
        int l = 2, r = 5;
        System.out.println(unsetBitsInGivenRange(n, l, r));
    }
}
  
// Contributed by Pramod Kumar

Python3

# python implementation to unset bits
# in the given range
  
# Function to toggle bits in the
# given range
def toggleBitsFromLToR(n, l, r):
      
    # calculating a number 'num' 
    # having 'r' number of bits
    # and bits in the range l to
    # r are the only set bits
    num = (((1 << r) - 1) ^ 
           ((1 << (l - 1)) - 1))
  
    # toggle the bits in the range 
    # l to r in 'n' and return the 
    # number
    return (n ^ num)
      
# Function to unset bits in the
# given range
def unsetBitsInGivenRange(n, l, r):
      
    # 'num' is the highest positive
    # integer number all the bits
    # of 'num' are set
    num = (1 << (4 * 8 - 1)) - 1
  
    # toggle the bits in the range 
    # l to r in 'num'
    num = toggleBitsFromLToR(num, l, r)
  
    # unset the bits in the range 
    # l to r in 'n' and return the
    # number
    return (n & num)
  
# Driver code    
n = 42
l = 2
r = 5
print(unsetBitsInGivenRange(n, l, r))
  
# This code is contributed by Sam007.

C#

// C#  implementation to unset
// bits in the given range 
using System;
   
class GFG {
       
    // Function to toggle bits in the given range 
    static int toggleBitsFromLToR(int n, int l, int r) 
    { 
        // calculating a number 'num'
        // having 'r' number of bits 
        // and bits in the range l 
        // to r are the only set bits 
        int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1); 
   
        // toggle the bits in the
        // range l to r in 'n' 
        // and return the number 
        return (n ^ num); 
    } 
       
    // Function to unset bits in the given range 
    static int unsetBitsInGivenRange(int n, int l, int r) 
    { 
        // 'num' is the highest 
        // positive integer number 
        // all the bits of 'num'
        // are set 
        int num = (1 << (2 * 8 - 1)) - 1; 
   
        // toggle the bits in
        // the range l to r in 'num' 
        num = toggleBitsFromLToR(num, l, r); 
   
        // unset the bits in 
        // the range l to r in 'n' 
        // and return the number 
        return (n & num); 
    } 
       
// Driver Code
static public void Main() {
      
    int n = 42; 
    int l = 2, r = 5; 
    Console.WriteLine(unsetBitsInGivenRange(n, l, r)); 
} 
} 
   
// This Code is  Contributed by akt_mit

输出：

32