计算范围内的未设置位

给定一个非负数n和两个值l和r 。问题是要对n的二进制表示形式中的l到r范围内的未设置位数进行计数，即对从最右边的第l位到最右边的r位进行计数。
例子：

Input : n = 42, l = 2, r = 5
Output : 2
(42)10 = (101010)2
There are '2' unset bits in the range 2 to 5.

Input : n = 80, l = 1, r = 4
Output : 4

方法：以下是步骤：

1. 计算num =(((1 << r)– 1)^((1 <<(l-1))– 1)。这将产生一个数字num，它具有r个位数，并且范围l至r的位数是唯一的设置位。
2. 计算数字中的设置位数(n和num) 。请参阅这篇文章。算吧。
3. 计算ans =(r – l + 1)–计数。
4. 返回ans 。

C++

// C++ implementation to count unset bits in the
// given range
#include 
using namespace std;
 
// Function to get no of set bits in the
// binary representation of 'n'
unsigned int countSetBits(int n)
{
    unsigned int count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// function to count unset bits
// in the given range
unsigned int countUnsetBitsInGivenRange(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);
 
    // returns number of unset bits in the range
    // 'l' to 'r' in 'n'
    return (r - l + 1) - countSetBits(n & num);
}
 
// Driver program to test above
int main()
{
    unsigned int n = 80;
    unsigned int l = 1, r = 4;
    cout << countUnsetBitsInGivenRange(n, l, r);
    return 0;
}

Java

// Java implementation to count unset bits in the
// given range
class GFG {
     
    // Function to get no of set bits in the
    // binary representation of 'n'
    static int countSetBits(int n)
    {
        int count = 0;
         
        while (n > 0) {
            n &= (n - 1);
            count++;
        }
         
        return count;
    }
 
    // function to count unset bits
    // in the given range
    static int countUnsetBitsInGivenRange(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);
 
        // returns number of unset bits in the range
        // 'l' to 'r' in 'n'
        return (r - l + 1) - countSetBits(n & num);
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int n = 80;
        int l = 1, r = 4;
         
        System.out.print(
            countUnsetBitsInGivenRange(n, l, r));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 implementation to count
# unset bits in the given range
 
# Function to get no of set bits in
# the binary representation of 'n'
def countSetBits (n):
    count = 0
    while n:
        n &= (n - 1)
        count += 1
    return count
 
# function to count unset bits
# in the given range
def countUnsetBitsInGivenRange (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)
     
    # returns number of unset bits
    # in the range 'l' to 'r' in 'n'
    return (r - l + 1) - countSetBits(n & num)
 
# Driver code to test above
n = 80
l = 1
r = 4
print(countUnsetBitsInGivenRange(n, l, r))
 
# This code is contributed by "Sharad_Bhardwaj"

C#

// C# implementation to count unset bits in the
// given range
using System;
 
class GFG {
     
    // Function to get no of set bits in the
    // binary representation of 'n'
    static int countSetBits(int n)
    {
        int count = 0;
         
        while (n > 0) {
            n &= (n - 1);
            count++;
        }
         
        return count;
    }
      
    // function to count unset bits
    // in the given range
    static int countUnsetBitsInGivenRange(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);
      
        // returns number of unset bits in the range
        // 'l' to 'r' in 'n'
        return (r - l + 1) - countSetBits(n & num);
    }
     
    //Driver code
    public static void Main()
    {
        int n = 80;
        int l = 1, r = 4;
         
        Console.Write(countUnsetBitsInGivenRange(n, l, r));
    }
}
 
//This code is contributed by Anant Agarwal.

Javascript

 
// Javascript implementation to count unset bits in the
// given range
 
// Function to get no of set bits in the
// binary representation of 'n'
function countSetBits(n)
{
    var count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// function to count unset bits
// in the given range
function countUnsetBitsInGivenRange(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
    var num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
 
    // returns number of unset bits in the range
    // 'l' to 'r' in 'n'
    return (r - l + 1) - countSetBits(n & num);
}
 
// Driver program to test above
var n = 80;
var l = 1, r = 4;
document.write( countUnsetBitsInGivenRange(n, l, r));
 

输出：

4