检查所有位是否在给定范围内未设置

给定一个非负数n和两个值l和r 。问题是检查在n的二进制表示形式中的l到r范围内，所有位是否都未设置。这些位从右到左编号，即，最低有效位被认为在第一位置。
约束： 1 <= l <= r <= n的二进制表示形式的位数。
例子：

Input : n = 17, l = 2, r = 4
Output : Yes
(17)10 = (10001)2
The bits in the range 2 to 4 are all unset.

Input : n = 39, l = 4, r = 6
Output : No
(39)10 = (100111)2
The bits in the range 4 to 6 are all not unset.

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new_num = n＆num。
3. 如果new_num == 0，则返回“是”(在给定范围内所有位均未设置)。
4. 否则返回“ No”(在给定范围内所有位均未设置)。

C++

// C++ implementation to check whether all the bits
// are unset in the given range or not
#include 
 
using namespace std;
 
// function to check whether all the bits
// are unset in the given range or not
bool allBitsSetInTheGivenRange(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);
 
    // new number which could only have one or more
    // set bits in the range l to r and nowhere else
    int new_num = n & num;
 
    // if true, then all bits are unset
    // in the given range
    if (new_num == 0)
        return true;
 
    // else all bits are not unset
    // in the given range
    return false;
}
 
// Driver program to test above
int main()
{
    unsigned int n = 17;
    unsigned int l = 2, r = 4;
    if (allBitsSetInTheGivenRange(n, l, r))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java implementation to check
// whether all the bits are
// unset in the given range or not
class GFG
{
     
// function to check whether
// all the bits are unset in
// the given range or not
static boolean allBitsSetInTheGivenRange(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);
 
    // new number which could only
    // have one or more set bits in
    // the range l to r and nowhere else
    int new_num = n & num;
 
    // if true, then all bits are
    // unset in the given range
    if (new_num == 0)
        return true;
 
    // else all bits are not
    // unset in the given range
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 17;
    int l = 2, r = 4;
    if (allBitsSetInTheGivenRange(n, l, r))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed
// by Smitha

Python3

# Python3 implementation to
# check whether all the bits
# are unset in the given
# range or not
 
# function to check whether
# all the bits are unset in
# the given range or not
def allBitsSetInTheGivenRange(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))
 
    # new number which could only
    # have one or more set bits in
    # the range l to r and nowhere else
    new_num = n & num
 
    # if true, then all bits are
    # unset in the given range
    if (new_num == 0):
        return True
 
    # else all bits are not
    # unset in the given range
    return false
 
# Driver Code
n = 17
l = 2
r = 4
if (allBitsSetInTheGivenRange(n, l, r)):
    print("Yes")
else:
    print("No")
 
# This code is contributed
# by Smitha

C#

// C# implementation to check
// whether all the bits are
// unset in the given range or not
using System;
 
class GFG
{
     
// function to check whether
// all the bits are unset in
// the given range or not
static bool allBitsSetInTheGivenRange(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);
 
    // new number which could 
    // only have one or more
    // set bits in the range
    // l to r and nowhere else
    int new_num = n & num;
 
    // if true, then all
    // bits are unset
    // in the given range
    if (new_num == 0)
        return true;
 
    // else all bits are not
    // unset in the given range
    return false;
}
 
// Driver Code
public static void Main()
{
    int n = 17;
    int l = 2, r = 4;
    if (allBitsSetInTheGivenRange(n, l, r))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed
// by Smitha

Javascript

 
// Javascript implementation to
// check whether all the bits
// are unset in the given range or not
 
// function to check whether all the bits
// are unset in the given range or not
function allBitsSetInTheGivenRange(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
    let num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
 
    // new number which could only have one or more
    // set bits in the range l to r and nowhere else
    let new_num = n & num;
 
    // if true, then all bits are unset
    // in the given range
    if (new_num == 0)
        return true;
 
    // else all bits are not unset
    // in the given range
    return false;
}
 
// Driver program to test above
    let n = 17;
    let l = 2, r = 4;
    if (allBitsSetInTheGivenRange(n, l, r))
        document.write("Yes");
    else
        document.write("No");
 

输出：

Yes