检查给定范围内是否所有位都没有置位
给定一个非负数 n 和两个值 l 和 r 。问题是检查在 n 的二进制表示中 l 到 r 的范围内是否所有的位都没有置位。这些位从右向左编号，即最低有效位被认为在第一个位置。 约束: 1 < = l < = r < =二进制表示中的位数 n 。 举例:
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.
进场:以下为步骤:
- 计算num=((1<<r)–1)^((1<<(l-1))–1)。这将产生一个数字号，其具有 r 数量的比特，并且范围 l 到 r 中的比特是唯一设置的比特。
- 计算新 _ 数 = n &数。
- 如果 new_num == 0，则返回“是”(在给定范围内，所有位都不设置)。
- 否则返回“否”(在给定的范围内，所有的位不会被取消设置)。
C++
// C++ implementation to check whether all the bits
// are unset in the given range or not
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP 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
$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))
echo "Yes";
else
echo "No";
// This code is contributed by Smitha
?>
java 描述语言
<script>
// 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");
</script>
输出:
Yes
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