检查两个数字从L到R的位是否彼此互补

本文概述

• C ++
• Java
• Python 3
• C#
• 的PHP

给定两个非负数a和b，以及两个值l和r，问题是检查两个给定数中l到r范围内对应位置的所有位是否互为补位。

这些比特位从右到左编号, 即, 最低有效比特wei被认为在第一位置。

例子:

Input: a = 10, b = 5
       l = 1, r = 3
Output: Yes
(10)10 = (1010)2
(5)10 = (101)2 = (0101)2
All the bits in the range 1 to 3 are complement of each other.

Input: a = 21, b = 13
       l = 2, r = 4
Output: No
(21)10 = (10101)2
(13)10 = (1101)2 = (1101)2
All the bits in the range 2 to 4 are not complement of each other.

方法：以下是解决问题的步骤

• 计算异或值= a ^ b。
• 检查所有位是否都设置在范围内升to[Rin异或值。参考这个发布。

下面是上述方法的实现。

C ++

//C++ implementation to check 
//whether all the bits in the given range
//of two numbers are complement of each other
#include <bits/stdc++.h>
using namespace std;
  
//function to check whether all the bits
//are set 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 will only have one or more
     //set bits in the range l to r and nowhere else
     int new_num = n & num;
  
     //if both are equal, then all bits are set
     //in the given range
     if (num == new_num)
         return true ;
  
     //else all bits are not set
     return false ;
}
  
//function to check whether all the bits in the given range
//of two numbers are complement of each other
bool bitsAreComplement(unsigned int a, unsigned int b, unsigned int l, unsigned int r)
{
     unsigned int xor_value = a ^ b;
     return allBitsSetInTheGivenRange(xor_value, l, r);
}
  
//Driver Code
int main()
{
     unsigned int a = 10, b = 5;
     unsigned int l = 1, r = 3;
  
     if (bitsAreComplement(a, b, l, r))
         cout <<"Yes" ;
     else
         cout <<"No" ;
  
     return 0;
}

Java

//Java implementation to check 
//whether all the bits in the 
//given range of two numbers 
//are complement of each other
class GFG
{
//function to check whether 
//all the bits are set 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 will only 
     //have one or more set bits 
     //in the range l to r and 
     //nowhere else
     int new_num = n & num;
  
     //if both are equal, //then all bits are set
     //in the given range
     if (num == new_num)
         return true ;
  
     //else all bits are not set
     return false ;
}
  
//function to check whether all 
//the bits in the given range
//of two numbers are complement 
//of each other
static boolean bitsAreComplement( int a, int b, int l, int r)
{
     int xor_value = a ^ b;
     return allBitsSetInTheGivenRange(xor_value, l, r);
}
  
//Driver Code
public static void main(String []args)
{
     int a = 10 , b = 5 ;
     int l = 1 , r = 3 ;
  
     if (bitsAreComplement(a, b, l, r))
         System.out.println( "Yes" );
     else
         System.out.println( "No" );
}
}
  
//This code is contributed by Smitha

Python 3

# Python 3 implementation to check whether 
# all the bits in the given range of two 
# numbers are complement of each other
  
# function to check whether all the bits
# are set 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 will only have one 
     # or more set bits in the range l to r
     # and nowhere else
     new_num = n & num
  
     # if both are equal, then all bits 
     # are set in the given range
     if (num = = new_num):
         return True
  
     # else all bits are not set
     return False
  
# function to check whether all the bits 
# in the given range of two numbers are 
# complement of each other
def bitsAreComplement(a, b, l, r):
     xor_value = a ^ b
     return allBitsSetInTheGivenRange(xor_value, l, r)
  
# Driver Code
if __name__ = = "__main__" :
      
     a = 10
     b = 5
     l = 1
     r = 3
  
     if (bitsAreComplement(a, b, l, r)):
         print ( "Yes" )
     else :
         print ( "No" )
  
# This code is contributed by ita_c

C#

//C# implementation to check 
//whether all the bits in the 
//given range of two numbers 
//are complement of each other 
using System;
  
class GFG
{
//function to check whether 
//all the bits are set 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 will only 
     //have one or more set bits 
     //in the range l to r and 
     //nowhere else 
     int new_num = n & num; 
  
     //if both are equal, //then all bits are set 
     //in the given range 
     if (num == new_num) 
         return true ; 
  
     //else all bits are not set 
     return false ; 
} 
  
//function to check whether all 
//the bits in the given range 
//of two numbers are complement 
//of each other 
static bool bitsAreComplement( int a, int b, int l, int r) 
{ 
     int xor_value = a ^ b; 
     return allBitsSetInTheGivenRange(xor_value, l, r); 
} 
  
//Driver Code 
static public void Main ()
{
     int a = 10, b = 5; 
     int l = 1, r = 3; 
      
     if (bitsAreComplement(a, b, l, r)) 
         Console.WriteLine( "Yes" ); 
     else
         Console.WriteLine( "No" ); 
} 
} 
  
//This code is contributed 
//by Ajit Deshpal.

的PHP

<?php
//PHP implementation to check 
//whether all the bits in the
//given range of two numbers
//are complement of each other 
  
//function to check whether
//all the bits are set 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 will 
     //only have one or more 
     //set bits in the range
     //l to r and nowhere else 
     $new_num = ( $n & $num ); 
  
     //if both are equal, //then all bits are set 
     //in the given range 
     if ( $num == $new_num ) 
         return true; 
  
     //else all bits
     //are not set 
     return false; 
} 
  
//function to check whether
//all the bits in the given range 
//of two numbers are complement 
//of each other 
function bitsAreComplement( $a , $b , $l , $r ) 
{ 
     $xor_value = $a ^ $b ; 
     return allBitsSetInTheGivenRange( $xor_value , $l , $r ); 
} 
  
//Driver Code 
$a = 10;
$b = 5; 
$l = 1;
$r = 3; 
if (bitsAreComplement( $a , $b , $l , $r )) 
     echo "Yes" ; 
else
     echo "No" ; 
  
//This Code is Contributed by ajit
?>

输出如下：

Yes

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