两个数字的XNOR

如果为二进制位，则XNOR给出XOR的反方向。

First bit | Second bit | XNOR  
0             0           1
0             1           0
1             0           0
1             1           1
It gives 1 if bits are same else if  
bits are different it gives 0.

它是XOR的反向，但是我们不能直接实现它，所以这是实现XNOR的程序
例子：

Input  : 10 20
Output : 1
Binary of 20 is 10100
Binary of 10 is  1010
So the XNOR is  00001
So output is 1

Input  : 10 10
Output : 15
Binary of 10 is  1010
Binary of 10 is  1010
So the XNOR is   1111
So output is 15

第一种方法：-(O(logn))在此解决方案中，我们一次检查一位。如果两个位相同，则将结果放入1，否则将放入0。
让我们用下面的代码来理解它

C++

// CPP program to find
// XNOR of two numbers
#include 
using namespace std;
 
// log(n) solution
int xnor(int a, int b)
{
    // Make sure a is larger
    if (a < b)
        swap(a, b);
 
    if (a == 0 && b == 0)
        return 1;
 
    int a_rem = 0; // for last bit of a
    int b_rem = 0; // for last bit of b
 
    // counter for count bit
    // and set bit  in xnornum
    int count = 0;
 
    // to make new xnor number
    int xnornum = 0;
 
    // for set bits in new xnor number
    while (a)
    {
        // get last bit of a
        a_rem = a & 1;
 
        // get last bit of b
        b_rem = b & 1;
 
        // Check if current two
        // bits are same
        if (a_rem == b_rem)       
            xnornum |= (1 << count);
         
        // counter for count bit
        count++;
        a = a >> 1;
        b = b >> 1;
    }
    return xnornum;
}
 
// Driver code
int main()
{
    int a = 10, b = 50;
    cout << xnor(a, b);
    return 0;
}

Java

// Java program to find
// XNOR of two numbers
import java.util.*;
import java.lang.*;
 
public class GfG {
 
    public static int xnor(int a, int b)
    {
        // Make sure a is larger
        if (a < b) {
            // swapping a and b;
            int t = a;
            a = b;
            b = t;
        }
 
        if (a == 0 && b == 0)
            return 1;
 
        // for last bit of a
        int a_rem = 0;
 
        // for last bit of b
        int b_rem = 0;
 
        // counter for count bit
        // and set bit in xnornum
        int count = 0;
 
        // to make new xnor number
        int xnornum = 0;
 
        // for set bits in new xnor number
        while (true) {
            // get last bit of a
            a_rem = a & 1;
 
            // get last bit of b
            b_rem = b & 1;
 
            // Check if current two bits are same
            if (a_rem == b_rem)
                xnornum |= (1 << count);
 
            // counter for count bit
            count++;
            a = a >> 1;
            b = b >> 1;
            if (a < 1)
                break;
        }
        return xnornum;
    }
 
    // Driver function
    public static void main(String argc[])
    {
        int a = 10, b = 50;
        System.out.println(xnor(a, b));
    }
}

Python3

# Python3 program to find XNOR
# of two numbers
import math
 
def swap(a,b):
 
    temp=a
    a=b
    b=temp
 
# log(n) solution
def xnor(a, b):
     
    # Make sure a is larger
    if (a < b):
        swap(a, b)
 
    if (a == 0 and b == 0) :
        return 1;
     
    # for last bit of a
    a_rem = 0
     
    # for last bit of b
    b_rem = 0
 
    # counter for count bit and
    #  set bit in xnor num
    count = 0
     
    # for make new xnor number
    xnornum = 0
 
    # for set bits in new xnor
    # number
    while (a!=0) :
     
        # get last bit of a
        a_rem = a & 1
         
        # get last bit of b
        b_rem = b & 1
 
        # Check if current two
        # bits are same
        if (a_rem == b_rem):    
            xnornum |= (1 << count)
         
        # counter for count bit
        count=count+1
         
        a = a >> 1
        b = b >> 1
     
    return xnornum;
     
# Driver method
a = 10
b = 50
print(xnor(a, b))
 
# This code is contributed by Gitanjali

C#

// C# program to find
// XNOR of two numbers
using System;
 
public class GfG {
 
    public static int xnor(int a, int b)
    {
        // Make sure a is larger
        if (a < b) {
            // swapping a and b;
            int t = a;
            a = b;
            b = t;
        }
 
        if (a == 0 && b == 0)
            return 1;
 
        // for last bit of a
        int a_rem = 0;
 
        // for last bit of b
        int b_rem = 0;
 
        // counter for count bit
        // and set bit in xnornum
        int count = 0;
 
        // to make new xnor number
        int xnornum = 0;
 
        // for set bits in new xnor number
        while (true) {
            // get last bit of a
            a_rem = a & 1;
 
            // get last bit of b
            b_rem = b & 1;
 
            // Check if current two bits are same
            if (a_rem == b_rem)
                xnornum |= (1 << count);
 
            // counter for count bit
            count++;
            a = a >> 1;
            b = b >> 1;
            if (a < 1)
                break;
        }
        return xnornum;
    }
 
    // Driver function
    public static void Main()
    {
        int a = 10, b = 50;
        Console.WriteLine(xnor(a, b));
    }
}
 
// This code is contributed by vt_m

PHP

> 1;
        $b = $b >> 1;
    }
    return $xnornum;
}
 
    // Driver code
    $a = 10;
    $b = 50;
    echo xnor($a, $b);
 
// This code is contributed by mits.    
?>

Javascript

 
// javascript program to find
// XNOR of two numbers
    function xnor(a, b)
    {
     
        // Make sure a is larger
        if (a < b)
        {
         
            // swapping a and b;
            let t = a;
            a = b;
            b = t;
        }
   
        if (a == 0 && b == 0)
            return 1;
   
        // for last bit of a
        let a_rem = 0;
   
        // for last bit of b
        let b_rem = 0;
   
        // counter for count bit
        // and set bit in xnornum
        let count = 0;
   
        // to make new xnor number
        let xnornum = 0;
   
        // for set bits in new xnor number
        while (true) {
            // get last bit of a
            a_rem = a & 1;
   
            // get last bit of b
            b_rem = b & 1;
   
            // Check if current two bits are same
            if (a_rem == b_rem)
                xnornum |= (1 << count);
   
            // counter for count bit
            count++;
            a = a >> 1;
            b = b >> 1;
            if (a < 1)
                break;
        }
        return xnornum;
    }
 
// Driver Function
 
         let a = 10, b = 50;
        document.write(xnor(a, b));
            
        // This code is contributed by susmitakundugoaldanga.

输出：

7

第二种方法：-O(1)
1)找出最多两个给定的数字。
2)切换两个数字中较高者的所有位。
3)返回原始较小数字和修改较大数字的XOR。

C++

// CPP program to find XNOR of two numbers.
#include 
using namespace std;
 
// Please refer below post for details of this
// function
// https://www.geeksforgeeks.org/toggle-bits-significant-bit/
int togglebit(int n)
{
    if (n == 0)
        return 1;
 
    // Make a copy of n as we are
    // going to change it.
    int i = n;
 
    // Below steps set bits after
    // MSB (including MSB)
 
    // Suppose n is 273 (binary
    // is 100010001). It does following
    // 100010001 | 010001000 = 110011001
    n |= n >> 1;
 
    // This makes sure 4 bits
    // (From MSB and including MSB)
    // are set. It does following
    // 110011001 | 001100110 = 111111111
    n |= n >> 2;
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
 
    return i ^ n;
}
 
// Returns XNOR of num1 and num2
int XNOR(int num1, int num2)
{
    // if num2 is greater then
    // we swap this number in num1
    if (num1 < num2)
        swap(num1, num2);
    num1 = togglebit(num1);
   
    return num1 ^ num2;
}
 
// Driver code
int main()
{
    int num1 = 10, num2 = 20;
    cout << XNOR(num1, num2);
    return 0;
}

Java

// Java program to find XNOR
// of two numbers
import java.io.*;
 
class GFG {
     
    // Please refer below post for
    // details of this function
    // https://www.geeksforgeeks.org/toggle-bits-significant-bit/
    static int togglebit(int n)
    {
        if (n == 0)
            return 1;
 
        // Make a copy of n as we are
        // going to change it.
        int i = n;
 
        // Below steps set bits after
        // MSB (including MSB)
 
        // Suppose n is 273 (binary
        // is 100010001). It does following
        // 100010001 | 010001000 = 110011001
        n |= n >> 1;
 
        // This makes sure 4 bits
        // (From MSB and including MSB)
        // are set. It does following
        // 110011001 | 001100110 = 111111111
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
 
        return i ^ n;
    }
 
    // Returns XNOR of num1 and num2
    static int xnor(int num1, int num2)
    {
        // if num2 is greater then
        // we swap this number in num1 
        if (num1 < num2)
        {
            int temp = num1;
            num1 = num2;
            num2 = temp;
        }
     
        num1 = togglebit(num1);
     
        return num1 ^ num2;
    }
 
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int a = 10, b = 20;    
        System.out.println(xnor(a, b));
    }
}
 
// This code is contributed by Gitanjali

Python

# python program to find XNOR of two numbers
import math
 
# Please refer below post for details of this function
# https://www.geeksforgeeks.org/toggle-bits-significant-bit/
def togglebit( n):
 
    if (n == 0):
        return 1
 
    # Make a copy of n as we are
    # going to change it.
    i = n
 
    # Below steps set bits after
    # MSB (including MSB)
 
    # Suppose n is 273 (binary
    # is 100010001). It does following
    # 100010001 | 010001000 = 110011001
    n = n|(n >> 1)
 
    # This makes sure 4 bits
    # (From MSB and including MSB)
    # are set. It does following
    # 110011001 | 001100110 = 111111111
    n |= n >> 2
    n |= n >> 4
    n |= n >> 8
    n |= n >> 16
 
    return i ^ n
     
# Returns XNOR of num1 and num2
def xnor( num1, num2):
     
    # Make sure num1 is larger
    if (num1 < num2):
        temp = num1
        num1 = num2
        num2 = temp
    num1 = togglebit(num1)
     
    return num1 ^ num2
     
# Driver code
a = 10
b = 20
print (xnor(a, b))
 
# This code is contributed by 'Gitanjali'.

C＃

// C# program to find XNOR
// of two numbers
using System;
 
class GFG {
     
    // Please refer below post for
    // details of this function
    // https://www.geeksforgeeks.org/toggle-bits-significant-bit/
    static int togglebit(int n)
    {
        if (n == 0)
            return 1;
 
        // Make a copy of n as we are
        // going to change it.
        int i = n;
 
        // Below steps set bits after
        // MSB (including MSB)
 
        // Suppose n is 273 (binary
        // is 100010001). It does following
        // 100010001 | 010001000 = 110011001
        n |= n >> 1;
 
        // This makes sure 4 bits
        // (From MSB and including MSB)
        // are set. It does following
        // 110011001 | 001100110 = 111111111
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
 
        return i ^ n;
    }
 
    // Returns XNOR of num1 and num2
    static int xnor(int num1, int num2)
    {
        // if num2 is greater then
        // we swap this number in num1
        if (num1 < num2)
        {
            int temp = num1;
            num1 = num2;
            num2 = temp;
        }
     
        num1 = togglebit(num1);
     
        return num1 ^ num2;
    }
 
    // Driver program
    public static void Main()
    {
        int a = 10, b = 20;
        Console.WriteLine(xnor(a, b));
    }
}
 
// This code is contributed by vt_m

的PHP

> 1;
 
    // This makes sure 4 bits
    // (From MSB and including MSB)
    // are set. It does following
    // 110011001 | 001100110 = 111111111
    $n |= $n >> 2;
    $n |= $n >> 4;
    $n |= $n >> 8;
    $n |= $n >> 16;
 
    return $i ^ $n;
}
 
// Returns XNOR of num1 and num2
function XNOR($num1, $num2)
{
     
    // if num2 is greater then
    // we swap this number in num1
    if ($num1 < $num2)
        list($num1, $num2)=array($num2, $num1);
    $num1 = togglebit($num1);
 
    return $num1 ^ $num2;
}
 
// Driver code
$num1 = 10;
$num2 = 20;
echo XNOR($num1, $num2);
 
// This code is contributed by Smitha.
?>

Java脚本

// Javascript program to find XNOR of two numbers.
 
// Please refer below post for details of this
// function
// https://www.geeksforgeeks.org/toggle-bits-significant-bit/
function togglebit(n)
{
    if (n == 0)
        return 1;
 
    // Make a copy of n as we are
    // going to change it.
    let i = n;
 
    // Below steps set bits after
    // MSB (including MSB)
 
    // Suppose n is 273 (binary
    // is 100010001). It does following
    // 100010001 | 010001000 = 110011001
    n |= n >> 1;
 
    // This makes sure 4 bits
    // (From MSB and including MSB)
    // are set. It does following
    // 110011001 | 001100110 = 111111111
    n |= n >> 2;
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
 
    return i ^ n;
}
 
// Returns XNOR of num1 and num2
function XNOR(num1, num2)
{
    // if num2 is greater then
    // we swap this number in num1
    if (num1 < num2) {
        let temp = num1;
        num1 = num2;
        num2 = temp;
    }
    num1 = togglebit(num1);
   
    return num1 ^ num2;
}
 
// Driver code
    let num1 = 10, num2 = 20;
    document.write(XNOR(num1, num2));
 

输出 ：

1