XNOR给出了二进制位的异或相反。
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First bit | Second bit | XNOR 0 0 10 1 01 0 01 1 1It gives 1 if bits are same else if bits are different it gives 0.
它与XOR相反，但我们不能直接实现它，所以这是实现XNOR的程序 例如：
Input : 10 20Output : 1Binary of 20 is 10100Binary of 10 is 1010So the XNOR is 00001So output is 1Input : 10 10Output : 15Binary of 10 is 1010Binary of 10 is 1010So the XNOR is 1111So output is 15
第一种方法：-（O（logn）） 在这个解决方案中，我们一次检查一位。如果两位相同，我们在结果中放入1，否则放入0。 让我们用下面的代码来理解它
C++
// CPP program to find // XNOR of two numbers #include <iostream> 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
<?php // PHP program to find // XNOR of two numbers // log(n) solution function xnor( $a , $b ) { // Make sure a is larger if ( $a < $b ) list( $a , $b ) = array ( $b , $a ); if ( $a == 0 && $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 xnornum $count = 0; // to make new xnor number $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 $a = 10; $b = 50; echo xnor( $a , $b ); // This code is contributed by mits. ?> |
Javascript
<script> // 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. </script> |
输出：
7
第二种方法：-O（1） 1） 求两个给定数字中的最大值。 2) 切换所有位 以两个数字中的较高者为单位。 3） 返回原始较小数字和修改后较大数字的异或。
C++
// CPP program to find XNOR of two numbers. #include <iostream> using namespace std; // Please refer below post for details of this // function 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 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 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 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 |
PHP
<?php // PHP program to find XNOR // of two numbers. // Please refer below post // for details of this function function 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 >> 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. ?> |
Javascript
<script> // Javascript program to find XNOR of two numbers. // Please refer below post for details of this // function 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)); </script> |
输出：
1
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