Dado un número entero 
. Encuentre el número de soluciones de 
las cuales satisface la ecuación:
a = b + (a^b)
Ejemplos:
Input: a = 4 Output: 2 The only values of b are 0 and 4 itself. Input: a = 3 Output: 4
Una solución ingenua es iterar de 0 a y contar el número de valores que satisfacen la ecuación dada. Necesitamos atravesar solo hasta que cualquier valor mayor que dará el valor XOR> , por lo tanto, no puede satisfacer la ecuación.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find the number of values
// of b such that a = b + (a^b)
#include <bits/stdc++.h>
using namespace std;
// function to return the number of solutions
int countSolutions(int a)
{
int count = 0;
// check for every possible value
for (int i = 0; i <= a; i++) {
if (a == (i + (a ^ i)))
count++;
}
return count;
}
// Driver Code
int main()
{
int a = 3;
cout << countSolutions(a);
}
Java
// Java program to find the number of values
// of b such that a = b + (a^b)
import java.io.*;
class GFG {
// function to return the number of solutions
static int countSolutions(int a)
{
int count = 0;
// check for every possible value
for (int i = 0; i <= a; i++) {
if (a == (i + (a ^ i)))
count++;
}
return count;
}
// Driver Code
public static void main (String[] args) {
int a = 3;
System.out.println( countSolutions(a));
}
}
// This code is contributed by inder_verma
Python3
# Python 3 program to find # the number of values of b # such that a = b + (a^b) # function to return the # number of solutions def countSolutions(a): count = 0 # check for every possible value for i in range(a + 1): if (a == (i + (a ^ i))): count += 1 return count # Driver Code if __name__ == "__main__": a = 3 print(countSolutions(a)) # This code is contributed # by ChitraNayal
C#
// C# program to find the number of
// values of b such that a = b + (a^b)
using System;
class GFG
{
// function to return the
// number of solutions
static int countSolutions(int a)
{
int count = 0;
// check for every possible value
for (int i = 0; i <= a; i++)
{
if (a == (i + (a ^ i)))
count++;
}
return count;
}
// Driver Code
public static void Main ()
{
int a = 3;
Console.WriteLine(countSolutions(a));
}
}
// This code is contributed by inder_verma
PHP
<?php
// PHP program to find the number of
// values of b such that a = b + (a^b)
// function to return the
// number of solutions
function countSolutions($a)
{
$count = 0;
// check for every possible value
for ($i = 0; $i <= $a; $i++)
{
if ($a == ($i + ($a ^ $i)))
$count++;
}
return $count;
}
// Driver Code
$a = 3;
echo countSolutions($a);
// This code is contributed
// by inder_verma
?>
Javascript
<script>
// Javascript program to find the number of values
// of b such that a = b + (a^b)
// function to return the number of solutions
function countSolutions(a)
{
let count = 0;
// check for every possible value
for (let i = 0; i <= a; i++) {
if (a == (i + (a ^ i)))
count++;
}
return count;
}
// Driver Code
let a = 3;
document.write(countSolutions(a));
</script>
Producción:
4
Complejidad del Tiempo : O(a)
Espacio Auxiliar : O(1)
Un Enfoque Eficiente es observar un patrón de respuestas cuando escribimos las posibles soluciones para diferentes valores de . Solo los bits establecidos se utilizan para determinar el número de respuestas posibles. La respuesta al problema siempre será 2 ^ (número de bits establecidos), que se puede determinar mediante la observación.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find the number of values
// of b such that a = b + (a^b)
#include <bits/stdc++.h>
using namespace std;
// function to return the number of solutions
int countSolutions(int a)
{
int count = __builtin_popcount(a);
count = pow(2, count);
return count;
}
// Driver Code
int main()
{
int a = 3;
cout << countSolutions(a);
}
Java
// Java program to find the number of values
// of b such that a = b + (a^b)
import java.io.*;
class GFG
{
// function to return the number of solutions
static int countSolutions(int a)
{
int count = Integer.bitCount(a);
count =(int) Math.pow(2, count);
return count;
}
// Driver Code
public static void main (String[] args)
{
int a = 3;
System.out.println(countSolutions(a));
}
}
// This code is contributed by Raj
Python3
# Python3 program to find the number
# of values of b such that a = b + (a^b)
# function to return the number
# of solutions
def countSolutions(a):
count = bin(a).count('1')
return 2 ** count
# Driver Code
if __name__ == "__main__":
a = 3
print(countSolutions(a))
# This code is contributed by
# Rituraj Jain
C#
// C# program to find the number of
// values of b such that a = b + (a^b)
class GFG
{
// function to return the number
// of solutions
static int countSolutions(int a)
{
int count = bitCount(a);
count =(int) System.Math.Pow(2, count);
return count;
}
static int bitCount(int n)
{
int count = 0;
while (n != 0)
{
count++;
n &= (n - 1);
}
return count;
}
// Driver Code
public static void Main()
{
int a = 3;
System.Console.WriteLine(countSolutions(a));
}
}
// This code is contributed by mits
PHP
<?php
// PHP program to find the number of
// values of b such that a = b + (a^b)
// function to return the number
// of solutions
function countSolutions($a)
{
$count = bitCount($a);
$count = (int)pow(2, $count);
return $count;
}
function bitCount($n)
{
$count = 0;
while ($n != 0)
{
$count++;
$n &= ($n - 1);
}
return $count;
}
// Driver Code
$a = 3;
echo (countSolutions($a));
// This code is contributed by ajit
?>
Javascript
<script>
// Javascript program to find the number
// of values of b such that a = b + (a^b)
function bitCount(n)
{
let count = 0;
while (n != 0)
{
count++;
n &= (n - 1);
}
return count;
}
// Function to return the number of solutions
function countSolutions(a)
{
let count = bitCount(a);
count = Math.pow(2, count);
return count;
}
// Driver Code
let a = 3;
document.write(countSolutions(a));
// This code is contributed by subhammahato348
</script>
Producción:
4
Complejidad de tiempo: O(log N)
Espacio auxiliar : O(1)