Suma igual y XOR – Part 1

Dado un entero positivo n, encuentra el número de enteros positivos i tales que 0 <= i <= n y n+i = n^i 

Ejemplos: 

C++

/* C++ program to print count of values such
   that n+i = n^i */
#include <iostream>
using namespace std;
 
// function to count number of values less than
// equal to n that satisfy the given condition
int countValues (int n)
{
    int countV = 0;
 
    // Traverse all numbers from 0 to n and
    // increment result only when given condition
    // is satisfied.
    for (int i=0; i<=n; i++ )
        if ((n+i) == (n^i) )
            countV++;
 
    return countV;
}
 
// Driver program
int main()
{
    int n = 12;
    cout << countValues(n);
    return 0;
}

Java

/* Java program to print count of values
 such that n+i = n^i */
import java.util.*;
 
class GFG {
     
    // function to count number of values
    // less than equal to n that satisfy
    // the given condition
    public static int countValues (int n)
    {
        int countV = 0;
      
        // Traverse all numbers from 0 to n
        // and increment result only when
        // given condition is satisfied.
        for (int i = 0; i <= n; i++ )
            if ((n + i) == (n ^ i) )
                countV++;
      
        return countV;
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 12;
        System.out.println(countValues(n));
         
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to print count
# of values such that n+i = n^i
 
# function to count number
# of values less than
# equal to n that satisfy
# the given condition
def countValues (n):
    countV = 0;
 
    # Traverse all numbers
    # from 0 to n and
    # increment result only
    # when given condition
    # is satisfied.
    for i in range(n + 1):
        if ((n + i) == (n ^ i)):
            countV += 1;
 
    return countV;
 
# Driver Code
n = 12;
print(countValues(n));
 
# This code is contributed by mits

C#

/* C# program to print count of values
such that n+i = n^i */
using System;
 
class GFG {
     
    // function to count number of values
    // less than equal to n that satisfy
    // the given condition
    public static int countValues (int n)
    {
        int countV = 0;
     
        // Traverse all numbers from 0 to n
        // and increment result only when
        // given condition is satisfied.
        for (int i = 0; i <= n; i++ )
            if ((n + i) == (n ^ i) )
                countV++;
     
        return countV;
    }
     
    /* Driver program to test above function */
    public static void Main()
    {
        int n = 12;
        Console.WriteLine(countValues(n));
         
    }
}
 
// This code is contributed by anuj_67.

PHP

<?php
// PHP program to print count
// of values such that n+i = n^i
 
// function to count number
// of values less than
// equal to n that satisfy
// the given condition
function countValues ($n)
{
    $countV = 0;
 
    // Traverse all numbers
    // from 0 to n and
    // increment result only
    // when given condition
    // is satisfied.
    for ($i = 0; $i <= $n; $i++ )
        if (($n + $i) == ($n^$i) )
            $countV++;
 
    return $countV;
}
 
    // Driver Code
    $n = 12;
    echo countValues($n);
 
// This code is contributed by m_kit
?>

Javascript

<script>
 
/* JavaScript program to print count of values such
that n+i = n^i */
 
// function to count number of values less than
// equal to n that satisfy the given condition
function countValues (n)
{
    let countV = 0;
 
    // Traverse all numbers from 0 to n and
    // increment result only when given condition
    // is satisfied.
    for (let i=0; i<=n; i++ )
        if ((n+i) == (n^i) )
            countV++;
 
    return countV;
}
 
// Driver program
 
    let n = 12;
    document.write(countValues(n));
 
// This code is contributed by Surbhi Tyagi.
 
</script>

C++

/* c++ program to print count of values such
  that n+i = n^i */
#include <bits/stdc++.h>
using namespace std;
 
// function to count number of values less than
// equal to n that satisfy the given condition
int countValues(int n)
{
    // unset_bits keeps track of count of un-set
    // bits in binary representation of n
    int unset_bits=0;
    while (n)
    {
        if ((n & 1) == 0)
            unset_bits++;
        n=n>>1;
    }
 
    // Return 2 ^ unset_bits
    return 1 << unset_bits;
}
 
// Driver code
int main()
{
    int n = 12;
    cout << countValues(n);
    return 0;
}

Java

/* Java program to print count of values
  such that n+i = n^i */
import java.util.*;
 
class GFG {
     
    // function to count number of values
    // less than equal to n that satisfy
    // the given condition
    public static int countValues(int n)
    {
        // unset_bits keeps track of count
        // of un-set bits in binary
        // representation of n
        int unset_bits=0;
        while (n > 0)
        {
            if ((n & 1) == 0)
                unset_bits++;
            n=n>>1;
        }
      
        // Return 2 ^ unset_bits
        return 1 << unset_bits;
    }
     
    /* Driver program to test above
    function */
    public static void main(String[] args)
    {
        int n = 12;
        System.out.println(countValues(n));
           
    }
}
   
// This code is contributed by Arnav Kr. Mandal.

C#

/* C# program to print count of values
  such that n+i = n^i */
using System;
public class GFG {
     
    // function to count number of values
    // less than equal to n that satisfy
    // the given condition
    public static int countValues(int n)
    {
       
        // unset_bits keeps track of count
        // of un-set bits in binary
        // representation of n
        int unset_bits=0;
        while (n > 0)
        {
            if ((n & 1) == 0)
                unset_bits++;
            n=n>>1;
        }
      
        // Return 2 ^ unset_bits
        return 1 << unset_bits;
    }
     
    /* Driver program to test above
    function */
    public static void Main(String[] args)
    {
        int n = 12;
        Console.WriteLine(countValues(n));
           
    }
}
 
// This code is contributed by umadevi9616

Python3

# Python3 program to print count of values such
# that n+i = n^i
 
# function to count number of values less than
# equal to n that satisfy the given condition
def countValues(n):
     
    # unset_bits keeps track of count of un-set
    # bits in binary representation of n
    unset_bits = 0
     
    while(n):
        if n & 1 == 0:
            unset_bits += 1
        n = n >> 1
         
    # Return 2 ^ unset_bits    
    return 1 << unset_bits
 
# Driver code
if __name__=='__main__':
    n = 12
    print(countValues(n))
 
# This code is contributed by rutvik

PHP

<?php
/* PHP program to print count
of values such that n+i = n^i */
 
 
// function to count number of
// values less than equal to n
// that satisfy the given
// condition
function countValues( $n)
{
     
    // unset_bits keeps track
    // of count of un-set bits
    // in binary representation
    // of n
    $unset_bits = 0;
    while ($n)
    {
        if (($n & 1) == 0)
            $unset_bits++;
        $n = $n >> 1;
    }
 
    // Return 2 ^ unset_bits
    return 1 << $unset_bits;
}
 
// Driver code
 
    $n = 12;
    echo countValues($n);
 
// This code is contributed
// by Anuj_67.
?>

Javascript

<script>
 
// Javascript program to print count of values
// such that n+i = n^i
 
// Function to count number of values
// less than equal to n that satisfy
// the given condition
function countValues(n)
{
     
    // unset_bits keeps track of count
    // of un-set bits in binary
    // representation of n
    let unset_bits = 0;
     
    while (n > 0)
    {
        if ((n & 1) == 0)
            unset_bits++;
             
        n = n >> 1;
    }
   
    // Return 2 ^ unset_bits
    return 1 << unset_bits;
}
   
// Driver Code
let n = 12;
 
document.write(countValues(n));
     
// This code is contributed by susmitakundugoaldanga
     
</script>

Publicación traducida automáticamente

Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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