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