Dado un número N , la tarea es encontrar el OR bit a bit (|) de todos los números pares del 1 al N.
Ejemplos:
Entrada: 2
Salida: 2Entrada: 10
Salida: 14
Explicación: 2 | 4 | 6 | 8 | 10 = 14
Enfoque ingenuo: inicialice el resultado como 2. Itere el ciclo de 4 a n (para todos los números pares) y actualice el resultado encontrando bit a bit o ( | ).
A continuación se muestra la implementación del enfoque:
C++
// C++ implementation of the above approach
#include <iostream>
using namespace std;
// Function to return the bitwise OR
// of all the even numbers upto N
int bitwiseOrTillN(int n)
{
// Initialize result as 2
int result = 2;
for (int i = 4; i <= n; i = i + 2) {
result = result | i;
}
return result;
}
// Driver code
int main()
{
int n = 10;
cout << bitwiseOrTillN(n);
return 0;
}
Java
// Java implementation of the above approach
class GFG
{
// Function to return the bitwise OR
// of all the even numbers upto N
static int bitwiseOrTillN(int n)
{
// Initialize result as 2
int result = 2;
for (int i = 4; i <= n; i = i + 2)
{
result = result | i;
}
return result;
}
// Driver code
static public void main (String args[])
{
int n = 10;
System.out.println(bitwiseOrTillN(n));
}
}
// This code is contributed by AnkitRai01
Python3
# Python 3 implementation of the above approach # Function to return the bitwise OR # of all the even numbers upto N def bitwiseOrTillN ( n ): # Initialize result as 2 result = 2; for i in range(4, n + 1, 2) : result = result | i return result # Driver code n = 10; print(bitwiseOrTillN(n)); # This code is contributed by ANKITKUMAR34
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the bitwise OR
// of all the even numbers upto N
static int bitwiseOrTillN(int n)
{
// Initialize result as 2
int result = 2;
for (int i = 4; i <= n; i = i + 2)
{
result = result | i;
}
return result;
}
// Driver code
static public void Main ()
{
int n = 10;
Console.WriteLine(bitwiseOrTillN(n));
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// Javascript implementation of the above approach
// Function to return the bitwise OR
// of all the even numbers upto N
function bitwiseOrTillN(n)
{
// Initialize result as 2
var result = 2;
for(var i = 4; i <= n; i = i + 2)
{
result = result | i;
}
return result;
}
// Driver code
var n = 10;
document.write( bitwiseOrTillN(n));
// This code is contributed by noob2000
</script>
Producción:
14
Enfoque eficiente: calcule el número total de bits en N . En OR bit a bit, el bit más a la derecha será 0 y todos los demás bits serán 1. Por lo tanto, devuelve pow(2, número total de bits)-2. Dará el valor equivalente en decimal de OR bit a bit.
A continuación se muestra la implementación del enfoque:
C++
// C++ implementation of the above approach
#include <iostream>
#include <math.h>
using namespace std;
// Function to return the bitwise OR
// of all even numbers upto N
int bitwiseOrTillN(int n)
{
// For value less than 2
if (n < 2)
return 0;
// Count total number of bits in bitwise or
// all bits will be set except last bit
int bitCount = log2(n) + 1;
// Compute 2 to the power bitCount and subtract 2
return pow(2, bitCount) - 2;
}
// Driver code
int main()
{
int n = 10;
cout << bitwiseOrTillN(n);
return 0;
}
Java
// Java implementation of the above approach
class GFG
{
// Function to return the bitwise OR
// of all even numbers upto N
static int bitwiseOrTillN(int n)
{
// For value less than 2
if (n < 2)
return 0;
// Count total number of bits in bitwise or
// all bits will be set except last bit
int bitCount = (int)(Math.log(n)/Math.log(2)) + 1;
// Compute 2 to the power bitCount and subtract 2
return (int)Math.pow(2, bitCount) - 2;
}
// Driver code
public static void main (String[] args)
{
int n = 10;
System.out.println(bitwiseOrTillN(n));
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 implementation of the above approach from math import log2 # Function to return the bitwise OR # of all even numbers upto N def bitwiseOrTillN(n) : # For value less than 2 if (n < 2) : return 0; # Count total number of bits in bitwise or # all bits will be set except last bit bitCount = int(log2(n)) + 1; # Compute 2 to the power bitCount and subtract 2 return pow(2, bitCount) - 2; # Driver code if __name__ == "__main__" : n = 10; print(bitwiseOrTillN(n)); # This code is contributed by AnkitRai01
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the bitwise OR
// of all even numbers upto N
static int bitwiseOrTillN(int n)
{
// For value less than 2
if (n < 2)
return 0;
// Count total number of bits in bitwise or
// all bits will be set except last bit
int bitCount = (int)(Math.Log(n)/Math.Log(2)) + 1;
// Compute 2 to the power bitCount and subtract 2
return (int)Math.Pow(2, bitCount) - 2;
}
// Driver code
public static void Main()
{
int n = 10;
Console.WriteLine(bitwiseOrTillN(n));
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// Javascript implementation of the approach
// Function to return the bitwise OR
// of all even numbers upto N
function bitwiseOrTillN(n)
{
// For value less than 2
if (n < 2)
return 0;
// Count total number of bits in bitwise or
// all bits will be set except last bit
var bitCount = parseInt(Math.log2(n) + 1);
// Compute 2 to the power bitCount and subtract 2
return Math.pow(2, bitCount) - 2;
}
// Driver code
var n = 10;
document.write( bitwiseOrTillN(n));
//This code is contributed by SoumikMondal
</script>
Producción:
14