Número total de subconjuntos en los que el producto de los elementos es par

Dada una array arr[] de elementos enteros, la tarea es encontrar el número total de subconjuntos de arr[] en los que el producto de los elementos es par.

Ejemplos:  

Entrada: arr[] = {2, 2, 3} 
Salida: 6 
Todos los subconjuntos posibles son {2}, {2}, {2, 2}, {2, 3}, {2, 3} y {2 , 2, 3}

Entrada: arr[] = {3, 3, 3} 
Salida: 6

Enfoque: Ya sabemos que: 

• par * par = par
• Impar * Par = Par
• Impar * Impar = Impar

Ahora, necesitamos contar los subconjuntos totales en los que está presente al menos un elemento par para que el producto de los elementos sea par. 
Ahora, Número total de subconjuntos que tienen al menos un elemento par = Total de subconjuntos posibles de n – Total de subconjuntos que tienen todos los elementos impares, 
es decir , (2 ⁿ – 1) – (2 ^totalOdd – 1)
A continuación se muestra la implementación de el enfoque anterior:  

// C++ implementation of above approach
 
#include <iostream>
#include<bits/stdc++.h>
 
using namespace std;
 
// Function to find total number of subsets
// in which product of the elements is even
void find(int a[], int n)
{
    int count_odd = 0;
     
    for(int i = 0; i < n ; i++)
    {
        // counting number of odds elements
        if (i % 2 != 0)
        count_odd += 1;
    }
 
    int result = pow(2, n) - 1 ;
    result -= (pow(2, count_odd) - 1) ;
    cout << result << endl;
     
}
 
// Driver code
int main()
{
   int a[] = {2, 2, 3} ;
   int n = sizeof(a)/sizeof(a[0]) ;
    
   // function calling
   find(a,n);
    
   return 0;
   // This code is contributed by ANKITRAI1;
}

// Java implementation of above approach
 
class GFG {
 
// Function to find total number of subsets
// in which product of the elements is even
    static void find(int a[], int n) {
        int count_odd = 0;
 
        for (int i = 0; i < n; i++) {
            // counting number of odds elements
            if (i % 2 != 0) {
                count_odd += 1;
            }
        }
 
        int result = (int) (Math.pow(2, n) - 1);
        result -= (Math.pow(2, count_odd) - 1);
        System.out.println(result);
 
    }
 
// Driver code
    public static void main(String[] args) {
        int a[] = {2, 2, 3};
        int n = a.length;
 
// function calling
        find(a, n);
 
    }
}
//this code contributed by 29AJayKumar

# Python3 implementation of above approach
import math as ma
 
# Function to find total number of subsets
# in which product of the elements is even
def find(a):
    count_odd = 0
    for i in a:
 
        # counting number of odds elements
        if(i % 2 != 0):
            count_odd+= 1
 
    result = pow(2, len(a)) - 1
    result = result - (pow(2, count_odd) - 1)
    print(result)
 
# Driver code
a =[2, 2, 3]
find(a)

     
// C# implementation of above approach
using System;
public class GFG {
  
// Function to find total number of subsets
// in which product of the elements is even
    static void find(int []a, int n) {
        int count_odd = 0;
  
        for (int i = 0; i < n; i++) {
            // counting number of odds elements
            if (i % 2 != 0) {
                count_odd += 1;
            }
        }
  
        int result = (int) (Math.Pow(2, n) - 1);
        result -= (int)(Math.Pow(2, count_odd) - 1);
        Console.Write(result);
  
    }
  
// Driver code
    public static void Main() {
        int []a = {2, 2, 3};
        int n = a.Length;
  
// function calling
        find(a, n);
  
    }
}
//this code contributed by 29AJayKumar

<?php
// PHP implementation of above approach
  
// Function to find total number of subsets
// in which product of the elements is even
function find(&$a, $n)
{
    $count_odd = 0;
      
    for($i = 0; $i < $n ; $i++)
    {
        // counting number of odds elements
        if ($i % 2 != 0)
        $count_odd += 1;
    }
  
    $result = pow(2, $n) - 1 ;
    $result -= (pow(2, $count_odd) - 1) ;
    echo $result ."\n";
      
}
  
// Driver code
 
$a = array(2, 2, 3) ;
$n = sizeof($a)/sizeof($a[0]) ;
 
// function calling
find($a,$n);
 
return 0;
 
?>

<script>
 
// Javascript implementation of above approach
 
// Function to find total number of subsets
// in which product of the elements is even
function find(a, n)
{
    var count_odd = 0;
     
    for(var i = 0; i < n ; i++)
    {
        // counting number of odds elements
        if (i % 2 != 0)
            count_odd += 1;
    }
 
    var result = Math.pow(2, n) - 1 ;
    result -= (Math.pow(2, count_odd) - 1) ;
    document.write( result );
     
}
 
// Driver code
var a = [2, 2, 3];
var n = a.length;
 
// function calling
find(a,n);
 
</script>

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