Dada una array arr[] de tamaño N que representa los dígitos de un número y un entero r que es la base (base) del número dado, es decir , n = arr[n – 1] * r 0 + arr[n – 2] * r 1 + … + a[0] * r norte – 1 . La tarea es encontrar si el número dado es par o impar.
Ejemplos:
Entrada: arr[] = {1, 0}, r = 2
Salida: Par
(10) 2 = (2) 10
Entrada: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}, r = 10
Salida: Impar
Enfoque ingenuo: el enfoque más simple es calcular el número n multiplicando los dígitos con la potencia de base correspondiente. Pero como el número de dígitos puede ser del orden de 10 5 , este enfoque no funcionará para un n grande.
Enfoque eficiente: Hay dos casos posibles.
- Si r es par , la respuesta final depende del último dígito, es decir , arr[n – 1] .
- Si r es impar , entonces tenemos que contar el número de dígitos impares. Si el número de dígitos impares es par, entonces la suma es par. De lo contrario, la suma es impar.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function that returns true if the number
// represented by arr[] is even in base r
bool isEven(int arr[], int n, int r)
{
// If the base is even, then
// the last digit is checked
if (r % 2 == 0) {
if (arr[n - 1] % 2 == 0)
return true;
}
// If base is odd, then the
// number of odd digits are checked
else {
// To store the count of odd digits
int oddCount = 0;
for (int i = 0; i < n; ++i) {
if (arr[i] % 2 != 0)
oddCount++;
}
if (oddCount % 2 == 0)
return true;
}
// Number is odd
return false;
}
// Driver code
int main()
{
int arr[] = { 1, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
int r = 2;
if (isEven(arr, n, r))
cout << "Even";
else
cout << "Odd";
return 0;
}
Java
// Java implementation of the approach
import java.io.*;
class GFG
{
// Function that returns true if the number
// represented by arr[] is even in base r
static boolean isEven(int arr[], int n, int r)
{
// If the base is even, then
// the last digit is checked
if (r % 2 == 0)
{
if (arr[n - 1] % 2 == 0)
return true;
}
// If base is odd, then the
// number of odd digits are checked
else {
// To store the count of odd digits
int oddCount = 0;
for (int i = 0; i < n; ++i) {
if (arr[i] % 2 != 0)
oddCount++;
}
if (oddCount % 2 == 0)
return true;
}
// Number is odd
return false;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1, 0 };
int n = arr.length;
int r = 2;
if (isEven(arr, n, r))
System.out.println ("Even");
else
System.out.println("Odd");
}
}
// This code is contributed by jit_t.
Python3
# Python 3 implementation of the approach
# Function that returns true if the number
# represented by arr[] is even in base r
def isEven(arr, n, r):
# If the base is even, then
# the last digit is checked
if (r % 2 == 0):
if (arr[n - 1] % 2 == 0):
return True
# If base is odd, then the
# number of odd digits are checked
else:
# To store the count of odd digits
oddCount = 0
for i in range(n):
if (arr[i] % 2 != 0):
oddCount += 1
if (oddCount % 2 == 0):
return True
# Number is odd
return False
# Driver code
if __name__ == '__main__':
arr = [1, 0]
n = len(arr)
r = 2
if (isEven(arr, n, r)):
print("Even")
else:
print("Odd")
# This code is contributed by
# Surendra_Gangwar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true if the number
// represented by arr[] is even in base r
static bool isEven(int []arr, int n, int r)
{
// If the base is even, then
// the last digit is checked
if (r % 2 == 0)
{
if (arr[n - 1] % 2 == 0)
return true;
}
// If base is odd, then the
// number of odd digits are checked
else
{
// To store the count of odd digits
int oddCount = 0;
for (int i = 0; i < n; ++i)
{
if (arr[i] % 2 != 0)
oddCount++;
}
if (oddCount % 2 == 0)
return true;
}
// Number is odd
return false;
}
// Driver code
public static void Main ()
{
int []arr = { 1, 0 };
int n = arr.Length;
int r = 2;
if (isEven(arr, n, r))
Console.WriteLine ("Even");
else
Console.WriteLine("Odd");
}
}
// This code is contributed by anuj_67...
PHP
<?php
// PHP implementation of the approach
// Function that returns true if the number
// represented by arr[] is even in base r
function isEven($arr, $n, $r)
{
// If the base is even, then
// the last digit is checked
if ($r % 2 == 0)
{
if ($arr[$n - 1] % 2 == 0)
return true;
}
// If base is odd, then the
// number of odd digits are checked
else
{
// To store the count of odd digits
$oddCount = 0;
for ($i = 0; $i < $n; ++$i)
{
if ($arr[$i] % 2 != 0)
$oddCount++;
}
if ($oddCount % 2 == 0)
return true;
}
// Number is odd
return false;
}
// Driver code
$arr = array( 1, 0 );
$n = Count($arr);
$r = 2;
if (isEven($arr, $n, $r))
echo "Even";
else
echo "Odd";
// This code is contributed by andrew1234
?>
Javascript
<script>
// Javascript implementation of the approach
// Function that returns true if the number
// represented by arr[] is even in base r
function isEven(arr, n, r)
{
// If the base is even, then
// the last digit is checked
if (r % 2 == 0)
{
if (arr[n - 1] % 2 == 0)
return true;
}
// If base is odd, then the
// number of odd digits are checked
else
{
// To store the count of odd digits
let oddCount = 0;
for (let i = 0; i < n; ++i)
{
if (arr[i] % 2 != 0)
oddCount++;
}
if (oddCount % 2 == 0)
return true;
}
// Number is odd
return false;
}
let arr = [ 1, 0 ];
let n = arr.length;
let r = 2;
if (isEven(arr, n, r))
document.write("Even");
else
document.write("Odd");
</script>
Producción:
Even
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por AbhishekSharma32 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA