Comprobar si el número formado por las últimas cifras de N números es divisible por 10 o no

Dada una array arr[] de tamaño N que consta de enteros positivos distintos de cero. La tarea es determinar si el número que se forma seleccionando los últimos dígitos de todos los números es divisible por 10 o no. Si el número es divisible por 10, escriba Sí; de lo contrario, escriba No.
Ejemplos: 
 

Entrada: arr[] = {12, 65, 46, 37, 99} 
Salida: No 
25679 no es divisible por 10.
Entrada: arr[] = {24, 37, 46, 50} 
Salida: Sí 
4760 es divisible por 10 . 
 

Enfoque: para que un número entero sea divisible por 10, debe terminar en 0. Entonces, el último elemento de la array decidirá si el número formado será divisible por 10 o no. Si el último dígito del último elemento es 0, imprima Sí; de lo contrario, imprima No.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function that returns true if the
// number formed by the last digits of
// all the elements is divisible by 10
bool isDivisible(int arr[], int n)
{
    // Last digit of the last element
    int lastDigit = arr[n - 1] % 10;
 
    // Number formed will be divisible by 10
    if (lastDigit == 0)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int arr[] = { 12, 65, 46, 37, 99 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    if (isDivisible(arr, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
     
class GFG
{
 
// Function that returns true if the
// number formed by the last digits of
// all the elements is divisible by 10
static boolean isDivisible(int arr[], int n)
{
    // Last digit of the last element
    int lastDigit = arr[n - 1] % 10;
 
    // Number formed will be divisible by 10
    if (lastDigit == 0)
        return true;
 
    return false;
}
 
// Driver code
static public void main ( String []arg)
{
    int arr[] = { 12, 65, 46, 37, 99 };
    int n = arr.length;
 
    if (isDivisible(arr, n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by Rajput-Ji

# Python3 implementation of the approach
 
# Function that returns true if the
# number formed by the last digits of
# all the elements is divisible by 10
def isDivisible(arr, n) :
 
    # Last digit of the last element
    lastDigit = arr[n - 1] % 10;
 
    # Number formed will be divisible by 10
    if (lastDigit == 0) :
        return True;
 
    return False;
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 12, 65, 46, 37, 99 ];
    n = len(arr);
 
    if (isDivisible(arr, n)) :
        print("Yes");
    else :
        print("No");
 
# This code is contributed by AnkitRai01

// C# implementation of the approach
using System;
     
class GFG
{
 
// Function that returns true if the
// number formed by the last digits of
// all the elements is divisible by 10
static bool isDivisible(int []arr, int n)
{
    // Last digit of the last element
    int lastDigit = arr[n - 1] % 10;
 
    // Number formed will be divisible by 10
    if (lastDigit == 0)
        return true;
 
    return false;
}
 
// Driver code
static public void Main(String []arg)
{
    int []arr = { 12, 65, 46, 37, 99 };
    int n = arr.Length;
 
    if (isDivisible(arr, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
     
// This code is contributed by Rajput-Ji

<script>
// Javascript implementation of the approach
 
// Function that returns true if the
// number formed by the last digits of
// all the elements is divisible by 10
function isDivisible(arr, n)
{
    // Last digit of the last element
    let lastDigit = arr[n - 1] % 10;
 
    // Number formed will be divisible by 10
    if (lastDigit == 0)
        return true;
 
    return false;
}
 
// Driver code
    let arr = [ 12, 65, 46, 37, 99 ];
    let n = arr.length;
 
    if (isDivisible(arr, n))
        document.write("Yes");
    else
        document.write("No");
 
</script>

No

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)