Dado un número N, la tarea es verificar si el producto de los dígitos en los lugares pares de un número es divisible por K. Si es divisible, emita «SÍ», de lo contrario, emita «NO».
Ejemplos:
Input: N = 5478, K = 5 Output: YES Since, 5 * 7 = 35, which is divisible by 5 Input: N = 19270, K = 2 Output: NO
Acercarse:
- Encuentra el producto de dígitos en lugares pares de derecha a izquierda.
- Luego verifica la divisibilidad tomando su módulo con ‘K’
- Si módulo da 0, salida SÍ, de lo contrario, salida NO
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// below function checks whether
// product of digits at even places
// is divisible by K
bool productDivisible(int n, int k)
{
int product = 1, position = 1;
while (n > 0) {
// if position is even
if (position % 2 == 0)
product *= n % 10;
n = n / 10;
position++;
}
if (product % k == 0)
return true;
return false;
}
// Driver code
int main()
{
int n = 321922;
int k = 3;
if (productDivisible(n, k))
cout << "YES";
else
cout << "NO";
return 0;
}
Java
// JAVA implementation of the above approach
class GFG {
// below function checks whether
// product of digits at even places
// is divisible by K
static boolean productDivisible(int n, int k) {
int product = 1, position = 1;
while (n > 0) {
// if position is even
if (position % 2 == 0) {
product *= n % 10;
}
n = n / 10;
position++;
}
if (product % k == 0) {
return true;
}
return false;
}
// Driver code
public static void main(String[] args) {
int n = 321922;
int k = 3;
if (productDivisible(n, k)) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
}
Python3
# Python3 implementation of the
# above approach
# below function checks whether
# product of digits at even places
# is divisible by K
def productDivisible(n, k):
product = 1
position = 1
while n > 0:
# if position is even
if position % 2 == 0:
product *= n % 10
n = n / 10
position += 1
if product % k == 0:
return True
return False
# Driver code
n = 321922
k = 3
if productDivisible(n, k) == True:
print("YES")
else:
print("NO")
# This code is contributed
# by Shrikant13
C#
// C# implementation of the above approach
using System;
class GFG
{
// below function checks whether
// product of digits at even places
// is divisible by K
static bool productDivisible(int n, int k)
{
int product = 1, position = 1;
while (n > 0)
{
// if position is even
if (position % 2 == 0)
product *= n % 10;
n = n / 10;
position++;
}
if (product % k == 0)
return true;
return false;
}
// Driver code
public static void Main()
{
int n = 321922;
int k = 3;
if (productDivisible(n, k))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
PHP
<?php
// PHP implementation of the
// above approach
// Below function checks whether
// product of digits at even places
// is divisible by K
function productDivisible($n, $k)
{
$product = 1;
$position = 1;
while ($n > 0)
{
// if position is even
if ($position % 2 == 0)
$product *= $n % 10;
$n = (int)($n / 10);
$position++;
}
if ($product % $k == 0)
return true;
return false;
}
// Driver code
$n = 321922;
$k = 3;
if (productDivisible($n, $k))
echo "YES";
else
echo "NO";
// This code is contributed by mits
?>
Javascript
<script>
// Javascript implementation of the above approach
// below function checks whether
// product of digits at even places
// is divisible by K
function productDivisible(n, k)
{
var product = 1, position = 1;
while (n > 0) {
// if position is even
if (position % 2 == 0)
product *= n % 10;
n =parseInt(n / 10);
position++;
}
if (product % k == 0)
return true;
return false;
}
// Driver code
var n = 321922;
var k = 3;
if (productDivisible(n, k))
document.write( "YES");
else
document.write( "NO");
</script>
Producción:
YES
Complejidad de tiempo: O (log 10 n)
Espacio Auxiliar: O(1)
Método #2: Usando el método string():
- Convierta el número entero en una string, luego recorra la string y multiplique todos los índices pares almacenándolos en el producto.
- Si el producto es divisible por k, devuelva Verdadero, de lo contrario, Falso.
A continuación se muestra la implementación:
Python3
# Python3 implementation of the
# above approach
# Function checks whether
# product of digits at even places
# is divisible by K
def productDivisible(n, k):
product = 1
# Converting integer to string
num = str(n)
# Traversing the string
for i in range(len(num)):
if(i % 2 == 0):
product = product*int(num[i])
if product % k == 0:
return True
return False
# Driver code
n = 321922
k = 3
if productDivisible(n, k) == True:
print("YES")
else:
print("NO")
# This code is contributed by vikkycirus
Javascript
<script>
// JavaScript implementation of the
// above approach
// Function checks whether
// product of digits at even places
// is divisible by K
function productDivisible(n, k){
var product = 1 ;
// Converting integer to string
var num = n.toString()
// Traversing the string
for(let i = 0 ; i < num.length ; i++){
if(i % 2 == 0){
product = product * Number(num[i])
}
}
if (product % k == 0){
return true
}
else{
return false;
}
}
// Driver code
var n = 321922
var k = 3
if(productDivisible(n, k)){
document.write("YES")
}
else{
document.write("NO")
}
</script>
Producción:
YES
Publicación traducida automáticamente
Artículo escrito por Shashank_Sharma y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA