Dado un número N y números de dígitos en N, la tarea es verificar si el producto de dígitos en lugares pares de un número es divisible por la suma de dígitos en lugares impares. Si es divisible, emite «VERDADERO»; de lo contrario, emite «FALSO».
Ejemplos:
Input: N = 2157 Output: TRUE Since, 1 * 7 = 7, which is divisible by 2+5=7 Input: N = 1234 Output: TRUE Since, 2 * 4 = 8, which is divisible by 1 + 3 = 4
Acercarse:
- Encuentra el producto de dígitos en lugares pares de derecha a izquierda.
- Encuentra la suma de dígitos en lugares impares de derecha a izquierda.
- Luego verifique la divisibilidad del producto tomando su módulo con suma
- Si el módulo da 0, emite VERDADERO, de lo contrario, emite FALSO
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 sum of digits at odd places
bool productSumDivisible(int n, int size)
{
int sum = 0, product = 1;
while (n > 0) {
// if size is even
if (size % 2 == 0) {
product *= n % 10;
}
// if size is odd
else {
sum += n % 10;
}
n = n / 10;
size--;
}
if (product % sum == 0)
return true;
return false;
}
// Driver code
int main()
{
int n = 1234;
int len = 4;
if (productSumDivisible(n, len))
cout << "TRUE";
else
cout << "FALSE";
return 0;
}
Java
// JAVA implementation of the above approach
class GFG {
// below function checks whether
// product of digits at even places
// is divisible by sum of digits at odd places
static boolean productSumDivisible(int n, int size)
{
int sum = 0, product = 1;
while (n > 0) {
// if size is even
if (size % 2 == 0) {
product *= n % 10;
}
// if size is odd
else {
sum += n % 10;
}
n = n / 10;
size--;
}
if (product % sum == 0) {
return true;
}
return false;
}
// Driver code
public static void main(String[] args)
{
int n = 1234;
int len = 4;
if (productSumDivisible(n, len)) {
System.out.println("TRUE");
}
else {
System.out.println("FALSE");
}
}
}
Python3
# Python 3 implementation of the above approach
# Below function checks whether product
# of digits at even places is divisible
# by sum of digits at odd places
def productSumDivisible(n, size):
sum = 0
product = 1
while (n > 0) :
# if size is even
if (size % 2 == 0) :
product *= n % 10
# if size is odd
else :
sum += n % 10
n = n // 10
size -= 1
if (product % sum == 0):
return True
return False
# Driver code
if __name__ == "__main__":
n = 1234
len = 4
if (productSumDivisible(n, len)):
print("TRUE")
else :
print("FALSE")
# This code is contributed by ChitraNayal
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 productSumDivisible(int n, int size)
{
int sum = 0, product = 1;
while (n > 0) {
// if size is even
if (size % 2 == 0) {
product *= n % 10;
}
// if size is odd
else {
sum += n % 10;
}
n = n / 10;
size--;
}
if (product % sum == 0) {
return true;
}
return false;
}
// Driver code
public static void Main()
{
int n = 1234;
int len = 4;
if (productSumDivisible(n, len))
Console.WriteLine("TRUE");
else
Console.WriteLine("FALSE");
}
}
PHP
<?php
// PHP implementation of the above approach
// Below function checks whether
// product of digits at even
// places is divisible by sum of
// digits at odd places
function productSumDivisible($n, $size)
{
$sum = 0; $product = 1;
while ($n > 0)
{
// if size is even
if ($size % 2 == 0)
{
$product *= $n % 10;
}
// if size is odd
else
{
$sum += $n % 10;
}
$n = $n / 10;
$size--;
}
if ($product % $sum == 0)
return true;
return false;
}
// Driver code
$n = 1234;
$len = 4;
if (productSumDivisible($n, $len))
echo "TRUE";
else
echo "FALSE";
// This code is contributed by anuj_67..
?>
Javascript
<script>
// Javascript implementation of the above approach
// below function checks whether
// product of digits at even places
// is divisible by sum of digits at odd places
function productSumDivisible(n, size)
{
var sum = 0, product = 1;
while (n > 0) {
// if size is even
if (size % 2 == 0) {
product *= n % 10;
}
// if size is odd
else {
sum += n % 10;
}
n = parseInt(n / 10);
size--;
}
if (product % sum == 0)
return true;
return false;
}
// Driver code
var n = 1234;
var len = 4;
if (productSumDivisible(n, len))
document.write("TRUE");
else
document.write("FALSE");
// This code is contributed by noob2000.
</script>
Producción:
TRUE
Complejidad de tiempo: O (logn)
Espacio Auxiliar: (1), ya que no se ha tomado ningún espacio extra.
Método #2: Usar el método string()
Convierta el entero en una string, luego recorra la string y realice dos operaciones
- Multiplique todos los índices impares y guárdelos en el producto
- Agregue todos los índices pares y guárdelos en suma
- Si el producto es divisible por la suma, devuelve Verdadero de lo contrario Falso
A continuación se muestra la implementación:
C++
// C++ implementation of the above approach
#include <iostream>
using namespace std;
#include<string>
// Below function checks whether product
// of digits at even places is divisible
// by sum of digits at odd places
bool productSumDivisible(int n)
{
int sum = 0;
int product = 1;
// Converting integer to string
string num = to_string(n);
// Traveersing the string
for(int i = 0 ; i < num.length() ; i++ ) {
if(i % 2 != 0){
product = product*(num[i]);
}
else{
sum = sum+int(num[i]);
}
}
if (product % sum == 0){
return true ;
}
else{
return false;
}
}
// Driver code
int main()
{
int n = 1234;
if (productSumDivisible(n))
{
cout<<"true";
}
else{
cout<<"false";
}
}
// This code is contributed by akshitsaxena07
Java
// Java implementation of the above approach
import java.io.*;
class GFG {
// Below function checks whether product
// of digits at even places is divisible
// by sum of digits at odd places
static boolean productSumDivisible(int n)
{
int sum = 0;
int product = 1;
// Converting integer to string
String num = String.valueOf(n);
// Traveersing the string
for(int i = 0 ; i < num.length() ; i++ ) {
if(i % 2 != 0){
product = product*Character.getNumericValue(num.charAt(i));
}
else{
sum = sum+Character.getNumericValue(num.charAt(i));
}
}
if (product % sum == 0){
return true ;
}
else{
return false;
}
}
// Driver code
public static void main (String[] args) {
int n = 1234;
if (productSumDivisible(n))
{
System.out.println("TRUE");
}
else{
System.out.println("FALSE");
}
}
}
// This code is contributed by rag2127.
Python3
# Python 3 implementation of the above approach
# Below function checks whether product
# of digits at even places is divisible
# by sum of digits at odd places
def productSumDivisible(n):
sum = 0
product = 1
# Converting integer to string
num = str(n)
# Traveersing the string
for i in range(len(num)):
if(i % 2 != 0):
product = product*int(num[i])
else:
sum = sum+int(num[i])
if (product % sum == 0):
return True
return False
# Driver code
if __name__ == "__main__":
n = 1234
if (productSumDivisible(n)):
print("TRUE")
else:
print("FALSE")
# This code is contributed by vikkycirus
C#
// C# implementation of the above approach
using System;
class GFG{
// Below function checks whether product
// of digits at even places is divisible
// by sum of digits at odd places
static bool productSumDivisible(int n)
{
int sum = 0;
int product = 1;
// Converting integer to string
string num = n.ToString();
// Traveersing the string
for(int i = 0; i < num.Length; i++ )
{
if (i % 2 != 0)
{
product = product*(int)Char.GetNumericValue(num[i]);
}
else
{
sum = sum+(int)Char.GetNumericValue(num[i]);
}
}
if (product % sum == 0)
{
return true;
}
else
{
return false;
}
}
// Driver code
static public void Main()
{
int n = 1234;
if (productSumDivisible(n))
{
Console.WriteLine("TRUE");
}
else
{
Console.WriteLine("FALSE");
}
}
}
// This code is contributed by avanitrachhadiya2155
Javascript
<script>
// JavaScript implementation of the above approach
// Below function checks whether product
// of digits at even places is divisible
// by sum of digits at odd places
function productSumDivisible(n){
var sum = 0
var product = 1
// Converting integer to string
var num = n.toString();
// Traveersing the string
for(i = 0 ; i < num.length ; i++ ) {
if(i % 2 != 0){
product = product*Number(num[i])
}
else{
sum = sum+Number(num[i])
}
}
if (product % sum == 0){
return true ;
}
else{
return false;
}
}
// Driver code
var n = 1234
if (productSumDivisible(n)){
document.write("TRUE")
}
else{
document.write("FALSE")
}
</script>
Producción:
TRUE
Complejidad temporal: O(d), donde d es el número de dígitos en n
Espacio Auxiliar: (1), ya que no se ha tomado ningún espacio extra.
Publicación traducida automáticamente
Artículo escrito por 29AjayKumar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA