Dados dos números enteros ‘N’ y ‘K’, la tarea es encontrar la suma de los dígitos de ‘N’ en sus lugares impares (de derecha a izquierda) y verificar si la suma es divisible por ‘K’. Si es divisible, emite SÍ , de lo contrario, emite NO .
Ejemplos:
Entrada: N = 4325, K = 4
Salida: SI
Ya que, 3 + 5 = 8, que es divisible por 4.Entrada: N = 1209, K = 3
Salida: NO
Acercarse:
- Encuentra la suma de los dígitos de ‘N’ en lugares impares (de derecha a izquierda).
- Luego verifica la divisibilidad de la suma tomando su módulo con ‘K’.
- Si es divisible, emita ‘SÍ’; de lo contrario, emita ‘NO’.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// function that checks the
// divisibility of the sum
// of the digits at odd places
// of the given number
bool SumDivisible(int n, int k)
{
int sum = 0, position = 1;
while (n > 0) {
// if position is odd
if (position % 2 == 1)
sum += n % 10;
n = n / 10;
position++;
}
if (sum % k == 0)
return true;
return false;
}
// Driver code
int main()
{
int n = 592452;
int k = 3;
if (SumDivisible(n, k))
cout << "YES";
else
cout << "NO";
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class solution
{
// function that checks the
// divisibility of the sum
// of the digits at odd places
// of the given number
static boolean SumDivisible(int n, int k)
{
int sum = 0, position = 1;
while (n > 0) {
// if position is odd
if (position % 2 == 1)
sum += n % 10;
n = n / 10;
position++;
}
if (sum % k == 0)
return true;
return false;
}
// Driver code
public static void main(String arr[])
{
int n = 592452;
int k = 3;
if (SumDivisible(n, k))
System.out.println("YES");
else
System.out.println("NO");
}
}
//This code is contributed by Surendra_Gangwar
Python 3
# Python 3 implementation of the approach
# function that checks the divisibility
# of the sum of the digits at odd places
# of the given number
def SumDivisible(n, k):
sum = 0
position = 1
while (n > 0) :
# if position is odd
if (position % 2 == 1):
sum += n % 10
n = n // 10
position += 1
if (sum % k == 0):
return True
return False
# Driver code
if __name__ =="__main__":
n = 592452
k = 3
if (SumDivisible(n, k)):
print("YES")
else:
print("NO")
# This code is contributed
# by ChitraNayal
C#
// C# implementation of the approach
using System;
class GFG
{
// function that checks the
// divisibility of the sum
// of the digits at odd places
// of the given number
static bool SumDivisible(int n, int k)
{
int sum = 0, position = 1;
while (n > 0)
{
// if position is odd
if (position % 2 == 1)
sum += n % 10;
n = n / 10;
position++;
}
if (sum % k == 0)
return true;
return false;
}
// Driver code
static public void Main ()
{
int n = 592452;
int k = 3;
if (SumDivisible(n, k))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
// This code is contributed by Sachin
PHP
<?php
// PHP implementation of the approach
// function that checks the divisibility
// of the sum of the digits at odd places
// of the given number
function SumDivisible($n, $k)
{
$sum = 0;
$position = 1;
while ($n > 0)
{
// if position is odd
if ($position % 2 == 1)
$sum += $n % 10;
$n = (int)$n / 10;
$position++;
}
if ($sum % $k == 0)
return true;
return false;
}
// Driver code
$n = 592452;
$k = 3;
if (SumDivisible($n, $k))
echo "YES";
else
echo "NO";
// This code is contributed
// by Sach_Code
?>
Javascript
<script>
// JavaScript implementation of the approach
// function that checks the
// divisibility of the sum
// of the digits at odd places
// of the given number
function SumDivisible(n, k)
{
let sum = 0, position = 1;
while (n > 0) {
// if position is odd
if (position % 2 == 1)
sum += n % 10;
n = Math.floor(n / 10);
position++;
}
if (sum % k == 0)
return true;
return false;
}
// Driver code
let n = 592452;
let k = 3;
if (SumDivisible(n, k))
document.write("YES");
else
document.write("NO");
// This code is contributed by Surbhi Tyagi.
</script>
Producción:
YES
Complejidad Temporal: O(log 10 n), ya que cada vez el valor de n se reduce a n/10.
Espacio Auxiliar: (1), ya que no se ha ocupado ningún espacio extra.
Método #2: Usando el método string():
- Convierta el número entero en una string, luego recorra la string y encuentre la suma de todos los índices impares almacenándolos en sum.
- Si la suma es divisible por k, devuelve Verdadero o Falso.
A continuación se muestra la implementación:
C++
// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;
bool sumDivisible(int n, int k)
{
int sum = 0;
// Converting integer to string
string num = to_string(n);
int i;
// Traversing the string
for (i = 0; i < num.size(); i++) {
if (i % 2 != 0) {
sum = sum + (num[i] - '0');
}
}
if (sum % k == 0) {
return true;
}
else {
return false;
}
}
// Driver code
int main()
{
int n = 592452;
int k = 3;
if (sumDivisible(n, k)) {
cout << ("YES");
}
else {
cout << ("NO");
}
return 0;
}
// This code is contributed by gauravrajput1
Java
// Java implementation of the
// above approach
import java.io.*;
class GFG
{
static boolean sumDivisible(int n, int k)
{
int sum = 0;
// Converting integer to string
String num = Integer.toString(n);
int i;
// Traversing the string
for (i = 0; i < num.length(); i++) {
if (i % 2 != 0) {
sum = sum + (num.charAt(i) - '0');
}
}
if (sum % k == 0) {
return true;
}
else {
return false;
}
}
// Driver code
public static void main(String[] args)
{
int n = 592452;
int k = 3;
if (sumDivisible(n, k)) {
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
}
// This code is contributed by shivanisinghss2110
Python3
# Python3 implementation of the
# above approach
def sumDivisible(n, k):
sum = 0
# Converting integer to string
num = str(n)
# Traversing the string
for i in range(len(num)):
if(i % 2 != 0):
sum = sum+int(num[i])
if sum % k == 0:
return True
return False
# Driver code
n = 592452
k = 3
if sumDivisible(n, k) == True:
print("YES")
else:
print("NO")
# This code is contributed by vikkycirus
C#
// C# implementation of the
// above approach
using System;
class GFG
{
static bool sumDivisible(int n, int k)
{
int sum = 0;
// Converting integer to string
string num = n.ToString();
int i;
// Traversing the string
for (i = 0; i < num.Length; i++) {
if (i % 2 != 0) {
sum = sum + (num[i] - '0');
}
}
if (sum % k == 0) {
return true;
}
else {
return false;
}
}
// Driver code
static public void Main ()
{
int n = 592452;
int k = 3;
if (sumDivisible(n, k)) {
Console.Write("YES");
}
else {
Console.Write("NO");
}
}
}
// This code is contributed by shivanisinghss2110
Javascript
<script>
// javascript implementation of the
// above approach
function sumDivisible(n, k){
var sum = 0
// Converting integer to string
var num = n.toString()
// Traversing the string
for(var i=0 ; i < num.length ; i++) {
if(i % 2 != 0) {
sum = sum + Number(num[i])
}
}
if (sum % k == 0){
return true;
}
else{
return false;
}
}
// Driver code
var n = 592452
var k = 3
if(sumDivisible(n, k)){
document.write("YES");
}
else{
document.write("NO");
}
</script>
Producción:
Yes
Complejidad temporal: O(d), donde d es el número de dígitos en n.
Espacio Auxiliar: O(d), donde d es el número de dígitos en n.
Publicación traducida automáticamente
Artículo escrito por Shashank_Sharma y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA