Dado un número entero N donde 
. La tarea es verificar si el número no es divisible por ninguno de sus dígitos. Si el número N dado es divisible por cualquiera de sus dígitos, escriba «SÍ», de lo contrario, escriba «NO».
Ejemplos:
Input : N = 5115 Output : YES Explanation: 5115 is divisible by both 1 and 5. So print YES. Input : 27 Output : NO Explanation: 27 is not divisible by 2 or 7
Planteamiento: La idea para resolver el problema es extraer los dígitos del número uno por uno y verificar si el número es divisible por alguno de sus dígitos. Si es divisible por cualquiera de sus dígitos, imprima SÍ; de lo contrario, imprima NO.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if given number is divisible
// by any of its digits
string isDivisible(long long int n)
{
long long int temp = n;
// check if any of digit divides n
while (n) {
int k = n % 10;
// check if K divides N
if (temp % k == 0)
return "YES";
n /= 10;
}
return "NO";
}
// Driver Code
int main()
{
long long int n = 9876543;
cout << isDivisible(n);
return 0;
}
Java
// Java implementation of above approach
class GFG
{
// Function to check if given number is divisible
// by any of its digits
static String isDivisible(int n)
{
int temp = n;
// check if any of digit divides n
while (n > 0)
{
int k = n % 10;
// check if K divides N
if (temp % k == 0)
{
return "YES";
}
n /= 10;
}
return "NO";
}
// Driver Code
public static void main(String[] args)
{
int n = 9876543;
System.out.println(isDivisible(n));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python program implementation of above approach # Function to check if given number is # divisible by any of its digits def isDivisible(n): temp = n # check if any of digit divides n while(n): k = n % 10 # check if K divides N if(temp % k == 0): return "YES" n /= 10; # Number is not divisible by # any of digits return "NO" # Driver Code n = 9876543 print(isDivisible(n)) # This code is contributed by # Sanjit_Prasad
C#
// C# implementation of above approach
using System;
class GFG
{
// Function to check if given number is divisible
// by any of its digits
static String isDivisible(int n)
{
int temp = n;
// check if any of digit divides n
while (n > 0)
{
int k = n % 10;
// check if K divides N
if (temp % k == 0)
{
return "YES";
}
n /= 10;
}
return "NO";
}
// Driver Code
public static void Main(String[] args)
{
int n = 9876543;
Console.WriteLine(isDivisible(n));
}
}
// This code is contributed by PrinciRaj1992
PHP
<?php
// PHP implementation of above approach
// Function to check if given number
// is divisible by any of its digits
function isDivisible($n)
{
$temp = $n;
// check if any of digit divides n
while ($n)
{
$k = $n % 10;
// check if K divides N
if ($temp % $k == 0)
return "YES";
$n = floor($n / 10);
}
return "NO";
}
// Driver Code
$n = 9876543;
echo isDivisible($n);
// This code is contributed by Ryuga
?>
Javascript
<script>
// Javascript implementation of above approach
// Function to check if given number
// is divisible by any of its digits
function isDivisible(n)
{
temp = n;
// Check if any of digit divides n
while (n)
{
k = n % 10;
// Check if K divides N
if (temp % k == 0)
return "YES";
n = Math.floor(n / 10);
}
return "NO";
}
// Driver Code
let n = 9876543;
document.write(isDivisible(n));
// This code is contributed by sravan kumar
</script>
Producción:
YES
Complejidad de Tiempo : O(log(N))
Espacio Auxiliar : O(1), ya que no se ha requerido espacio extra.
Método #2: Usando una string:
- Tenemos que convertir el número dado en string tomando una nueva variable.
- Atraviesa la cuerda,
- Convertir carácter a entero (dígito)
- Verifique si el número es divisible por cualquiera de sus dígitos, luego imprima SÍ, de lo contrario, imprima NO.
A continuación se muestra la implementación del enfoque anterior:
C++
//C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
string getResult(int n)
{
// Converting integer to string
string st = to_string(n);
// Traversing the string
for (int i = 0; i < st.length(); i++)
{
//find the actual digit
int d = st[i] - 48;
// If the number is divisible by
// digits then return yes
if(n % d == 0)
{
return "Yes";
}
}
// If no digits are dividing the
// number then return no
return "No";
}
// Driver Code
int main()
{
int n = 9876543;
// passing this number to get result function
cout<<getResult(n);
}
// this code is contributed by SoumiikMondal
Java
// JAva implementation of above approach
import java.io.*;
class GFG{
static String getResult(int n)
{
// Converting integer to string
String st = Integer.toString(n);
// Traversing the string
for (int i = 0; i < st.length(); i++)
{
//find the actual digit
int d = st.charAt(i) - 48;
// If the number is divisible by
// digits then return yes
if(n % d == 0)
{
return "Yes";
}
}
// If no digits are dividing the
// number then return no
return "No";
}
// Driver Code
public static void main(String[] args)
{
int n = 9876543;
// passing this number to get result function
System.out.println(getResult(n));
}
}
// this code is contributed by shivanisinghss2110
Python3
# Python implementation of above approach def getResult(n): # Converting integer to string st = str(n) # Traversing the string for i in st: # If the number is divisible by # digits then return yes if(n % int(i) == 0): return 'Yes' # If no digits are dividing the # number then return no return 'No' # Driver Code n = 9876543 # passing this number to get result function print(getResult(n)) # this code is contributed by vikkycirus
C#
// C# implementation of above approach
using System;
public class GFG{
static String getResult(int n)
{
// Converting integer to string
string st = n.ToString();
// Traversing the string
for (int i = 0; i < st.Length; i++)
{
//find the actual digit
int d = st[i] - 48;
// If the number is divisible by
// digits then return yes
if(n % d == 0)
{
return "Yes";
}
}
// If no digits are dividing the
// number then return no
return "No";
}
// Driver Code
public static void Main(String[] args)
{
int n = 9876543;
// passing this number to get result function
Console.Write(getResult(n));
}
}
// this code is contributed by shivanisinghss2110
Javascript
<script>
// JavaScript implementation of above approach
function getResult(n)
{
// Converting integer to string
let st = n.toString();
// Traversing the string
for (let i = 0; i < st.length; i++)
{
//find the actual digit
let d = st[i].charCodeAt(0) - 48;
// If the number is divisible by
// digits then return yes
if(n % d == 0)
{
return "Yes";
}
}
// If no digits are dividing the
// number then return no
return "No";
}
// Driver Code
let n = 9876543;
// passing this number to get result function
document.write(getResult(n));
// This code is contributed by unknown2108
</script>
Producción:
Yes
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(n)
Publicación traducida automáticamente
Artículo escrito por Sanjit_Prasad y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA