Dado un número que puede estar en octal o en decimal. Si el número no es octal, conviértalo en octal y luego verifique si es palíndromo o no. Si es palíndromo, imprima 1; de lo contrario, imprima 0. Si el número ya está en octal, compruebe si el número es palíndromo o no.
Ejemplos:
Input :n = 111 Output : Yes Explanation: all digits of 111 are in 0-7 so it is a octal number. Read 111 the result is same backward and forward so it is a palindrome number. Input : 68 Output : No Explanation: 68 is not a octal number because it's all digits are not in 0-7. So first we need to convert it into octal (68)base10(Decimal) = (104)base8(octal) 104 is not palindrome. Input : 97 Output : Yes Explanation: 97 is not a octal number because it's all digits are not in 0-7 so first we need to convert it into decimal to octal (97)base10(Decimal) = (141)base8(octal) 141 is palindrome so output = 1.
Número octal: El sistema numérico octal, o oct para abreviar, es el sistema numérico de base 8 y utiliza los dígitos del 0 al 7.
Cómo convertir un número decimal en un número octal:
C++
// C++ program to check if octal
// representation of a number is prime
#include <iostream>
using namespace std;
const int MAX_DIGITS = 20;
/* Function to Check no is in octal
or not */
bool isOctal(long n)
{
while (n)
{
if ((n % 10) >= 8)
return false;
else
n = n / 10;
}
return true;
}
/* Function To check no is palindrome
or not*/
int isPalindrome(long n)
{
// If number is already in octal, we traverse
// digits using repeated division with 10. Else
// we traverse digits using repeated division
// with 8
int divide = (isOctal(n) == false)? 8 : 10;
// To store individual digits
int octal[MAX_DIGITS];
// Traversing all digits
int i = 0;
while (n != 0)
{
octal[i++] = n % divide;
n = n / divide;
}
// checking if octal no is palindrome
for (int j = i - 1, k = 0; k <= j; j--, k++)
if (octal[j] != octal[k])
return false;
return true;
}
// Driver code
int main()
{
long n = 97;
if (isPalindrome(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java program to check if
// octal representation of
// a number is prime
import java.io.*;
import java.util.*;
import java.lang.*;
class GFG
{
static int MAX_DIGITS = 20;
/* Function to Check no
is in octal or not */
static int isOctal(int n)
{
while (n > 0)
{
if ((n % 10) >= 8)
return 0;
else
n = n / 10;
}
return 1;
}
/* Function To check no
is palindrome or not*/
static int isPalindrome(int n)
{
// If number is already in
// octal, we traverse digits
// using repeated division
// with 10. Else we traverse
// digits using repeated
// division with 8
int divide = (isOctal(n) == 0) ? 8 : 10;
// To store individual digits
int octal[] = new int[MAX_DIGITS];
// Traversing all digits
int i = 0;
while (n != 0)
{
octal[i++] = n % divide;
n = n / divide;
}
// checking if octal
// no is palindrome
for (int j = i - 1,
k = 0; k <= j; j--, k++)
if (octal[j] != octal[k])
return 0;
return 1;
}
// Driver Code
public static void main(String[] args)
{
int n = 97;
if (isPalindrome(n) > 0)
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
Python3
# Python3 program to check if
# octal representation of
# a number is prime
MAX_DIGITS = 20;
# Function to Check no
# is in octal or not
def isOctal(n):
while(n):
if((n % 10) >= 8):
return False
else:
n = int(n / 10)
return True
# Function To check no
# is palindrome or not
def isPalindrome(n):
# If number is already in
# octal, we traverse digits
# using repeated division
# with 10. Else we traverse
# digits using repeated
# division with 8
divide = 8 if(isOctal(n) == False) else 10
# To store individual digits
octal=[]
# Traversing all digits
while (n != 0):
octal.append(n % divide)
n = int(n / divide)
# checking if octal
# no is palindrome
j = len(octal)-1
k = 0
while(k <= j):
if(octal[j] != octal[k]):
return False
j-=1
k+=1
return True
# Driver Code
if __name__=='__main__':
n = 97;
if (isPalindrome(n)):
print("Yes")
else:
print("No")
# This code is contributed by mits
C#
// C# program to check if
// octal representation of
// a number is prime
using System;
class GFG
{
static long MAX_DIGITS = 20;
/* Function to Check no
is in octal or not */
static long isOctal(long n)
{
while (n > 0)
{
if ((n % 10) >= 8)
return 0;
else
n = n / 10;
}
return 1;
}
/* Function To check no
is palindrome or not*/
static long isPalindrome(long n)
{
// If number is already in
// octal, we traverse digits
// using repeated division
// with 10. Else we traverse
// digits using repeated
// division with 8
long divide = (isOctal(n) == 0) ? 8 : 10;
// To store individual digits
long[] octal = new long[MAX_DIGITS];
// Traversing all digits
long i = 0;
while (n != 0)
{
octal[i++] = n % divide;
n = n / divide;
}
// checking if octal
// no is palindrome
for (long j = i - 1,
k = 0; k <= j; j--, k++)
if (octal[j] != octal[k])
return 0;
return 1;
}
// Driver Code
static int Main()
{
long n = 97;
if (isPalindrome(n) > 0)
Console.Write("Yes");
else
Console.Write("No");
return 0;
}
}
// This code is contributed
// by mits
PHP
<?php
// PHP program to check if
// octal representation of
// a number is prime
$MAX_DIGITS = 20;
// Function to Check no
// is in octal or not
function isOctal($n)
{
while ($n)
{
if (($n % 10) >= 8)
return false;
else
$n = (int)$n / 10;
}
return true;
}
// Function To check no
// is palindrome or not
function isPalindrome($n)
{
global $MAX_DIGITS;
// If number is already in
// octal, we traverse digits
// using repeated division
// with 10. Else we traverse
// digits using repeated
// division with 8
$divide = (isOctal($n) ==
false) ? 8 : 10;
// To store individual digits
$octal;
// Traversing all digits
$i = 0;
while ($n != 0)
{
$octal[$i++] = $n % $divide;
$n = (int)$n / $divide;
}
// checking if octal
// no is palindrome
for ($j = $i - 1, $k = 0;
$k <= $j; $j--, $k++)
if ($octal[$j] != $octal[$k])
return -1;
return 0;
}
// Driver Code
$n = 97;
if (isPalindrome($n))
echo "Yes";
else
echo "No";
// This code is contributed by ajit
?>
Javascript
<script>
// Javascript program to check if octal
// representation of a number is prime
const MAX_DIGITS = 20;
/* Function to Check no is in octal
or not */
function isOctal(n)
{
while (n)
{
if ((n % 10) >= 8)
return false;
else
n = Math.floor(n / 10);
}
return true;
}
/* Function To check no is palindrome
or not*/
function isPalindrome(n)
{
// If number is already in octal, we traverse
// digits using repeated division with 10. Else
// we traverse digits using repeated division
// with 8
var divide = (isOctal(n) == false)? 8 : 10;
// To store individual digits
var octal = new Array(MAX_DIGITS);
// Traversing all digits
var i = 0;
while (n != 0)
{
octal[i++] = n % divide;
n = Math.floor(n / divide);
}
// checking if octal no is palindrome
for (var j = i - 1, k = 0; k <= j; j--, k++)
if (octal[j] != octal[k])
return false;
return true;
}
// Driver code
var n = 97;
if (isPalindrome(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by Mayank Tyagi
</script>
Producción :
Yes
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA