检查编号是否为回文数(八进制)

给定一个数字，可以是八进制或十进制。如果该数字不是八进制，则将其转换为八进制，然后检查其是否为回文。如果是回文，则打印1，否则打印0。如果数字已经是八进制的，则检查该数字是否是回文。
例子 ：

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.

八进制数字：八进制数字系统(简称oct)是以8为底的数字系统，使用0到7之间的数字。
如何将十进制数转换为八进制数：

C++

// C++ program to check if octal
// representation of a number is prime
#include 
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

= 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

 
// 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
 

输出 ：

Yes