// C++ Program to print all palindromic primes // smaller than or equal to n. #include   using namespace std; // A function that returns true only if num // contains one digit int oneDigit(int num) {  // comparison operation is faster than  // division operation. So using following  // instead of 'return num / 10 == 0;'  return (num >= 0 && num < 10); } // A recursive function to find out whether // num is palindrome or not. Initially dupNum // contains address of a copy of num. bool isPalUtil(int num int* dupNum) {  // Base case (needed for recursion termination):  // This statement/ mainly compares the first  // digit with the last digit  if (oneDigit(num))  return (num == (*dupNum) % 10);  // This is the key line in this method. Note  // that all recursive/ calls have a separate  // copy of num but they all share same copy  // of *dupNum. We divide num while moving up  // the recursion tree  if (!isPalUtil(num/10 dupNum))  return false;  // The following statements are executed when  // we move up the recursion call tree  *dupNum /= 10;  // At this point if num%10 contains i'th  // digit from beginning then (*dupNum)%10  // contains i'th digit from end  return (num % 10 == (*dupNum) % 10); } // The main function that uses recursive function // isPalUtil() to find out whether num is palindrome // or not int isPal(int num) {  // If num is negative make it positive  if (num < 0)  num = -num;  // Create a separate copy of num so that  // modifications made to address dupNum don't  // change the input number.  int *dupNum = new int(num); // *dupNum = num  return isPalUtil(num dupNum); } // Function to generate all primes and checking // whether number is palindromic or not void printPalPrimesLessThanN(int n) {  // Create a boolean array 'prime[0..n]' and  // initialize all entries it as true. A value  // in prime[i] will finally be false if i is  // Not a prime else true.  bool prime[n+1];  memset(prime true sizeof(prime));  for (int p=2; p*p<=n; p++)  {  // If prime[p] is not changed then it is  // a prime  if (prime[p] == true)  {  // Update all multiples of p  for (int i=p*2; i<=n; i += p)  prime[i] = false;  }  }  // Print all palindromic prime numbers  for (int p=2; p<=n; p++)  // checking whether the given number is  // prime palindromic or not  if (prime[p] && isPal(p))  cout << p << ' '; } // Driver Program int main() {  int n = 100;  printf('Palindromic primes smaller than or '  'equal to %d are :n' n);  printPalPrimesLessThanN(n); } 

// Java Program to print all palindromic primes // smaller than or equal to n. import java.util.*; class GFG {    // A function that returns true only if num  // contains one digit  static boolean oneDigit(int num)  {  // comparison operation is faster than  // division operation. So using following  // instead of 'return num / 10 == 0;'  return (num >= 0 && num < 10);  }    // A recursive function to find out whether  // num is palindrome or not. Initially dupNum  // contains address of a copy of num.  static boolean isPalUtil(int num int dupNum)  {  // Base case (needed for recursion termination):  // This statement/ mainly compares the first  // digit with the last digit  if (oneDigit(num))  return (num == (dupNum) % 10);    // This is the key line in this method. Note  // that all recursive/ calls have a separate  // copy of num but they all share same copy  // of dupNum. We divide num while moving up  // the recursion tree  if (!isPalUtil(num/10 dupNum))  return false;    // The following statements are executed when  // we move up the recursion call tree  dupNum /= 10;    // At this point if num%10 contains ith  // digit from beginning then (dupNum)%10  // contains ith digit from end  return (num % 10 == (dupNum) % 10);  }    // The main function that uses recursive function  // isPalUtil() to find out whether num is palindrome  // or not  static boolean isPal(int num)  {  // If num is negative make it positive  if (num < 0)  num = -num;    // Create a separate copy of num so that  // modifications made to address dupNum don't  // change the input number.  int dupNum = num; // dupNum = num    return isPalUtil(num dupNum);  }    // Function to generate all primes and checking  // whether number is palindromic or not  static void printPalPrimesLessThanN(int n)  {  // Create a boolean array 'prime[0..n]' and  // initialize all entries it as true. A value  // in prime[i] will finally be false if i is  // Not a prime else true.  boolean prime[] = new boolean[n+1];    Arrays.fill(prime true);    for (int p = 2; p*p <= n; p++)  {  // If prime[p] is not changed then it is  // a prime  if (prime[p])  {  // Update all multiples of p  for (int i = p*2; i <= n; i += p){  prime[i] = false;}  }  }    // Print all palindromic prime numbers  for (int p = 2; p <= n; p++){    // checking whether the given number is  // prime palindromic or not  if (prime[p] && isPal(p)){  System.out.print(p + ' ');  }  }  }    // Driver function  public static void main(String[] args)   {  int n = 100;  System.out.printf('Palindromic primes smaller than or '  +'equal to %d are :n' n);  printPalPrimesLessThanN(n);  }  }   // This code is contributed by Arnav Kr. Mandal. 

# Python3 Program to print all palindromic  # primes smaller than or equal to n.  # A function that returns true only if  # num contains one digit  def oneDigit(num): # comparison operation is faster than # division operation. So using following # instead of 'return num / 10 == 0;' return (num >= 0 and num < 10); # A recursive function to find out whether  # num is palindrome or not. Initially dupNum  # contains address of a copy of num. def isPalUtil(num dupNum): # Base case (needed for recursion termination):  # This statement/ mainly compares the first  # digit with the last digit  if (oneDigit(num)): return (num == (dupNum) % 10); # This is the key line in this method. Note  # that all recursive/ calls have a separate  # copy of num but they all share same copy  # of dupNum. We divide num while moving up  # the recursion tree  if (not isPalUtil(int(num / 10) dupNum)): return False; # The following statements are executed  # when we move up the recursion call tree  dupNum =int(dupNum/10); # At this point if num%10 contains ith  # digit from beginning then (dupNum)%10  # contains ith digit from end  return (num % 10 == (dupNum) % 10); # The main function that uses recursive  # function isPalUtil() to find out whether  # num is palindrome or not  def isPal(num): # If num is negative make it positive  if (num < 0): num = -num; # Create a separate copy of num so that  # modifications made to address dupNum  # don't change the input number.  dupNum = num; # dupNum = num  return isPalUtil(num dupNum); # Function to generate all primes and checking  # whether number is palindromic or not  def printPalPrimesLessThanN(n): # Create a boolean array 'prime[0..n]' and  # initialize all entries it as true. A value  # in prime[i] will finally be false if i is  # Not a prime else true.  prime = [True] * (n + 1); p = 2; while (p * p <= n): # If prime[p] is not changed  # then it is a prime  if (prime[p]): # Update all multiples of p  for i in range(p * 2 n + 1 p): prime[i] = False; p += 1; # Print all palindromic prime numbers  for p in range(2 n + 1): # checking whether the given number  # is prime palindromic or not  if (prime[p] and isPal(p)): print(p end = ' '); # Driver Code  n = 100; print('Palindromic primes smaller' 'than or equal to' n 'are :'); printPalPrimesLessThanN(n); # This code is contributed by chandan_jnu 

// C# Program to print all palindromic  // primes smaller than or equal to n. using System;  class GFG {    // A function that returns true only  // if num contains one digit  static bool oneDigit(int num)  {  // comparison operation is faster than  // division operation. So using following  // instead of 'return num / 10 == 0;'  return (num >= 0 && num < 10);  }    // A recursive function to find out whether  // num is palindrome or not. Initially dupNum  // contains address of a copy of num.  static bool isPalUtil(int num int dupNum)  {  // Base case (needed for recursion termination):  // This statement/ mainly compares the first  // digit with the last digit  if (oneDigit(num))  return (num == (dupNum) % 10);    // This is the key line in this method. Note  // that all recursive/ calls have a separate  // copy of num but they all share same copy  // of dupNum. We divide num while moving up  // the recursion tree  if (!isPalUtil(num/10 dupNum))  return false;    // The following statements are executed when  // we move up the recursion call tree  dupNum /= 10;    // At this point if num%10 contains ith  // digit from beginning then (dupNum)%10  // contains ith digit from end  return (num % 10 == (dupNum) % 10);  }    // The main function that uses recursive   // function isPalUtil() to find out   // whether num is palindrome or not  static bool isPal(int num)  {  // If num is negative make it positive  if (num < 0)  num = -num;    // Create a separate copy of num so that  // modifications made to address dupNum don't  // change the input number.  int dupNum = num; // dupNum = num    return isPalUtil(num dupNum);  }    // Function to generate all primes and checking  // whether number is palindromic or not  static void printPalPrimesLessThanN(int n)  {  // Create a boolean array 'prime[0..n]' and  // initialize all entries it as true. A value  // in prime[i] will finally be false if i is  // Not a prime else true.  bool []prime = new bool[n+1];    for (int i=0;i<n+1;i++)  prime[i]=true;    for (int p = 2; p*p <= n; p++)  {  // If prime[p] is not changed  // then it is a prime  if (prime[p])  {  // Update all multiples of p  for (int i = p*2; i <= n; i += p){  prime[i] = false;}  }  }    // Print all palindromic prime numbers  for (int p = 2; p <= n; p++){    // checking whether the given number is  // prime palindromic or not  if (prime[p] && isPal(p)){  Console.Write(p + ' ');  }  }  }    // Driver function  public static void Main()   {  int n = 100;  Console.Write('Palindromic primes smaller than or ' +   'equal to are :n' n);  printPalPrimesLessThanN(n);  } }   // This code is contributed by nitin mittal. 

// javascript Program to print all palindromic primes // smaller than or equal to n. // A function that returns true only if num // contains one digit function oneDigit(num) {  // comparison operation is faster than  // division operation. So using following  // instead of 'return num / 10 == 0;'  return (num >= 0 && num < 10); } // A recursive function to find out whether // num is palindrome or not. Initially dupNum // contains address of a copy of num. function isPalUtil(num dupNum) {  // Base case (needed for recursion termination):  // This statement/ mainly compares the first  // digit with the last digit  if (oneDigit(num))  return (num == (dupNum) % 10);  // This is the key line in this method. Note  // that all recursive/ calls have a separate  // copy of num but they all share same copy  // of dupNum. We divide num while moving up  // the recursion tree  if (!isPalUtil(parseInt(num / 10) dupNum))  return false;  // The following statements are executed when  // we move up the recursion call tree  dupNum = parseInt(dupNum / 10);  // At this point if num%10 contains ith  // digit from beginning then (dupNum)%10  // contains ith digit from end  return (num % 10 == (dupNum) % 10); } // The main function that uses recursive function // isPalUtil() to find out whether num is palindrome // or not function isPal(num) {  // If num is negative make it positive  if (num < 0)  num = -num;  // Create a separate copy of num so that  // modifications made to address dupNum don't  // change the input number.  var dupNum = num; // dupNum = num  return isPalUtil(num dupNum); } // Function to generate all primes and checking // whether number is palindromic or not function printPalPrimesLessThanN(n) {  // Create a boolean array 'prime[0..n]' and  // initialize all entries it as true. A value  // in prime[i] will finally be false if i is  // Not a prime else true.  var prime  = Array.from({length : n + 1} (_ i) => true);  for (p = 2; p * p <= n; p++) {  // If prime[p] is not changed then it is  // a prime  if (prime[p]) {  // Update all multiples of p  for (i = p * 2; i <= n; i += p) {  prime[i] = false;  }  }  }  // Print all palindromic prime numbers  for (p = 2; p <= n; p++) {  // checking whether the given number is  // prime palindromic or not  if (prime[p] && isPal(p)) {  console.log(p);  }  } } // Driver function var n = 100; console.log('Palindromic primes smaller than or equal to ' + n + ' are :'); printPalPrimesLessThanN(n); 

 // PHP Program to print all palindromic  // primes smaller than or equal to n.  // A function that returns true only  // if num contains one digit  function oneDigit($num) { // comparison operation is faster than  // division operation. So using following  // instead of 'return num / 10 == 0;'  return ($num >= 0 && $num < 10); } // A recursive function to find out whether  // num is palindrome or not. Initially  // dupNum contains address of a copy of num.  function isPalUtil($num $dupNum) { // Base case (needed for recursion termination):  // This statement/ mainly compares the first  // digit with the last digit  if (oneDigit($num)) return ($num == ($dupNum) % 10); // This is the key line in this method. Note  // that all recursive/ calls have a separate  // copy of num but they all share same copy  // of dupNum. We divide num while moving up  // the recursion tree  if (!isPalUtil((int)($num/10) $dupNum)) return false; // The following statements are executed  // when we move up the recursion call tree  $dupNum = (int)($dupNum / 10); // At this point if num%10 contains ith  // digit from beginning then (dupNum)%10  // contains ith digit from end  return ($num % 10 == ($dupNum) % 10); } // The main function that uses recursive  // function isPalUtil() to find out whether // num is palindrome or not  function isPal($num) { // If num is negative make it positive  if ($num < 0) $num = -$num; // Create a separate copy of num so that  // modifications made to address dupNum  // don't change the input number.  $dupNum = $num; // dupNum = num  return isPalUtil($num $dupNum); } // Function to generate all primes and checking  // whether number is palindromic or not  function printPalPrimesLessThanN($n) { // Create a boolean array 'prime[0..n]' and  // initialize all entries it as true. A value  // in prime[i] will finally be false if i is  // Not a prime else true.  $prime = array_fill(0 $n + 1 true); for ($p = 2; $p * $p <= $n; $p++) { // If prime[p] is not changed then  // it is a prime  if ($prime[$p]) { // Update all multiples of p  for ($i = $p * 2; $i <= $n; $i += $p) { $prime[$i] = false; } } } // Print all palindromic prime numbers  for ($p = 2; $p <= $n; $p++) { // checking whether the given number  // is prime palindromic or not  if ($prime[$p] && isPal($p)) { print($p . ' '); } } } // Driver Code $n = 100; print('Palindromic primes smaller ' . 'than or equal to '.$n.' are :n'); printPalPrimesLessThanN($n); // This code is contributed by mits  ?>