Write a program in C which studies the number of integers wh
Write a program in C which studies the number of integers which are palindromic in base 10 and another base. Your program should go through each base, b from 2-10. It should then count how many numbers between 1 and min(1,000,000, b^10-1) are palindromic in both base 10 and base b.
Solution
// A C++ program to generate palindromic numbers
// less than n.
#include <iostream>
using namespace std;
// A utility for creating palindrome
int createPalindrome(int input, int b, bool isOdd)
{
int n = input;
int palin = input;
// checks if number of digits is odd or even
// if odd then neglect the last digit of input in
// finding reverse as in case of odd number of
// digits middle element occur once
if (isOdd)
n /= b;
// Creates palindrome by just appending reverse
// of number to itself
while (n > 0)
{
palin = palin * b + (n % b);
n /= b;
}
return palin;
}
// Fruition to print decimal palindromic number
void generatePaldindromes(int n)
{
int number;
// Run two times for odd and even length palindromes
for (int j = 0; j < 2; j++)
{
// Creates palindrome numbers with first half as i.
// Value of j decided whether we need an odd length
// of even length palindrome.
int i = 1;
while ((number = createPalindrome(i, 10, j % 2)) < n)
{
cout << number << \" \";
i++;
}
}
}
// Driver Program to test above function
int main()
{
int n = 104;
generatePaldindromes(n);
return 0;
}
| // A C++ program to generate palindromic numbers // less than n. #include <iostream> using namespace std; // A utility for creating palindrome int createPalindrome(int input, int b, bool isOdd) { int n = input; int palin = input; // checks if number of digits is odd or even // if odd then neglect the last digit of input in // finding reverse as in case of odd number of // digits middle element occur once if (isOdd) n /= b; // Creates palindrome by just appending reverse // of number to itself while (n > 0) { palin = palin * b + (n % b); n /= b; } return palin; } // Fruition to print decimal palindromic number void generatePaldindromes(int n) { int number; // Run two times for odd and even length palindromes for (int j = 0; j < 2; j++) { // Creates palindrome numbers with first half as i. // Value of j decided whether we need an odd length // of even length palindrome. int i = 1; while ((number = createPalindrome(i, 10, j % 2)) < n) { cout << number << \" \"; i++; } } } // Driver Program to test above function int main() { int n = 104; generatePaldindromes(n); return 0; } |