If a number's a palindrom and divisible by all of its digits

Question

I have a problem where in a range of numbers L to R I should check how many of them are both palindroms (read the same forward and backward) OR divisible by all of its digits. Now, I have started a program which doesn't really work well, I made a few functions but I have a problem where in line 30 the compiler calls an error saying:

Main.cpp:30:41: error: in evaluation of 'operator%=(int, __gnu_cxx::__promote_2::__type {aka double})'
   palindrom_check %= pow(10, upper);
   ^

Here are examples of input and output given in the problem:

• Input: 1 15
• Output: 12

Explanation: In the interval of 1 to 15, the numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15 - therefore the output is 12.

• Input: 303 304
• Output: 1

I am pretty sure I'm not using the pow() correctly. I hope anyone can give me a way to continue or edit my code to fix it and make it work.

#include <iostream>
#include <cmath>

using namespace std;

int digits = 1;
int check_dig = 0;

int Digit(int n) {
    digits = 1;
    check_dig = n;
    while (check_dig > 10) {
        check_dig /= 10;
        digits++;
    }
}

int exponent = 1;
int front_check = 0;
int back_check = 0;
bool palindrom = false;
int palindrom_check = 0;
int upper = digits - 1;

int Palindrom(int n) {
    palindrom = false;
    exponent = (int)pow(10, upper);
    front_check = n;
    palindrom_check = n;
    while (palindrom_check > 0) {
        back_check += (palindrom_check % 10) * exponent;
        palindrom_check %= exponent;
        exponent /= 10;
    }
    if (back_check == front_check) {
        palindrom = true;
    }
}

bool divisible = false;
int divisible_check = 0;

int Divisibility(int n) {
    divisible_check = n;
    while (divisible_check > 0) {
        if (divisible_check % (divisible_check % 10) == 0) {
            divisible_check /= 10;
            divisible = true;
            continue;
        } else {
            divisible = false;
            break;
        }
    }
}

int main() {

    int L, R;
    cin >> L >> R;
    int result = 0;

    for (int i = L; i <= R; i++) {
        Digit(i);
        Palindrom(i);
        Divisibility(i);
        if (palindrom || divisible) {
            result++;
        }
    }

    cout << result;

    return 0;
}

Edit:

#include <iostream>
#include <cmath>

using namespace std;

int digits = 1;
int check_dig = 0;

void Digit(int n) {
    digits = 1;
    check_dig = n;
    while (check_dig > 10) {
        check_dig /= 10;
        digits++;
    }
}

int front_check = 0;
int back_check = 0;
int palindrom_check = 0;
int upper = digits - 1;

bool palindrom(int n) {
    front_check = n;
    back_check = 0;
    palindrom_check = n;
    while (palindrom_check > 0) {
        back_check = (back_check * 10) + (palindrom_check % 10);
        palindrom_check /= 10;
    }
    if (back_check == front_check) {
        return true;
    }
}

int divisible_check = 0;

bool divisible(int n) {
    divisible_check = n;
    while (divisible_check > 0) {
        if (divisible_check % (divisible_check % 10) == 0) {
            divisible_check /= 10;
            return true;
        } else {
            return false;
            break;
        }
    }
}

int main() {

    int L, R;
    cin >> L >> R;
    int result = 0;

    for (int i = L; i <= R; i++) {
        Digit(i);
        palindrom(i);
        divisible(i);
        if (palindrom || divisible) {
            result++;
        }
    }

    cout << result;

    return 0;
}

Answer 1

It looks to me as though this line:

palindrom_check %= pow(10, upper);

will be unhappy as pow(10,upper) is coming out as a double, not an int. The modulus operator % doesn't really make sense for double precision numbers. The error you have can be fixed by typecasting this to an integer:

palindrom_check %= (int)pow(10, upper);

Answer 2

In C++, a function defined as

bool divisible(int n) { /* … */ }

Is a function that accepts an integer value passed as an argument when called and returns a bool. A properly instructed compiler would complain (among other things) if there exist at least one path that the execution could take which doesn't return anything.

bool divisible(int n) {
    divisible_check = n;           // No need for that variable to be global.
    while (divisible_check > 0) {  // If it's 0, the loop (which isn't a loop) is not executed.
        if ( /* ... */ ) {
            // ...
            return true;
        } else {
            return false;
            break;                 // <- Useless.
        }
    }
    // And now what?
}

The same applies to the other function, that could be implemented like so:

bool is_divisible_by_all_of_its_digits(int n)
{
    int tmp = n;
    while (tmp > 0)
    {
        int digit = tmp % 10;
        // We want to prevent n % 0 which would cuase a floating point exception
        if ( digit == 0  ||  n % digit != 0)
        {
            return false;
        }
        tmp /= 10;         
    }
    return true;
}

The returned value can be used directly in the condition

// ...
for (int i = L; i <= R; i++) {
    if ( is_palindrome(i)  ||  is_divisible_by_all_of_its_digits(i)) {
        result++;
    }
}