136B - Ternary Logic - CodeForces Solution

#include <iostream>
#include <vector>
#include <string>
#include <utility>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <stack>
#include <algorithm>
#include <cmath>

using namespace std;

int main(int argc, char** argv){

    int a, c;
    cin >> a >> c;

    string a3, b3, c3;
    while (a) {
        a3 = to_string(a%3) + a3;
        a /= 3;
    }   

    while (c) {
        c3 = to_string(c%3) + c3;
        c /= 3;
    }   

    while (a3.length() < c3.length()) {
        a3 = '0' + a3;
    }
    while (c3.length() < a3.length()) {
        c3 = '0' + c3;
    }

    for (int i = 0; i < a3.length() || i < c3.length(); i++) {
        if (i < a3.length()) {
            if (c3[i] >= a3[i]) {
                b3 += (c3[i] - a3[i] + '0');
            } else {
                if (a3[i] == '2' && c3[i] == '1') {
                    b3 += '2';
                } else if (a3[i] == '1' && c3[i] == '0') {
                    b3 += '2';
                } else {
                    b3 += '1';
                }
            }
        }
    }

    int sum = 0;
    for (int i = b3.length()-1; i >= 0; i--) {
        sum += (b3[i] - '0') * pow(3, b3.length() - i - 1);
    }

    cout << sum << endl;

    return 0;
}