This algorithm takes a positive integer as input and calculates its square using a recursive approach. It divides the input into three parts and recursively calculates the squares of these parts. Using the values of the squared parts, it calculates the square of the original number and returns the result. This algorithm is more efficient than traditional multiplication-based methods of finding the square of a number.

Time Complexity: O(n^log₃5)

            long double square_fast(long long num)
            {
                if (num < 0)
                {
                    num = num * -1;
                }

                // base case
                if (to_string(num).length() == 1)
                {
                    return num * num;
                }
                // divide to three parts
                string numStr = to_string(num);
                int numDigits = numStr.length();
                int numPerPart = (numDigits) / 3;
                int extraDigits = numDigits % 3;
                vector<string> parts;
                parts.resize(3, "");
                int powerM;
                if (numDigits == 2)
                {
                    parts.at(0) = "0";
                    parts.at(1) = to_string(num / 10);
                    parts.at(2) = to_string(num % 10);
                    powerM = 1;
                }
                else
                {
                    if (extraDigits == 0)
                    {
                        parts.at(0) = numStr.substr(0, numPerPart);
                        parts.at(1) = numStr.substr(numPerPart, numPerPart);
                        parts.at(2) = numStr.substr(numPerPart * 2, numPerPart);
                        powerM = numPerPart;
                    }
                    else if (extraDigits == 1)
                    {
                        parts.at(0) = numStr.substr(0, numPerPart + 1);
                        parts.at(1) = numStr.substr(numPerPart + 1, numPerPart);
                        parts.at(2) = numStr.substr(numPerPart * 2 + 1, numPerPart);
                        powerM = numPerPart;
                    }
                    else if (extraDigits == 2)
                    {
                        parts.at(0) = numStr.substr(0, numPerPart);
                        parts.at(1) = numStr.substr(numPerPart + 2, numPerPart + 1);
                        parts.at(2) = numStr.substr(numPerPart * 2 + 1, numPerPart + 1);
                        powerM = numPerPart + 1;
                    }
                }
                long double a = parts.at(0) == "" ? 0 : stoi(parts.at(0));
                long double b = parts.at(1) == "" ? 0 : stoi(parts.at(1));
                long double c = parts.at(2) == "" ? 0 : stoi(parts.at(2));

                long double m = square_fast(c);
                long double n = square_fast(a);
                long double p = square_fast(a + b + c);
                long double q = square_fast(a - b + c);
                long double s = square_fast(4 * a - 2 * b + c);

                long double answer = pow(10, powerM * 4) * n + pow(10, powerM * 3) * (2 * n + (p + 3 * q - s) / 6 - m / 2) + pow(10, powerM * 2) * (-n + (p + q) / 2 - m) + pow(10, powerM * 1) * ((p / 3) - (2 * n) - q + (s / 6) + (m / 2)) + m;
                return answer;
            }