/**
 * @file
 * @brief C++ Program for Modular Exponentiation Iteratively.
 * Calculate the value of an integer a raised to an integer exponent b
 * under modulo c.
 * @note The time complexity of this approach is O(log b).
 *
 * Example:
 * (4^3) % 5 (where ^ stands for exponentiation and % for modulo)
 * (4*4*4) % 5
 * (4 % 5) * ( (4*4) % 5 )
 * 4 * (16 % 5)
 * 4 * 1
 * 4
 * We can also verify the result as 4^3 is 64 and 64 modulo 5 is 4
 */

#include <iostream>

/// Iterative Function to calculate a raised to exponent b
/// under modulo c in O(log b) using modular exponentiation.
int power(int a, unsigned int b, int c) {
    int ans = 1;  /// Initialize the answer to be returned

    a = a % c;  /// Update a if it is more than or
                /// equal to c

    if (a == 0) {
        return 0;  /// In case a is divisible by c;
    }

    while (b > 0) {
        /// If b is odd, multiply a with answer
        if (b & 1) {
            ans = (ans * a) % c;
        }

        /// b must be even now
        b = b >> 1;  // b = b/2
        a = (a * a) % c;
    }

    return ans;
}

/**
 * @brief Main function
 * @returns 0 on exit
 */
int main() {
    /// Give two numbers num1, num2 and modulo m
    int num1 = 2;
    int num2 = 5;
    int m = 13;

    std::cout << "The value of " << num1 << " raised to exponent " << num2
              << " under modulo " << m << " is " << power(num1, num2, m);
    /// std::cout << "The value of "<<num1<<" raised to exponent "<<num2<<"
    /// " under modulo "<<m<<" is " << power(num1, num2, m);
}