## C++, math, modular_division.cpp

``````/**
* @file
* @brief An algorithm to divide two numbers under modulo p [Modular
* Division](https://www.geeksforgeeks.org/modular-division)
* @details To calculate division of two numbers under modulo p
* Modulo operator is not distributive under division, therefore
* we first have to calculate the inverse of divisor using
* [Fermat's little
theorem](https://en.wikipedia.org/wiki/Fermat%27s_little_theorem)
* Now, we can multiply the dividend with the inverse of divisor
* and modulo is distributive over multiplication operation.
* Let,
* We have 3 numbers a, b, p
* To compute (a/b)%p
* (a/b)%p ≡ (a*(inverse(b)))%p ≡ ((a%p)*inverse(b)%p)%p
* NOTE: For the existence of inverse of 'b', 'b' and 'p' must be coprime
* For simplicity we take p as prime
* Time Complexity: O(log(b))
* Example: ( 24 / 3 ) % 5 => 8 % 5 = 3 --- (i)
Now the inverse of 3 is 2
(24 * 2) % 5 = (24 % 5) * (2 % 5) = (4 * 2) % 5 = 3 --- (ii)
(i) and (ii) are equal hence the answer is correct.
* @see modular_inverse_fermat_little_theorem.cpp, modular_exponentiation.cpp
* @author [Shubham Yadav](https://github.com/shubhamamsa)
*/

#include <cassert>   /// for assert
#include <iostream>  /// for IO operations

/**
* @namespace math
* @brief Mathematical algorithms
*/
namespace math {
/**
* @namespace modular_division
* @brief Functions for [Modular
* Division](https://www.geeksforgeeks.org/modular-division) implementation
*/
namespace modular_division {
/**
* @brief This function calculates a raised to exponent b under modulo c using
* modular exponentiation.
* @param a integer base
* @param b unsigned integer exponent
* @param c integer modulo
* @return a raised to power b modulo c
*/
uint64_t power(uint64_t a, uint64_t b, uint64_t c) {
uint64_t 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 % c) * (a % c)) % c;
}
/// b must be even now
b = b >> 1;  /// b = b/2
a = ((a % c) * (a % c)) % c;
}
return ans;
}

/**
* @brief This function calculates modular division
* @param a integer dividend
* @param b integer divisor
* @param p integer modulo
* @return a/b modulo c
*/
uint64_t mod_division(uint64_t a, uint64_t b, uint64_t p) {
uint64_t inverse = power(b, p - 2, p) % p;  /// Calculate the inverse of b
uint64_t result =
((a % p) * (inverse % p)) % p;  /// Calculate the final result
return result;
}
}  // namespace modular_division
}  // namespace math

/**
* Function for testing power function.
* test cases and assert statement.
* @returns `void`
*/
static void test() {
uint64_t test_case_1 = math::modular_division::mod_division(8, 2, 2);
assert(test_case_1 == 0);
std::cout << "Test 1 Passed!" << std::endl;
uint64_t test_case_2 = math::modular_division::mod_division(15, 3, 7);
assert(test_case_2 == 5);
std::cout << "Test 2 Passed!" << std::endl;
uint64_t test_case_3 = math::modular_division::mod_division(10, 5, 2);
assert(test_case_3 == 0);
std::cout << "Test 3 Passed!" << std::endl;
uint64_t test_case_4 = math::modular_division::mod_division(81, 3, 5);
assert(test_case_4 == 2);
std::cout << "Test 4 Passed!" << std::endl;
uint64_t test_case_5 = math::modular_division::mod_division(12848, 73, 29);
assert(test_case_5 == 2);
std::cout << "Test 5 Passed!" << std::endl;
}

/**
* @brief Main function
* @param argc commandline argument count (ignored)
* @param argv commandline array of arguments (ignored)
* @returns 0 on exit
*/
int main(int argc, char *argv[]) {
test();  // execute the tests
return 0;
}
``````