How to calculate (A*B)%C?

Question

Can someone help me how to calculate (A*B)%C, where 1<=A,B,C<=10^18 in C++, without big-num, just a mathematical approach.

Answer 1

Off the top of my head (not extensively tested)

typedef unsigned long long BIG;
BIG mod_multiply( BIG A, BIG B, BIG C )
{
    BIG mod_product = 0;
    A %= C;

    while (A) {
        B %= C;
        if (A & 1) mod_product = (mod_product + B) % C;
        A >>= 1;
        B <<= 1;
    }

    return mod_product;
}

This has complexity O(log A) iterations. You can probably replace most of the % with a conditional subtraction, for a bit more performance.

typedef unsigned long long BIG;
BIG mod_multiply( BIG A, BIG B, BIG C )
{
    BIG mod_product = 0;
    // A %= C; may or may not help performance
    B %= C;

    while (A) {
        if (A & 1) {
            mod_product += B;
            if (mod_product > C) mod_product -= C;
        }
        A >>= 1;
        B <<= 1;
        if (B > C) B -= C;
    }

    return mod_product;
}

This version has only one long integer modulo -- it may even be faster than the large-chunk method, depending on how your processor implements integer modulo.

• Live demo: https://ideone.com/1pTldb -- same result as Yakk's.

Answer 2

An implementation of this stack overflow answer prior:

#include <stdint.h>
#include <tuple>
#include <iostream>

typedef std::tuple< uint32_t, uint32_t > split_t;
split_t split( uint64_t a )
{
  static const uint32_t mask = -1;
  auto retval = std::make_tuple( mask&a, ( a >> 32 ) );
  // std::cout << "(" << std::get<0>(retval) << "," << std::get<1>(retval) << ")\n";
  return retval;
}

typedef std::tuple< uint64_t, uint64_t, uint64_t, uint64_t > cross_t;
template<typename Lambda>
cross_t cross( split_t lhs, split_t rhs, Lambda&& op )
{
  return std::make_tuple( 
    op(std::get<0>(lhs), std::get<0>(rhs)),
    op(std::get<1>(lhs), std::get<0>(rhs)),
    op(std::get<0>(lhs), std::get<1>(rhs)),
    op(std::get<1>(lhs), std::get<1>(rhs))
  );
}

// c must have high bit unset:
uint64_t a_times_2_k_mod_c( uint64_t a, unsigned k, uint64_t c )
{
  a %= c;
  for (unsigned i = 0; i < k; ++i)
  {
    a <<= 1;
    a %= c;
  }
  return a;
}

// c must have about 2 high bits unset:
uint64_t a_times_b_mod_c( uint64_t a, uint64_t b, uint64_t c )
{
  // ensure a and b are < c:
  a %= c;
  b %= c;
  
  auto Z = cross( split(a), split(b), [](uint32_t lhs, uint32_t rhs)->uint64_t {
    return (uint64_t)lhs * (uint64_t)rhs;
  } );
  
  uint64_t to_the_0;
  uint64_t to_the_32_a;
  uint64_t to_the_32_b;
  uint64_t to_the_64;
  std::tie( to_the_0, to_the_32_a, to_the_32_b, to_the_64 ) = Z;
  
  // std::cout << to_the_0 << "+ 2^32 *(" << to_the_32_a << "+" << to_the_32_b << ") + 2^64 * " << to_the_64 << "\n";
  
  // this line is the one that requires 2 high bits in c to be clear
  // if you just add 2 of them then do a %c, then add the third and do
  // a %c, you can relax the requirement to "one high bit must be unset":
  return
    (to_the_0
    + a_times_2_k_mod_c(to_the_32_a+to_the_32_b, 32, c) // + will not overflow!
    + a_times_2_k_mod_c(to_the_64, 64, c) )
  %c;
}

int main()
{
  uint64_t retval = a_times_b_mod_c( 19010000000000000000, 1011000000000000, 1231231231231211 );
  std::cout << retval << "\n";
}

The idea here is to split your 64-bit integer into a pair of 32-bit integers, which are safe to multiply in 64-bit land.

We express a*b as (a_high * 2^32 + a_low) * (b_high * 2^32 + b_low), do the 4-fold multiplication (keeping track of the 2³² factors without storing them in our bits), then note that doing a * 2^k % c can be done via a series of k repeats of this pattern: ((a*2 %c) *2%c).... So we can take this 3 to 4 element polynomial of 64-bit integers in 2³² and reduce it without having to worry about things.

The expensive part is the a_times_2_k_mod_c function (the only loop).

You can make it go many times faster if you know that c has more than one high bit clear.

You could instead replace the a %= c with subtraction a -= (a>=c)*c;

Doing both isn't all that practical.

Live example