How to calculate (A*B)%C? [duplicate] - c++

Question

This question already has answers here:
Multiply two overflowing integers modulo a third
(2 answers)
Closed 9 years ago.
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