Modulo of multiplication of large numbers

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Need help in mod 1000000007 questions 3 answers

How can I calculate (A*B)%C for A,B,C <= 10^18, in C++? 2 answers

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Basic idea here is to first define a non-overflowing addmod function which takes advantage of negative numbers in its arithmetic. Then define timesmod in terms of it also using bit operations. The time complexity is O(N) where N is the number of bits used (64 in this case).

#include <iostream>
using namespace std;

typedef long long BigInt; // must be signed, to detect overflow

BigInt A = 0x7fffffffffffff01;
BigInt B = 0x7fffffffffffff02;
BigInt M = 0x7fffffffffffff03;

// For simplicity it is assumed x, y, and m are all positive.
BigInt addmod( BigInt x, BigInt y, BigInt m )
{
  x %= m;
  y %= m;
  BigInt sum = x-m+y; // -m <= sum < m-1
  return sum < 0 ? sum + m : sum;
}

BigInt timesmod( BigInt x, BigInt y, BigInt m )
{
  x %= m;
  y %= m;
  BigInt a = x < y ? x : y; // min
  BigInt b = x < y ? y : x; // max
  BigInt product = 0;
  for (; a != 0; a >>= 1, b = addmod(b,b,m) )
    if (a&1) product = addmod(product,b,m);
  return product;
}

int main()
{
  cout << "A = " << A << endl;
  cout << "B = " << B << endl;
  cout << "M = " << M << endl;
  cout << "A*B mod M = " << timesmod(A,B,M) << endl;
  return 0;
}

Output:

A = 9223372036854775553
B = 9223372036854775554
M = 9223372036854775555
A*B mod M = 2

This is easily confirmed since A=-2 and B=-1 mod M .

Note: this code is not optimized.

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I don't have time to answer this right now, so I'll give a pointer and come back to edit this answer later. Look up "Schrage multiplication" in your favorite algorithms textbook (or search engine). The basic idea is to split both A and B into pieces, process the pieces separately, then combine the pieces to complete the calculation.

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I think you can form the 128-bit product in two pieces (high 64 bits and low 64 bits) and reduce each piece modulo p. Supposing p is around 4^k , you can then figure out roughly how many p's are in this number by dividing hi64 / (p>>k) ; this should give you about k-1 bits of the right answer. Subtract off that many p 's from the whole thing and now hi64 has about k-1 fewer bits. Do this again, but compute (hi64 << k-1) / (p >> k) . Then do it again, computing (hi64 << k+k-2) / (p >> k) .

Schrage's trick, suggested by another poster, sounds like a better deal, but I don't understand it. Hopefully that poster returns and completes his answer!

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