Question

有人可以帮助我如何计算(A*B)%C，其中1<=A,B,C<=10^18在C ++中，没有big-num，只是一种数学方法。

Answer 1

脱离我的头顶（未经过广泛测试）

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;
}

这具有复杂度O(log A)次迭代。您可以使用条件减法替换大多数%，以获得更高的性能。

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;
}

这个版本只有一个长整数模 - 它甚至可能比大块方法更快，这取决于你的处理器如何实现整数模数。

现场演示：https://ideone.com/1pTldb - 与Yakk的结果相同。

Answer 2

执行（this）[http://stackoverflow.com/a/14859713/256138]堆栈溢出回答之前：

#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";
}

这里的想法是将64位整数分成一对32位整数，这些整数可以安全地在64位域中相乘。

我们将a*b表示为(a_high * 2^32 + a_low) * (b_high * 2^32 + b_low)，进行4倍乘法（跟踪2^32因子而不将它们存储在我们的位中），然后注意执行{{1}可以通过此模式的一系列a * 2^k % c重复来完成：k。所以我们可以在((a*2 %c) *2%c)...中取这个64位整数的3到4元素多项式并减少它而不必担心事情。

昂贵的部分是2^32函数（唯一的循环）。

如果您知道a_times_2_k_mod_c有多个高位清除，您可以加快速度。

您可以使用减法c

替换a %= c

两者都不是那么实用。

Live example

如何计算（A * B）％C？

2 个答案: