931 lines
28 KiB
C++
931 lines
28 KiB
C++
/**
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* @file int2048.cpp --- 2048-bit integer class implementation
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*
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* @details This file contains the implementation of the 2048-bit integer class.
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*
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* Codesytle: This file is written in a sytle mainly based on Google C++ Style
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* Guide. As I use Clang-format to format my code, so the code style may be a
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* little bit strange sometimes, in that case I'll manually format the
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* code.What's sepecial is the comment:
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* 1. Multi-line comments are always before the code they comment on.
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* Usually the code they comment on is a complex procedure,like the definition
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* of a function,a class or a variable with complex operation. If a multi-line
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* comment is in one line, it will start with "/*" instead of "/**",otherwise it
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* will start with "/**" and in the format of Doxygen.
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* 2. Single-line comments are always after the code they comment on.
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* Usually they are in the same line with the code they comment on,but sometimes
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* they may come in the next lines. single-line comments shouldn't exceed 3
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* lines as they are intended to be short and easy to understand.
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* 3. Temporary disabled code will be marked with "//" in the front of each
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* 4. Some comments have special meanings,like "//TODO", "//FIXME", "//XXX","//
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* clang-format off" and "// clang-format on". They are not controlled by the
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* previous rules.
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*/
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#include "int2048.h"
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#include <cassert>
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#include <cstdio>
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#include <cstring>
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#include <iostream>
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#include <string>
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#include <utility>
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static_assert(sizeof(int) == 4, "sizeof(int) != 4");
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static_assert(sizeof(long long) == 8, "sizeof(long long)!=8");
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namespace ZYM {
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// 构造函数
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int2048::int2048() {
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// 实现构造函数逻辑
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buf_length = kDefaultLength;
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val = new int[buf_length]();
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flag = 1;
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num_length = 1;
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}
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int2048::~int2048() {
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// 实现析构函数逻辑
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if (val != nullptr) delete[] val;
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}
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int2048::int2048(long long input_value) {
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// 实现构造函数逻辑
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buf_length = kDefaultLength;
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val = new int[buf_length]();
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if (input_value < 0) {
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flag = -1;
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input_value = -input_value;
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} else
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flag = 1;
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if (input_value == 0) {
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num_length = 1;
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return;
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}
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num_length = 0;
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const static int kPow10[9] = {1, 10, 100, 1000, 10000,
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100000, 1000000, 10000000, 100000000};
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while (input_value > 0) {
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val[num_length / kNum] += (input_value % 10) * kPow10[num_length % kNum];
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input_value /= 10;
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num_length++;
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}
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}
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int2048::int2048(const std::string &input_value) {
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// 实现构造函数逻辑
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buf_length = (input_value.length() + kNum - 1) / kNum * kMemAdditionScalar;
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val = new int[buf_length]();
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flag = 1;
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num_length = 0;
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if (input_value[0] == '-') {
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flag = -1;
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}
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const static int kPow10[9] = {1, 10, 100, 1000, 10000,
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100000, 1000000, 10000000, 100000000};
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int read_highest_pos = (flag > 0 ? 0 : 1);
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while (input_value[read_highest_pos] == '0' &&
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read_highest_pos + 1 < input_value.length())
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read_highest_pos++;
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for (int i = input_value.length() - 1; i >= read_highest_pos; i--) {
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val[num_length / kNum] +=
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(input_value[i] - '0') * kPow10[num_length % kNum];
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num_length++;
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}
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if (num_length == 1 && val[0] == 0) flag = 1;
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}
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int2048::int2048(const int2048 &input_value) {
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buf_length = input_value.buf_length;
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val = new int[buf_length]();
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memcpy(val, input_value.val, buf_length * sizeof(int));
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flag = input_value.flag;
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num_length = input_value.num_length;
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}
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/**
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* @brief Move constructor
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*
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* @param input_value
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*
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* @details Note that after moving, the input_value will be set to 0.
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*/
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int2048::int2048(int2048 &&input_value) noexcept {
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buf_length = input_value.buf_length;
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val = input_value.val;
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flag = input_value.flag;
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num_length = input_value.num_length;
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input_value.buf_length = kDefaultLength;
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input_value.flag = 1;
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input_value.num_length = 1;
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input_value.val = new int[input_value.buf_length]();
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}
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// 读入一个大整数
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void int2048::read(const std::string &input_value) {
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delete[] val;
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buf_length = (input_value.length() + kNum - 1) / kNum * kMemAdditionScalar;
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val = new int[buf_length]();
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flag = 1;
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num_length = 0;
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if (input_value[0] == '-') {
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flag = -1;
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}
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const static int kPow10[9] = {1, 10, 100, 1000, 10000,
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100000, 1000000, 10000000, 100000000};
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int read_highest_pos = (flag > 0 ? 0 : 1);
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while (input_value[read_highest_pos] == '0' &&
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read_highest_pos + 1 < input_value.length())
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read_highest_pos++;
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for (int i = input_value.length() - 1; i >= read_highest_pos; i--) {
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val[num_length / kNum] +=
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(input_value[i] - '0') * kPow10[num_length % kNum];
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num_length++;
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}
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if (num_length == 1 && val[0] == 0) flag = 1;
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}
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// 输出储存的大整数，无需换行
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void int2048::print() {
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// 实现输出逻辑
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char *buf = new char[num_length + 5];
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char *p = buf;
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if (flag == -1) *p++ = '-';
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const static int kPow10[9] = {1, 10, 100, 1000, 10000,
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100000, 1000000, 10000000, 100000000};
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for (int i = num_length - 1; i >= 0; i--)
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*p++ = char('0' + val[i / int2048::kNum] / kPow10[i % int2048::kNum] % 10);
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*p++ = 0;
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std::cout << buf;
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delete[] buf;
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}
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void int2048::print_debug() {
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// 实现输出逻辑
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char *buf = new char[num_length + 5];
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char *p = buf;
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if (flag == -1) *p++ = '-';
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const static int kPow10[9] = {1, 10, 100, 1000, 10000,
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100000, 1000000, 10000000, 100000000};
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for (int i = num_length - 1; i >= 0; i--)
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*p++ = char('0' + val[i / int2048::kNum] / kPow10[i % int2048::kNum] % 10);
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*p++ = 0;
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std::cerr << buf << std::endl;
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delete[] buf;
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}
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/**
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* @brief Claim memory for the number
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*
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* @param number_length The length of the number
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*
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* @details If the number_length is smaller than the current buf_length/4 or
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* bigger than current buf_length, then val will be reset to accomodate
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* number_length*2 numbers. Something like std::vector.
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*/
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void int2048::ClaimMem(size_t number_length) {
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size_t new_number_blocks = (number_length + kNum - 1) / kNum;
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if (new_number_blocks > buf_length) {
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int *new_val = new int[new_number_blocks * kMemAdditionScalar]();
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memcpy(new_val, val, buf_length * sizeof(int));
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delete[] val;
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val = new_val;
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buf_length = new_number_blocks * kMemAdditionScalar;
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} else if (new_number_blocks * kMemDeleteScalar < buf_length) {
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int *new_val = new int[new_number_blocks * kMemAdditionScalar]();
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memcpy(new_val, val, new_number_blocks * sizeof(int));
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delete[] val;
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val = new_val;
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buf_length = new_number_blocks * kMemAdditionScalar;
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}
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}
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/**
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* @brief Compare two unsigned numbers
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*
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* @param A The first number
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* @param B The second number
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*
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* @return -1 if A < B, 0 if A == B, 1 if A > B
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*
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* @details This function is used to compare two numbers wittout considering the
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* sign.
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*/
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inline int UnsignedCmp(const int2048 &A, const int2048 &B) {
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if (A.num_length != B.num_length) return A.num_length < B.num_length ? -1 : 1;
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int number_of_blocks = (A.num_length + int2048::kNum - 1) / int2048::kNum;
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for (int i = number_of_blocks - 1; i >= 0; i--)
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if (A.val[i] != B.val[i]) return A.val[i] < B.val[i] ? -1 : 1;
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return 0;
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}
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inline void UnsignedAdd(int2048 &A, const int2048 *const pB) {
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if (&A == pB) throw "UnsignedAdd: A and B are the same object";
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A.ClaimMem(std::max(A.num_length, pB->num_length) + 2);
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for (int i = 0;
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i < (std::max(A.num_length, pB->num_length) + int2048::kNum - 1) /
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int2048::kNum;
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i++) {
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if (i < (pB->num_length + int2048::kNum - 1) / int2048::kNum)
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A.val[i] += pB->val[i];
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if (i + 1 < A.buf_length) A.val[i + 1] += A.val[i] / int2048::kStoreBase;
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A.val[i] %= int2048::kStoreBase;
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}
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A.num_length = std::max(A.num_length, pB->num_length);
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const static int kPow10[9] = {1, 10, 100, 1000, 10000,
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100000, 1000000, 10000000, 100000000};
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if (A.val[A.num_length / int2048::kNum] /
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kPow10[A.num_length % int2048::kNum] >
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0)
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A.num_length++;
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}
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// 加上一个大整数
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int2048 &int2048::add(const int2048 &B) {
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// 实现加法逻辑
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const int2048 *pB = &B;
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if (this->flag == pB->flag) {
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if (this == &B) pB = new int2048(B);
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UnsignedAdd(*this, pB);
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} else if (this->flag == 1 && pB->flag == -1) {
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int cmp = UnsignedCmp(*this, *pB);
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if (cmp >= 0) {
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if (this == &B) pB = new int2048(B);
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UnsignedMinus(*this, pB);
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this->flag = 1;
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} else {
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int2048 new_B = std::move(*this);
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*this = B;
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UnsignedMinus(*this, &new_B);
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this->flag = -1;
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}
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} else if (this->flag == -1 && pB->flag == 1) {
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int cmp = UnsignedCmp(*this, *pB);
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if (cmp >= 0) {
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if (this == &B) pB = new int2048(B);
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UnsignedMinus(*this, pB);
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this->flag = -1;
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if (this->num_length == 1 && this->val[0] == 0) this->flag = 1;
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} else {
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int2048 new_B = std::move(*this);
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*this = B;
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UnsignedMinus(*this, &new_B);
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this->flag = 1;
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}
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}
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if (pB != &B) delete pB;
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return *this;
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}
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// 返回两个大整数之和
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int2048 add(int2048 A, const int2048 &B) {
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// 实现加法逻辑
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return std::move(A.add(B));
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}
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inline void UnsignedMinus(int2048 &A, const int2048 *const pB) {
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if (&A == pB) throw "UnsignedMinus: A and B are the same object";
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for (int i = 0; i < (pB->num_length + int2048::kNum - 1) / int2048::kNum;
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i++) {
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A.val[i] -= pB->val[i];
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if (A.val[i] < 0) {
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A.val[i] += int2048::kStoreBase;
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A.val[i + 1]--;
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}
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}
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const static int kPow10[9] = {1, 10, 100, 1000, 10000,
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100000, 1000000, 10000000, 100000000};
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int new_length = 0;
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for (int i = 0; i < A.num_length; i++)
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if (A.val[i / int2048::kNum] / kPow10[i % int2048::kNum] > 0)
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new_length = i + 1;
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A.num_length = new_length;
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if (A.num_length == 0) A.num_length = 1;
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A.ClaimMem(A.num_length);
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}
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// 减去一个大整数
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int2048 &int2048::minus(const int2048 &B) {
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// 实现减法逻辑
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const int2048 *pB = &B;
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if (this->flag == B.flag) {
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int cmp = UnsignedCmp(*this, *pB);
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if (cmp >= 0) {
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if (this == &B) pB = new int2048(B);
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UnsignedMinus(*this, pB);
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if (this->num_length == 1 && this->val[0] == 0) this->flag = 1;
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} else {
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int2048 new_B = std::move(*this);
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*this = B;
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UnsignedMinus(*this, &new_B);
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this->flag = -this->flag;
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if (this->num_length == 1 && this->val[0] == 0) this->flag = 1;
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}
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} else {
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if (this == &B) pB = new int2048(B);
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UnsignedAdd(*this, pB);
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}
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if (pB != &B) delete pB;
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return *this;
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}
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// 返回两个大整数之差
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int2048 minus(int2048 A, const int2048 &B) {
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// 实现减法逻辑
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return std::move(A.minus(B));
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}
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// 运算符重载
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int2048 int2048::operator+() const {
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// 实现一元加法逻辑
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return std::move(int2048(*this));
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}
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int2048 int2048::operator-() const {
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// 实现一元减法逻辑
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int2048 ret(*this);
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if (!(ret.num_length == 1 && ret.val[0] == 0)) ret.flag = -ret.flag;
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return std::move(ret);
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}
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int2048 &int2048::operator=(const int2048 &B) {
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// 实现赋值运算符逻辑
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// similar to int2048::int2048(const int2048 &input_value)
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if (this == &B) return *this;
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delete[] val;
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buf_length = B.buf_length;
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val = new int[buf_length]();
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memcpy(val, B.val, buf_length * sizeof(int));
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flag = B.flag;
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num_length = B.num_length;
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return *this;
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}
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int2048 &int2048::operator=(int2048 &&B) noexcept {
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// 实现移动赋值运算符逻辑
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if (this == &B) return *this;
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delete[] val;
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buf_length = B.buf_length;
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val = B.val;
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flag = B.flag;
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num_length = B.num_length;
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B.buf_length = kDefaultLength;
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B.flag = 1;
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B.num_length = 1;
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B.val = new int[B.buf_length]();
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return *this;
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}
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int2048 &int2048::operator+=(const int2048 &B) {
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// 实现复合加法逻辑
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return this->add(B);
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}
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int2048 operator+(int2048 A, const int2048 &B) {
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// 实现加法逻辑
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A.add(B);
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return std::move(A);
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}
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int2048 &int2048::operator-=(const int2048 &B) {
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// 实现复合减法逻辑
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return this->minus(B);
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}
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int2048 operator-(int2048 A, const int2048 &B) {
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// 实现减法逻辑
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A.minus(B);
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return std::move(A);
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}
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__int128_t int2048::QuickPow(__int128_t v, long long q) {
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__int128_t ret = 1;
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v %= int2048::kNTTMod;
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while (q > 0) {
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if (q & 1) (ret *= v) %= int2048::kNTTMod;
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(v *= v) %= int2048::kNTTMod;
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q >>= 1;
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}
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return ret;
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}
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/**
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* @brief Move the number to the left by L digits. That is, v'=v*(10^L)
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*/
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void int2048::LeftMoveBy(int L) {
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const static long long kPow10[9] = {1, 10, 100,
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1000, 10000, 100000,
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1000000, 10000000, 100000000};
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int big_move = L / int2048::kNum;
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int small_move = L % int2048::kNum;
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this->ClaimMem(this->num_length + L);
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for (int i = this->buf_length - 1; i >= big_move; i--) {
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this->val[i] = this->val[i - big_move];
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}
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for (int i = 0; i < big_move; i++) {
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this->val[i] = 0;
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}
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this->num_length += big_move * int2048::kNum;
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if (small_move == 0) return;
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for (int i = this->buf_length - 1; i >= 0; i--) {
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this->val[i] =
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((long long)this->val[i] * kPow10[small_move]) % int2048::kStoreBase;
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if (i - 1 >= 0) {
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this->val[i] += this->val[i - 1] / kPow10[int2048::kNum - small_move];
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}
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}
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this->val[big_move] =
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(this->val[big_move] / kPow10[small_move]) * kPow10[small_move];
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this->num_length += small_move;
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}
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/**
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* @brief Move the number to the right by L digits. That is, v'=v//(10^L)
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*/
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void int2048::RightMoveBy(int L) {
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if (L >= this->num_length) {
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this->num_length = 1;
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this->val[0] = 0;
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return;
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}
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int big_move = L / int2048::kNum;
|
|
int small_move = L % int2048::kNum;
|
|
for (int i = 0; i < this->buf_length - big_move; i++) {
|
|
this->val[i] = this->val[i + big_move];
|
|
}
|
|
for (int i = this->buf_length - big_move; i < this->buf_length; i++) {
|
|
this->val[i] = 0;
|
|
}
|
|
this->num_length -= big_move * int2048::kNum;
|
|
if (small_move == 0) return;
|
|
const static int kPow10[9] = {1, 10, 100, 1000, 10000,
|
|
100000, 1000000, 10000000, 100000000};
|
|
for (int i = 0; i < this->buf_length; i++) {
|
|
this->val[i] /= kPow10[small_move];
|
|
if (i + 1 < this->buf_length) {
|
|
this->val[i] += this->val[i + 1] % kPow10[small_move] *
|
|
kPow10[int2048::kNum - small_move];
|
|
}
|
|
}
|
|
this->num_length -= small_move;
|
|
}
|
|
|
|
/**
|
|
* @brief the definition of NTTTransform
|
|
*
|
|
* @param a the array to be transformed
|
|
* @param NTT_blocks the length of the array
|
|
* @param inverse whether to do inverse transform
|
|
*
|
|
* @details the array a will be transformed to a new array a', which is the NTT.
|
|
* Note that the base in NTT is 10000.
|
|
*/
|
|
void int2048::NTTTransform(__int128_t *a, int NTT_blocks,
|
|
bool inverse = false) {
|
|
for (int i = 1, j = 0; i < NTT_blocks; i++) {
|
|
int bit = NTT_blocks >> 1;
|
|
while (j >= bit) {
|
|
j -= bit;
|
|
bit >>= 1;
|
|
}
|
|
j += bit;
|
|
if (i < j) std::swap(a[i], a[j]);
|
|
}
|
|
for (int len = 2; len <= NTT_blocks; len <<= 1) {
|
|
__int128_t wlen = QuickPow(int2048::kNTTRoot, (int2048::kNTTMod - 1) / len);
|
|
if (inverse) wlen = QuickPow(wlen, int2048::kNTTMod - 2);
|
|
for (int i = 0; i < NTT_blocks; i += len) {
|
|
__int128_t w = 1;
|
|
for (int j = 0; j < len / 2; j++) {
|
|
__int128_t u = a[i + j], v = a[i + j + len / 2] * w % int2048::kNTTMod;
|
|
a[i + j] = (u + v) % int2048::kNTTMod;
|
|
a[i + j + len / 2] = (u - v + int2048::kNTTMod) % int2048::kNTTMod;
|
|
(w *= wlen) %= int2048::kNTTMod;
|
|
}
|
|
}
|
|
}
|
|
if (inverse) {
|
|
__int128_t inv = QuickPow(NTT_blocks, int2048::kNTTMod - 2);
|
|
for (int i = 0; i < NTT_blocks; i++) (a[i] *= inv) %= int2048::kNTTMod;
|
|
}
|
|
}
|
|
|
|
/**
|
|
* @brief Unsigned multiply two numbers
|
|
*
|
|
* @param A The first number
|
|
* @param pB The second number
|
|
*
|
|
* @details This function will multiply two numbers without considering the
|
|
* sign.
|
|
*/
|
|
inline void UnsignedMultiply(int2048 &A, const int2048 *pB) {
|
|
if (&A == pB) throw "UnsignedMultiply: A and B are the same object";
|
|
int blocks_of_A = ((A.num_length + int2048::kNum - 1) / int2048::kNum);
|
|
int blocks_of_B = ((pB->num_length + int2048::kNum - 1) / int2048::kNum);
|
|
int max_blocks = blocks_of_A + blocks_of_B;
|
|
int NTT_blocks = 1;
|
|
while (NTT_blocks < (max_blocks << 1)) NTT_blocks <<= 1;
|
|
__int128_t *pDA = new __int128_t[NTT_blocks]();
|
|
__int128_t *pDB = new __int128_t[NTT_blocks]();
|
|
__int128_t *pDC = new __int128_t[NTT_blocks]();
|
|
for (int i = 0; i < blocks_of_A; i++) {
|
|
pDA[i << 1] = A.val[i] % int2048::kNTTBlcokBase;
|
|
pDA[(i << 1) | 1] = A.val[i] / int2048::kNTTBlcokBase;
|
|
}
|
|
for (int i = 0; i < blocks_of_B; i++) {
|
|
pDB[i << 1] = pB->val[i] % int2048::kNTTBlcokBase;
|
|
pDB[(i << 1) | 1] = pB->val[i] / int2048::kNTTBlcokBase;
|
|
}
|
|
A.NTTTransform(pDA, NTT_blocks);
|
|
A.NTTTransform(pDB, NTT_blocks);
|
|
for (int i = 0; i < NTT_blocks; i++)
|
|
pDC[i] = (pDA[i] * pDB[i]) % int2048::kNTTMod;
|
|
A.NTTTransform(pDC, NTT_blocks, true);
|
|
for (int i = 0; i < NTT_blocks - 1; i++) {
|
|
pDC[i + 1] += pDC[i] / int2048::kNTTBlcokBase;
|
|
pDC[i] %= int2048::kNTTBlcokBase;
|
|
}
|
|
if (pDC[NTT_blocks - 1] >= int2048::kNTTBlcokBase)
|
|
throw "UnsignedMultiply: NTT result overflow";
|
|
int flag_store = A.flag;
|
|
A.ClaimMem(NTT_blocks * 4);
|
|
memset(A.val, 0, A.buf_length * sizeof(int));
|
|
for (int i = 0; i < NTT_blocks / 2; i++) {
|
|
A.val[i] = pDC[(i << 1) | 1] * int2048::kNTTBlcokBase + pDC[i << 1];
|
|
}
|
|
A.num_length = NTT_blocks * 4;
|
|
const static int kPow10[9] = {1, 10, 100, 1000, 10000,
|
|
100000, 1000000, 10000000, 100000000};
|
|
while (A.val[(A.num_length - 1) / int2048::kNum] /
|
|
kPow10[(A.num_length - 1) % int2048::kNum] ==
|
|
0) {
|
|
A.num_length--;
|
|
if (A.num_length == 0) {
|
|
A.num_length = 1;
|
|
break;
|
|
}
|
|
}
|
|
delete[] pDA;
|
|
delete[] pDB;
|
|
delete[] pDC;
|
|
}
|
|
|
|
int2048 &int2048::Multiply(const int2048 &B) {
|
|
// 实现复合乘法逻辑
|
|
const int2048 *pB = &B;
|
|
if (this == &B) pB = new int2048(B);
|
|
if ((this->num_length == 1 && this->val[0] == 0) ||
|
|
(pB->num_length == 1 && pB->val[0] == 0)) {
|
|
*this = std::move(int2048(0));
|
|
if (pB != &B) delete pB;
|
|
return *this;
|
|
}
|
|
this->flag = this->flag * pB->flag;
|
|
UnsignedMultiply(*this, pB);
|
|
if (pB != &B) delete pB;
|
|
return *this;
|
|
}
|
|
|
|
int2048 Multiply(int2048 A, const int2048 &B) {
|
|
// 实现乘法逻辑
|
|
return std::move(A.Multiply(B));
|
|
}
|
|
|
|
int2048 &int2048::operator*=(const int2048 &B) {
|
|
// 实现复合乘法逻辑
|
|
return this->Multiply(B);
|
|
}
|
|
|
|
int2048 operator*(int2048 A, const int2048 &B) {
|
|
// 实现乘法逻辑
|
|
A.Multiply(B);
|
|
return std::move(A);
|
|
}
|
|
|
|
/**
|
|
* @brief Unsigned multiply a number by an integer
|
|
*
|
|
* @param v The integer
|
|
*
|
|
* @details This function will multiply a number by an integer without
|
|
* considering the sign.
|
|
*/
|
|
void int2048::UnsignedMultiplyByInt(int v) {
|
|
if (v < 0) throw "UnsignedMultiplyByInt: v < 0";
|
|
if (v == 0) {
|
|
*this = std::move(int2048(0));
|
|
return;
|
|
}
|
|
if (v == 1) return;
|
|
int blocks = (this->num_length + int2048::kNum - 1) / int2048::kNum;
|
|
this->ClaimMem(this->num_length + int2048::kNum * 2);
|
|
blocks += 2;
|
|
long long c = 0;
|
|
for (int i = 0; i < blocks; i++) {
|
|
long long tmp = (long long)this->val[i] * v + c;
|
|
c = tmp / int2048::kStoreBase;
|
|
this->val[i] = tmp % int2048::kStoreBase;
|
|
}
|
|
this->num_length = blocks * int2048::kNum;
|
|
const static int kPow10[9] = {1, 10, 100, 1000, 10000,
|
|
100000, 1000000, 10000000, 100000000};
|
|
while (this->num_length > 1 &&
|
|
this->val[(this->num_length - 1) / int2048::kNum] /
|
|
kPow10[(this->num_length - 1) % int2048::kNum] ==
|
|
0)
|
|
this->num_length--;
|
|
}
|
|
|
|
/**
|
|
* @brief Estimate the inverse of B, which is the result of [10^2n/B]
|
|
*/
|
|
int2048 GetInv(const int2048 &B, int n) {
|
|
const static int kPow10[9] = {1, 10, 100, 1000, 10000,
|
|
100000, 1000000, 10000000, 100000000};
|
|
int total_blocks = (B.num_length + int2048::kNum - 1) / int2048::kNum;
|
|
if (n <= 16) {
|
|
/**
|
|
* This code block is used to calculate the inverse of B when n is small.
|
|
*/
|
|
long long b = 0;
|
|
for (int j = B.num_length - 1, i = 0; i < n; i++, j--)
|
|
b = b * 10 + (B.val[j / int2048::kNum] / kPow10[j % int2048::kNum]) % 10;
|
|
int2048 res;
|
|
res.ClaimMem((2 * n) + 1);
|
|
res.val[2 * n / int2048::kNum] = kPow10[2 * n % int2048::kNum];
|
|
__uint128_t c = 0;
|
|
int tmp_B = (((2 * n) + 1) + int2048::kNum - 1) / int2048::kNum;
|
|
for (int i = tmp_B - 1; i >= 0; i--) {
|
|
c = c * int2048::kStoreBase + res.val[i];
|
|
res.val[i] = c / b;
|
|
c %= b;
|
|
assert(res.val[i] < int2048::kStoreBase);
|
|
}
|
|
res.num_length = (2 * n) + 1;
|
|
while (res.num_length > 1 &&
|
|
res.val[(res.num_length - 1) / int2048::kNum] /
|
|
kPow10[(res.num_length - 1) % int2048::kNum] ==
|
|
0)
|
|
res.num_length--;
|
|
return std::move(res);
|
|
}
|
|
int k = (n + 2) >> 1;
|
|
int2048 sub_soluton = std::move(GetInv(B, k));
|
|
int2048 sub_soluton_copy_1(sub_soluton);
|
|
int2048 sub_soluton_copy_2(sub_soluton);
|
|
sub_soluton_copy_1.UnsignedMultiplyByInt(2);
|
|
sub_soluton_copy_1.LeftMoveBy((n - k));
|
|
// Now sub_soluton_copy_1 is 2*sub_soluton*10^(n-k)
|
|
/*now we get the current B*/
|
|
int2048 current_B;
|
|
current_B.ClaimMem(n);
|
|
for (int i = B.num_length - 1, j = n - 1; j >= 0; j--, i--)
|
|
current_B.val[j / int2048::kNum] +=
|
|
(B.val[i / int2048::kNum] / kPow10[i % int2048::kNum] % 10) *
|
|
kPow10[j % int2048::kNum];
|
|
// current_B is the highest n numbers of B
|
|
current_B.num_length = n;
|
|
// std::cerr << "n=" << n << " while B=";
|
|
// int2048 new_B = B;
|
|
// new_B.print_debug();
|
|
// std::cerr << "current_B: ";
|
|
// current_B.print_debug();
|
|
|
|
UnsignedMultiply(sub_soluton_copy_2, ¤t_B);
|
|
UnsignedMultiply(sub_soluton_copy_2, &sub_soluton);
|
|
sub_soluton_copy_2.RightMoveBy(2 * k);
|
|
// Now sub_soluton_copy_2 is sub_soluton^2*current_B/(10^2k)
|
|
UnsignedMinus(sub_soluton_copy_1, &sub_soluton_copy_2);
|
|
int2048 res = sub_soluton_copy_1;
|
|
// Now we get an estimation for [10^2n/current_B]
|
|
/*Now begin doing some ajustments*/
|
|
int2048 remain;
|
|
remain.ClaimMem((2 * n) + 1);
|
|
remain.val[2 * n / int2048::kNum] = kPow10[2 * n % int2048::kNum];
|
|
remain.num_length = (2 * n) + 1;
|
|
UnsignedMultiply(sub_soluton_copy_1, ¤t_B);
|
|
int cmp = UnsignedCmp(remain, sub_soluton_copy_1);
|
|
if (cmp == 0) return std::move(res);
|
|
if (cmp > 0) {
|
|
UnsignedMinus(remain, &sub_soluton_copy_1);
|
|
for (int i = 64; i > 0; i >>= 1) {
|
|
int2048 tmp_B(current_B);
|
|
tmp_B.UnsignedMultiplyByInt(i);
|
|
if (UnsignedCmp(remain, tmp_B) >= 0) {
|
|
res += i;
|
|
UnsignedMinus(remain, &tmp_B);
|
|
}
|
|
}
|
|
return std::move(res);
|
|
} else {
|
|
UnsignedMinus(sub_soluton_copy_1, &remain);
|
|
for (int i = 64; i > 0; i >>= 1) {
|
|
int2048 tmp_B(current_B);
|
|
tmp_B.UnsignedMultiplyByInt(i);
|
|
if (UnsignedCmp(sub_soluton_copy_1, tmp_B) >= 0) {
|
|
res -= i;
|
|
UnsignedMinus(sub_soluton_copy_1, &tmp_B);
|
|
}
|
|
}
|
|
int2048 tmp_zero(0);
|
|
if (UnsignedCmp(sub_soluton_copy_1, tmp_zero) != 0) res -= 1;
|
|
return std::move(res);
|
|
}
|
|
}
|
|
|
|
/**
|
|
* @brief Unsigned divide two numbers
|
|
*
|
|
* @param A The first number
|
|
* @param pB The second number
|
|
*
|
|
* @details This function will divide two numbers without considering the sign.
|
|
* reference:
|
|
* <https://github.com/lzyrapx/Competitive-Programming-Docs/blob/master/WC%E8%AE%B2%E8%AF%BE%E8%B5%84%E6%96%99/%E7%90%86%E6%80%A7%E6%84%89%E6%82%A6%E2%80%94%E2%80%94%E9%AB%98%E7%B2%BE%E5%BA%A6%E6%95%B0%E5%80%BC%E8%AE%A1%E7%AE%97%EF%BC%882012WC%EF%BC%89.pdf>
|
|
*/
|
|
inline void UnsignedDivide(int2048 &A, const int2048 *pB) {
|
|
int2048 B(*pB);
|
|
int L1 = A.num_length;
|
|
int L2 = B.num_length;
|
|
if (L2 <= 16) {
|
|
long long b = B.val[0] + (long long)B.val[1] * int2048::kStoreBase;
|
|
__uint128_t c = 0;
|
|
int blocks_of_A = (A.num_length + int2048::kNum - 1) / int2048::kNum;
|
|
for (int i = blocks_of_A - 1; i >= 0; i--) {
|
|
c = c * int2048::kStoreBase + A.val[i];
|
|
A.val[i] = c / b;
|
|
c %= b;
|
|
assert(A.val[i] < int2048::kStoreBase);
|
|
}
|
|
A.num_length = blocks_of_A * int2048::kNum;
|
|
const static int kPow10[9] = {1, 10, 100, 1000, 10000,
|
|
100000, 1000000, 10000000, 100000000};
|
|
while (A.num_length > 1 &&
|
|
A.val[(A.num_length - 1) / int2048::kNum] /
|
|
kPow10[(A.num_length - 1) % int2048::kNum] ==
|
|
0)
|
|
A.num_length--;
|
|
return;
|
|
}
|
|
if (UnsignedCmp(A, *pB) < 0) {
|
|
A = std::move(int2048(0));
|
|
return;
|
|
}
|
|
if (2 * L2 < L1) {
|
|
int delta = L1 - 2 * L2;
|
|
A.LeftMoveBy(delta);
|
|
L1 += delta;
|
|
B.LeftMoveBy(delta);
|
|
L2 += delta;
|
|
}
|
|
assert(L1 == A.num_length);
|
|
assert(L2 == B.num_length);
|
|
assert(2 * L2 >= L1);
|
|
/*Now we calculate [10^2n/B]*/
|
|
int2048 res_hat = std::move(GetInv(B, L2));
|
|
UnsignedMultiply(res_hat, &A);
|
|
res_hat.RightMoveBy(2 * L2);
|
|
// now res_hat is [[10^2n/B]*A/(10^2L2)]
|
|
/*Now begin doing ajustments*/
|
|
int2048 remain(A), tmp_B(B);
|
|
UnsignedMultiply(tmp_B, &res_hat);
|
|
UnsignedMinus(remain, &tmp_B);
|
|
// now remain is A-res_hat*B
|
|
for (int i = 8; i > 0; i >>= 1) {
|
|
tmp_B = B;
|
|
tmp_B.UnsignedMultiplyByInt(i);
|
|
if (UnsignedCmp(remain, tmp_B) >= 0) {
|
|
res_hat += i;
|
|
UnsignedMinus(remain, &tmp_B);
|
|
}
|
|
}
|
|
A = std::move(res_hat);
|
|
}
|
|
int2048 &int2048::Divide(const int2048 &B) {
|
|
if (B.num_length == 1 && B.val[0] == 0) {
|
|
*this = std::move(int2048(0));
|
|
return *this;
|
|
// throw "Divide: divide by zero";
|
|
}
|
|
if (this == &B) {
|
|
*this = std::move(int2048(1));
|
|
return *this;
|
|
}
|
|
int2048 origin_A(*this);
|
|
int flag_store = this->flag * B.flag;
|
|
UnsignedDivide(*this, &B);
|
|
this->flag = flag_store;
|
|
if (this->flag == -1) {
|
|
if (origin_A != (*this) * B) {
|
|
*this -= 1;
|
|
}
|
|
}
|
|
if (this->num_length == 1 && this->val[0] == 0) this->flag = 1;
|
|
return *this;
|
|
}
|
|
int2048 Divide(int2048 A, const int2048 &B) {
|
|
A.Divide(B);
|
|
return std::move(A);
|
|
}
|
|
|
|
int2048 &int2048::operator/=(const int2048 &B) {
|
|
// 实现复合除法逻辑
|
|
return this->Divide(B);
|
|
}
|
|
|
|
int2048 operator/(int2048 A, const int2048 &B) {
|
|
// 实现除法逻辑
|
|
A.Divide(B);
|
|
return std::move(A);
|
|
}
|
|
|
|
int2048 &int2048::operator%=(const int2048 &B) {
|
|
// 实现复合取模逻辑
|
|
int2048 C(*this);
|
|
C.Divide(B);
|
|
this->minus(C.Multiply(B));
|
|
return *this;
|
|
}
|
|
|
|
int2048 operator%(int2048 A, const int2048 &B) {
|
|
// 实现取模逻辑
|
|
int2048 C(A);
|
|
C.Divide(B);
|
|
A.minus(C.Multiply(B));
|
|
return std::move(A);
|
|
}
|
|
|
|
std::istream &operator>>(std::istream &stream, int2048 &V) {
|
|
// 实现输入运算符逻辑
|
|
std::string v_str;
|
|
stream >> v_str;
|
|
V.read(v_str);
|
|
return stream;
|
|
}
|
|
|
|
std::ostream &operator<<(std::ostream &stream, const int2048 &v) {
|
|
// 实现输出运算符逻辑
|
|
char *buf = new char[v.num_length + 5];
|
|
char *p = buf;
|
|
if (v.flag == -1) *p++ = '-';
|
|
const static int kPow10[9] = {1, 10, 100, 1000, 10000,
|
|
100000, 1000000, 10000000, 100000000};
|
|
for (int i = v.num_length - 1; i >= 0; i--)
|
|
*p++ =
|
|
char('0' + v.val[i / int2048::kNum] / kPow10[i % int2048::kNum] % 10);
|
|
*p++ = 0;
|
|
stream << buf;
|
|
delete[] buf;
|
|
return stream;
|
|
}
|
|
|
|
bool operator==(const int2048 &A, const int2048 &B) {
|
|
// 实现等于运算符逻辑
|
|
if (A.flag != B.flag) return false;
|
|
return UnsignedCmp(A, B) == 0;
|
|
}
|
|
|
|
bool operator!=(const int2048 &A, const int2048 &B) {
|
|
// 实现不等于运算符逻辑
|
|
if (A.flag != B.flag) return true;
|
|
return UnsignedCmp(A, B) != 0;
|
|
}
|
|
|
|
bool operator<(const int2048 &A, const int2048 &B) {
|
|
// 实现小于运算符逻辑
|
|
if (A.flag != B.flag) return A.flag < B.flag;
|
|
int cmp = UnsignedCmp(A, B);
|
|
if (A.flag == 1)
|
|
return cmp < 0;
|
|
else
|
|
return cmp > 0;
|
|
}
|
|
|
|
bool operator>(const int2048 &A, const int2048 &B) {
|
|
// 实现大于运算符逻辑
|
|
if (A.flag != B.flag) return A.flag > B.flag;
|
|
int cmp = UnsignedCmp(A, B);
|
|
if (A.flag == 1)
|
|
return cmp > 0;
|
|
else
|
|
return cmp < 0;
|
|
}
|
|
|
|
bool operator<=(const int2048 &A, const int2048 &B) {
|
|
// 实现小于等于运算符逻辑
|
|
if (A.flag != B.flag) return A.flag < B.flag;
|
|
int cmp = UnsignedCmp(A, B);
|
|
if (A.flag == 1)
|
|
return cmp <= 0;
|
|
else
|
|
return cmp >= 0;
|
|
}
|
|
|
|
bool operator>=(const int2048 &A, const int2048 &B) {
|
|
// 实现大于等于运算符逻辑
|
|
if (A.flag != B.flag) return A.flag > B.flag;
|
|
int cmp = UnsignedCmp(A, B);
|
|
if (A.flag == 1)
|
|
return cmp >= 0;
|
|
else
|
|
return cmp <= 0;
|
|
}
|
|
} // namespace ZYM
|