C++中两个大十六进制数的mod [复制]

q0qdq0h2 于 2023-11-19 发布在其他

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Mod of two large numbers in C++（1个答案）
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我有两个大的十六进制数，每个至少有70位（十进制），我试图计算值hex_a%hex_b：

string hex_a="1133A5DCDEF3216A63EB879A82F5A1DC4490CCF6412492CF1B242DB";
string hex_b="AAB3A5DCDEF3216A6AAA2F5A1DC4490CCF6412492CF1B242DB";

字符串
我尝试用stoi将两个十六进制数转换为十进制数，并将值存储在字符串中：

string a=to_string(stoi(hex_a, 0, 16));

型
但是我在使用stoi方法时得到一个错误：

terminate called after throwing an instance of 'std::out_of_range'
  what():  stoi
Aborted (core dumped)

型
如何在C++中计算两个大数字的模？

c++

来源：https://stackoverflow.com/questions/77466422/mod-of-two-big-hex-numbers-in-c

2条答案

按热度按时间

blmhpbnm1#

您可以使用boost/multiprecision/cpp_int：

#include <boost/multiprecision/cpp_int.hpp>
#include <iostream>
#include <sstream>
#include <string>
using boost::multiprecision::cpp_int;
int main() {
  std::string hex_a = "0x1133A5DCDEF3216A63EB879A82F5A1DC4490CCF6412492CF1B242DB";
  std::string hex_b = "0xAAB3A5DCDEF3216A6AAA2F5A1DC4490CCF6412492CF1B242DB";
  cpp_int a(hex_a), b(hex_b);
  cpp_int result = a % b;
  std::cout << "Result as int: " << result << '\n';
  std::stringstream ss;
  ss << std::hex << result;
  std::string result_hex = ss.str();
  std::cout << "Result as hex: 0x" << result_hex << '\n';
  return 0;
}

字符串
或者GNU Multiple Precision Arithmetic Library (GMP)，这可能更快：

#include <gmp.h>
int main() {
  mpz_t a, b, result;
  mpz_init(a);
  mpz_init(b);
  mpz_init(result);
  mpz_set_str(a, "1133A5DCDEF3216A63EB879A82F5A1DC4490CCF6412492CF1B242DB", 16);
  mpz_set_str(b, "AAB3A5DCDEF3216A6AAA2F5A1DC4490CCF6412492CF1B242DB", 16);
  mpz_mod(result, a, b);
  gmp_printf("Result as int: %Zd\n", result);
  gmp_printf("Result as hex: Ox%Zx\n", result);
  mpz_clear(a);
  mpz_clear(b);
  mpz_clear(result);
  return 0;
}

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输出：

Result as int: 887778964514700662847592530576646032246936850244260028835776
Result as hex: 0x8d6e6cdeb792c9d42b257481c738989f678484c94fd6b567c0

型
在Python中确认相同的结果：

>>> hex_a = "0x1133A5DCDEF3216A63EB879A82F5A1DC4490CCF6412492CF1B242DB"
>>> hex_b = "0xAAB3A5DCDEF3216A6AAA2F5A1DC4490CCF6412492CF1B242DB"
>>> a = int(hex_a, 16)
>>> b = int(hex_b, 16)
>>> result = a % b
>>> print(f'Result as int: {result}')
Result as int: 887778964514700662847592530576646032246936850244260028835776
>>> print(f'Result as hex: 0x{result:x}')
Result as hex: 0x8d6e6cdeb792c9d42b257481c738989f678484c94fd6b567c0

型

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赞(0）回复(0）举报 2023-11-19

l2osamch2#

首先，你不能把这么大的数字转换成“int”、“long”、“long long”、“double”等。
第一种方法是在字符串上实现基本的数学运算（+，-，乘16或移位），然后自己实现除法算法。
第二种方法是使用和“任意精度算术库”：https://en.wikipedia.org/wiki/List_of_arbitrary-precision_arithmetic_software
例如https://www.boost.org/doc/libs/1_83_0/libs/multiprecision/doc/html/index.html

赞(0）回复(0）举报 2023-11-19

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C++中两个大十六进制数的mod [复制]

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