如何使用C++设计一种数据类型， 名称为int20。 仅限于3.12号完成

Question

使用C++设计一种数据类型， 名称为int20。名思义它是一个20（16+4）位的int整形，要求使用类来设计，位于std命名空间，重载加减乘除取余输入输出等基本运算，使用方法与int无异。
要求提交一个头文件以及源文件，头文件中包括类的声明以及功能注释，源文件中包含方法实现等。我们的测试将如下，新建一个main.cpp文件写入以下内容，将提交的头文件与源文件同时放入同级目录之后运行。

```c++
#include "int20.h"
int main() {
    int20 first, second;
    std::cin >> first >> second;
    long long sum1 = 0 ,sum2 = 0;
    for (long long loop = 0; loop < 1048560; loop += 1)
    {
        first++;
        second--;
        sum1 += first;
        sum2 -= second;
    }
    std::cout << first + second << sum1 << sum2 << std::endl;
    return 0;
}

```

Answer 1

参考GPT和自己的思路，以下是int20的头文件和源文件实现：

头文件：int20.h：

#ifndef INT20_H
#define INT20_H

#include <iostream>

namespace std {

    class int20 {
        friend std::ostream& operator<<(std::ostream& os, const int20& num);
        friend std::istream& operator>>(std::istream& is, int20& num);

    public:
        int20();
        int20(int num);

        int20 operator+(const int20& other) const;
        int20 operator-(const int20& other) const;
        int20 operator*(const int20& other) const;
        int20 operator/(const int20& other) const;
        int20 operator%(const int20& other) const;

        int20& operator++();
        int20 operator++(int);
        int20& operator--();
        int20 operator--(int);

    private:
        int digits[20];
        bool is_negative;
    };

}

#endif // INT20_H

源文件：int20.cpp：

#include "int20.h"

namespace std {

    int20::int20() {
        for (int i = 0; i < 20; i++) {
            digits[i] = 0;
        }
        is_negative = false;
    }

    int20::int20(int num) {
        if (num < 0) {
            is_negative = true;
            num = -num;
        } else {
            is_negative = false;
        }
        for (int i = 0; i < 20; i++) {
            digits[i] = num % 10;
            num /= 10;
        }
    }

    int20 int20::operator+(const int20& other) const {
        int20 result;
        int carry = 0;
        for (int i = 0; i < 20; i++) {
            int sum = digits[i] + other.digits[i] + carry;
            result.digits[i] = sum % 10;
            carry = sum / 10;
        }
        result.is_negative = (is_negative && other.is_negative);
        return result;
    }

    int20 int20::operator-(const int20& other) const {
        int20 result;
        int borrow = 0;
        for (int i = 0; i < 20; i++) {
            int diff = digits[i] - other.digits[i] - borrow;
            if (diff < 0) {
                diff += 10;
                borrow = 1;
            } else {
                borrow = 0;
            }
            result.digits[i] = diff;
        }
        if (is_negative && other.is_negative) {
            result.is_negative = !result.is_negative;
        } else if (is_negative && !other.is_negative) {
            result.is_negative = true;
        } else if (!is_negative && other.is_negative) {
            result.is_negative = false;
        } else {
            int cmp = (*this) < other;
            if (cmp == 0) {
                result.is_negative = false;
            } else if (cmp < 0) {
                result.is_negative = true;
            } else {
                result.is_negative = false;
            }
        }
        return result;
    }

    int20 int20::operator*(const int20& other) const {
        int20 result;
        for (int i = 0; i < 20; i++) {
            for (int j = 0; j < 20; j++) {
                int k = i + j;
                int product = digits[i] * other.digits[j];
            result.digits[k] += product % 10;
            result.digits[k+1] += product / 10;
        }
    }
    for (int i = 0; i < 19; i++) {
        result.digits[i+1] += result.digits[i] / 10;
        result.digits[i] %= 10;
    }
    result.is_negative = (is_negative != other.is_negative);
    return result;
}

int20 int20::operator/(const int20& other) const {
    int20 result;
    int20 dividend(*this);
    int20 divisor(other);
    dividend.is_negative = false;
    divisor.is_negative = false;
    if (dividend < divisor) {
        return result;
    }
    int20 quotient;
    for (int i = 19; i >= 0; i--) {
        quotient = quotient * 10;
        int digit = 0;
        while (dividend >= divisor) {
            dividend = dividend - divisor;
            digit++;
        }
        quotient = quotient + digit;
        divisor = divisor / 10;
    }
    result = quotient;
    result.is_negative = (is_negative != other.is_negative);
    return result;
}

int20 int20::operator%(const int20& other) const {
    int20 dividend(*this);
    int20 divisor(other);
    dividend.is_negative = false;
    divisor.is_negative = false;
    if (dividend < divisor) {
        return dividend;
    }
    for (int i = 19; i >= 0; i--) {
        while (dividend >= divisor) {
            dividend = dividend - divisor;
        }
        divisor = divisor / 10;
    }
    dividend.is_negative = is_negative;
    return dividend;
}

int20& int20::operator++() {
    *this = *this + 1;
    return *this;
}

int20 int20::operator++(int) {
    int20 old_value(*this);
    *this = *this + 1;
    return old_value;
}

int20& int20::operator--() {
    *this = *this - 1;
    return *this;
}

int20 int20::operator--(int) {
    int20 old_value(*this);
    *this = *this - 1;
    return old_value;
}

std::ostream& operator<<(std::ostream& os, const int20& num) {
    if (num.is_negative) {
        os << '-';
    }
    bool leading_zero = true;
    for (int i = 19; i >= 0; i--) {
        if (num.digits[i] != 0) {
            leading_zero = false;
        }
        if (!leading_zero) {
            os << num.digits[i];
        }
    }
    if (leading_zero) {
        os << '0';
    }
    return os;
}

std::istream& operator>>(std::istream& is, int20& num) {
    std::string input;
    is >> input;
    int len = input.length();
    num = int20();
    int pos = 0;
    if (input[pos] == '-') {
        num.is_negative = true;
        pos++;
    }
    while (pos < len) {
        num = num * 10 + (input[pos] - '0');
        pos++;
    }
    return is;
}

然后，在main.cpp中，我们可以这样使用int20类：

#include "int20.h"

int main
() {
int20 first, second;
std::cin >> first >> second;
long long sum1 = 0 ,sum2 = 0;
for (long long loop = 0; loop < 1048560; loop += 1)
{
first++;
second--;
sum1 += first;
sum2 -= second;
}
std::cout << first + second << ' ' << sum1 << ' ' << sum2 << std::endl;
return 0;
}

这个测试程序会从标准输入中读取两个int20类型的整数，然后对它们进行1048560次加减运算，并统计结果的和。最后输出两个数之和，sum1和sum2的值。
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