编程求解，如内容所示题

Question

每月工作日只考虑周末情况，即只有周六、周日放假。每月第一个工作日如果是星期一则该月是A类月，每月最后一个工作日如果是星期五则该月是B类月。一个月可能是A类月也可能是B类月。
如2006 10 2是该月第一个工作日为星期一是A类月，如2006 9 29是该月最后一个工作日为星期五则是B类月。
输入一个时间段如：2006 6 2006 6（闭区间），求出该时间段内的A类月和B类月数量。python或c/c++。

Answer 1

#include <iostream>
#include <stdexcept>
#include <ctime>

class YearMonth
{
public:
    YearMonth() {}
    YearMonth(int year, int month) : _year(year), _month(month)
    {
        if (year < 0 || month < 1 || month > 12)
            throw std::invalid_argument("invalid year or month");
    }

    int year() const { return _year; }
    int month() const { return _month; }

    bool isLeapYear() const
    {
        return ((_year % 4 == 0) && (_year % 100 != 0)) || (_year % 400 == 0);
    }
    bool isTypeA() const;
    bool isTypeB() const;

    YearMonth &operator++()
    {
        if (++_month > 12)
        {
            _month -= 12;
            ++_year;
        }
        return *this;
    }

    bool operator<=(const YearMonth &other) const
    {
        return _year < other._year || (_year == other._year && _month <= other._month);
    }

private:
    int _year = 0;
    int _month = 1;
};

bool YearMonth::isTypeA() const
{
    std::tm tm = {.tm_mday = 1, .tm_mon = _month - 1, .tm_year = _year - 1900};
    std::time_t time = std::mktime(&tm);
    const std::tm *t = std::localtime(&time);
    return t->tm_wday == 0 || t->tm_wday == 1 || t->tm_wday == 6;
}

bool YearMonth::isTypeB() const
{
    static int days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int lastDay = days[_month - 1];
    if (isLeapYear() && _month == 2)
        lastDay++;
    std::tm tm = {.tm_mday = lastDay, .tm_mon = _month - 1, .tm_year = _year - 1900};
    std::time_t time = std::mktime(&tm);
    const std::tm *t = std::localtime(&time);
    return t->tm_wday == 0 || t->tm_wday == 5 || t->tm_wday == 6;
}

std::istream &operator>>(std::istream &is, YearMonth &yearMonth)
{
    int year, month;
    is >> year >> month;
    yearMonth = YearMonth(year, month);
    return is;
}

int main()
{
    YearMonth start, end;
    std::cin >> start >> end;
    int typeA = 0, typeB = 0;
    while (start <= end)
    {
        if (start.isTypeA())
            typeA++;
        if (start.isTypeB())
            typeB++;
        ++start;
    }
    std::cout << "Type A: " << typeA << '\n'
              << "Type B: " << typeB << '\n';
    return 0;
}

$ g++ -Wall main.cpp
$ ./a.out
2006 6 2006 6
Type A: 0
Type B: 1
$ ./a.out
2006 1 2006 6
Type A: 3
Type B: 3