1.编写程序，解决两个方向日期转换问题，实现要求如下： (1)屏幕菜单提示信息如下： 1.year/month/day->yearday 2. year/yearday->year/month/day 3.Exit 根据以上输入1.2.3选项分别完成不同功能，如果输入错误，应提示相应的出错信息。 (2)实现unsigned int Dayofyear(unsigned int year, unsigned int month, unsigned int day)函数，完成将年、月、日信息转换为该年中的总天数，即参数为年月日，返回值为天数。 (3)实现 void MonthandDay (unsigned int year, unsigned int totolday, unsigned int * month, unsigned int * day）函数，完成将年、总天数转换为对应的月、日，月份和日期信息通过指针返回。

好的，这是一个比较复杂的问题，需要进行程序设计和编写。我会逐步为您解答。

首先，我们需要设计一个菜单，让用户选择需要进行的操作。可以使用switch语句来实现。以下是一个基本的框架：

#include <iostream>
using namespace std;

unsigned int Dayofyear(unsigned int year, unsigned int month, unsigned int day);
void MonthandDay(unsigned int year, unsigned int totolday, unsigned int* month, unsigned int* day);

int main()
{
    int choice = 0;
    while (choice != 3)
    {
        cout << "Please choose an option:" << endl;
        cout << "1. year/month/day->yearday" << endl;
        cout << "2. year/yearday->year/month/day" << endl;
        cout << "3. Exit" << endl;
        cin >> choice;

        switch (choice)
        {
            case 1:
                // year/month/day->yearday
                break;
            case 2:
                // year/yearday->year/month/day
                break;
            case 3:
                // Exit
                break;
            default:
                cout << "Invalid choice. Please try again." << endl;
                break;
        }
    }

    return 0;
}

接下来，我们需要实现Dayofyear函数和MonthandDay函数。以下是Dayofyear函数的代码实现：

unsigned int Dayofyear(unsigned int year, unsigned int month, unsigned int day)
{
    bool isLeapYear = false;
    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
    {
        isLeapYear = true;
    }

    unsigned int daysOfMonth[] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    if (isLeapYear)
    {
        daysOfMonth[1] = 29;
    }

    unsigned int dayOfYear = 0;
    for (int i = 0; i < month - 1; i++)
    {
        dayOfYear += daysOfMonth[i];
    }
    dayOfYear += day;

    return dayOfYear;
}

Dayofyear函数接收年、月、日三个参数，返回该年中的总天数。首先判断是否为闰年，然后根据每个月的天数计算总天数。

接下来是MonthandDay函数的代码实现：

void MonthandDay(unsigned int year, unsigned int totolday, unsigned int* month, unsigned int* day)
{
    bool isLeapYear = false;
    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
    {
        isLeapYear = true;
    }

    unsigned int daysOfMonth[] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    if (isLeapYear)
    {
        daysOfMonth[1] = 29;
    }

    unsigned int i = 0;
    while (totolday > daysOfMonth[i])
    {
        totolday -= daysOfMonth[i];
        i++;
    }

    *month = i + 1;
    *day = totolday;
}

MonthandDay函数接收年、总天数两个参数，返回该年中对应的月和日。首先判断是否为闰年，然后根据每个月的天数计算月和日。

最后，我们需要在菜单中调用Dayofyear函数和MonthandDay函数，根据用户选择的不同选项执行不同的操作。以下是完整的代码实现：

#include <iostream>
using namespace std;

unsigned int Dayofyear(unsigned int year, unsigned int month, unsigned int day);
void MonthandDay(unsigned int year, unsigned int totolday, unsigned int* month, unsigned int* day);

int main()
{
    int choice = 0;
    while (choice != 3)
    {
        cout << "Please choose an option:" << endl;
        cout << "1. year/month/day->yearday" << endl;
        cout << "2. year/yearday->year/month/day" << endl;
        cout << "3. Exit" << endl;
        cin >> choice;

        switch (choice)
        {
            case 1:
            {
                unsigned int year, month, day;
                cout << "Please enter year/month/day:" << endl;
                cin >> year >> month >> day;

                unsigned int dayOfYear = Dayofyear(year, month, day);
                cout << "Day of year: " << dayOfYear << endl;
                break;
            }
            case 2:
            {
                unsigned int year, dayOfYear;
                cout << "Please enter year and day of year:" << endl;
                cin >> year >> dayOfYear;

                unsigned int month, day;
                MonthandDay(year, dayOfYear, &month, &day);
                cout << "Month: " << month << ", Day: " << day << endl;
                break;
            }
            case 3:
                cout << "Goodbye!" << endl;
                break;
            default:
                cout << "Invalid choice. Please try again." << endl;
                break;
        }
    }

    return 0;
}

unsigned int Dayofyear(unsigned int year, unsigned int month, unsigned int day)
{
    bool isLeapYear = false;
    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
    {
        isLeapYear = true;
    }

    unsigned int daysOfMonth[] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    if (isLeapYear)
    {
        daysOfMonth[1] = 29;
    }

    unsigned int dayOfYear = 0;
    for (int i = 0; i < month - 1; i++)
    {
        dayOfYear += daysOfMonth[i];
    }
    dayOfYear += day;

    return dayOfYear;
}

void MonthandDay(unsigned int year, unsigned int totolday, unsigned int* month, unsigned int* day)
{
    bool isLeapYear = false;
    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
    {
        isLeapYear = true;
    }

    unsigned int daysOfMonth[] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    if (isLeapYear)
    {
        daysOfMonth[1] = 29;
    }

    unsigned int i = 0;
    while (totolday > daysOfMonth[i])
    {
        totolday -= daysOfMonth[i];
        i++;
    }

    *month = i + 1;
    *day = totolday;
}

希望这个问题的解答对您有所帮助！