C语言。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，以下是代码实现：

#include <stdio.h>

// 计算该年中的总天数
unsigned int Dayofyear(unsigned int year, unsigned int month, unsigned int day) {
    unsigned int dayTab[2][13] = {
        {0,31,59,90,120,151,181,212,243,273,304,334,365},
        {0,31,60,91,121,152,182,213,244,274,305,335,366}
    };
    int leap = (year%4==0 && year%100!=0) || year%400==0;
    return dayTab[leap][month-1] + day;
}

// 将总天数转换为月份和日期
void MonthandDay(unsigned int year, unsigned int totolday, unsigned int *month, unsigned int *day) {
    unsigned int dayTab[2][13] = {
        {0,31,59,90,120,151,181,212,243,273,304,334,365},
        {0,31,60,91,121,152,182,213,244,274,305,335,366}
    };
    int leap = (year%4==0 && year%100!=0) || year%400==0;
    int i;
    for (i = 1; i <= 12; i++) {
        if (totolday <= dayTab[leap][i]) {
            *month = i;
            *day = totolday - dayTab[leap][i-1];
            break;
        }
    }
}

int main() {
    unsigned int year, month, day, yearday, option;
    unsigned int totolday, ret_month, ret_day;
    
    while (1) {
        printf("1.year/month/day->yearday\n");
        printf("2.year/yearday->year/month/day\n");
        printf("3.Exit\n");
        printf("Please enter option: ");
        scanf("%u", &option);
        
        if (option == 1) {
            printf("Please enter year, month, day (separated by spaces): ");
            scanf("%u %u %u", &year, &month, &day);
            yearday = Dayofyear(year, month, day);
            printf("%u/%u/%u is the %u day of %u\n", year, month, day, yearday, year);
        } else if (option == 2) {
            printf("Please enter year, yearday: ");
            scanf("%u %u", &year, &totolday);
            MonthandDay(year, totolday, &ret_month, &ret_day);
            printf("%u year %u day is %u/%u/%u\n", year, totolday, year, ret_month, ret_day);
        } else if (option == 3) {
            break;
        } else {
            printf("Invalid option!\n");
        }
    }
    return 0;
}

这个程序实现了一个菜单，让用户选择日期转换的功能。当用户选择 1 时，程序会要求用户输入年、月、日，然后计算出该日期是该年的第几天。当用户选择 2 时，程序会要求用户输入年和该年的第几天，然后计算出对应的月份和日期。当用户选择 3 时，程序会退出。

注意，为了方便起见，这里假设所有输入都是正确的，没有做输入检查。如果需要更加健壮的程序，需要加上输入检查的代码。