ex6_3.c

自己做的常用库和实现的数据结构。public domain.

/*
  Ex6_3
  Calculate the number of days between two dates and the weekday of the two
  days.
*/
struct date_type
{
    int year, month, day;
};
char *week[7] = {"SUN", "MON", "TUE", "WED", "THU", "FRI", "SAT"};

/* function to calculate the N value
*/
long n_value(date)
struct date_type date;
{
    long n, f, g;

    if (date.month <= 2)
    {
	f = date.year - 1;
	g = date.month + 13;
    } else
    {
	f = date.year;
	g = date.month + 1;
    }
    n = 1461 * f / 4 + 153 * g / 5 + date.day;
    if (date.year < 1700 || date.year == 1700 && date.month < 3)
	printf("THE DATE IS TOO LATE.\n");
    else if (date.year < 1800 || date.year == 1800 && date.month < 3)
	n += 2;
    else if (date.year < 1900 || date.year == 1900 && date.month < 3)
	n ++;
    return(n);
}   /* N_VALUE */

/* main function to calculate the number of days between two dates
*/
main()
{
    struct date_type date1, date2;
    long n1, n2;
    int d1,d2;

    printf("Please input the first date! (dd/mm/yyyy)\n");
    scanf("%d/%d/%d", &date1.day, &date1.month, &date1.year);
    printf("Please input the second date! (dd/mm/yyyy)\n");
    scanf("%d/%d/%d", &date2.day, &date2.month, &date2.year);

    n1 = n_value(date1);
    n2 = n_value(date2);

    printf("THE NUMBER OF DAYS BETWEEN TWO DATES IS %-d.\n",n1>n2?n1-n2:n2-n1);

    d1 = (int)((n1-621049)%7);
    d2 = (int)((n2-621049)%7);

    printf("THE DAY OF THE DATE1 IS %s.\n", week[d1]);
    printf("THE DAY OF THE DATE2 IS %s.\n", week[d2]);
}   /* MAIN */