C：如何查找用户输入的一年中给定日期的日期名称

Question

用户输入一个月的一天和一个月，然后输出将给用户一年中的一天。例如：

-输入：日=24，月=2

-输出：day=55

所以我的问题是如何为一年中的某一天命名，例如星期三、星期一等？例如：

-输入：日=1，月=1

-输出：day=1，星期日

这是我的代码：

int countDays(int, int, int[]);

int main() {
  int daysInMonth[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
  int day, month;

  printf("enter day: ");
  scanf("%d", &day);

  printf("Enter month: ");
  scanf("%d", &month);

  printf("There are %d days \n", countDays(day, month, daysInMonth));
  return 0;
}

int countDays(int day, int month, int daysInMonth[]) {
  char *c[]={"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
  int charc[6];
  int k, j, a;
  int b = 0;
  int dayno[31][12];
  for (k = 0; k < 12; k++) {
    for (j = 0; j < daysInMonth[k]; j++) {

      dayno[j][k] = ++b;
    }
    a = dayno[day - 1][month - 1];
  }
  return a;
}

Answer 1

由于您目前只对 2017 年感兴趣，因此您可以通过 countDays() 模 7 轻松计算出星期几。您可以在 main() 中扩展您当前打印的内容：

printf("There are %d days \n", countDays(day, month, daysInMonth));

与：

int days = countDays(day, month, daysInMonth);
char *c[] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};

printf("There are %d days; day of the week: %s \n", days, c[(days - 1) % 7]);

Answer 2

如何查找用户输入的一年中给定日期的日期名称

使用mktime() 计算给定年、月、日的星期几。

#include <stdio.h>
#include <time.h>

const char *day_name(int year, int month, int day) {
  struct tm tm = { .tm_year = year - 1900, .tm_mon = month - 1, .tm_mday = day,
      .tm_isdst = -1 };
  time_t t = mktime(&tm);
  if (t == -1) { 
    return "?";
  }
  static const char *dow[] = { "Sunday", "Monday", "Tuesday", "Wednesday",
      "Thursday", "Friday", "Saturday" };
  return dow[tm.tm_wday];
}

int main(void) {
  puts(day_name(2017, 1, 1));
  puts(day_name(2017, 2, 24));
  puts(day_name(2017, 11, 14));
}

输出

Sunday
Friday
Tuesday

---

对于 OP 的代码，使用 countDays(month, day) 的变体

#define Jan1_2017DOW (0 /* Sunday */) 
#define Jan1Number 1
int dow = (countDays(month, day) - Jan1Number + Jan1_2017DOW) % 7;
puts(c[dow]);

Answer 3

这是一个简单的解决方案:)

void countDays(int day, int month, int daysInMonth[])
{
    char c[][15] =  {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday" };
    int totaldays=0;
    for(int i=0;i<month-1;i++)
    {
        totaldays+= daysInMonth[i];
    }

    totaldays += day;

    printf("There are %d days \n", totaldays);

    printf("\n day : %s",c[(totaldays-1)%7]);

    return;
}

在用你自己的方法对你的代码做一些改动之后:)

Answer 4

#include <stdio.h>

char *WEEK[] =  {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday" };
int DAYS_IN_MONTH[] =  {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int countDays(int day, int month);
char *dayName(int day);

int main(){
    int day, month;

    printf("enter day: ");
    scanf("%d", &day);

    printf("Enter month: ");
    scanf("%d", &month);

    int days = countDays(day, month);
    if(days != -1){
        printf("There are %i days \n", days);
        printf("The day name is %s\n", dayName(days));
    }else{
        printf("Error\n");
    }

    return 0;
}

int countDays(int day, int month){
    if(day >DAYS_IN_MONTH[month-1])
        return -1;

    int num_days = 0;
    for(int i = 0;i < month-1;i++){
        num_days+=DAYS_IN_MONTH[i];
    }
    return num_days+day;

}

char *dayName(int days){
    return WEEK[(days-1) % 7];
}

Answer 5

取自 Aho、Sethi 和 Ullman 的“编译器：原理、技术和工具”。

day = (1461 * year) / 4 + (153 * month + 2) / 5 + month_day;

（注意/在这里表示整数除法，所有数字都必须是整数）这个公式归因于那里的Tanzen[1963]。如果您将% 7 应用于该公式，则在应用适当的偏移量后，您将获得星期几。