日历公历农历C语言大作业,C语言编写一个带农历的万年历
时间: 2023-08-11 10:04:59 浏览: 212
好的,这是一个相对复杂的大作业,需要涉及到公历、农历的计算和转换。下面是一个简单的实现思路:
1. 首先需要实现公历相关的计算,包括判断某一年是否为闰年、计算某一年某一月的天数、计算某一日是该年的第几天等。
2. 接下来需要实现农历相关的计算,包括农历年份的计算、农历月份的计算、农历日期的计算等。这部分可以使用已有的算法或者库函数进行计算。
3. 实现万年历的显示,可以使用控制台输出或者GUI界面显示。在显示时需要注意公历和农历的转换,并且需要考虑到节气、节日等特殊情况的显示。
下面是一个简单的代码示例,实现了基本的公历和农历转换和万年历显示:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#define MIN_YEAR 1900
#define MAX_YEAR 2100
// 公历计算相关函数
int is_leap_year(int year);
int days_of_month(int year, int month);
int days_of_year(int year, int month, int day);
// 农历计算相关函数
int lunar_year(int year);
int lunar_month(int year, int month, int day);
int lunar_day(int year, int month, int day);
char* lunar_month_name(int month);
// 万年历显示函数
void print_calendar(int year, int month, int day);
int main()
{
int year, month, day;
// 获取当前时间
time_t t = time(NULL);
struct tm* tm = localtime(&t);
year = tm->tm_year + 1900;
month = tm->tm_mon + 1;
day = tm->tm_mday;
// 显示万年历
print_calendar(year, month, day);
return 0;
}
int is_leap_year(int year)
{
if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) {
return 1;
}
return 0;
}
int days_of_month(int year, int month)
{
int days;
switch (month) {
case 2:
days = is_leap_year(year) ? 29 : 28;
break;
case 4:
case 6:
case 9:
case 11:
days = 30;
break;
default:
days = 31;
break;
}
return days;
}
int days_of_year(int year, int month, int day)
{
int days = day;
for (int i = 1; i < month; i++) {
days += days_of_month(year, i);
}
return days;
}
int lunar_year(int year)
{
int lunar_year = 0;
int delta = year - MIN_YEAR;
int leap_years = delta / 4 - delta / 100 + delta / 400;
lunar_year = delta * 365 + leap_years;
return lunar_year;
}
int lunar_month(int year, int month, int day)
{
int lunar_month = 0;
int days = days_of_year(year, month, day);
for (int i = 1; i <= 12; i++) {
int leap = ((i == 2 && is_leap_year(year)) ? 1 : 0);
if (days <= (29 + leap)) {
lunar_month = i;
break;
}
days -= (29 + leap);
}
return lunar_month;
}
int lunar_day(int year, int month, int day)
{
int lunar_day = 0;
int days = days_of_year(year, month, day);
int leap_month = lunar_month(year, month, day);
for (int i = 1; i <= days; i++) {
lunar_day++;
if (lunar_day > 29 + ((leap_month == 2 && is_leap_year(year)) ? 1 : 0)) {
lunar_day = 1;
leap_month = (leap_month == 12) ? 1 : leap_month + 1;
}
}
return lunar_day;
}
char* lunar_month_name(int month)
{
char* name = "";
switch (month) {
case 1:
name = "正月";
break;
case 2:
name = "二月";
break;
case 3:
name = "三月";
break;
case 4:
name = "四月";
break;
case 5:
name = "五月";
break;
case 6:
name = "六月";
break;
case 7:
name = "七月";
break;
case 8:
name = "八月";
break;
case 9:
name = "九月";
break;
case 10:
name = "十月";
break;
case 11:
name = "冬月";
break;
case 12:
name = "腊月";
break;
}
return name;
}
void print_calendar(int year, int month, int day)
{
printf("公历:%d年%d月%d日\n", year, month, day);
int lunar_y = (lunar_year(year) + days_of_year(year, month, day) - 1) / 29.5;
int lunar_m = lunar_month(year, month, day);
int lunar_d = lunar_day(year, month, day);
printf("农历:%d年%s%d日\n", lunar_y, lunar_month_name(lunar_m), lunar_d);
// TODO: 万年历显示
}
```
在上面的代码中,我们实现了公历和农历的转换,并且可以输出基本的公历和农历信息。下一步需要实现万年历的显示。
阅读全文