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📄 time_from_mjd.c

📁 最大似然估计算法
💻 C
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#include <stdio.h>#include "utc_time.h"utc_time time_from_mjd(mjd)double mjd;{	int j, leap;	int year, month, day, hour, minute;	double second;	utc_time Time;	Time.mjd = mjd;	Time.julian = Time.mjd + 2400000.5;	Time.sec85 = (Time.julian - gregorian_to_julian(1985,1,1,0,0,0))*86400.0;	julian_to_gregorian(&year,&month,&day,&hour,&minute,&second,Time.julian);	Time.year = year;	Time.month = month;	Time.day = day;	Time.hour = hour;	Time.minute = minute;	Time.second = second;	Time.decimal_day = (double)hour / 24.0 + (double)minute / 1440.0 + second / 86400.0;	if (Time.decimal_day >= 0.5) strcpy(Time.ampm,"PM");	else strcpy(Time.ampm,"AM");	if (Time.year >= 0) strcpy(Time.adbc,"AD");	else strcpy(Time.adbc,"BC");	strcpy(Time.monthname,months[month-1]);	strcpy(Time.smonth,short_months[month-1]);	strcpy(Time.weekday,days[(int)(Time.julian+1.5)%7]);	j = (int)(Time.julian - 2444244.5);	Time.gweek[0] = j/7;	Time.gweek[1] = j%7;	leap = year%4 == 0 && year%100 != 0 || year%400 == 0;	Time.doy = mth[leap][month-1] + day;	Time.date = (double)Time.year + (double)(Time.doy - 1) / (365.0 + (double)leap) +		Time.decimal_day / (365.0 + (double)leap);	Time.j2000 = (Time.julian - 2451545.0) / 36525.0;	return Time;				}

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