24. Conversion Formula for solar term and calendar date of any year

There is a certain error in the formula, but it will be within several minutes between and...
Many calendar scripts use the hour () as the start of a day, so that a day will be added after...

 

# Include <stdio. h> <br/> # include <stdlib. h> <br/> static const double x_1900_1_6_2_5 = 693966.08680556; <br/> double get_solar_term (INT y, int N) <br/>{< br/> static const int terminfo [] ={< br/> 0, 21208,424 67, 63836,853 37, 107014, <br/> 128867,150921, 173149,195551, 218072,240693, <br/> 263343,285989, 308563,331033, 353350,375494, <br/> average, minimum 795,462224, 482.16,504758 <br/>}; <br/> RET Urn x_1900_1_6_2_5 + 365.2422 * (y-1900) + terminfo [N]/(60. * 24); <br/>}< br/> int format_date (unsigned _ days, char * result); <br/> int main (INT argc, char * argv []) <br/>{< br/> static const char * solar_term_name [] ={< br/> "Xiao Han", "da Han", "Li Chun ", "Rain", <br/> "stunned", "Spring Equinox", "Qingming", "Gu Yu", <br/> "lixia", "xiaoman", "Mango ", "Summer Solstice", <br/> "Summer heat", "Summer heat", "Autumn Heat", "Summer heat", <br/> "White Dew", "Autumn Equinox", "Cold Dew ", "Frost down", <br/> "Winter", "light snow", "heavy snow", "Winter Solstice" <br/>}; <br/> char Str _ D [2, 100]; <br/> int year = 2008, I; <br/> If (argc = 2) <br/>{< br/> year = atoi (argv [1]); <br/> If (year <1900 | year> 2099) <br/> year = 2008; <br/>}< br/> for (I = 0; I <24; ++ I) <br/>{< br/> format_date (unsigned) get_solar_term (year, I), str_d); <br/> printf ("% s: % s/n ", solar_term_name [I], str_d); <br/>}</P> <p> return 0; <br/>}< br/> int format_date (unsigned _ days, Cha R * result) <br/>{< br/> static const int mdays [] = {0, 31, 59, 90,120,151,181,212,243,273,304,334,365}; <br/> int y, m, d, diff; <br/> unsigned days; <br/> days = 100 * (_ days-_ days/(3652425l/(3652425l-3652400l ))); <br/> Y = days/36524; days % = 36524; <br/> M = 1 + days/3044;/* [1 .. 12] */<br/> d = 1 + (days % 3044)/100;/* [1 .. 31] */<br/> diff = y * 365 + Y/4-y/100 + Y/400 + mdays [M-1] + d -(M <= 2 & (Y & 3) = 0) & (Y % 100 )! = 0 | Y % 400 = 0)-_ days; <br/> If (diff> 0 & diff> = d )/*~ 0.5% */<br/>{< br/> If (M = 1) <br/>{< br/> -- y; M = 12; <br/> d = 31-(diff-D ); <br/>}< br/> else <br/> {<br/> d = mdays [M-1]-(diff-D ); <br/> If (-- M = 2) <br/> D + = (Y & 3) = 0) & (Y % 100 )! = 0 | Y % 400 = 0 ); <br/>}< br/> else <br/> {<br/> If (D-= diff)> mdays [m]) /*~ 1.6% */<br/> {<br/> If (M = 2) <br/> {<br/> If (Y & 3) = 0) & (Y % 100 )! = 0 | Y % 400 = 0) <br/>{< br/> If (D! = 29) <br/> M = 3, D-= 29; <br/>}< br/> else <br/>{< br/> M = 3, d-= 28; <br/>}< br/> else <br/>{< br/> D-= mdays [m]; <br/> If (M ++ = 12) <br/> ++ y, M = 1; <br/>}< br/> return sprintf (result, "% 04d-% 02d-% 02d", Y, m, d); <br/>}< br/>