日出日落时间计算程序(C语言)

1、/日出日落时间计算C语言程序#define PI 3.1415926 #include<math.h> #include<iostream> using namespace std; int days_of_month_1=31,28,31,30,31,30,31,31,30,31,30,31; int days_of_month_2=31,29,31,30,31,30,31,31,30,31,30,31; long double h=-0.833; /定义全局变量 void input_date(int c) int i; cout<<"Ente

2、r the date (form: 2009 03 10):"<<endl; for(i=0;i<3;i+) cin>>ci; /输入日期 void input_glat(int c) int i; cout<<"Enter the degree of latitude(range: 0°- 60°,form: 40 40 40 (means 40°4040):"<<endl; for(i=0;i<3;i+) cin>>ci; /输入纬度 void input_g

3、long(int c) int i; cout<<"Enter the degree of longitude(west is negativ,form: 40 40 40 (means 40°4040):"<<endl; for(i=0;i<3;i+) cin>>ci; /输入经度 int leap_year(int year) if(year%400=0) | (year%100!=0) && (year%4=0) return 1; else return 0; /判断是否为闰年:若为闰年,返回1；若

4、非闰年,返回0 int days(int year, int month, int date) int i,a=0; for(i=2000;i<year;i+) if(leap_year(i) a=a+366; else a=a+365; if(leap_year(year) for(i=0;i<month-1;i+) a=a+days_of_month_2i; else for(i=0;i<month-1;i+) a=a+days_of_month_1i; a=a+date; return a; /求从格林威治时间公元2000年1月1日到计算日天数days long dou

5、ble t_century(int days, long double UTo) return (long double)days+UTo/360)/36525; /求格林威治时间公元2000年1月1日到计算日的世纪数t long double L_sun(long double t_century) return (280.460+36000.770*t_century); /求太阳的平黄径 long double G_sun(long double t_century) return (357.528+35999.050*t_century); /求太阳的平近点角 long double

6、ecliptic_longitude(long double L_sun,long double G_sun) return (L_sun+1.915*sin(G_sun*PI/180)+0.02*sin(2*G_sun*PI/180); /求黄道经度 long double earth_tilt(long double t_century) return (23.4393-0.0130*t_century); /求地球倾角 long double sun_deviation(long double earth_tilt, long double ecliptic_longitude) ret

7、urn (180/PI*asin(sin(PI/180*earth_tilt)*sin(PI/180*ecliptic_longitude); /求太阳偏差 long double GHA(long double UTo, long double G_sun, long double ecliptic_longitude) return (UTo-180-1.915*sin(G_sun*PI/180)-0.02*sin(2*G_sun*PI/180)+2.466*sin(2*ecliptic_longitude*PI/180)-0.053*sin(4*ecliptic_longitude*PI

8、/180); /求格林威治时间的太阳时间角GHA long double e(long double h, long double glat, long double sun_deviation) return 180/PI*acos(sin(h*PI/180)-sin(glat*PI/180)*sin(sun_deviation*PI/180)/(cos(glat*PI/180)*cos(sun_deviation*PI/180); /求修正值e long double UT_rise(long double UTo, long double GHA, long double glong,

9、long double e) return (UTo-(GHA+glong+e); /求日出时间 long double UT_set(long double UTo, long double GHA, long double glong, long double e) return (UTo-(GHA+glong-e); /求日落时间 long double result_rise(long double UT, long double UTo, long double glong, long double glat, int year, int month, int date) long

10、double d; if(UT>=UTo) d=UT-UTo; else d=UTo-UT; if(d>=0.1) UTo=UT; UT=UT_rise(UTo,GHA(UTo,G_sun(t_century(days(year,month,date),UTo),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo),G_sun(t_century(days(year,month,date),UTo),glong,e(h,glat,sun_deviation(earth_tilt(t_century(days(ye

11、ar,month,date),UTo),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo),G_sun(t_century(days(year,month,date),UTo); result_rise(UT,UTo,glong,glat,year,month,date); return UT; /判断并返回结果（日出） long double result_set(long double UT, long double UTo, long double glong, long double glat, int year,

12、 int month, int date) long double d; if(UT>=UTo) d=UT-UTo; else d=UTo-UT; if(d>=0.1) UTo=UT; UT=UT_set(UTo,GHA(UTo,G_sun(t_century(days(year,month,date),UTo),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo),G_sun(t_century(days(year,month,date),UTo),glong,e(h,glat,sun_deviation(ea

13、rth_tilt(t_century(days(year,month,date),UTo),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo),G_sun(t_century(days(year,month,date),UTo); result_set(UT,UTo,glong,glat,year,month,date); return UT; /判断并返回结果（日落） int Zone(long double glong) if(glong>=0) return (int)(int)(glong/15.0)+1);

14、 else return (int)(int)(glong/15.0)-1); /求时区 void output(long double rise, long double set, long double glong) if(int)(60*(rise/15+Zone(glong)-(int)(rise/15+Zone(glong)<10) cout<<"The time at which the sun rises is "<<(int)(rise/15+Zone(glong)<<":0"<<(i

15、nt)(60*(rise/15+Zone(glong)-(int)(rise/15+Zone(glong)<<" .n" else cout<<"The time at which the sun rises is "<<(int)(rise/15+Zone(glong)<<":"<<(int)(60*(rise/15+Zone(glong)-(int)(rise/15+Zone(glong)<<" .n" if(int)(60*(set/15

16、+Zone(glong)-(int)(set/15+Zone(glong)<10) cout<<"The time at which the sun sets is "<<(int)(set/15+Zone(glong)<<": "<<(int)(60*(set/15+Zone(glong)-(int)(set/15+Zone(glong)<<" .n" else cout<<"The time at which the sun sets is &

17、quot;<<(int)(set/15+Zone(glong)<<":"<<(int)(60*(set/15+Zone(glong)-(int)(set/15+Zone(glong)<<" .n" /打印结果 int main() long double UTo=180.0; int year,month,date; long double glat,glong; int c3; input_date(c); year=c0; month=c1; date=c2; input_glat(c); glat=c

18、0+c1/60+c2/3600; input_glong(c); glong=c0+c1/60+c2/3600; long double rise,set; rise=result_rise(UT_rise(UTo,GHA(UTo,G_sun(t_century(days(year,month,date),UTo),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo),G_sun(t_century(days(year,month,date),UTo),glong,e(h,glat,sun_deviation(earth_tilt(t_century(days(year,month,date