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  • 利用日期、经纬度求日出日落时间 C语言程序代码（zz）


    先贴在这了，后面应该用得着
    http://zhidao.baidu.com/link?url=iw-hcd_tLpRtf4r2Kh-NmDPaQ10UdlunBQUWaz14J-eNEq5fw-y83ydgOTn7KX9Z_t5rGywtlFGaygRJeEYBPa  

    1. #define PI 3.1415926 
    2.  
    3.  #include<math.h> 
    4.  
    5.  #include<iostream> 
    6.  
    7.  using namespace std; 
    8.  
    9.  
    10.  int days_of_month_1[]={31,28,31,30,31,30,31,31,30,31,30,31}; 
    11.  
    12.  int days_of_month_2[]={31,29,31,30,31,30,31,31,30,31,30,31}; 
    13.  
    14.  long double h=-0.833; 
    15.  
    16.  //定义全局变量 
    17.  
    18.  
    19. void input_date(int c[]){ 
    20.  
    21.      int i; 
    22.  
    23.      cout<<"Enter the date (form: 2009 03 10):"<<endl; 
    24.  
    25.      for(i=0;i<3;i++){ 
    26.  
    27.          cin>>c[i]; 
    28.  
    29.      } 
    30.  
    31.  } 
    32.  
    33.  //输入日期 
    34.  
    35.  
    36. void input_glat(int c[]){ 
    37.  
    38.      int i; 
    39.  
    40.      cout<<"Enter the degree of latitude(range: 0°- 60°,form: 40 40 40 (means 40°40′40″)):"<<endl; 
    41.  
    42.      for(i=0;i<3;i++){ 
    43.  
    44.          cin>>c[i]; 
    45.  
    46.      } 
    47.  
    48.  } 
    49.  
    50.  //输入纬度 
    51.  
    52.  
    53. void input_glong(int c[]){ 
    54.  
    55.      int i; 
    56.  
    57.      cout<<"Enter the degree of longitude(west is negativ,form: 40 40 40 (means 40°40′40″)):"<<endl; 
    58.  
    59.      for(i=0;i<3;i++){ 
    60.  
    61.          cin>>c[i]; 
    62.  
    63.      } 
    64.  
    65.  } 
    66.  
    67.  //输入经度 
    68.  
    69.  
    70. int leap_year(int year){ 
    71.  
    72.      if(((year%400==0) || (year%100!=0) && (year%4==0))) return 1; 
    73.  
    74.      else return 0; 
    75.  
    76.  } 
    77.  
    78.  //判断是否为闰年:若为闰年，返回1；若非闰年，返回0 
    79.  
    80.  
    81.  int days(int year, int month, int date){ 
    82.  
    83.      int i,a=0; 
    84.  
    85.      for(i=2000;i<year;i++){ 
    86.  
    87.          if(leap_year(i)) a=a+366; 
    88.  
    89.          else a=a+365; 
    90.  
    91.      } 
    92.  
    93.      if(leap_year(year)){ 
    94.  
    95.          for(i=0;i<month-1;i++){ 
    96.  
    97.              a=a+days_of_month_2[i]; 
    98.  
    99.          } 
    100.  
    101.      } 
    102.  
    103.      else { 
    104.  
    105.          for(i=0;i<month-1;i++){ 
    106.  
    107.              a=a+days_of_month_1[i]; 
    108.  
    109.          } 
    110.  
    111.      } 
    112.  
    113.      a=a+date; 
    114.  
    115.      return a; 
    116.  
    117.  } 
    118.  
    119.  //求从格林威治时间公元2000年1月1日到计算日天数days 
    120.  
    121.  
    122.  long double t_century(int days, long double UTo){ 
    123.  
    124.      return ((long double)days+UTo/360)/36525; 
    125.  
    126.  } 
    127.  
    128.  //求格林威治时间公元2000年1月1日到计算日的世纪数t 
    129.  
    130.  
    131.  long double L_sun(long double t_century){ 
    132.  
    133.      return (280.460+36000.770*t_century); 
    134.  
    135.  } 
    136.  
    137.  //求太阳的平黄径 
    138.  
    139.  
    140. long double G_sun(long double t_century){ 
    141.  
    142.      return (357.528+35999.050*t_century); 
    143.  
    144.  } 
    145.  
    146.  //求太阳的平近点角 
    147.  
    148.  
    149. long double ecliptic_longitude(long double L_sun,long double G_sun){ 
    150.  
    151.      return (L_sun+1.915*sin(G_sun*PI/180)+0.02*sin(2*G_sun*PI/180)); 
    152.  
    153.  } 
    154.  
    155.  //求黄道经度 
    156.  
    157.  
    158. long double earth_tilt(long double t_century){ 
    159.  
    160.      return (23.4393-0.0130*t_century); 
    161.  
    162.  } 
    163.  
    164.  //求地球倾角 
    165.  
    166.  
    167. long double sun_deviation(long double earth_tilt, long double ecliptic_longitude){ 
    168.  
    169.      return (180/PI*asin(sin(PI/180*earth_tilt)*sin(PI/180*ecliptic_longitude))); 
    170.  
    171.  } 
    172.  
    173.  //求太阳偏差 
    174.  
    175.  
    176. long double GHA(long double UTo, long double G_sun, long double ecliptic_longitude){ 
    177.  
    178.      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/180)); 
    179.  
    180.  } 
    181.  
    182.  //求格林威治时间的太阳时间角GHA 
    183.  
    184.  
    185.  long double e(long double h, long double glat, long double sun_deviation){ 
    186.  
    187.      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))); 
    188.  
    189.  } 
    190.  
    191.  //求修正值e 
    192.  
    193.  
    194.  long double UT_rise(long double UTo, long double GHA, long double glong, long double e){ 
    195.  
    196.      return (UTo-(GHA+glong+e)); 
    197.  
    198.  } 
    199.  
    200.  //求日出时间 
    201.  
    202.  
    203. long double UT_set(long double UTo, long double GHA, long double glong, long double e){ 
    204.  
    205.      return (UTo-(GHA+glong-e)); 
    206.  
    207.  } 
    208.  
    209.  //求日落时间 
    210.  
    211.  
    212. long double result_rise(long double UT, long double UTo, long double glong, long double glat, int year, int month, int date){ 
    213.  
    214.      long double d; 
    215.  
    216.      if(UT>=UTo) d=UT-UTo; 
    217.  
    218.      else d=UTo-UT; 
    219.  
    220.      if(d>=0.1) { 
    221.  
    222.          UTo=UT; 
    223.  
    224.          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(year,month,date),UTo)),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo)),G_sun(t_century(days(year,month,date),UTo)))))); 
    225.  
    226.          result_rise(UT,UTo,glong,glat,year,month,date); 
    227.  
    228.      } 
    229.  
    230.      return UT; 
    231.  
    232.  } 
    233.  
    234.  //判断并返回结果（日出） 
    235.  
    236.  
    237. long double result_set(long double UT, long double UTo, long double glong, long double glat, int year, int month, int date){ 
    238.  
    239.      long double d; 
    240.  
    241.      if(UT>=UTo) d=UT-UTo; 
    242.  
    243.      else d=UTo-UT; 
    244.  
    245.      if(d>=0.1){ 
    246.  
    247.          UTo=UT; 
    248.  
    249.          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(earth_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)))))); 
    250.  
    251.          result_set(UT,UTo,glong,glat,year,month,date); 
    252.  
    253.      } 
    254.  
    255.      return UT; 
    256.  
    257.  } 
    258.  
    259.  //判断并返回结果（日落） 
    260.  
    261.  
    262. int Zone(long double glong){ 
    263.  
    264.      if(glong>=0) return (int)((int)(glong/15.0)+1); 
    265.  
    266.      else return (int)((int)(glong/15.0)-1); 
    267.  
    268.  } 
    269.  
    270.  //求时区 
    271.  
    272.  
    273. void output(long double rise, long double set, long double glong){ 
    274.  
    275.      if((int)(60*(rise/15+Zone(glong)-(int)(rise/15+Zone(glong))))<10) 
    276.  
    277.          cout<<"The time at which the sun rises is "<<(int)(rise/15+Zone(glong))<<":0"<<(int)(60*(rise/15+Zone(glong)-(int)(rise/15+Zone(glong))))<<" .
"; 
    278.  
    279.      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))))<<" .
"; 
    280.  
    281.      if((int)(60*(set/15+Zone(glong)-(int)(set/15+Zone(glong))))<10) 
    282.  
    283.          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))))<<" .
"; 
    284.  
    285.      else 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))))<<" .
"; 
    286.  
    287.  } 
    288.  
    289.  //打印结果 
    290.  
    291.           
    292.  
    293. int main(){ 
    294.  
    295.      long double UTo=180.0; 
    296.  
    297.      int year,month,date; 
    298.  
    299.      long double glat,glong; 
    300.  
    301.      int c[3]; 
    302.  
    303.      input_date(c); 
    304.  
    305.      year=c[0]; 
    306.  
    307.      month=c[1]; 
    308.  
    309.      date=c[2]; 
    310.  
    311.      input_glat(c); 
    312.  
    313.      glat=c[0]+c[1]/60+c[2]/3600; 
    314.  
    315.      input_glong(c); 
    316.  
    317.      glong=c[0]+c[1]/60+c[2]/3600; 
    318.  
    319.      long double rise,set; 
    320.  
    321.      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),UTo)),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo)),G_sun(t_century(days(year,month,date),UTo)))))),UTo,glong,glat,year,month,date); 
    322.  
    323.      set=result_set(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(earth_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)))))),UTo,glong,glat,year,month,date); 
    324.  
    325.      output(rise,set,glong); 
    326.  
    327.      system("pause"); 
    328.  
    329.      return 0; 
    330.  
    331.  }
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  • 原文地址：https://www.cnblogs.com/strinkbug/p/5942419.html
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