利用日期、经纬度求日出日落时间 C语言程序代码（zz）

时间：2022-05-05 22:04:02

先贴在这了，后面应该用得着

    

  1. #define PI 3.1415926 
    2. 

  2.  
    3. 

  3.  #include<math.h> 
    4. 

  4.  
    5. 

  5.  #include<iostream> 
    6. 

  6.  
    7. 

  7.  using namespace std; 
    8. 

  8.  
    9. 

  9.  
    10. 

  10.  int days_of_month_1[]={31,28,31,30,31,30,31,31,30,31,30,31}; 
    11. 

  11.  
    12. 

  12.  int days_of_month_2[]={31,29,31,30,31,30,31,31,30,31,30,31}; 
    13. 

  13.  
    14. 

  14.  long double h=-0.833; 
    15. 

  15.  
    16. 

  16.  //定义全局变量 
    17. 

  17.  
    18. 

  18.  
    19. 

  19. void input_date(int c[]){ 
    20. 

  20.  
    21. 

  21.      int i; 
    22. 

  22.  
    23. 

  23.      cout<<"Enter the date (form: 2009 03 10):"<<endl; 
    24. 

  24.  
    25. 

  25.      for(i=0;i<3;i++){ 
    26. 

  26.  
    27. 

  27.          cin>>c[i]; 
    28. 

  28.  
    29. 

  29.      } 
    30. 

  30.  
    31. 

  31.  } 
    32. 

  32.  
    33. 

  33.  //输入日期 
    34. 

  34.  
    35. 

  35.  
    36. 

  36. void input_glat(int c[]){ 
    37. 

  37.  
    38. 

  38.      int i; 
    39. 

  39.  
    40. 

  40.      cout<<"Enter the degree of latitude(range: 0°- 60°,form: 40 40 40 (means 40°40′40″)):"<<endl; 
    41. 

  41.  
    42. 

  42.      for(i=0;i<3;i++){ 
    43. 

  43.  
    44. 

  44.          cin>>c[i]; 
    45. 

  45.  
    46. 

  46.      } 
    47. 

  47.  
    48. 

  48.  } 
    49. 

  49.  
    50. 

  50.  //输入纬度 
    51. 

  51.  
    52. 

  52.  
    53. 

  53. void input_glong(int c[]){ 
    54. 

  54.  
    55. 

  55.      int i; 
    56. 

  56.  
    57. 

  57.      cout<<"Enter the degree of longitude(west is negativ,form: 40 40 40 (means 40°40′40″)):"<<endl; 
    58. 

  58.  
    59. 

  59.      for(i=0;i<3;i++){ 
    60. 

  60.  
    61. 

  61.          cin>>c[i]; 
    62. 

  62.  
    63. 

  63.      } 
    64. 

  64.  
    65. 

  65.  } 
    66. 

  66.  
    67. 

  67.  //输入经度 
    68. 

  68.  
    69. 

  69.  
    70. 

  70. int leap_year(int year){ 
    71. 

  71.  
    72. 

  72.      if(((year%400==0) || (year%100!=0) && (year%4==0))) return 1; 
    73. 

  73.  
    74. 

  74.      else return 0; 
    75. 

  75.  
    76. 

  76.  } 
    77. 

  77.  
    78. 

  78.  //判断是否为闰年:若为闰年，返回1；若非闰年，返回0 
    79. 

  79.  
    80. 

  80.  
    81. 

  81.  int days(int year, int month, int date){ 
    82. 

  82.  
    83. 

  83.      int i,a=0; 
    84. 

  84.  
    85. 

  85.      for(i=2000;i<year;i++){ 
    86. 

  86.  
    87. 

  87.          if(leap_year(i)) a=a+366; 
    88. 

  88.  
    89. 

  89.          else a=a+365; 
    90. 

  90.  
    91. 

  91.      } 
    92. 

  92.  
    93. 

  93.      if(leap_year(year)){ 
    94. 

  94.  
    95. 

  95.          for(i=0;i<month-1;i++){ 
    96. 

  96.  
    97. 

  97.              a=a+days_of_month_2[i]; 
    98. 

  98.  
    99. 

  99.          } 
    100. 

  100.  
    101. 

  101.      } 
    102. 

  102.  
    103. 

  103.      else { 
    104. 

  104.  
    105. 

  105.          for(i=0;i<month-1;i++){ 
    106. 

  106.  
    107. 

  107.              a=a+days_of_month_1[i]; 
    108. 

  108.  
    109. 

  109.          } 
    110. 

  110.  
    111. 

  111.      } 
    112. 

  112.  
    113. 

  113.      a=a+date; 
    114. 

  114.  
    115. 

  115.      return a; 
    116. 

  116.  
    117. 

  117.  } 
    118. 

  118.  
    119. 

  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. 

  120.  
    121. 

  121.  
    122. 

  122.  long double t_century(int days, long double UTo){ 
    123. 

  123.  
    124. 

  124.      return ((long double)days+UTo/360)/36525; 
    125. 

  125.  
    126. 

  126.  } 
    127. 

  127.  
    128. 

  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. 

  129.  
    130. 

  130.  
    131. 

  131.  long double L_sun(long double t_century){ 
    132. 

  132.  
    133. 

  133.      return (280.460+36000.770*t_century); 
    134. 

  134.  
    135. 

  135.  } 
    136. 

  136.  
    137. 

  137.  //求太阳的平黄径 
    138. 

  138.  
    139. 

  139.  
    140. 

  140. long double G_sun(long double t_century){ 
    141. 

  141.  
    142. 

  142.      return (357.528+35999.050*t_century); 
    143. 

  143.  
    144. 

  144.  } 
    145. 

  145.  
    146. 

  146.  //求太阳的平近点角 
    147. 

  147.  
    148. 

  148.  
    149. 

  149. long double ecliptic_longitude(long double L_sun,long double G_sun){ 
    150. 

  150.  
    151. 

  151.      return (L_sun+1.915*sin(G_sun*PI/180)+0.02*sin(2*G_sun*PI/180)); 
    152. 

  152.  
    153. 

  153.  } 
    154. 

  154.  
    155. 

  155.  //求黄道经度 
    156. 

  156.  
    157. 

  157.  
    158. 

  158. long double earth_tilt(long double t_century){ 
    159. 

  159.  
    160. 

  160.      return (23.4393-0.0130*t_century); 
    161. 

  161.  
    162. 

  162.  } 
    163. 

  163.  
    164. 

  164.  //求地球倾角 
    165. 

  165.  
    166. 

  166.  
    167. 

  167. long double sun_deviation(long double earth_tilt, long double ecliptic_longitude){ 
    168. 

  168.  
    169. 

  169.      return (180/PI*asin(sin(PI/180*earth_tilt)*sin(PI/180*ecliptic_longitude))); 
    170. 

  170.  
    171. 

  171.  } 
    172. 

  172.  
    173. 

  173.  //求太阳偏差 
    174. 

  174.  
    175. 

  175.  
    176. 

  176. long double GHA(long double UTo, long double G_sun, long double ecliptic_longitude){ 
    177. 

  177.  
    178. 

  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. 

  179.  
    180. 

  180.  } 
    181. 

  181.  
    182. 

  182.  //求格林威治时间的太阳时间角GHA 
    183. 

  183.  
    184. 

  184.  
    185. 

  185.  long double e(long double h, long double glat, long double sun_deviation){ 
    186. 

  186.  
    187. 

  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. 

  188.  
    189. 

  189.  } 
    190. 

  190.  
    191. 

  191.  //求修正值e 
    192. 

  192.  
    193. 

  193.  
    194. 

  194.  long double UT_rise(long double UTo, long double GHA, long double glong, long double e){ 
    195. 

  195.  
    196. 

  196.      return (UTo-(GHA+glong+e)); 
    197. 

  197.  
    198. 

  198.  } 
    199. 

  199.  
    200. 

  200.  //求日出时间 
    201. 

  201.  
    202. 

  202.  
    203. 

  203. long double UT_set(long double UTo, long double GHA, long double glong, long double e){ 
    204. 

  204.  
    205. 

  205.      return (UTo-(GHA+glong-e)); 
    206. 

  206.  
    207. 

  207.  } 
    208. 

  208.  
    209. 

  209.  //求日落时间 
    210. 

  210.  
    211. 

  211.  
    212. 

  212. long double result_rise(long double UT, long double UTo, long double glong, long double glat, int year, int month, int date){ 
    213. 

  213.  
    214. 

  214.      long double d; 
    215. 

  215.  
    216. 

  216.      if(UT>=UTo) d=UT-UTo; 
    217. 

  217.  
    218. 

  218.      else d=UTo-UT; 
    219. 

  219.  
    220. 

  220.      if(d>=0.1) { 
    221. 

  221.  
    222. 

  222.          UTo=UT; 
    223. 

  223.  
    224. 

  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. 

  225.  
    226. 

  226.          result_rise(UT,UTo,glong,glat,year,month,date); 
    227. 

  227.  
    228. 

  228.      } 
    229. 

  229.  
    230. 

  230.      return UT; 
    231. 

  231.  
    232. 

  232.  } 
    233. 

  233.  
    234. 

  234.  //判断并返回结果（日出） 
    235. 

  235.  
    236. 

  236.  
    237. 

  237. long double result_set(long double UT, long double UTo, long double glong, long double glat, int year, int month, int date){ 
    238. 

  238.  
    239. 

  239.      long double d; 
    240. 

  240.  
    241. 

  241.      if(UT>=UTo) d=UT-UTo; 
    242. 

  242.  
    243. 

  243.      else d=UTo-UT; 
    244. 

  244.  
    245. 

  245.      if(d>=0.1){ 
    246. 

  246.  
    247. 

  247.          UTo=UT; 
    248. 

  248.  
    249. 

  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. 

  250.  
    251. 

  251.          result_set(UT,UTo,glong,glat,year,month,date); 
    252. 

  252.  
    253. 

  253.      } 
    254. 

  254.  
    255. 

  255.      return UT; 
    256. 

  256.  
    257. 

  257.  } 
    258. 

  258.  
    259. 

  259.  //判断并返回结果（日落） 
    260. 

  260.  
    261. 

  261.  
    262. 

  262. int Zone(long double glong){ 
    263. 

  263.  
    264. 

  264.      if(glong>=0) return (int)((int)(glong/15.0)+1); 
    265. 

  265.  
    266. 

  266.      else return (int)((int)(glong/15.0)-1); 
    267. 

  267.  
    268. 

  268.  } 
    269. 

  269.  
    270. 

  270.  //求时区 
    271. 

  271.  
    272. 

  272.  
    273. 

  273. void output(long double rise, long double set, long double glong){ 
    274. 

  274.  
    275. 

  275.      if((int)(60*(rise/15+Zone(glong)-(int)(rise/15+Zone(glong))))<10) 
    276. 

  276.  
    277. 

  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))))<<" .\n"; 
    278. 

  278.  
    279. 

  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))))<<" .\n"; 
    280. 

  280.  
    281. 

  281.      if((int)(60*(set/15+Zone(glong)-(int)(set/15+Zone(glong))))<10) 
    282. 

  282.  
    283. 

  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))))<<" .\n"; 
    284. 

  284.  
    285. 

  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))))<<" .\n"; 
    286. 

  286.  
    287. 

  287.  } 
    288. 

  288.  
    289. 

  289.  //打印结果 
    290. 

  290.  
    291. 

  291.           
    292. 

  292.  
    293. 

  293. int main(){ 
    294. 

  294.  
    295. 

  295.      long double UTo=180.0; 
    296. 

  296.  
    297. 

  297.      int year,month,date; 
    298. 

  298.  
    299. 

  299.      long double glat,glong; 
    300. 

  300.  
    301. 

  301.      int c[3]; 
    302. 

  302.  
    303. 

  303.      input_date(c); 
    304. 

  304.  
    305. 

  305.      year=c[0]; 
    306. 

  306.  
    307. 

  307.      month=c[1]; 
    308. 

  308.  
    309. 

  309.      date=c[2]; 
    310. 

  310.  
    311. 

  311.      input_glat(c); 
    312. 

  312.  
    313. 

  313.      glat=c[0]+c[1]/60+c[2]/3600; 
    314. 

  314.  
    315. 

  315.      input_glong(c); 
    316. 

  316.  
    317. 

  317.      glong=c[0]+c[1]/60+c[2]/3600; 
    318. 

  318.  
    319. 

  319.      long double rise,set; 
    320. 

  320.  
    321. 

  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. 

  322.  
    323. 

  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. 

  324.  
    325. 

  325.      output(rise,set,glong); 
    326. 

  326.  
    327. 

  327.      system("pause"); 
    328. 

  328.  
    329. 

  329.      return 0; 
    330. 

  330.  
    331. 

  331.  }
    332.

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