四分历python实现-阿里云开发者社区

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四分历python实现

简介: 根据一个新加坡人的c代码改写成python代码   1 ''' 四分历''' 2 3 # 4 zq = 0 5 month = 0 6 7 def main(): 8 global month 9 year = 1 ...

根据一个新加坡人的c代码改写成python代码

 

  1 ''' 四分历'''
  2 
  3 # 
  4 zq = 0
  5 month = 0
  6 
  7 def main():
  8     global month
  9     year = 1
 10     rb_year = 0
 11     moon = 0  # number of new moon since beginning of ru bu year.
 12     tmonth = 0
 13     continues = False
 14 
 15     while year != 0:
 16         year = int(input("\nPlease Enter a year to do computation (range:85~236, 0 to exit):"))
 17 
 18         if year == 0: 
 19             return
 20         if year < 85 or year > 236:
 21             print("\nCalculation of Si Fen Li doesn't apply to your input value.")
 22             input("\nPress Enter to continue.")
 23             #getch(continues);
 24             continue
 25 
 26         rb_year = (year + 9281) % 76
 27         tmonth = 14 + Leap_y(rb_year)
 28 
 29         print("月\t朔\t望\t长度\t中气\t时间\t时间\t时间")
 30         print("- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -")
 31 
 32         month = 11
 33         #zq = 0
 34         for moon in range(1, tmonth + 1):
 35             if moon % 8 == 0:
 36                 #print("\n")
 37                 pass
 38             print_s(rb_year, moon)
 39             month += 1
 40             if month > 12:
 41                 month = 1
 42 
 43         input("\nPress Enter to continue.")
 44 
 45 # 润年
 46 def Leap_y( rbyear):
 47     isLeap = 0
 48     if ((rbyear - 1) * 235 ) % 19 >= 12:
 49         isLeap = 1
 50     return isLeap
 51 
 52 # 润月
 53 def Leap_m(rbyear, month):
 54     isLeap = 0
 55     completed_month = (rbyear - 1) * (235.0/19.0) + (month - 1)
 56     acd_first = (completed_month * (29.0+499.0/940.0) )
 57     acd_last  = ((completed_month + 1.0) * (29.0+499.0/940.0) )
 58     qi_first  = int(acd_first)/(30.0+14.0/32.0)
 59     qi_last   = int(acd_last) / (30.0+14.0/32.0)
 60 
 61     if qi_first - int(qi_first) != 0:
 62         if int(qi_first) == int(qi_last) or qi_last == int(qi_last):
 63             isLeap = 1
 64 
 65     return isLeap
 66 
 67 
 68 # 打印结果
 69 def print_s(rbyear, moon):
 70     global zq
 71     global month
 72     name_s = acd_shuo(rbyear, moon)
 73     name_w = acd_wang(rbyear, moon)
 74     time_s = ( acd_shuo(rbyear, moon) - float(name_s) ) * 24.0
 75     time_w = ( acd_wang(rbyear, moon) - float(name_w) ) * 24.0
 76     length = int(acd_shuo(rbyear, moon + 1)) - name_s
 77 
 78     if Leap_m(rbyear, moon) == 0:    #if is not a leap month.
 79         zq += 1
 80         name_q = acd_qi(rbyear, zq);
 81         time_q = ( acd_qi(rbyear, zq) - (float)(name_q) ) * 24.0
 82         print("{}\t1\t{:.1f}\t{:.3f}\t{:.1f}\t{:.3f}\t{:.3f}\t{:.1f}".format(month, time_s, name_w - name_s + 1, time_w, length, name_q - name_s + 1, time_q))
 83     else:
 84         month -= 1
 85         print("{}(Leap)\t1\t{:.1f}\t{:.3f}\t{:.1f}\t{:.3f}".format(month, time_s, name_w - name_s + 1, time_w, length))
 86 
 87 
 88 #
 89 def acd_shuo(rbyear, moon):
 90     completed_month = ((rbyear - 1) * 235 / 19) + moon - 1
 91     return completed_month * (29.0+499.0/940.0)
 92 
 93 #
 94 def acd_wang(rbyear, moon):
 95     completed_month = ((rbyear - 1) * 235 / 19) + moon - 1
 96     return (completed_month + 0.5) * (29.0+499.0/940.0)
 97 
 98 #
 99 def acd_qi(rbyear, qi):
100     completed_qi = ((rbyear - 1) * 12) + qi - 1
101     return completed_qi * (30.0+14.0/32.0)
102 
103 
104 if __name__ == '__main__':
105     main()

 

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