練習(xí)題（阿拉伯?dāng)?shù)字轉(zhuǎn)化為羅馬數(shù)字）

yestreenstars

回復(fù) 30# Herowinter

寫成多行腳本會(huì)好些~
   

yestreenstars

回復(fù) 27# jason680

不錯(cuò)的思路，但那些單字符的變量看得有點(diǎn)暈，單行命令用單字符變量還好~
   

jason680

回復(fù) 32# yestreenstars

$ awk '
BEGIN{
  # 0. prepare roman and value array
  # rom[1] = I, val[1] = 1
  # rom[2] = V, val[2] = 5
  # rom[3] = X, val[3] = 10
  # ...
  # rom[7] = X, val[7] = 1000
  max = split("IVXLCDM",rom,"");
  for(n=1; n <= max; n++){
    val[n] = 10**int(n/2) * (n%2?1:.5);
  }
}
{
  num = $0;
  # 1. check number ( 1 <= number <= 3999 )
  if(num < 1 || 3999 < num){
    print "out of range 1~3999: " num;
    next
  }
  rom_str = "";
  # 2. check number with 10 times only (1000, 100, 10, and 1)
  #    max=7, and V[7] = 1000, V[5]=100, ...
  for(n = max; n > 0; n -= 2){
    if(num >= val[n]){
      multi = int(num / val[n]);
      num %= val[n];
      # 3. for special number 4 and 9
      #   4xx = 100 and 500  = CD  (Note: multi=4, rom[5] and rom[5+1], n=5)
      #   9xx = 100 and 1000 = CM  (Note: multi=9, rom[5] and rom[5+2], n=5)
      #   rom[5] = C, val[5] = 100   , n =5
      #   rom[6] = D, val[6] = 500   , n+1=6
      #   rom[7] = M, val[7] = 1000 , n+2=7
      if(multi %5 == 4){
        rom_str = rom_str rom[n] rom[n+(multi>5?2:1)];
        continue;
      }
     #4. more than number 5
     #  6x = 50 and 1x = LX   (Note: multi=6 ==> rom[3+1] and multi=6-5=1)
     #  7x = 50 and 2x = LXX  (Note: Multi=7 ==> rom[3+1] and multi=7-5=2)
     #  rom[3] = X, val[3] = 10,  n=3
     #  rom[4] = L, val[4] = 50,  n+1=4
     if(multi >= 5){
       rom_str = rom_str rom[n+1];
       multi -= 5;
     }
     # for normal number
     for(c=1; c <= multi; c++)
       rom_str = rom_str rom[n]
    }
  }
  print rom_str
}' FILE
VI
LXXVI
CDXCIX
MMMDCCCLXXXVIII
IX
XX
out of range 1~3999: 4000
out of range 1~3999: 0
out of range 1~3999: -1

$ cat FILE
6
76
499
3888
9
20
4000
0
-1

   

substr函數(shù)

這是我的python2代碼，比較幼稚有些地方很不嚴(yán)謹(jǐn)高手請(qǐng)蛋定

1. #!/usr/bin/python2
   
2. 
   
3. ROM = [['I', 'V'], ['X', 'L'], ['C', 'D'], ['M']]
   
4. GEN = lambda a, b, c, d, e, f, r: a * ROM[b][c] + d * ROM[e][f] + r
   
5. LAM = [[3, lambda m, i, r: (0, i, 0, m, i, 0, r)],
   
6.        [5, lambda m, i, r: (1, i, 0, 5 - m, i, 1, r)],
   
7.        [8, lambda m, i, r: (1, i, 1, m - 5, i, 0, r)],
   
8.        [9, lambda m, i, r: (10 - m, i, 0, 1, i + 1, 0, r)]]
   
9. 
   
10. def A2R(n, i=0, R=''):
    
11.     if n is 0: return R
    
12.     n, m = divmod(n, 10)
    
13.     return A2R(n, i + 1, GEN(*next(l(m, i, R) for v, l in LAM if m <= v)))
    
14. 
    
15. 
    
16. test = [6, 9, 10, 11, 76, 499, 3888]
    
17. good = ['VI', 'IX', 'X', 'XI', 'LXXVI', 'CDXCIX', 'MMMDCCCLXXXVIII']
    
18. 
    
19. for I, N in enumerate(test):
    
20.     rom = A2R(N)
    
21.     print N, '=', rom, ':', rom == good[I]
    
22. 
    

復(fù)制代碼

ichenxiaohu

dfgsdg hsdfgh

lichmama

1. #-*- coding:utf8 -*-
   
2. 
   
3. def arab_to_roma(targ):
   
4.     nums = [1000, 500, 100, 50, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
   
5.     roma = {1000:'M', 500:'D', 100:'C', 50:'L', 10:'X',
   
6.             9:'IX', 8:'VIII', 7:'VII', 6:'VI', 5:'V',
   
7.             4:'IV',3:'III', 2:'II', 1:'I'}
   
8.     answ = {}
   
9. 
   
10.     for item in nums:
    
11.         while targ >= item:
    
12.             if not answ.has_key(item):
    
13.                 answ[item] = 1
    
14.             else:
    
15.                 answ[item] += 1
    
16.             targ -= item
    
17. 
    
18.     return (''.join([roma[key]*answ[key] for key in sorted(answ, reverse=True)]))
    
19. 
    
20. if __name__ == '__main__':
    
21.     arab = [6, 76, 499, 3888]
    
22.     for item in arab:
    
23.         print arab_to_roma(item)

復(fù)制代碼

yestreenstars

回復(fù) 34# substr函數(shù)

這腳本有問(wèn)題哈~5的時(shí)候有問(wèn)題~
   

yestreenstars

回復(fù) 36# lichmama

你這腳本也有問(wèn)題哦~不只是499時(shí)有問(wèn)題，其他很多轉(zhuǎn)化都有問(wèn)題~
   

lichmama

回復(fù) 38# yestreenstars


    是的，沒(méi)注意到還有其他規(guī)則。

hjfeng1988

回復(fù) 16# ly5066113


    好猥瑣的枚舉法