罗马数字和数字的转化和一个puzzle

1. 罗马数字和数字的转化
下面两段代码出自http://rosettacode.org/wiki/Roman_numerals/Encode#Python

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1. def to_roman(x):
   
2.     anums = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
   
3.     rnums = "M CM D CD C XC L XL X IX V IV I".split()
   
4.     ret=[]
   
5.     for a,r in zip(anums,rnums):
   
6.         n,x=divmod(x,a)
   
7.         ret.append(r*n)
   
8.     return "".join(ret)
   
9. 
   
10. def test_to_roman():
    
11.     test=(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,25,30,40,
    
12.         50,60,69,70,80,90,99,100,200,300,400,500,600,666,700,800,900,
    
13.         1000,1009,1444,1666,1945,1997,1999,2000,2008,2010,2011,2500,
    
14.         3000,3999)
    
15. 
    
16.     for i in test:
    
17.         print("number {} to roman is {}".format(i,to_roman(i)))
    
18. 
    
19. #test_to_roman()

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1. def to_arabic(x):
   
2.     decoder=dict(zip('MDCLXVI',(1000,500,100,50,10,5,1)))
   
3.     result=0
   
4.     for r,r1 in zip(x,x[1:]):
   
5.         rd,rd1=decoder[r],decoder[r1]
   
6.         if rd < rd1:
   
7.             result+=-rd
   
8.         else:
   
9.             result+=rd
   
10.     return result+decoder[x[-1]]
    
11. 
    
12. def test_to_arabic():
    
13.     ll='MCMXC MMVIII MDCLXVI'.split()
    
14.     for i in ll:
    
15.         print(to_arabic(i))
    
16. 
    
17. test_to_arabic()

下面的代码是我自己写的。 和别人的比，果然是差很多了。

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1. def to_arabic2(s):
   
2.     #s: sequence
   
3.     decoder=dict(zip('MDCLXVI',(1000,500,100,50,10,5,1)))
   
4.     result=0
   
5.     for i in range(len(s)-1):
   
6.         value=decoder.get(s[i])
   
7.         vnext=decoder.get(s[i+1])
   
8.         if value < vnext:
   
9.             result+=-value
   
10.         else:
    
11.             result+=value
    
12.     return result+decoder.get(s[-1])
    
13. 
    
14. def test_to_arabic2():
    
15.     ll='MCMXC MMVIII MDCLXVI'.split()
    
16.     for i in ll:
    
17.         print(to_arabic2(i))
    
18. 
    
19. test_to_arabic2()

2. 一个puzzle ,出自： http://stackoverflow.com/questions/10388060/solving-puzzle-in-python
Puzzle:
A merchant has a 40 kg weight which he used in his shop. Once, it fell from his hands and was broken into 4 pieces. But surprisingly, now he can weigh any weight between 1 kg to 40 kg with the combination of these 4 pieces.
So question is, what are weights of those 4 pieces?
这个哥们的代码写的太通用了，还加上pieces. 即N个pieces，sum(n)==L, 同时可以表示1->L，很有趣的题目

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1. import itertools as it
   
2. def find_weights(weight,pieces):
   
3.     full=range(1,weight+1)
   
4.     all_nums=set(full)
   
5.     comb=[x for x in it.combinations(full,pieces) if sum(x)==40]
   
6.     funcs=(lambda x: 0, lambda x: x, lambda x: -x)
   
7.     for c in comb:
   
8.         sums=set()
   
9.         for fmap in it.product(funcs,repeat=pieces):
   
10.             s=sum(f(x) for x,f in zip(c,fmap))
    
11.             if s>0:
    
12.                 sums.add(s)
    
13.                 if sums==all_nums:
    
14.                     return c
    
15. 
    
16. 
    
17. print(find_weights(40,5))