题目122：找出计算幂的最高效方法

jerryxjr1220

解答：

1. #coding:utf-8
   
2. L = 201
   
3. ways = [[[0]*L] for i in range(L)]
   
4. ways[2] = [[2,1]]
   
5. for i in range(2,L):
   
6.     for eachway in ways[i]:
   
7.         for p in eachway:
   
8.             n = i + p
   
9.             if n < L:
   
10.                 if min([len(each) for each in ways[n]]) > min([len(each) for each in ways[i]]) + 1:
    
11.                     ways[n] = [[n] + eachway]
    
12.                 elif min([len(each) for each in ways[n]]) == min([len(each) for each in ways[i]]) + 1:
    
13.                     ways[n] += [[n] + eachway]
    
14. 
    
15. print (sum([(min([len(each) for each in ways[x]])-1) for x in range(2,201)]))

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答案：
1582