所以我终于为foobar挑战写了一个递归函数＆＃34;特殊的平衡＆＃34;。它与我当地的蟒蛇运行良好，但foobar给了我＆＃34;时间限制超过＆＃34;错误。我以为我会分享这些代码并询问是否有人知道问题所在。
以下是说明和解决方案：
我们能救他们吗？ Beta兔子试图进入一个包含唯一已知僵尸治疗方法的实验室 - 但这是一个障碍。只有正确解决了挑战，门才会打开。僵尸兔群的未来受到威胁，因此Beta读取了挑战：左侧有一个物体，其质量以一定数量的单位给出。可以预见，任务是平衡双方。但有一个问题：你只有这个特殊的重量集，有1,3,3,27 ......单位的质量。也就是说，每个3的幂都是一个。作为一个出色的数学家，Beta Rabbit很快发现任何数量的质量单位都可以使用这个集合进行精确平衡。
为了帮助Beta进入房间，编写一个名为answer（x）的方法，该方法输出一个表示权重放置位置的字符串列表，以便双方平衡，假设左边的权重有质量x单位。
输出列表的第一个元素应对应于1个单位的权重，第二个元素对应于3个单位的权重，依此类推。每个字符串都是以下之一：
＆＃34; L＆＃34; ：把重量放在左手边 ＆＃34; R＆＃34; ：把重量放在右手边 ＆＃34; - ＆＃34; ：不要使用重量
为了确保输出尽可能小，列表的最后一个元素不能是＆＃34; - ＆＃34;。
x将始终为正整数，不大于1000000000。
要提供Python解决方案，请编辑solution.py 要提供Java解决方案，请编辑solution.java
输入： （int）x = 2 输出： （字符串列表）[＆＃34; L＆＃34;，＆＃34; R＆＃34;]
输入： （int）x = 8 输出： （字符串列表）[＆＃34; L＆＃34;，＆＃34; - ＆＃34;，＆＃34; R＆＃34;]
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import math
def calc_weights(x, A, B):
# For a given x, A, B, use the closest integer which is a power of 3 and see if it works to balance the scale
# Finds the closest power of 3 number to x, then subtracts it from x and calls itself again. x will reduce at each step
# A keeps a list of numbers used e.g. [1, 9, 27]
# B keeps a list of side for each element in A. e.g. ['R', 'L', 'R']
# x, A, B are updated everytime this function is called
if abs(x) > 1:
xorig = x; Aorig=list(A); Borig=list(B);
for lb in range(0, 2): # For each x, we have 2 combinations. E.g. x=45: use 3**3=27 or 3**4=81 (lower bound, and higher bound). The code will examine one number, if it fails it goes for the second number
#a = pow(3, int(math.log(abs(x), 3))) * (3**(1-lb)) # if lb=0 then use lower bound. If lb=1 then use the higher bound.
a = pow(3, int(math.log(abs(x), 3))) * (3**(lb))
while (a in A) and (a>0): # if a is used already, then use another number
a = int(a/3)
if (a==0) and (lb==0): # using the lower bound number didn't work
continue
if (a==0) and (lb==1): # using the higher bound didn't work either
A.append('0'); B.append('-')
return 10000000001, A, B # 10000000001 means it didn't work, return to previous step
A.append(a) # Move forward with a which is a power of 3 number
if (x>=0):
x-=a
B.append('R')
else:
x+=a
B.append('L')
[x, A, B] = calc_weights(x, A, B) # Recursion
if x==10000000001: # The last value added, didn't work
A = list(Aorig)
B = list(Borig)
x = xorig
if x==0:
return x, A, B
else:
if lb==1:
return 10000000001, A, B
else:
if x==0:
return x, A, B
else:
if 1 in A:
A.append(0)
B.append('-')
#print 'return 1'
return 10000000001, A, B
elif x==1:
A.append(1)
B.append('R')
return 0, A, B
elif x==-1:
A.append(1)
B.append('L')
return 0, A, B
def answer(x):
# your code here
[xp, A, B] = calc_weights(x, [], [])
if xp != 0:
print '**************************************** BAD', xp, A, B
i = 1
j = 0
out=[]
while i <= max(A):
out.append('-')
if i in A:
out[j] = B[A.index(i)]
j += 1
i = i * 3
return out
# Out is the required output of the code. I check again to see if the functions worked fine.
# If there's no input on screen, everything worked fine. Here I check x=1 to 20000
for x in range(1, 20000):
out = answer(x)
xp = x
for i in range(0, len(out)):
if out[i]=='R':
xp -= 3**i
if out[i]=='L':
xp += 3**i
#print xp
if (xp!=0):
print x, xp, out
#print out