优化Project Euler #5的Python代码(从1到20所有数字的LCM)

Question

这是我对欧拉计划的第五个问题的解决方案，它要求从1到20 (含)的数字的最小公共倍数。是否有任何方法来改进while循环条件而不是使用sum？

table = range(1, 21)
result = 1
pf = 2
while sum(table) > len(table):
   flag = False
   for x, y in enumerate(table):
      if y % pf == 0:
         table[x] = y/pf 
         flag = True
   if flag:
      result *= pf
   else:
      pf += 1
print result

Answer 1

您的目的是在table中的所有条目变为1时终止循环。

while sum(table) > len(table):

是一种相当模糊的表达意图的方式。我建议你

while max(table) > 1:

或者更明显的短路

while any(x > 1 for x in table):

Answer 2

用常量。当你想回去尝试不同的价值时，就更容易理解了。

复制我的答案从堆栈溢出，也改变了周围的事情，因为这是代码审查。

MAX_FACTOR = 20 #The largest divisor we want to be able to divide by

#should have a general outline of the algorithm here
table = range(1,MAX_FACTOR + 1)
result = 1 #final result is stored here
currentFactor = 2 #what we're currently trying to divide by
while currentFactor <= MAX_FACTOR:
   isDivisor = False #is the currentFactor a divisor
   for x,y in enumerate(table):
      if y % currentFactor == 0:
     table[x] = y/currentFactor 
     isDivisor = True
   if isDivisor:
      result *= currentFactor
   else:
      currentFactor += 1
print result