我有这个代码,它将一个数字分成5个组,将它们放入一个列表中,然后将它们相乘。如果你感到困惑,这就是Project Euler中的问题8。它还没有完成,因为我需要找到其他可能的5个连续整数。
def split_number(number, n):
line = str(number)
split = [line[i:i+n] for i in range(0, len(line), n)]
return split
splitnum = split_number((extremely long number), 5)
for x in enumerate(splitnum[:-1]):
split5 = split_number(splitnum[x], 1)
for n in split5:
splitproduct = reduce(lambda x, y: x*y, splitnum[n])
if (splitproduct > solution):
solution = splitproduct
print solution
当我尝试运行时,我收到错误
TypeError: list indices must be integers, not tuple
我想当我遍历splitnum时,x是一个元组。我需要它是一个整数,所以我可以正确使用split5()。
新代码:
def split_number(number, n):
line = str(number)
split = [line[i:i+n] for i in range(1, len(line)-n+1, n)]
return split
number =
while len(split_number(number,1)) is not 0:
splitnum = split_number((number), 5)
solution = 0
for x in splitnum[:-1]:
split5 = split_number(x, 1)
for n in split5:
splitproduct = reduce(lambda x, y: x*y, n)
if (splitproduct > solution):
solution = splitproduct
number = split_number(number, 1)
del number[0]
print solution
现在我在函数split_number中的'split'行上出现内存错误。这可能是因为数字非常长。但这不是主题问题,我只是想让你们看看我是如何实现他们的解决方案的(这是有效的,因为程序实际运行)。 :)
答案 0 :(得分:2)
Enumerate返回元组,其中元组的第一个元素是序列的索引。
我认为您只想迭代splitnum
,而不是使用枚举,因为您没有在任何地方使用索引(for x in splitnum[:-1]
)。
答案 1 :(得分:2)
您需要做的就是将字符串的所有5个长子串相乘,看看哪个是最大的:
import operator
n = map(int, '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450')
print max(reduce(operator.mul, n[i:i + 5]) for i in range(len(n) - 5))