我正在研究Project Euler Problem 50,它指出：

The prime 41,can be written as the sum of six consecutive primes:

41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a prime,contains 21 terms,and is equal to 953.

Which prime,below one-million,can be written as the sum of the most consecutive primes?

为了确定素数P中的项(如果它可以写成素数的总和),我使用所有素数的滑动窗口(按递增顺序)直到(但不包括)P,并计算所有的总和这些窗口,如果总和等于考虑的素数,我算一下窗口的长度……

这适用于1000以下的所有素数,但是对于高达10 ** 6的素数来说它非常慢,所以我希望备忘录会有所帮助;在计算滑动窗口的总和时,做了很多双重工作……(对吧？)

所以我在网上找到了标准的memoizaton实现,并将其粘贴在我的代码中,这是正确的吗？ (我不知道应该怎么在这里工作……)

primes = tuple(n for n in range(1,10**6) if is_prime(n)==True)

count_best = 0


##http://docs.python.org/release/2.3.5/lib/itertools-example.html:
## Slightly modified (first for loop)
from itertools import islice
    def window(seq):
    for n in range(2,len(seq) + 1):

        it = iter(seq)
        result = tuple(islice(it,n))
        if len(result) == n:
            yield result    
        for elem in it:
            result = result[1:] + (elem,)
            yield result   

def memoize(function):
    cache = {}
    def decorated_function(*args):
        if args in cache:
            return cache[args]
        else:
            val = function(*args)
            cache[args] = val
            return val
    return decorated_function


@memoize 


def find_lin_comb(prime):
    global count_best

    for windows in window(primes[0 : primes.index(prime)]):
        if sum(windows) == prime and len(windows) > count_best:
            count_best = len(windows)
            print('Prime: ',prime,'Terms: ',count_best)


##Find them:
for x in primes[::-1]: find_lin_comb(x)

(顺便说一句,素数的元组是“正常”快速生成的)

所有的输入都很受欢迎,我只是一个业余爱好程序员,所以请不要对我有所了解.

谢谢！

编辑：这是一个没有破坏缩进的工作代码粘贴：
http://pastebin.com/R1NpMqgb