Question

我正在与Euler Problem 12挣扎。我不希望答案或确切的代码更改会让我在那里。我只是想指向正确的方向。我运行此代码大约10分钟，并得出一个不正确的答案。这让我相信我的假设是一个带有＆gt;的三角形数字。 500除数不会有任何因素＆gt; 10000不正确。我想我需要使用更快的素数生成器并让程序停止迭代列表。我不知道怎么做后者。

def eratosthenes_sieve(limit):
    primes = {}
    listofprimes = []
    for i in range(2, limit + 1):
        primes[i] = True

    for i in primes:
        factors = range(i, limit + 1, i)
        for f in factors[1:limit + 1]:
            primes[f] = False

    for i in primes:
        if primes[i] == True:
            listofprimes.append(i)
    return listofprimes


def prime_factorization(n):
    global primal
    prime_factors = {}
    for i in primal:
        if n < i:
            i = primal[0]
        if n % i == 0:
            if i not in prime_factors.keys():
                prime_factors[i] = 1
            else:
                prime_factors[i] += 1
            n = n / i
        if n in primal:
            if n not in prime_factors.keys():
                prime_factors[n] = 1
            else:
                prime_factors[n] += 1
            return prime_factors
    return prime_factors    

def divisor_function(input):
    x = 1
    for exp in input.values():
        x *= exp + 1
    return x

def triangle(th):
    terms = []
    for each in range(1, th+1):
        terms.append(each)
    return sum(terms)

z = 1
primal = eratosthenes_sieve(10000)
found = False
while found == False:
triz = triangle(z)
number_of_divisors = divisor_function(prime_factorization(triz))

if number_of_divisors > 300:
    print "GETTING CLOSE!! ********************************"
if number_of_divisors > 400:
    print "SUPER DUPER CLOSE!!! *********************************************************"
if number_of_divisors < 501:
    print "Nope. Not %s...Only has %s divisors." % (triz, number_of_divisors)

    z += 1
else:
    found = True
    print "We found it!"
    print "The first triangle number with over 500 divisors is %s!" % triangle(z)

Answer 1

当然，我在发布后几分钟就知道了。

在我的prime_factorization函数中。

如果n％i == 0：应该是n％i == 0。

这导致程序遗漏因素并贯穿整个素数列表。

与欧拉12，慢代码 - Python挣扎

1 个答案: