我试图通过解决Project Euler中的问题来学习python。我陷入了问题58.问题就这样说了:
Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18 5 4 3 12 29
40 19 6 1 2 11 28
41 20 7 8 9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49
It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?
以下是我为解决此问题而编写的代码。我利用一个素数来检查素数,但我不知道设置素数的限制。所以当我需要增加限制时,我让代码告诉我。代码运行良好,最高限制= 10 ^ 8,但当我将其设置为10 ^ 9时,代码冻结我的PC,我必须重新启动。不知道我做错了什么。如果您需要其他信息,请与我们联系。谢谢!
def primesieve(limit):
primelist=[]
for i in xrange(limit):
primelist.append(i)
primelist[1]=0
for i in xrange(2,limit):
if primelist[i]>0:
ctr=2
while (primelist[i]*ctr<limit):
a=primelist[i]*ctr
primelist[a]=0
ctr+=1
primelist=filter(lambda x: x!=0, primelist)
return primelist
limit=10**7
plist=primesieve(limit)
pset=set(plist)
diagnumbers=5.0
primenumbers=3.0
sidelength=3
lastnumber=9
while (primenumbers/diagnumbers)>=0.1:
sidelength+=2
for i in range(3):
lastnumber+=(sidelength-1)
if lastnumber in pset:
primenumbers+=1
diagnumbers+=4
lastnumber+=(sidelength-1)
if lastnumber>plist[-1]:
print lastnumber,"Need to increase limit"
break
print "sidelength",sidelength," last number",lastnumber,(primenumbers/diagnumbers)
答案 0 :(得分:0)
即使您正在使用xrange,在制作初级版本时,仍然会生成大小为10 ** 9的列表。使用大量内存,可能是你的问题。
相反,您可以考虑通过检查(2,N ** .5)之间的任何数字来均匀地编写一个检查数字N是否为素数的函数。然后,您可以开始生成角数,然后执行素性测试。
答案 1 :(得分:0)
您可以采取以下措施使您的主要生成器更有效:
def primesieve(limit):
primelist=[]
# Don't create a list of all your numbers up front.
# And even if you do, at least skip the even numbers!
#for i in xrange(limit):
# primelist.append(i)
# Skip counting - no even number > 3 is prime!
for i in xrange(3, limit, 2):
# You only need to check up to the square root of a number:
# I *thought* that there was some rule that stated that a number
# was prime if it was not divisible by all primes less than it,
# but I couldn't find that for certain. That would make this go
# a lot faster if you only had to check primes and numbers greater
# than the greatest prime found so far up to the square root of
# the number
for divisor in xrange(3, int(i**0.5)+1, 2):
if not i % divisor: # no remainder, so sad
break
else:
# loop exited naturally, number has no divisors hooray!
primelist.append(i)
# Need to put the number 2 back, though
primelist.insert(0, 2)
return primelist
这使用了我的CPU中的 mess (100%或更多,万岁!)但几乎没有使用任何内存(比如7分钟内几个MB RAM)。我的CPU只有2.something GHz,到目前为止,10**8
作为最大素数花费了7分钟。
如果你看一下我在评论中链接的帖子,有一些更好的方法来生成素数,但这些是一些简单的改进。