是! 我做到了继续分数！
我从生成2的平方根的数字中找到了这些代码
def z(contfrac, a=1, b=0, c=0, d=1): for x in contfrac: while a > 0 and b > 0 and c > 0 and d > 0: t = a // c t2 = b // d if not t == t2: break yield t a = (10 * (a - c*t)) b = (10 * (b - d*t)) # continue with same fraction, don't pull new x a, b = x*a+b, a c, d = x*c+d, c for digit in rdigits(a, c): yield digitdef rdigits(p, q): while p > 0: if p > q:d = p // qp = p - q * d else:d = (10 * p) // qp = 10 * p - q * d yield d
我制作了连续分数生成器：
def e_cf_expansion(): yield 1 k = 0 while True: yield k k += 2 yield 1 yield 1
并将它们放在一起：
def e_dec(): return z(e_cf_expansion())
然后：
>>> gen = e_dec()>>> e = [str(gen.next()) for i in xrange(1000)]>>> e.insert(1, '.')>>> print ''.join(e)2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019011573834187930702154089149934884167509244761460668082264800168477411853742345442437107539077744992069551702761838606261331384583000752044933826560297606737113200709328709127443747047230696977209310141692836819025515108657463772111252389784425056953696770785449969967946864454905987931636889230098793127736178215424999229576351482208269895193668033182528869398496465105820939239829488793320362509443117301238197068416140397019837679320683282376464804295311802328782509819455815301756717361332069811250996181881593041690351598888519345807273866738589422879228499892086805825749279610484198444363463244968487560233624827041978623209002160990235304369941849146314093431738143640546253152096183690888707016768396424378140592714563549061303107208510383750510115747704171898610687396965521267154688957035035
奖励：生成sqrt（n）的连续分数的代码，其中n是一个正整数，而sqrt（n）是不合理的：
def sqrt_cf_expansion(S): """Generalized generator to compute continued fraction representation of sqrt(S)""" m = 0 d = 1 a = int(math.sqrt(S)) a0 = a while True: yield a m = d*a-m d = (S-m**2)//d a = (a0+m)//d
