擅长:python、mysql、java
<p>实现可以是:</p>
<pre><code>from itertools import count
def fast_fib_generator():
F = [1, 1]
yield 1
yield 1
for k in count(1):
F.append(F[k] ** 2 + F[k - 1] ** 2)
yield F[-1]
F.append(F[k] * (2 * F[k + 1] - F[k]))
yield F[-1]
for x in fast_fib_generator():
print x
</code></pre>
<p>初步结果:</p>
^{pr2}$