python 算法：Product of consecutive Fib numbers

The Fibonacci numbers are the numbers in the following integer sequence (Fn):
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...
such as：F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

Given a number, say prod (for product), we search two Fibonacci numbers F(n) and F(n+1) verifying
F(n) * F(n+1) = prod.
Your function productFib takes an integer (prod) and returns an array:
[F(n), F(n+1), true] or {F(n), F(n+1), 1} or (F(n), F(n+1), True)
depending on the language if F(n) * F(n+1) = prod.

If you don't find two consecutive F(m) verifying F(m) * F(m+1) = prodyou will return
[F(m), F(m+1), false] or {F(n), F(n+1), 0} or (F(n), F(n+1), False)

F(m) being the smallest one such as F(m) * F(m+1) > prod.

注意两点：

1.while循环没范围，用条件划范围，range总需要先给出个范围，这就是用起来要注意的区别和优化

2.对错的返回，完全可以不用去if定义，直接列出一个运算的式子，满足就自然对，不满足自然错，这是if的优化

【fib数列到200个，数位可达到20+位以上以上以上】

我的解法：（如此智障的解法昂）

def productFib(prod):
    a,b=1,1
    l=[0,1,1]#数列初始的几个数
    ll=[]
    for i in range(200):
        b,a=a,a+b  #累加的一个变换
        l.append(a)
    for j in range (len(l)-2):#防止出范围
        if prod==l[j]*l[j+1]:
            ll.append(l[j])
            ll.append(l[j+1])
            ll.append(True)
        elif l[j]*l[j+1]<prod<l[j+1]*l[j+2]:
            ll.append(l[j+1])
            ll.append(l[j+2])
            ll.append(False)
    return ll

大神解法：

def productFib(prod):
  a, b = 0, 1
  while prod > a * b:
    a, b = b, a + b
  return [a, b, prod == a * b]

【打脸不？？！！跟大神的解法一比，我的逻辑思维幼稚的一笔啊~~~整那么多五五六六七七八八有啥用！！】

python 算法：Product of consecutive Fib numbers

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