倒数斐波那契常数

Question

我正在开发一个程序，该程序计算倒数斐波那契常数（斐波那契数的无穷总和。）它计算每个项直到错误为止：

我有一个程序，但只用了1474个学期，我需要达到10000个学期。它返回一个错误：

Traceback (most recent call last):

  File "/Users/jaddvirji/Desktop/PapaTechChallenges/Challenge2/Part1/main.py", line 23, in        <module>

curf.write(str(Decimal(fibConstant(x))))

  File "/Users/jaddvirji/Desktop/PapaTechChallenges/Challenge2/Part1/main.py", line 18, in     fibConstant

return (1.0 / fib(n)) + fibConstant(n - 1.0)

File "/Users/jaddvirji/Desktop/PapaTechChallenges/Challenge2/Part1/main.py", line 12, in   fib

return long(((phi**n) - (1-phi)**n) / 5**0.5)

OverflowError: (34, 'Result too large')

我的代码是：

#!/usr/bin/env python

print "(C) COPYRIGHT JADD VIRJI 2013. ALL RIGHTS RESERVED."

from decimal import *

import time as t

import sys

sys.setrecursionlimit(10000)

phi = (1+(5**0.5))/2

def fib(n):

   return long(((phi**n) - (1-phi)**n) / 5**0.5)

def fibConstant(n):

  if(n == 1):

      return (1.0 / fib(n))

else:

  return (1.0 / fib(n)) + fibConstant(n - 1.0)

x = 1

while True:

  curf = open(str(x)+" term.txt","w")

  curf.write(str(Decimal(fibConstant(x))))

  curf.close()

  x = x+1

  print Decimal(x)

print "DONE. THANKS FOR USING."

另外，上述大约200个术语的每个结果都是相同的（而且是错误的）。

有人知道如何解决这些问题吗？

Answer 1

尝试将的值存储fibConstant在列表中。然后，对于每个后续的计算，您只需要调用列表的最后一个值即可，而无需重新计算。例如：from math import sqrtphi = (1 + sqrt(5)) / 2.def fib(n):&nbsp; &nbsp; return (phi**n - (1-phi)**n) / sqrt(5)fib_constant_list = [1./fib(1)]def fib_constant(n):&nbsp; &nbsp; new_fib_c = (1./fib(n) + fib_constant_list[-1])&nbsp; &nbsp; fib_constant_list.append(new_fib_c)&nbsp; &nbsp; return new_fib_cn = 2N_MAX = 1000while n < N_MAX:&nbsp; &nbsp; &nbsp;print fib_constant(n)