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     fibConstantreturn (1.0 / fib(n)) + fibConstant(n - 1.0)File "/Users/jaddvirji/Desktop/PapaTechChallenges/Challenge2/Part1/main.py", line 12, in   fibreturn long(((phi**n) - (1-phi)**n) / 5**0.5)OverflowError: (34, 'Result too large')我的代码是：#!/usr/bin/env pythonprint "(C) COPYRIGHT JADD VIRJI 2013. ALL RIGHTS RESERVED."from decimal import *import time as timport syssys.setrecursionlimit(10000)phi = (1+(5**0.5))/2def 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 = 1while 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 sqrt

phi = (1 + sqrt(5)) / 2.

def fib(n):

    return (phi**n - (1-phi)**n) / sqrt(5)

fib_constant_list = [1./fib(1)]

def fib_constant(n):

    new_fib_c = (1./fib(n) + fib_constant_list[-1])

    fib_constant_list.append(new_fib_c)

    return new_fib_c

n = 2

N_MAX = 1000

while n < N_MAX:

     print fib_constant(n)