Euler 25项目的非蛮力解决方案 [英] Non Brute Force Solution to Project Euler 25

问题描述

项目欧拉问题25 :

斐波那契数列由递归关系定义:

The Fibonacci sequence is defined by the recurrence relation:

F _n = F _n-1 + F _n-2 ，其中F _{1 = 1} F ₂ =1.因此前12个项 将是F ₁ = 1，F ₂ = 1，F ₃ = 2，F ₄ = 3， F ₅ = 5，F ₆ = 8，F ₇ = 13，F ₈ = 21，F ₉ = 34，F ₁₀ = 55，F ₁₁ = 89，F ₁₂ = 144

F_n = F_n−1 + F_n−2, where F_{1 = 1} and F₂ = 1. Hence the first 12 terms will be F₁ = 1, F₂ = 1, F₃ = 2, F₄ = 3, F₅ = 5, F₆ = 8, F₇ = 13, F₈ = 21, F₉ = 34, F₁₀ = 55, F₁₁ = 89, F₁₂ = 144

第12个术语F ₁₂ 是第一个包含三个数字的术语.

The 12th term, F₁₂, is the first term to contain three digits.

斐波纳契数列中第一个包含1000的项是什么 数字?

What is the first term in the Fibonacci sequence to contain 1000 digits?

我在Python中制作了一个蛮力解决方案，但是计算实际解决方案绝对需要永远.谁能建议非暴力解决方案?

I made a brute force solution in Python, but it takes absolutely forever to calculate the actual solution. Can anyone suggest a non brute force solution?

def Fibonacci(NthTerm):
    if NthTerm == 1 or NthTerm == 2:
        return 1 # Challenge defines 1st and 2nd term as == 1
    else:  # recursive definition of Fib term
        return Fibonacci(NthTerm-1) + Fibonacci(NthTerm-2)

FirstTerm = 0 # For scope to include Term in scope of print on line 13
for Term in range(1, 1000): # Arbitrary range
    FibValue = str(Fibonacci(Term)) # Convert integer to string for len()
    if len(FibValue) == 1000:
        FirstTerm = Term
        break # Stop there
    else:
        continue # Go to next number
print "The first term in the\nFibonacci sequence to\ncontain 1000 digits\nis the", FirstTerm, "term."

推荐答案

您可以编写一个fibonacci函数，该函数以线性时间运行并且具有恒定的内存占用，因此不需要列表即可保留它们. 这是一个递归版本(但是，如果n足够大，它将只是 stackoverflow )

You can write a fibonacci function that runs in linear time and with constant memory footprint, you don't need a list to keep them. Here's a recursive version (however, if n is big enough, it will just stackoverflow)

def fib(a, b, n):
    if n == 1:
        return a
    else: 
        return fib(a+b, a, n-1)


print fib(1, 0, 10) # prints 55

此函数仅调用一次(导致大约N调用参数N)，而您的解决方案则调用两次(大约2 ^ N调用参数N).

This function calls itself only once (resulting in around N calls for a parameter N), in contrast with your solution which calls itself twice (around 2^N calls for a parameter N).

这是一个永远不会堆栈溢出的版本，它使用循环而不是递归:

Here's a version that won't ever stackoverflow and uses a loop instead of recursion:

def fib(n):
    a = 1
    b = 0
    while n > 1:
        a, b = a+b, a
        n = n - 1
    return a

print fib(100000)

那已经足够快了:

$ time python fibo.py 
3364476487643178326662161200510754331030214846068006390656476...

real    0m0.869s

但是调用fib直到获得足够大的结果并不完美:意甲联赛的第一个数字被多次计算. 您可以计算下一个斐波那契数，并在同一循环中检查其大小:

But calling fib until you get a result big enough isn't perfect: the first numbers of the serie are calculated multiple times. You can calculate the next fibonacci number and check its size in the same loop:

a = 1
b = 0
n = 1
while len(str(a)) != 1000:
    a, b = a+b, a
    n = n + 1
print "%d has 1000 digits, n = %d" % (a, n)

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