斐波那契加法器 - 计算斐波那契数列中的数字

Question

我正在开发一个程序，该程序将计算具有某些数字限制的斐波那契数（即第一个具有 100 位数字的斐波那契数）。下面的代码总体上正在运行，但我遇到了一个让我难过的逻辑错误。

代码的目标是以类似于二进制加法的方式计算斐波那契数。使用数组，每个元素都保存一个从 0 到 9 的数字，因此每个数组索引代表一个 10 的位置。

它开始工作并正常循环，但由于处理循环的方式，它在 13 到 21 之间停止。它将数字与 10 的位置相加，然后保存一个 31 数字。

有没有办法打破或阻止它把我没有看到的那些加在一起？

num1 = [0]*100
num2 = [0]*100
num2[len(num2)-1] = 1
carry = 0
flag = True

while (flag):
    #Main for loop to work through the array
    for count in range (1, len(num2)):
        carry = num2[len(num2) - count] + num1[len(num1) - count]
        if carry > 9:
            num2[len(num2)- (count + 1)] = num2[len(num2)- (count + 1)] + 1
            carry = carry % 10
            num1[len(num1) - count] = num2[len(num2) - count]
            num2[len(num2) - count] = carry
        else:
            num1[len(num1) - count] = num2[len(num2) - count]
            num2[len(num2) - count] = carry
    print(num2)
    if num2[0] != 0:
        flag = False

每次它通过我希望看到的主 while 循环

[0,0,...,0,1]
[0,0,...,0,2]
[0,0,...,0,3]
[0,0,...,0,5]
[0,0,...,0,8]
[0,0,...,1,3]
[0,0,...,2,1]
...

但在它到达 [...,2,1] 循环后，它会移动到 [...,3,1]

Answer 1

这是对您的代码的更正。请注意，Python 具有无限精度整数，因此我添加了一个斐波那契生成器来检查答案。

num1 = [0]*100
num2 = [0]*100
num2[len(num2)-1] = 1
flag = True

# Fibonacci generator for verification of answer
def fib():
    a,b = 0,1
    while True:
        a,b = b,a+b
        yield a

# Instance of the generator
f = fib()

# convert a list of single-digit integers to a real integer for display
def value(L):
    assert all(n < 10 for n in L) # bug checking for invalid list values.
    return int(''.join([str(i) for i in L]))

while (flag):
    #Main for loop to work through the array

    # Start with zero carry for first digit
    carry = 0

    for count in range (1,len(num2)+1): # originally off-by-1.

        # compute the sum plus the carry of previous sum
        temp = num2[len(num2) - count] + num1[len(num1) - count] + carry

        # shift num2 digit to num1 digit
        num1[len(num1) - count] = num2[len(num2) - count]

        # new num2 digit is the one's place of temp sum.
        num2[len(num2) - count] = temp % 10

        # remember the carry (10's place) for next sum.
        carry = temp // 10

    # Check for bugs...compare the answer with the next Fibonacci number
    assert value(num1) == next(f)

    if num1[0] != 0:
        flag = False

print(value(num1))

请注意，您可以通过记住负偏移量从末尾访问数组（num2[-1] 是数组中的最后一项）和range 可以倒数来使 for 循环更简单：

for count in range(-1,-len(num2)-1,-1):
    temp = num2[count] + num1[count] + carry
    num1[count] = num2[count]
    num2[count] = temp % 10
    carry = temp // 10 # remember carry for next digit

Answer 2

它比我的 cmets 更棘手，但这个版本可以正常工作：

num1 = [0]*10
num2 = [0]*10
num2[len(num2)-1] = 1
sum = 0
carry = 0
flag = True

while (flag):
    #Main for loop to work through the array
    for count in range (1, len(num2)):
        sum = num2[len(num2) - count] + num1[len(num1) - count] + carry
        num1[len(num1) - count] = num2[len(num2) - count]
        if sum > 9:
            sum = sum % 10
            carry = 1
        else:
            carry = 0
        num2[len(num2) - count] = sum
    if carry == 1:
        num2[0] = num2[0] + 1
    print(num2)
    if num2[0] != 0:
        flag = False

您还必须在应用进位之前复制到 new1，即使是在下一个较高位执行此操作时...

Answer 3

这是我认为您想要了解的内容的一个更简洁的版本。

#Init of Fib variables 
a = 0
b = 1
num = 10  #Change this to the number of fib calculation loops.
x = 0
output_arr_len = 100 #Change this to 10 for testing, as it's easier to see.    

while x < num:
    #Calculate Fib Sequence
    c = a + b 
    a = b
    b = c 
    x += 1

    #Output Array
    print([0] * (output_arr_len - len(str(c))) + [int(i) for i in str(c)])

下面是前 20 个循环的输出，output_arr_len 设置为 10。

[0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 2]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 3]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 5]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 8]
[0, 0, 0, 0, 0, 0, 0, 0, 1, 3]
[0, 0, 0, 0, 0, 0, 0, 0, 2, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 3, 4]
[0, 0, 0, 0, 0, 0, 0, 0, 5, 5]
[0, 0, 0, 0, 0, 0, 0, 0, 8, 9]
[0, 0, 0, 0, 0, 0, 0, 1, 4, 4]
[0, 0, 0, 0, 0, 0, 0, 2, 3, 3]
[0, 0, 0, 0, 0, 0, 0, 3, 7, 7]
[0, 0, 0, 0, 0, 0, 0, 6, 1, 0]
[0, 0, 0, 0, 0, 0, 0, 9, 8, 7]
[0, 0, 0, 0, 0, 0, 1, 5, 9, 7]
[0, 0, 0, 0, 0, 0, 2, 5, 8, 4]
[0, 0, 0, 0, 0, 0, 4, 1, 8, 1]
[0, 0, 0, 0, 0, 0, 6, 7, 6, 5]
[0, 0, 0, 0, 0, 1, 0, 9, 4, 6]