在 Python 中生成斐波那契级数

Question

我想用Python简单地生成fibonacci序列。但不知怎么的，我看不出正确的系列。例如，如果我输入了3，那么正确的答案应该是：1 1 2 3

下面是我的代码。有人能指出这有什么问题吗：

def genfibonacci(no):
    if no <= 1:
        return no
    else:
        sum = genfibonacci(no - 1) + genfibonacci(no - 2)
    print (sum)
    return(sum)

number = int(input())
genfibonacci(number)

提前谢谢。

Answer 1

这里最简单的斐波纳契数列代表：

#Fibonacci series in Short Code

#initilize the base list
ls2=[0,1]

#user input: How many wants to print
c=int(input('Enter required numbers:'))

#fibonacci Function to add last two elements of list
ls2.extend((ls2[i-1]+ls2[i-2]) for i in range(2,c))

#This will print recuired numbers of list
print(ls2[0:c])

如果您想创建功能，那么：

#Fibonacci series in Short Code

#initilize the base list
ls2=[0,1]
#user input: How many wants to print
c=int(input('Enter required numbers:'))

#fibonacci Function to add last two elements of list
def fibonacci(c):    

    ls2.extend((ls2[i-1]+ls2[i-2]) for i in range(2,c))    
    #This will print required numbers of list
    print(ls2[0:c])

fibonacci(c)

Answer 2

在python中有很多方法可以计算斐波那契数列..

def fib(n):
 a,b = 1,1
 for i in range(n-1):
  a,b = b,a+b
 return a
print fib(5)

def fibR(n):
 if n==1 or n==2:
  return 1
 return fib(n-1)+fib(n-2)
print fibR(5)

a,b = 0,1
def fibI():
 global a,b
 while True:
  a,b = b, a+b
  yield a
f=fibI()
f.next()
f.next()
f.next()
f.next()
print f.next()

def memoize(fn, arg):
 memo = {}
 if arg not in memo:
  memo[arg] = fn(arg)
  return memo[arg]

fibm = memoize(fib,5)
print fibm

class Memoize:
 def __init__(self, fn):
  self.fn = fn
  self.memo = {}
 def __call__(self, arg):
  if arg not in self.memo:
   self.memo[arg] = self.fn(arg)
   return self.memo[arg]

@Memoize
def fib(n):
 a,b = 1,1
 for i in range(n-1):
  a,b = b,a+b
 return a
print fib(5)

Answer 3

部分问题是在计算时打印（除了if＜code＞no＜/code＞

def genfibonacci(no):
    if no <= 1:
        sum = no
    else:
        sum = genfibonacci(no-1) + genfibonacci(no-2)
    return sum

>>> [genfibonacci(i) for i in range(4)]
[0, 1, 1, 2]
>>> [genfibonacci(i) for i in range(5)]
[0, 1, 1, 2, 3]

此范围从 0 开始，因此您可以根据需要将其删除。

因为4的< code>genfibonacci将调用32和2，而32和2又将调用2和1，所以对于同一个数字，您的print语句将出现不止一次。而对于< code>no则完全不是