def fib2(n): # return Fibonacci series up to n """Return a list containing the Fibonacci series up to n.""" result = [] a, b = 0, 1 while a < n: result.append(a) # see below a, b = b, a+b print(b) return result f100 = fib2(100)
时间: 2024-04-26 11:22:05 浏览: 6
这段代码定义了一个函数fib2,它接受一个整数n作为参数,并返回一个列表,其中包含了小于n的所有斐波那契数列元素。在函数内部,定义了一个空列表result和两个变量a和b,它们的初始值分别为0和1。接着,使用while循环,当a小于n时,将a添加到result列表中,然后更新a和b的值,使它们变为b和a+b。最后,函数返回result列表,并将其赋值给变量f100。在函数执行过程中,还会输出每个斐波那契数列元素的后一个元素。
相关问题
修改一下def fib(n): if n <= 1: return n else: return fib(n-1) + fib(n-2) def main(): n=input() print(fib(n))
你需要将输入的字符串类型转换为整型,否则会出现类型错误。修改后的代码如下所示:
```python
def fib(n):
if n <= 1:
return n
else:
return fib(n-1) + fib(n-2)
def main():
n = int(input())
print(fib(n))
```
这样,当你运行 `main()` 函数并输入一个整数时,程序就会计算斐波那契数列的第 n 项并输出结果。
def fib(n=5):
This is a Python function that generates the first n Fibonacci numbers. Here is the code:
```python
def fib(n=5):
a, b = 0, 1
for _ in range(n):
print(a)
a, b = b, a + b
```
The function initializes two variables, `a` and `b`, to 0 and 1 respectively. It then enters a loop that iterates `n` times. In each iteration, it prints the current value of `a` and updates `a` and `b` to be the next two Fibonacci numbers. Finally, the function returns `None` (implicitly, since there is no `return` statement). By default, the function generates the first 5 Fibonacci numbers, but you can pass a different value of `n` to generate more or fewer numbers. For example, `fib(10)` would generate the first 10 Fibonacci numbers.