斐波那契数列

# 方法一：
def fib(n):
    if n <= 2:
        return 1
    else:
        return fib(n-1) + fib(n-2)
for i in range(1,10):
    print(f‘fib({i}) = {fib(i)}‘)

# 方法二：
def fab(max):
    n, a, b = 0, 0, 1
    while n < max:
        print(b)
        a, b = b, a + b
        n += 1
fab(9)

# 方法二改良
def fab(max):
    n, a, b = 0, 0, 1
    L = []
    while n < max:
        L.append(b)
        a, b = b, a + b
        n += 1
    return L
fab(9)

# 方法一改良：使用生成器，节约内存空间
def fab(max):
    n, a, b = 0, 0, 1
    while n < max:
        yield b
        a, b = b, a + b
        n += 1
for n in fab(9):
    print(n)