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)
原文地址:https://www.cnblogs.com/wintalau/p/11622271.html