斐波那契数(python)

斐波那契数(python)

def f(x):
    if x ==0 or x == 1:
        return 1
    else:
        return f(x-1)+f(x-2)

@cal_time  #装饰器 确定运行时间
def fib(n):
    return f(n)
print(fib(30))
#给递归函数加装饰器,会出现重复打印
#重新定义一个函数,调用,这样就不会重复打印

fib running time: 0.2029705047607422 secs.
1346269

@cal_time
def f1(n):
    l = [1,1]
    for i in range(2,n+1):
        l.append(l[-1]+l[-2])
    return l[n]
print(f1(5))

f1 running time: 0.0 secs.
8

@cal_time
def f2(n):
    a, b, c = 1, 1, 1
    for i in range(2,n+1):
        c = a + b
        a = b
        b = c
    return(c)
print(f2(5))

f2 running time: 0.0 secs.
8

#确定运行时间
import time

def cal_time(func):
    def wrapper(*args, **kwargs):
        t1 = time.time()
        result = func(*args, **kwargs)
        t2 = time.time()
        print("%s running time: %s secs." % (func.__name__, t2 - t1))
        return result

    return wrapper

一段有n个台阶组成的楼梯,小明从楼梯的最底层向最高处前进,它可以选择一次迈一级台阶或者一次迈两级台阶.问:他有多少种不同的走法?

#从1 开始的斐波那契数列,1,2,3,5,8,13...
def taijie(n):
    a = 1
    b = 2
    c = 0
    for i in range(3,n+1):#从1开始
        c = a + b
        a = b
        b = c
    return c
print(taijie(5))