Python程序计算到达第n个楼梯的方式

有n个楼梯，站在底部的人想爬到顶部。此人一次可以爬 1 个楼梯或 2 个楼梯。数一数路数，人能登顶。

考虑图中所示的示例。 n 的值为 3。有 3 种方法可以到达顶部。该图取自 Easier Fibonacci 谜题

Python3

# A program to count the number of ways to reach n'th stair
 
# Recurssive program to find n'th fibonacci number
def fib(n):
    if n <= 1:
        return n
    return fib(n-1) + fib(n-2)
 
# returns no. of ways to reach s'th stair
def countWays(s):
    return fib(s + 1)
 
# Driver program
 
s = 4
print ("Number of ways =",countWays(s))
 
# Contributed by Harshit Agrawal

输出：

Number of ways =  5

上述实现的时间复杂度是指数级的（黄金比例提高到 n 次方）。可以使用前面讨论的斐波那契函数优化来优化它以在 O(Logn) 时间内工作。

Python3

# A program to count the number of ways to reach n'th stair
 
# Recursive function used by countWays
def countWaysUtil(n, m):
    if n <= 1:
        return n
    res = 0
    i = 1
    while i<= m and i<= n:
        res = res + countWaysUtil(n-i, m)
        i = i + 1
    return res
     
# Returns number of ways to reach s'th stair   
def countWays(s, m):
    return countWaysUtil(s + 1, m)
     
 
# Driver program
s, m = 4, 2
print ("Number of ways =", countWays(s, m))
 
# Contributed by Harshit Agrawal

输出：

Number of ways = 5

Python3

# A program to count the number of ways to reach n'th stair
 
# Recursive function used by countWays
def countWaysUtil(n, m):
    res = [0 for x in range(n)] # Creates list res with all elements 0
    res[0], res[1] = 1, 1
     
    for i in range(2, n):
        j = 1
        while j<= m and j<= i:
            res[i] = res[i] + res[i-j]
            j = j + 1
    return res[n-1]
 
# Returns number of ways to reach s'th stair
def countWays(s, m):
    return countWaysUtil(s + 1, m)
     
# Driver Program
s, m = 4, 2
print ("Number of ways =", countWays(s, m))
     
# Contributed by Harshit Agrawal

输出：

Number of ways = 5

有关详细信息，请参阅有关到达第 n 级楼梯的计数方式的完整文章！