优化此递归功能(动态编程)

Question

我正在求解一个非常简单的算法问题，要求递归和记忆。下面的代码正常运行，但不符合时间限制。有人建议我优化尾部递归，但这不是尾部递归。这只是一种研究材料，而不是作业。

问题

•如果下雨，蜗牛可以每天攀升2m，否则可以攀升1M。

•每天下雨的可能性为75％。

•给定天数(＆lt; = 1000(和高度(＆lt; = 1000(，计算蜗牛可以从井中取出的概率(比高度攀升更多(

此Python代码是通过递归和回忆实现的。

import sys
sys.setrecursionlimit(10000)
# Probability of success that snails can climb 'targetHeight' within 'days'
def successRate(days, targetHeight):
    global cache
    # edge case
    if targetHeight <= 1:
        return 1
    if days == 1:
        if targetHeight > 2:
            return 0
        elif targetHeight == 2:
            return 0.75
        elif targetHeight == 1:
            return 0.25
    answer = cache[days][targetHeight]
    # if the answer is not previously calculated
    if answer == -1:
        answer = 0.75 * (successRate(days - 1, targetHeight - 2)) + 0.25 * (successRate(days - 1, targetHeight - 1))
        cache[days][targetHeight] = answer
    return answer

height, duration = map(int, input().split())
cache = [[-1 for j in range(height + 1)] for i in range(duration + 1)] # cache initialized as -1
print(round(successRate(duration, height),7))

Answer 1

这很简单。所以这只是一个提示。对于intal部分集：

# suppose cache is allocated
cache[1][1] = 0.25
cache[1][2] = 0.75
for i in range(3,targetHeight+1):
    cache[1][i] = 0
for i in range(days+1):
    cache[i][1] = 1
    cache[i][0] = 1

，然后尝试使用初始化值重写递归部分(您应该迭代自下而上，喜欢以下(。最后，返回cache[days][targetHeight]的值。

for i in range(2, days+1):
    for j in range(2, targetHeight+1):
        cache[i][j] = 0.75 * cache[i-1][j-2] + 0.25 * cache[i-1][j-1]