作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/
目录
- 动态规划
题目地址：https://leetcode.com/problems/new-21-game/description/
题目描述
Alice plays the following game, loosely based on the card game “21”.
Alice starts with 0 points, and draws numbers while she has less than K points. During each draw, she gains an integer number of points randomly from the range [1, W], where W is an integer. Each draw is independent and the outcomes have equal probabilities.
Alice stops drawing numbers when she gets K or more points. What is the probability that she has N or less points?
Example 1:
Input: N = 10, K = 1, W = 10
Output: 1.00000
Explanation: Alice gets a single card, then stops.
Example 2:
Input: N = 6, K = 1, W = 10
Output: 0.60000
Explanation: Alice gets a single card, then stops.
In 6 out of W = 10 possibilities, she is at or below N = 6 points.
Example 3:
Input: N = 21, K = 17, W = 10
Output: 0.73278
Note:
- 0 <= K <= N <= 10000
- 1 <= W <= 10000
- Answers will be accepted as correct if they are within 10^-5 of the correct answer.
The judging time limit has been reduced for this question.
题目大意
刚开始的时候，有0分，她会已知在[1,W]中随机选数字，直到有K分或者K分以上停止。问她能够正好得到N分或者更少分的概率。
解题方法
动态规划
类似爬楼梯的问题，每次可以跨[1,W]个楼梯，当一共爬了K个和以上的台阶时停止，问这个时候总台阶数<=N的概率。
使用动态规划，dp[i]表示得到点数i的概率，只有当现在的总点数少于K的时候，才会继续取数。那么状态转移方程可以写成：
- 当
i <= K时，dp[i] = （前W个dp的和）/ W；(爬楼梯得到总楼梯数为i的概率） - 当
K < i < K + W时，那么在这次的前一次的点数范围是[i - W, K - 1]。我们的dp数组表示的是得到点i的概率，所以dp[i]=(dp[K-1]+dp[K-2]+…+dp[i-W])/W.（可以从前一次的基础的上选[1,W]个数字中的一个） - 当i>=K+W时，这种情况下无论如何不都应该存在的，所以dp[i]=0.
时间复杂度是O(N)，空间复杂度是O(N).
class Solution(object):
def new21Game(self, N, K, W):
"""
:type N: int
:type K: int
:type W: int
:rtype: float
"""
if K == 0: return 1
dp = [1.0] + [0] * N
tSum = 1.0
for i in range(1, N + 1):
dp[i] = tSum / W
if i < K:
tSum += dp[i]
if 0 <= i - W < K:
tSum -= dp[i - W]
return sum(dp[K:])
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日期
2018 年 11 月 1 日 —— 小光棍节
