目錄

• 題目描述
• 題目大意
• 解題方法
• 動態規劃
• 相似題目
• 參考資料
• 日期

題目描述

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:

1. 0 <= K <= N <= 10000
2. 1 <= W <= 10000
3. 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.

題目大意

剛開始的時候，有0分，她會已知在[1,W]中隨機選數字，直到有K分或者K分以上停止。問她能夠正好得到N分或者更少分的概率。

解題方法

動態規劃

類似爬樓梯的問題，每次可以跨[1,W]個樓梯，當一共爬了K個和以上的臺階時停止，問這個時候總檯階數<=N的概率。

使用動態規劃，dp[i]表示得到點數i的概率，只有當現在的總點數少於K的時候，才會繼續取數。那麼狀態轉移方程可以寫成：

1. 當i <= K時，dp[i] = （前W個dp的和）/ W；(爬樓梯得到總樓梯數為i的概率）
2. 當K < i < K + W時，那麼在這次的前一次的點數範圍是[i - W, K - 1]。我們的dp陣列表示的是得到點i的概率，所以dp[i]=(dp[K-1]+dp[K-2]+…+dp[i-W])/W.（可以從前一次的基礎的上選[1,W]個數字中的一個）
3. 當i>=K+W時，這種情況下無論如何不都應該存在的，所以dp[i]=0.

時間複雜度是O(N)，空間複雜度是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:])

相似題目

參考資料

日期

2018 年 11 月 1 日 —— 小光棍節