LeetCode[Hard]------837. New 21 Game

问题描述

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.
4. The judging time limit has been reduced for this question.

简单翻译一下，此题从21点游戏演变过来，W表示每次能从[1,W]区间中取点, N表示总点数最大不能超过N, K表示取点一旦超过K就停手。题目要求停手且不爆掉的概率。

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思路：

此题看成一个dp问题。概率dp[i]应该等于前W个dp的和除以W,我们所要的结果就是dp[K]到dp[N]的和。注意当i>K是，由于停手的关系，前一个的概率只能是dp[K-1]而不是dp[i-1]。

代码:

public double new21Game(int N, int K, int W) {
		if (K == 0 || N >= K + W)
			return 1;
		double dp[] = new double[N + 1], Wsum = 1, res = 0;
		dp[0] = 1;
		for (int i = 1; i <= N; ++i) {
			dp[i] = Wsum / W;
			if (i < K)
				Wsum += dp[i];
			else
				res += dp[i];
			if (i - W >= 0)
				Wsum -= dp[i - W];
		}
		return res;
	}