837. New 21 Game

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.

Dynamic programming用到一個DP array，dp[i]紀錄點數和為i的機率。 dp[i] = dp[i - W] * (1 / W) + dp[i - (W - 1)] * (1 / W) + ... + dp[i - 2] * (1 / W) + dp[i - 1] * (1 / W) = (dp[i - W] + dp[i - (W - 1)] + ... + dp[i - 1]) * (1 / W) = Wsum / W 再用一個變數Wsum紀錄一個大小最大為W的sliding window，紀錄前K個點數和的機率。

// Dynamic Programming
double new21Game(int N, int K, int W) { // time: O(N); space: O(N)
    if (K == 0 || N >= K + W) return 1.0;
    double res = 0.0, Wsum = 1.0;
    vector<double> dp(N + 1, 0.0);
    dp[0] = 1.0;
    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;
}