LeetCode[Hard]------837. New 21 Game
問題描述
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.
簡單翻譯一下，此題從21點遊戲演變過來，W表示每次能從[1,W]區間中取點, N表示總點數最大不能超過N, K表示取點一旦超過K就停手。題目要求停手且不爆掉的概率。
思路：
此題看成一個dp問題。概率dp[i]應該等於前W個dp的和除以W,我們所要的結果就是dp[K]到dp[N]的和。注意當i>K是，由於停手的關係，前一個的概率只能是dp[K-1]而不是dp[i-1]。
程式碼:
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;
}