原题链接在这里：

题目：

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 = 10Output: 1.00000Explanation:  Alice gets a single card, then stops.

Example 2:

Input: N = 6, K = 1, W = 10Output: 0.60000Explanation:  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 = 10Output: 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.

题解：

When the draws sum up to K, it stops, calculate the possibility K<=sum<=N.

Think about one step earlier, sum = K-1, game is not ended and draw largest card W. K-1+W is the maximum sum could get when game is ended. If it is <= N, then for sure the possiblity when games end ans sum <= N is 1.

Because the maximum is still <= 1.

Otherwise calculate the possibility sum between K and N. 

Let dp[i] denotes the possibility of that when game ends sum up to i.

i is a number could be got equally from i - m and draws value m card.

Then dp[i] should be sum of dp[i-W] + dp[i-W+1] + ... + dp[i-1], devided by W.

We only need to care about previous W value sum, accumlate winSum, reduce the possibility out of range.

Time Complexity: O(N).

Space: O(N).

AC Java:   

1 class Solution { 2     public double new21Game(int N, int K, int W) { 3         if(K == 0 || K-1+W <= N){ 4             return 1; 5         } 6          7         if(K > N){ 8             return 0; 9         }10         11         double [] dp = new double[N+1];12         dp[0] = 1.0;13         double winSum = 1;14         15         double res = 0.0;16         for(int i = 1; i<=N; i++){17             dp[i] = winSum/W;18             19             if(i
     
      = W){26                 winSum -= dp[i-W];27             }28         }29         30         return res;31     }32 }
     

类似.