Description：
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.
思路：
题目的意思是，在已有点数不超过K的情况下，可以从[1, W]中任选一个数，保证和不超过N。求最后当点数超过K时，这个点数小于N的概率。这个问题其实类似于爬楼梯，每次可跨上[1, W]阶楼梯。
采用动态规划，dp[i]表示点数为i的概率，只有当已有点数小于K时，才能再取数，所以有：
当i<=K时，dp[i] = (前W个dp之和) / W
当K< i< K+W时，前一次最多只能是K-1点，最少为i-W点，所以dp[i] = (dp[K-1]+dp[K-2]+…+dp[i-W])/W
当i >= K+W时，无论如何都无法取到，dp[i] = 0
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] + [0.0] * N
# Wsum表示(前W个dp之和)/W
Wsum = 1.0000
for i in range(1, N + 1):
dp[i] = Wsum / W
# 因为当i>=K时，不能再取数，因此后面的概率不能累加
if i < K: Wsum += dp[i]
# 因为只需要计算前w个dp之和，所以当i>=W时，减去最前面的dp。
if 0 <= i - W < K: Wsum -= dp[i - W]
return sum(dp[K:])