leetcode 837. New 21 Game
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 = 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.
解法:
这个题大致的意思是在点数不超过K
的情况下,每次从[1,W]
中任选一个数字,求出当和点数大于等于K
时,小于等于N
的概率。
这个题我们可以看作是给定N个台阶,每次上升[1,W]
个台阶的问题,可以分三种情况,设当前已走步数为 test
,当前步数test
下的概率 dp[test]
- 当
i <= K
时,dp[i] = (前W个dp的和)/ W
;(爬楼梯得到总楼梯数为i的概率) - 当
K < i < K + W
时,那么在这次的前一次的点数范围是[i - W, K - 1]
。我们的dp数组表示的是得到点i的概率,所以dp[i]=(dp[K-1]+dp[K-2]+…+dp[i-W])/W
.(可以从前一次的基础的上选[1,W]个数字中的一个) - 当
i>=K+W
时,这种情况下无论如何不都应该存在的,所以dp[i]=0.
代码:
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] * N
tSum = 1.0
for i in range(1, N + 1):
dp[i] = tSum / W
if i < K:
tSum += dp[i]
#减去不符合题意的情况
if 0 <= i - W < K:
tSum -= dp[i - W]
return sum(dp[K:])