LeetCode #837 - New 21 Game_leetcode 837 c++-优快云博客

本文链接：https://blog.youkuaiyun.com/LawFile/article/details/86697575

本文介绍了一种基于动态规划的方法来解决类似于卡牌游戏“21”的概率计算问题。通过构建一个概率分布模型，我们能够精确计算出玩家在获得特定点数范围内的概率。这种方法适用于各种参数设置，包括不同的目标点数、起始点数和每轮可能获得的最大点数。

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题目描述：

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:

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.

class Solution {
public:
    double new21Game(int N, int K, int W) {
        if(K==0) return 1;
        double preWSum=1.0;
        vector<double> dp(N+1,0); //dp[i]表示点数为i的概率，每个dp[i]都由前W个dp转移过来
        dp[0]=1.0; //初始条件是0，所以一开始0的概率是1
        for(int i=1;i<=N;i++)
        {
            dp[i]=preWSum/W; //表示前W个dp之和
            if(i<K) preWSum+=dp[i]; // 当i大于等于K时，dp[i]不能转移
            if(i-W>=0&&i-W<K) preWSum-=dp[i-W];
        }
        double result=0.0;
        for(int i=K;i<=N;i++) result+=dp[i];
        return result;
    }
};