leetcode 837

问题

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:

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Input: N = 10, K = 1, W = 10
Output: 1.00000
Explanation:  Alice gets a single card, then stops.

Example 2:

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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:

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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.

分析

分析可知获得任意一个点的大小与前面的点的大小是有关系的。因此，我们可以自顶向下递归地求出每一个点的概率，然后将 大于等于 K，小于等于 N 的点的概率相加就可以得到答案。但如果不进行优化答案自然就会TLE，如代码1。分析可知，每个点依赖于其前面的W个点，在我们依次求出 $1-N$ 这些点的概率的时候其实是可以重复利用之前的结果，也就是我们一直维护待求点前的$W$ 个点的概率和，这样求解每一个点的概率的时间复杂度就是 $O(1)$，而不是 $O(W)$ 。

代码1

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class Solution {
public:
    double new21Game(int N, int K, int W) {
        vector<double> memo(K+W);
        // K .. K+i ... N ... W
        if (N >= K+W-1 || K == 0)  // corner case
            return 1;
        double ans = 0.0f;
        memo[0] = 1;
        for (int i = N; i >= K; --i) {
            ans += dp(i, K, W, memo);
        }
        return ans;
    }
    double dp(int x, int K, int W, vector<double> &memo) {
        if (memo[x] != 0) return memo[x];
        double ans = 0;
        for (int i = 1; i <= W; ++i) {
            int y = x-i;
            if (y >= K) continue;
            if (y < 0) break;
            ans += dp(y, K, W, memo);
        }
        ans /= W;
        return memo[x] = ans;
    }
};

代码2

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class Solution {
public:
    double new21Game(int N, int K, int W) {
        vector<double> dp(N+1);
        // K .. K+i ... N ... W
        if (N >= K+W-1 || K == 0) return 1;  // corner case
        double ans = 0.0f;
        double sumW = 1;
        dp[0] = 1;
        for (int i = 1; i <= N; ++i) {
            dp[i] = sumW/W;
            if (i < K)
                sumW += dp[i];
            else
                ans  += dp[i];
            if (i-W >= 0)
                sumW -= dp[i-W];
        }
        return ans;
    }
};