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.

暴力DFS

count = 1.0
count, count, count, ..., ..., count += 0.1

count = 0.1
count, count, count, ..., ..., count += 0.01

代码

class Solution {
public double new21Game(int N, int K, int W) {
if (K == 0 && N == 0) return (double) 1.0;
if (N == 0) return (double) 0.0;
if (K == 0) return (double) 1.0;
n = N;
k = K;
w = W;
max = K - 1 + W;
count = new double[max + 1];
dfs(0, (double)1.0);
double sumProb = 0.0;
for (int i = K; i <= N; i++) sumProb += count[i];
return sumProb;
}

private int n;
private int k;
private int w;
private int max;
private double[] count;

private void dfs(int num, double prob) {
if (num < k) {
for (int i = 1; i <= w; i++) dfs(num + i, prob / w);
} else {
count[num] += prob;
}
}
}

结果 动态规划

代码

class Solution {
public double new21Game(int N, int K, int W) {
if (K == 0 && N == 0) return (double) 1.0;
if (N == 0) return (double) 0.0;
if (K == 0) return (double) 1.0;
int max = K - 1 + W;
double[] dp = new double[max + 1];
dp = 1.0;
for (int i = 1; i <= max; i++) {
for (int j = i - W; j < i; j++) {
if (j >= 0 && j < K) dp[i] += dp[j];
}
dp[i] /= W;
}
double sumProb = 0.0;
for (int i = K; i <= N; i++) sumProb += dp[i];
return sumProb;
}
}

结果 dp[i]本身就记录跳转概率总和

• dp = 1.0
• dp[i]: probability of get points i
• dp[i] = sum(last W dp values) / W
• Wsum = sum(last W dp values)

到达[1~10]点的概率总和 / 10

17开始，跳转结束。比如点20

到达[10~16]点的概率总和 / 10

17, 18, 19三点不跳转了。他们应该被计入到最后的输出结果中。

代码

class Solution {
public double new21Game(int N, int K, int W) {
if (K == 0 && N == 0) return (double) 1.0;
if (N == 0) return (double) 0.0;
if (K == 0) return (double) 1.0;
double[] dp = new double[N + 1];
dp = 1.0;
double Wsum = 1.0, res = 0.0;
for (int i = 1; i <= N; i++) {
dp[i] = Wsum / W;
if (i < K) {
Wsum += dp[i];
} else {
res += dp[i];
}
if (i - W >= 0) Wsum -= dp[i - W];
}
return res;
}
}

结果 