剑指 Offer 60. n个骰子的点数

题目描述

把n个骰子扔在地上，所有骰子朝上一面的点数之和为s。输入n，打印出s的所有可能的值出现的概率。

你需要用一个浮点数数组返回答案，其中第 i 个元素代表这 n 个骰子所能掷出的点数集合中第 i 小的那个的概率。

示例1：

输入: 1
输出: [0.16667,0.16667,0.16667,0.16667,0.16667,0.16667]

示例2：

输入: 2
输出: [0.02778,0.05556,0.08333,0.11111,0.13889,0.16667,0.13889,0.11111,0.08333,0.05556,0.02778]

限制：

1 <= n <= 11

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/nge-tou-zi-de-dian-shu-lcof

思路解析

记\(n\)个骰子，点数和为\(s\)的概率为\(P(n,s)\)，易得以下递推关系：

\[P(n,s) = \frac{1}{6} \cdot \sum_{i = 1}^6 P(n-1,s-i) \]

上式为动态规划状态转移方程，使用二维矩阵存储\(P(n,s)\)的值。

代码实现

class Solution {
public:
    vector<double> twoSum(int n) {
        int s = n * 6;
        vector<vector<double>> probability_mat(n + 1, vector<double>(s + 1, 0));
        double base_p = 1.0 / 6.0;
        for(int i = 1; i <= 6; i++) 
            probability_mat[1][i] = base_p;
        for(int dice = 2; dice <= n; dice++) {
            for(int ts = 1; ts <= s; ts++) {
                for(int i = 1; i <= 6; i++)
                    if(i <= ts)
                        probability_mat[dice][ts] += probability_mat[dice - 1][ts - i] / 6.0;
            }
        }
        vector<double> result;
        for(int i = 0; i < s + 1; i++) {
            if(probability_mat[n][i] > 0)
                result.push_back(probability_mat[n][i]);
        }
        return result;
    }
};