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;
}
};