[剑指 Offer] 60. n个骰子的点数

[frdmu]

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
   

解法：
动态规划。搞一个二维数组dp[i][j]，表示i个骰子点数为j时的概率。图示如下：

代码如下：

class Solution {
public:
    vector<double> dicesProbability(int n) {
        vector<vector<double>>  dp(n+1, vector<double>(6*n+1, 0));
        vector<double> res;

        for (int i = 1; i < 7; i++) {
            dp[1][i] = 1.0 / 6;
        }   
        for (int i = 1; i < n; i++) {
            for (int j = i; j <= 6*i; j++) {
                for (int s = 1; s < 7; s++) {
                    dp[i+1][j+s] += dp[i][j] * 1.0 / 6;
                }    
            }
        }
        for (int i = n; i <= 6*n; i++) {
            res.push_back(dp[n][i]);
        }

        return res;
    }
};

Refer:
剑指 Offer 60. n 个骰子的点数（动态规划，清晰图解)