[剑指 Offer 60]n个骰子的点数

//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 
// 👍 167 👎 0

/*
* 剑指 Offer 60 n个骰子的点数
* 2021-02-18 11:47:26
* @author oxygenbytes
*/ 
#include "leetcode.h" 
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public:
    vector<double> dicesProbability(int n) {
        vector<vector<int>> dp(n + 1, vector<int>(6 * n + 1, 0));

        dp[0][0] = 1;
        for(int i = 1;i <= n;i++){
            for(int j = 1;j <= 6 * n;j++){
                for(int k = 1; k <= min(6, j);k++){
                    dp[i][j] += dp[i-1][j-k];
                }
            }
        }

        vector<double> ans;
        int sum = 0;
        for(int i = n;i <= 6 * n;i++) sum += dp[n][i];
        for(int i = n;i <= 6 * n;i++) ans.push_back((double)dp[n][i] / sum);
        return ans;
    }
};
//leetcode submit region end(Prohibit modification and deletion)