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
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代码:

class Solution {
public:
    vector<double> twoSum(int n) {
        vector<double> ans;
        double d = 1;
        long long f[12][67] = {1};
        for(int i = 1;i <= n;i ++) {
            d *= 6;
            for(int j = i;j <= i * 6;j ++) {
                for(int k = 1;k <= min(6,j - i + 1);k ++) {
                    f[i][j] += f[i - 1][j - k];
                }
            }
        }
        for(int i = n;i <= n * 6;i ++) {
            ans.push_back(f[n][i] / d);
        }
        return ans;
    }
};