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

动态规划
n个骰子，一共有6**n种情况
n=1, 和为s的情况有 F(n,s)=1 s=1,2,3,4,5,6
n>1 , F(n,s) = F(n-1,s-1)+F(n-1,s-2) +F(n-1,s-3)+F(n-1,s-4)+F(n-1,s-5)+F(n-1,s-6)
可以看作是从前(n-1)个骰子投完之后的状态转移过来。
其中F（N,S）表示投第N个骰子时，点数和为S的次数

class Solution:
    def twoSum(self, n: int) -> List[float]:

        dp = [ [0 for _ in range(6*n+1)] for _ in range(n+1)]
        for i in range(1,7):
            dp[1][i] = 1

        for i in range(2,n+1):
            for j in range(i,i*6+1):
                for k in range(1,7):
                    dp[i][j] +=dp[i-1][j-k]
        res = []
        for i in range(n,n*6+1):
            res.append(dp[n][i]*1.0/6**n)
        return res