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]

感觉难点在于怎么表达不可能的情况。
class Solution {
    public double[] dicesProbability(int n) {
        double[][] dp = new double[n][6 * n + 1];
        for (int i = 1; i < 7; i++) {
            dp[0][i] = 1;
        }
        for (int i = 1; i < n; i++) {
            for (int j = i; j <= 6 * n; j++) {
                for (int k = 0; k < 6; k++) {
                    if (j - k - 1 >= 0 && dp[i - 1][j - k - 1] != 0) {
                        dp[i][j] += dp[i - 1][j - k - 1];
                    }
                }
            }
        }
        double total = 0;
        for (int i = 0; i < dp[0].length; i++) {
            total += dp[dp.length - 1][i];
        }
        for (int i = 0; i < dp[0].length; i++) {
            dp[dp.length - 1][i] /= total;
        }
        return Arrays.copyOfRange(dp[dp.length - 1], n, dp[dp.length - 1].length);
    }
}