【剑指offer】【动态规划】60. n个骰子的点数

题目链接：https://leetcode-cn.com/problems/nge-tou-zi-de-dian-shu-lcof/

朴素版

状态表示：f[i][j]表示投掷wani枚骰子，j点数的出现次数；
状态计算：f[i][j] = f[i - 1][j - 1 ] + f[i - 1][j - 1 ] + ... + f[i - 1][j - 6]；投掷完i枚，点数j出现的次数，等于投掷完i - 1枚, 点数j - 1, j - 2， ... , j - 6出现的次数之和；
状态初始化：f[1][1 ~ 6] = 1

投掷n枚骰子，出现的数字个最大为n * 6；不用坐标0，因此初始化的时候，f的大小为[n + 1, 6 * n + 1]

class Solution {
public:
    vector<double> twoSum(int n) {
        vector<vector<int>> f(n + 1, vector<int>(6 * n + 1, 0)); 
        f[0][0] = 1;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= 6 * i; j++)
                for(int cur = 1; cur <= 6; cur++)
                    if(j - cur >= 0) 
                        f[i][j] += f[i - 1][j - cur];
                
        double total = 0;
        for(int i = n; i <= 6 * n; i++)
            total += f[n][i];

        vector<double> res(6 * n + 1);
        for(int i = n; i <= 6 * n; i++)
            res[i] = f[n][i] / total;

        return vector<double> (res.begin() + n, res.end());
    }
};