扔骰子，求概率

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。
ANSWER:
All the possible values includes n to 6n. All the event number is 6^n.
For n<=S<=6n, the number of events is f(S, n)
f(S,n) = f(S-6, n-1) + f(S-5, n-1) + … + f(S-1, n-1)
number of events that all dices are 1s is only 1, and thus f(k, k) = 1, f(1-6, 1) = 1, f(x, 1)=0 where x<1 or x>6, f(m, n)=0 where m<n 
Can do it in DP.

using System;using System.Collections.Generic;using System.IO;using System.Diagnostics;namespace ConsoleApp{    class RunClass    {        public static void Main()        {            int sum = 28;            int N = 27;            Stopwatch sw = new Stopwatch();            sw.Start();            int r1 = CalcBruteForce(sum, N);//too too too slow            sw.Stop();            Console.WriteLine("" + r1 + ":" +                 sw.ElapsedMilliseconds);            sw.Reset();            int r2 = CalcDP(sum, N);            sw.Stop();            Console.WriteLine("" + r2 + ":" +                 sw.ElapsedMilliseconds);        }        public static int CalcDP(int sum, int n)        {            int[,] arr = new int[sum + 1, n + 1];            for (int j = 1; j < arr.GetLongLength(1); j++)//n                for (int i = 0; i < arr.GetLongLength(0); i++)//sum                {                    if (i <= 0)                        arr[i, j] = 0;                    else if (i > 6 * j)                        arr[i, j] = 0;                    else if (j == 1)                        arr[i, j] = 1;                    //dp                    else                    {                        int m = 0;                        for (int t = 1; t <= 6; t++)                        {                            if (i - t >= 0)                                m += arr[i - t, j - 1];                        }                        arr[i, j] = m;                    }                }            return arr[sum,n];        }        public static int CalcBruteForce(int sum, int n)        {            if (sum <= 0)                return 0;            if (sum == n)                return 1;            if (sum > 6 * n)                return 0;            if(n==1)                return 1;            int rst = CalcBruteForce(sum - 1, n - 1) +                         CalcBruteForce(sum - 2, n - 1) +                        CalcBruteForce(sum - 3, n - 1) +                        CalcBruteForce(sum - 4, n - 1) +                        CalcBruteForce(sum - 5, n - 1) +                        CalcBruteForce(sum - 6, n - 1);            return rst;        }    }}