[Programming question] points of n dice

67. Two games (Computing ).
2. n dice points.
Place n dice on the ground, and the sum of all dice toward the point above is S. Input N,
Print the probability that all possible values of S appear.

 

Idea: recursively convert the possible condition variables of each dice, and record the total number of occurrences of various and S as probability.

/* 67. Two playmates (Computing ). 2. n dice points. Place n dice on the ground, and the sum of all dice toward the point above is S. Input n to print the probability of all possible values of S. */# Include <stdio. h> # include <stdlib. h> # include <string. h> # include <math. h> void getallarr (int n, int * record) // recursively obtain each possible occurrence and Occurrence Count {static int sum = 0; static int num = N; if (0 = N) {record [Sum-num] ++; return ;}for (INT I = 1; I <= 6; I ++) {sum + = I; getallarr (n-1, record); sum-= I ;}} void printall (INT N) {If (0 = N) {printf ("input error");} int * record = (int *) malloc (5 * n + 1) * sizeof (INT); memset (record, 0, (5 * n + 1) * sizeof (INT); getallarr (n, record); For (INT I = N; I <= 6 * n; I ++) {printf ("% d: % d/% d \ n", I, record [I-n], INT (POW (6.0, n )));}
Free (record);} int main () {printall (8); Return 0 ;}

 

Looking at the answers on the Internet, we found that we can use dynamic planning.

Http://blog.csdn.net/whuqin/article/details/6639187

1. The current variables include: number of dice, points, and. When there are K dice and the sum of points is N, the number of occurrences is recordedF (k, n). What is the relationship with the K-1 dice stage?

2. When I have a K-1 dice and then add a dice, the dice points can only be 1, 2, 3, 4, 5 or 6. The K dice get the sum of points and N:

(K-1, n-1): Number 1 for the K dice

(K-1, N-2): K dice shot points 2

(K-1, n-3): K dice shot points 3

....

(K-1, n-6): K dice shot 6 points

On the basis of the K-1 dice, then add a dice appear points and N results only these 6 cases!

Therefore:F (k, n) = f (K-1, n-1) + f (K-1, N-2) + f (K-1, n-3) + f (K-1, n-4) + f (K-1, n-5 (K-1, n-6) + f)

3. There is 1 dice, F () = f () = 1.