67. 两个闲玩娱乐

1.扑克牌的顺子

从扑克牌中随机抽5 张牌，判断是不是一个顺子，即这5 张牌是不是连续的。2-10 为数字本身，A 为1，J 为11，Q 为12，K 为13，而大小王可以看成任意数字。

HANDWRITING:

大小王当0，5张牌排好序

bool straight (int poker[]) {int jokers = 0, i;for (i = 0; poker[i] == 0; ++i) ++jokers;int start = poker[i];for(; i < 5; ++i) {if (poker[i] == start) ++start;else if (jokers > 0) --jokers;else return false;}return true;}

1、没有考虑到当ooker[i] != start时，i不能向后移

改正：

bool straight (int poker[]) {int jokers = 0, i;for (i = 0; poker[i] == 0; ++i) ++jokers;int start = poker[i];while (i < 5) {if (poker[i] == start) ++start, ++i;else if (jokers > 0) --jokers, ++start;else return false;}return true;}

2.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。

HANDWRITING:

计算出各个S出现的次数，概率即S[ i ] / 6^N

void dice (int n) {int *s = new int[6 * n + 1];for (int i = 1; i <= 6; ++i) s[i] = 1;for (int c = 2; c <= n; ++c) {for (int d = c * d; d >= c ; --d) {s[d] = s[d-1] + s[d-2] + s[d-3] + s[d-4] + s[d-5] + s[d-6];}}for (int i = 1; i <= 6 * n; ++i) cout<<s[i]<<" ";cout<<endl;}

1、没考虑d-6是否大于0

2、没释放new出来的空间

改正：

void dice (int n) {int *s = new int[6 * n + 1];for (int i = 1; i <= 6; ++i) s[i] = 1;for (int c = 2; c <= n; ++c) {for (int d = c * 6; d >= c ; --d) {for (int i = 1; i < d && i <= 6; ++i)s[d] += s[d - i];}}for (int i = 1; i <= 6 * n; ++i) cout<<s[i]<<" ";cout<<endl;delete [] s;}

ANSWER:

FROM:http://blog.csdn.net/v_july_v/article/details/6870251
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.

void listAllProbabilities(int n) {  int[][] f = new int[6*n+1][];  for (int i=0; i<=6*n; i++) {    f[i] = new int[n+1];  }  for (int i=1; i<=6; i++) {    f[i][1] = 1;  }  for (int i=1; i<=n; i++) {    f[i][i] = 1;  }  for (int i=2; i<=n; i++) {    for (int j=i+1; j<=6*i; j++) {      for (int k=(j-6<i-1)?i-1:j-6; k<j-1; k++)        f[j][i] += f[k][i-1];    }  }  double p6 = Math.power(6, n);  for (int i=n; i<=6*n; i++) {    System.out.println(“P(S=”+i+”)=”+((double)f[i][n] / p6));  }}