[ACM_HDU_1297递归推导]Children’s Queue
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Children’s Queue
There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?
Input
There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)
Output
For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.
Sample Input
123
Sample Output
124
题目大概意思是n个小该排队,女孩不能单独一个人站着,所以要么没有女生,要么>=2位女孩站在一起,问一共有多少种排法?
分析:
用F(n)表示n个人的合法队列,作如下分析:
按照最后一个人的性别分析,他要么是男,要么是女,所以可以分两大类讨论:
Case 1:
如果n个人的合法队列的最后一个人是男,则对前面n-1个人的队列没有任何限制,他只要站在最后即可,所以,这种情况一共有F(n-1);
Case 2:
如果n个人的合法队列的最后一个人是女,则要求队列的第n-1个人务必也是女生,这就是说,限定了最后两个人必须都是女生,这又可以分两种情况:
2.1、如果队列的前n-2个人是合法的队列,则显然后面再加两个女生,也一定是合法的,这种情况有F(n-2);2.2、但是,难点在于,即使前面n-2个人不是合法的队列,加上两个女生也有可能是合法的,当然,这种长度为n-2的不合法队列,不合法的地方必须是尾巴,就是说,这里说的长度是n-2的不合法串的形式必须是“F(n-4)+男+女”,这种情况一共有F(n-4).
所以,通过以上的分析,可以得到递推的通项公式:F(n)=F(n-1)+F(n-2)+F(n-4) (n>3)
然后就是对n<=3 的一些特殊情况的处理了,显然:
F(0)=1 (没有人也是合法的,这个可以特殊处理,就像0的阶乘定义为1一样)
F(1)=1
F(2)=2
F(3)=4
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