HDOJ1297 Children’s Queue[递推+大数]

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6000    Accepted Submission(s): 1887

Problem Description

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

1 2 3

 

Sample Output

1 2 4

 

Author

SmallBeer (CML)

 

Source

杭电ACM集训队训练赛（VIII）

 

Recommend

lcy

 

 

 别人的解法：

 F(n)表示n个人的合法队列 
 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：
 
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);
 
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)

 

 

知道递推公式就不难了- -

code:

import java.util.*;
 import java.math.*;
 
 public class Main
 {
     public static void main(String []args)
     {
         int n;
         int i;
         Scanner cin=new Scanner(System.in);
         BigInteger []result=new BigInteger[1002];
         result[1]=BigInteger.valueOf(1);
         result[2]=BigInteger.valueOf(2);
         result[3]=BigInteger.valueOf(4);
         result[4]=BigInteger.valueOf(7);
         for(i=5;i<=1000;i++)
             result[i]=result[i-1].add(result[i-2].add(result[i-4]));
         while(cin.hasNext())
         {
             n=cin.nextInt();
             System.out.println(result[n]);
         }
     }
 }