【HDU】1297 - Children’s Queue（BigDecimal & 递推 & 思维）

点击打开题目

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13723 Accepted Submission(s): 4496

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）

这个递推也有点难想（其实是容易想错），之前有一个LELE的RPG难题和这个类似。

现在说一下这个题的推理：我们考虑当前情况的最后一个人

①如果是男生，那么就是 f（i - 1）；

②如果是女生，而f（i - 1）的最后一个有可能是男生，这样就违反了题目的要求。我们考虑f（i - 2）在后面直接放两个女生就满足了条件。而想到这里并没有结束（就wa在这里了），还有一种情况，如果f（i - 2）的最后只有一女生（这种情况在f（i - 2）中不存在，但是在f（i）中可能存在，即连续三个女生。）这个时候就应该加上f（i - 4）（即f（i - 4）后面放一个男生 + 一个女生 + 两个女生）。

所以递推公式为 ：

f（n）= f（n - 1）+ f（n - 2）+ f（n - 4）

这个数还是非常大的，所以我选择了用java的BigDecimal来做。

代码如下：

import java.math.BigDecimal;
import java.util.Scanner;

public class Main
{
	public static void main(String[] args)
	{
	    BigDecimal f[] = new BigDecimal [1011];
	    f[1] = new BigDecimal(1);
	    f[2] = new BigDecimal(2);
	    f[3] = new BigDecimal(4);
	    f[4] = new BigDecimal(7);
	    for (int i = 5 ; i <= 1000 ; i++)
	    {
	        f[i] = f[i-1].add(f[i-2]).add(f[i-4]);
	    }
	    Scanner sc = new Scanner(System.in);
	    while (sc.hasNext())
	    {
	        int n = sc.nextInt();
	        System.out.println(f[n]);
	    }
	}
}