Children’s Queue

本文探讨了在特定条件下排列儿童队列的问题，即女孩不能单独排队，通过动态规划方法解决该问题，并提供了Java实现。讨论了算法的时间和空间复杂度，以及如何利用递推公式来计算满足条件的队列总数。

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11915 Accepted Submission(s): 3874

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

题意：n个同学站队，女生不能落单，问有多少种排列方案

思路：dp[i][1]表示第i位是男生的方案数，dp[i][0]表示第i位是女生的方案数，转移方程为

dp[i][1]=dp[i-1][1]+dp[i-1][0];
dp[i][0]=dp[i-2][1]+dp[i-1][0];

Hint：注意数据范围，要用高精度，求方便直接用了Java的BigInteger

import java.util.Scanner;
import java.math.*;

public class HDU1297 {

	public static void main(String[] args) {
		int n;
		BigInteger dp[][]=new BigInteger [1010][2];
		dp[1][1]=new BigInteger("1");
		dp[1][0]=new BigInteger("0");
		dp[2][1]=new BigInteger("1");
		dp[2][0]=new BigInteger("1");
		for(int i=3;i<=1000;i++)
		{
			dp[i][1]=dp[i-1][1].add(dp[i-1][0]);
			dp[i][0]=dp[i-2][1].add(dp[i-1][0]);
		}
		Scanner in=new Scanner(System.in);
		while(in.hasNext())
		{
			n=in.nextInt();
			System.out.println(dp[n][1].add(dp[n][0]));
		}
	}

}