ACM刷题之HDU————Children’s Queue

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1170 Accepted Submission(s): 625 

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

123

 

Sample Output

124

 

递推+大数

假设最后一位是男，那么会有f(n-1)种情况

假设最后一位是女，则倒数第二位必定是女，那么会有两种情况：

1、倒数第三位是男，那么就f(n-2)种情况

2、倒数第三位是女，那么就有f(n-4）种情况。

所以f(n)=f(n-1)+f(n-2)+f(n-4)

再配合JAVA的大数，这题就好了

下面是ac代码：

import java.math.BigInteger;import java.util.Scanner;import java.math.*;import java.text.*;public class Main{public static void main(String[] args) {// TODO 自动生成的方法存根Scanner cin=new Scanner(System.in);BigInteger[] child = new BigInteger[1002];child[1] = new BigInteger("1");child[2] = new BigInteger("2");child[3] = new BigInteger("4");child[4] = new BigInteger("7");int i = 0;for(i=5;i<=1000;i++){child[i] = child[i-1].add(child[i-2].add(child[i-4])) ;}while(cin.hasNext())   //等同于!=EOF{i=cin.nextInt();System.out.println(child[i]);}}}