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： 就是n个人，站成一排。 有一个要求，(F)女生不能单独一个人站在男生之间。 可以没有女生。
输出有多少种站法； (不考虑人与人的不同，只考虑位置和男女区别) （如果一排以MF结尾是不合法的）
分析： 假如n个人的站法为db[n]; 由前面的推导出db[n]。 db[n-1]结尾添加一个M，是一定可以的。 db[n-2]结尾添加FF，也是一定可以的。 添加MF不可以，添加MM也是可以的（但是这个情况和db[n-1]中重复了），添加FM也是和db[n-1] M重复了。
在不可以序列后面加上FF(MF不可以,加上FF),成为合法, 所以db[n-4]后面 MFFF可以, 其实加一个F也能构成合法，但是这种情况包含在db[n-2]（相当与 FF）里面；
所以递推方程式db[n] =db[n-1] db[n-2] db[n-4];
db[i] 中保存的都是合法序列数。
Java大数秒A~~~
代码语言：javascript复制import java.math.BigInteger;
import java.util.Scanner;
public class Main{
static BigInteger db[] = new BigInteger[1001];
public static void main(String[] args) {
dabiao();
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
int n =sc.nextInt();
System.out.println(db[n]);
}
}
private static void dabiao() {
db[0]=new BigInteger("1");
db[1]=new BigInteger("1");
db[2]=new BigInteger("2");
db[3]=new BigInteger("4");
db[4]=new BigInteger("7");
for(int i=5;i<db.length;i ){
db[i]=db[i-1].add(db[i-2]).add(db[i-4]);
}
}
}
