Children’s Queue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10540 Accepted Submission(s): 3390
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
Sample Output
递推+大数：
分男女生考虑，设共有n个人。 //f(n)表示当有n个人的时候合法序列的个数，ak表示第k个人
1.当a1为男生时，有f(n-1)种可能。
2.当a1是女生时，a2必为女生.
1）当a1、a2都为女生时，a3为男生，有f(n-3)种可能。
2）当a1、a2、a3都为女生，a4为男生时，有f(n-4)种可能。
......
以此类推 // 下式中f(1)表示n全为女生时那种情况，规定f(0) = 1。
故有 f(n) = f(n-1) + f(n-3) + f(n-4) + f(n-5) + f(n-6) + ... + f(1) + f(0) + 1; //（1）式
又 f(n-2) = f(n-3) + f(n-5) + f(n-6) + ... + f(1) + f(0) + 1; //（2）式
由以上两式（（1）式-（2）式）得 f(n) = f(n-1) + f(n-2) + f(n-4)
import java.math.BigInteger;
import java.util.Scanner;
public class pp {
public static void main(String[] args)
{
Scanner cin=new Scanner(System.in);
BigInteger a[]=new BigInteger[1001];
a[0]=BigInteger.valueOf(1);
a[1]=BigInteger.valueOf(1);
a[2]=BigInteger.valueOf(2);
a[3]=BigInteger.valueOf(4);
for(int i=4;i<=1000;i++)
//a[i]=(a[i-1].multiply(BigInteger.valueOf(2))).subtract(a[i-3]);
a[i]=(a[i-1].add(a[i-2])).add(a[i-4]);
while(cin.hasNext())
{
int n=cin.nextInt();
System.out.println(a[n]);
}
}
}
hdu 1297 Children’s Queue,布布扣,bubuko.com
hdu 1297 Children’s Queue
原文：http://blog.csdn.net/fyxz1314/article/details/38339565
