HDOJ/HDU 1297 Children’s Queue(推导~大数)
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~~~
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]); } }
}
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文章来源: chenhx.blog.csdn.net,作者:谙忆,版权归原作者所有,如需转载,请联系作者。
原文链接:chenhx.blog.csdn.net/article/details/51472495
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