（递推 大整数） Children’s Queue hdu1297

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 16006    Accepted Submission(s): 5337

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

对于各种排列而言，用f(n)表示其长度为n时的可能数，有以下的情形：

　　长度为1时，只有1种可能，即“M”（女生不能单独站立，下同）；

　　长度为2时，有2种可能，即“FF”和“MM”;

　　长度为3时，有4种可能，即“FFF”、“FFM”、“MFF”和“MMM”;

　　长度为4时，有7种可能，即“FFFF”、“FFFM”、“FFMM”、“MFFM”、“MFFF”、“MMFF”、“MMMM”;

　　长度为n>4时，对于之前长度n-1的排列，加一个M是可行的（最好是男孩的情形）；对于之前长度n-2的排列，加加上FF是可行的（加MM有可能造成排列重复，这是最后是女孩的情形，并且前n-2个人是可行的排列）；另外一种情况是，前n-2个人是不可行的排列，即最后的两个人是“MF”，再加上两个“FF”就变成可行的排列，这种情况也就是在n-4人可行的排列基础上加上“MFFF”。

C++：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int s[][];
int main()
{
    s[][]=;
    s[][]=;
    s[][]=;
    s[][]=;
    int i,j,c,ans=;
    for(i=;i<=;i++)
    {
        for(j=,c=;j<;j++)
        {
            ans=s[i-][j]+s[i-][j]+s[i-][j]+c;
            c=ans/;
            s[i][j]=ans%;
        }
    }
    int n;
    while(cin>>n)
    {
        j=;
        while(!s[n][j])
            j--;
        cout<<s[n][j];
        for(i=j-;i>=;i--)
            printf("%08d",s[n][i]);
        cout<<endl;
    }
    return ;
}

用JAVA

import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner (System.in);
        BigInteger a[]=new BigInteger[1001];
        int n;
        while(in.hasNextInt()) {
            n=in.nextInt();
            a[1]=BigInteger.valueOf(1);
            a[2]=BigInteger.valueOf(2);
            a[3]=BigInteger.valueOf(4);
            a[4]=BigInteger.valueOf(7);
            for(int i=5;i<=n;i++) {
                a[i]=BigInteger.ZERO;
                a[i]=a[i].add(a[i-1]);
                a[i]=a[i].add(a[i-2]);
                a[i]=a[i].add(a[i-4]);
            }
            System.out.println(a[n]);
        }
    }
}