hdu 1008 Count 101 n个0-1串的种类，不能包括101

Count 101

Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 267    Accepted Submission(s): 106

Problem Description

You know YaoYao is fond of his chains. He has a lot of chains and each chain has n diamonds on it. There are two kinds of diamonds, labeled 0 and 1. We can write down the label of diamonds on a chain. So each chain can be written as a sequence consisting of 0 and 1.
We know that chains are different with each other. And their length is exactly n. And what’s more, each chain sequence doesn’t contain “101” as a substring.
Could you tell how many chains will YaoYao have at most?

 

Input

There will be multiple test cases in a test data. For each test case, there is only one number n(n<10000). The end of the input is indicated by a -1, which should not be processed as a case.

 

Output

For each test case, only one line with a number indicating the total number of chains YaoYao can have at most of length n. The answer should be print after module 9997.

 

SampleInput

3
4
-1

 

SampleOutput

7
12
Hint
We can see when the length equals to 4. We can have those chains:
0000,0001,0010,0011
0100,0110,0111,1000
1001,1100,1110,1111

 

 

/*a[n][0][0]=a[n-1][0][0]+a[n-1][1][0];
a[n][0][1]=a[n-1][0][0];
a[n][1][0]=a[n-1][0][1]+a[n-1][1][1];
a[n][1][1]=a[n-1][0][1]+a[n-1][1][1];*/
#include<iostream>
#include<cstdio>
using namespace std;
int a[10001][2][2];
int ans[10001];
int main()
{
    ans[1]=2;ans[2]=4;
    a[2][0][0]=1;a[2][0][1]=1;a[2][1][0]=1;a[2][1][1]=1;
    for(int n=3;n<=10000;n++)
    {
        a[n][0][0]=(a[n-1][0][0]+a[n-1][1][0])%9997;
        a[n][0][1]=a[n-1][0][0]%9997;
        a[n][1][0]=(a[n-1][0][1]+a[n-1][1][1])%9997;
        a[n][1][1]=(a[n-1][0][1]+a[n-1][1][1])%9997;
        ans[n]=(a[n][0][0]+a[n][0][1]+a[n][1][0]+a[n][1][1])%9997;
    }
    int n;
    while(scanf("%d",&n)==1)
    {
        if(n<0) break;
        printf("%d/n",ans[n]);
    }
    return 0;
}