HDOJ 3485 Count 101

数位DP。。。。

Count 101

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 912    Accepted Submission(s): 477

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.

 

Sample Input

   
   
   
   

    
    
    
    3
4
-1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    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 
    
    
    
    
     
   
   
   
   

 

Source

2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int mod=9997;

int dp[11000][2][2],n;
int ans[11000];

void init()
{
    dp[1][1][0]=dp[1][0][0]=1;
    dp[2][0][0]=dp[2][1][0]=1;
    dp[2][0][1]=dp[2][1][1]=1;

    ans[0]=0;ans[1]=2;ans[2]=4;
    for(int i=3;i<11000;i++)
    {
        for(int j=0;j<2;j++)
        {
            for(int k=0;k<2;k++)
            {
                if(j==0)
                {
                    dp[i][j][k]+=(dp[i-1][k][0]+dp[i-1][k][1])%mod;
                }
                else if(j==1)
                {
                    if(k==1)
                    {
                         dp[i][j][k]+=(dp[i-1][k][0]+dp[i-1][k][1])%mod;
                    }
                    else
                    {
                        dp[i][j][k]+=dp[i-1][k][0];
                    }
                }
                dp[i][j][k]=dp[i][j][k]%mod;
            }
        }
        ans[i]=(dp[i][1][0]+dp[i][1][1]+dp[i][0][1]+dp[i][0][0])%mod;
    }

}

int main()
{
    init();
while(scanf("%d",&n)!=EOF&&~n)
{
    printf("%d\n",ans[n]);
}
    return 0;
}