HDU 3485 Count 101(DP)

Count 101

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1114    Accepted Submission(s): 568

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

 

Recommend

zhengfeng

 

这个有点类似于数位DP，但是不是数位DP

设dp[i][j][k]表示长度为i串的最后两位数为i,j的不包含101的个数。

所以dp方程就有了：

dp[i][0][0]+=dp[i-1][0][1]+dp[i-1][0][0]
dp[i][0][1]+=dp[i-1][1][0]+dp[i-1][1][1]
dp[i][1][0]+=dp[i-1][0][0]
dp[i][1][1]+=dp[i-1][1][0]+dp[i-1][1][1]

最后取ans求总数就行了

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int mod=9997;
const int MAXN=10000+5;
int dp[MAXN][2][2],n,ans[MAXN];

void init()
{
    memset(dp,0,sizeof(dp));
    memset(ans,0,sizeof(ans));
    ans[0]=0;ans[1]=2;ans[2]=4;
    dp[2][0][0]=dp[2][0][1]=dp[2][1][0]=dp[2][1][1]=1;
    dp[1][1][0]=dp[1][0][0]=1;
    for(int i=3;i<MAXN;i++)
    {
        dp[i][0][0]=(dp[i][0][0]+(dp[i-1][0][1]+dp[i-1][0][0])%mod)%mod;
        dp[i][0][1]=(dp[i][0][1]+(dp[i-1][1][0]+dp[i-1][1][1])%mod)%mod;
        dp[i][1][0]=(dp[i][1][0]+(dp[i-1][0][0])%mod)%mod;
        dp[i][1][1]=(dp[i][1][1]+(dp[i-1][1][0]+dp[i-1][1][1])%mod)%mod;
        ans[i]=(dp[i][0][0]+dp[i][0][1]+dp[i][1][0]+dp[i][1][1])%mod;
    }
}

int main()
{
    init();
    while(scanf("%d",&n)&&n!=-1)
    {
        printf("%d
",ans[n]);
    }
    return 0;
}