hdu 3485 H - Count 101

问题描述

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?

 

输入

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.

 

输出

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.

 

样例输入

     
     
     
     

      
      
      
       3
4
-1 
     
     
     
     

 

样例输出

     
     
     
     

      
      
      
       7
12 
     
     
     
     

提示

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 

01串去掉101串的个数；
可以用dp[i]表示长度为i时的最多个数；
算dp[i]时，因为首位相比i-1新加了一个1，所以就应该是dp[i-1]*2,但是因为多加了以为所以前面三尾可能出现101的情况
，所以这个时候还要减去这种情况的个数，这种情况就是第i-1位是0，i-2位是1，即第i-1位取0的个数减去第i-2位取0的个数（这里需要思考一下）
所以就是dp[i-2]-dp[i-3];
所以状态转移方程为dp[i]=dp[i-1]*2-(dp[i-2]-dp[i-3]);

代码如下：
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <math.h>
using namespace std;

int dp[10005];

int main()
{
    dp[1]=2;dp[2]=4;dp[3]=7;
    for(int i=4;i<=10005;i++)
        dp[i]=(2*dp[i-1]-(dp[i-2]-dp[i-3]))%9997;
    int n;
    while(cin>>n,n!=-1)
        cout<<dp[n]<<endl;

}