題幹：

In how many ways can you tile a 3xn rectangle with 2x1 dominoes? 
Here is a sample tiling of a 3x12 rectangle. 

Input

Input consists of several test cases followed by a line containing -1. Each test case is a line containing an integer 0 <= n <= 30.

Output

For each test case, output one integer number giving the number of possible tilings.

Sample Input

2
8
12
-1

Sample Output

3
153
2131

解題報告：

AC程式碼：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 200 + 5;
ll dp[55][15];
int n;
int main()
{
	dp[0][7]=1;
	dp[1][4]=dp[1][5]=dp[1][8]=1;
	for(int i = 2; i<=35; i++) {
		dp[i][1] = dp[i-1][5];
		dp[i][2] = dp[i-1][6];
		dp[i][3] = dp[i-1][4];
		dp[i][4] = dp[i-1][3] + dp[i-1][7];
		dp[i][5] = dp[i-1][7] + dp[i-1][1];
		dp[i][6] = dp[i-1][2];
		dp[i][7] = dp[i-1][8] + dp[i-1][4] + dp[i-1][5];
		dp[i][8] = dp[i-1][7];
	}
	while(cin>>n&& n!=-1) {
		printf("%lld\n",dp[n][7]);
	}
	return 0 ;
}
 

雖然有很多狀態都是不可能用到的（比如2和6），但還是定義一下比較好。

狀態橫著寫一下：

100

010

001

110

011

101

111

000

一定別忘了000這種狀態哈~