题干：
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解题报告：
AC代码：
#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 ;
}虽然有很多状态都是不可能用到的（比如2和6），但还是定义一下比较好。
状态横着写一下：
100
010
001
110
011
101
111
000
一定别忘了000这种状态哈~
附一种错误的递推：
#include<bits/stdc++.h>using namespace std;
int dp[100000][3];
int n;
int main()
{dp[1][2] = 1;dp[1][1] = 0;dp[1][0] = 0;cin>>n;for(int i = 2; i<=n; i++) {dp[i][1] = dp[i-1][2] + 1;dp[i][0] = dp[i][1] + 1;dp[i][2] = dp[i-2][0] * 2 + dp[i-1][1];}cout << dp[n][0] << endl; return 0 ;}
