题意：我们可以用11或21的小矩形，横着或竖着去覆盖一个更大的矩形（可以横盖或竖盖），请问恰好完全覆盖n列两行的大矩形，有多少种盖法？

分四种情况，类似状态压缩dp。

#include <iostream>
#include <cstdio>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int
#define l(x) ((x)<<1)
#define r(x) ((x)<<1|1)
#define lowbit(x) ((x)&(-(x)))
#define abs(x) ((x)>=0?(x):(-(x)))
#define ms(a,b) memset(a,b,sizeof(a))

using namespace std;

int dp[111111][4],n;

int main(){
	ios::sync_with_stdio(false);
	dp[1][0] = 1;
	dp[1][1] = 1;
	dp[1][2] = 1;
	dp[1][3] = 1;
	cin >> n;
	for (int i = 2; i <= n; i++) {
		dp[i][0] = dp[i - 1][1] + dp[i - 1][2] + dp[i - 1][3];
		dp[i][1] = dp[i - 1][0] + dp[i - 1][2] + dp[i - 1][3];
		dp[i][2] = dp[i][0] + dp[i][1] - dp[i - 1][2];
		dp[i][3] = dp[i - 1][3] + dp[i - 1][2];
	}
	cout << dp[n][2] + dp[n][3] << endl;
	
	return 0;
}