解法:推匯出公式後,轉換成矩陣快速冪求解即可.

/*dfs暴力求解,便於找規律*/ 
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
bool book[7][1007];
int n, cnt;
bool findpos(int &x, int &y) {
    for(int i=1 ; i<=4 ; i++) {
        for(int j=1 ; j<=n ; j++) {
            if(!book[i][j]) {
                x = i, y = j;
                return false;
            }
        }
    }
    return true;
}
void dfs(int x, int y) {
    if(!book[x+1][y] && x<4) {
        book[x][y] = book[x+1][y] = true;//縱向 
        int newx, newy;
        if(findpos(newx, newy)) {//已經填滿 
            book[x][y] = book[x+1][y] = false;//登出掉 
            cnt++;
            return ;
        }
        dfs(newx, newy);//從未填滿的地方開始繼續dfs 
        book[x][y] = book[x+1][y] = false;//恢復 
    }
    if(!book[x][y+1] && y<n) {//橫向 
        book[x][y] = book[x][y+1] = true;
        int newx, newy;
        if(findpos(newx, newy)) {
            book[x][y] = book[x][y+1] = false;
            cnt++;
            return ;
        }
        dfs(newx, newy);
        book[x][y] = book[x][y+1] = false;
    }
}
int main(){
	while(~scanf("%d", &n)){
		cnt=0;
		dfs(1, 1);
		cout << cnt<<endl;
	}
}
 

/*
推導公式:F[i]   = F[i-1] + 5*F[i-2] + F[i-3] - F[i-4]
*/
#include<bits/stdc++.h>
#define rep(i, a, b) for(int i=(a); i<(b); ++i)
using namespace std;
typedef long long ll;
const int maxn = 4;
const ll mod= 1e9+7;
struct Matrix{  
    ll tmp[maxn][maxn];  
}a;  
void init()  
{  
	rep(i, 0, maxn){  
		rep(j, 0, maxn){
			a.tmp[i][j]=0;  
		} 
    }  
    a.tmp[0][0]=a.tmp[0][2]=1;  
    a.tmp[0][1]=5;  
    a.tmp[0][3]=-1;  
    a.tmp[1][0]=a.tmp[2][1]=a.tmp[3][2]=1;
    return;
}  
Matrix mul(Matrix a,Matrix b)  
{  
    Matrix ans;  
    rep(i, 0, maxn)
        rep(j, 0, maxn){  
            ans.tmp[i][j]=0;  
            rep(k, 0, maxn){  
                ans.tmp[i][j]+=(a.tmp[i][k]*b.tmp[k][j]+mod)%mod;
                ans.tmp[i][j]%=mod;  
            }  
        }  
    return ans;  
}  
void fun(Matrix ans,ll k)  
{  
    rep(i,0,maxn){
 		rep(j, 0,  maxn) {
 			a.tmp[i][j]=(i==j);	
		}     	
	}  
    while(k){  
        if(k%2)  a=mul(a,ans);  
        ans=mul(ans,ans);  
        k/=2;  
    }  
    return ;
}  
int main(){
	//freopen("in.txt", "r", stdin);
 	Matrix t;  
    ll n;  
    rep(i, 0, maxn){
    	rep(j, 0, maxn){
    		t.tmp[i][j]=0;
		}
	}  
    t.tmp[0][0]=36;  
    t.tmp[1][0]=11;  
    t.tmp[2][0]=5;  
    t.tmp[3][0]=1; 
    while(~scanf("%lld",&n))  
    {  
        init();  
        if(n<=4){  
            printf("%lld\n",t.tmp[4-n][0]);  
            continue;  
        }  
        fun(a,n-4);  
        a=mul(a,t);  
        printf("%lld\n",a.tmp[0][0]%mod);  
    }  
    return 0;  
}