Description
Cao Cao was hunted down by thousands of enemy soldiers when he escaped from Hua Rong Dao. Assuming Hua Rong Dao is a narrow aisle (one N*4 rectangle), while Cao Cao can be regarded as one 2*2 grid. Cross general can be regarded as one 1*2 grid.Vertical general can be regarded as one 2*1 grid. Soldiers can be regarded as one 1*1 grid. Now Hua Rong Dao is full of people, no grid is empty.
There is only one Cao Cao. The number of Cross general, vertical general, and soldier is not fixed. How many ways can all the people stand?
Input
There is a single integer T (T≤4) in the first line of the test data indicating that there are T test cases.
Then for each case, only one integer N (1≤N≤4) in a single line indicates the length of Hua Rong Dao.
Output
Sample Input
2
1
2
Sample Output
0
18
Hint
Here are 2 possible ways for the Hua Rong Dao 2*4.
/*/
题意：
华容道，N*4的格子，N=[1,4]，一个曹操 2*2 有 2*1、1*2和1*1的士兵，问有多少中放法。 一开始不知道自己怎么想的，看一下数据这么小，而且1，2，3很容易就推出来了，以为可以推出来，【MDZZ】结果在N==4这种情况下爆炸了。。 下面进入正题。直接DFS暴力。 如果怕超时可以打表，实际上不会。 AC代码：
/*/
#include"algorithm"
#include"iostream"
#include"cstring"
#include"cstdlib"
#include"cstdio"
#include"string"
#include"vector"
#include"stack"
#include"queue"
#include"cmath"
#include"map"
using namespace std;
typedef long long LL ;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define FK(x) cout<<"["<<x<<"]\n"
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(x))
#define bigfor(T) for(int qq=1;qq<= T ;qq++)
int n;
int ans,flag;
bool vis[10][10]; bool check(int x,int y) {
if(x<0||y<0||x>=n||y>=4||vis[x][y])return 0;
return 1;
} void DFS(int x) {
if(x==4*n&&flag==1) {
ans++;
return ;
}
if(x>=4*n) {
return ;
}
for(int i=0; i<4; i++) {
for(int j=0; j<4; j++) {
if(check(i,j)&&check(i+1,j)&&check(i,j+1)&&check(i+1,j+1)&&!flag) {
vis[i][j]=vis[i+1][j]=vis[i][j+1]=vis[i+1][j+1]=1;
flag=1;
DFS(x+4);
vis[i][j]=vis[i+1][j]=vis[i][j+1]=vis[i+1][j+1]=0;
flag=0;
}
if(check(i,j)&&check(i+1,j)) {
vis[i][j]=vis[i+1][j]=1;
DFS(x+2);
vis[i][j]=vis[i+1][j]=0;
}
if(check(i,j)&&check(i,j+1)) {
vis[i][j]=vis[i][j+1]=1;
DFS(x+2);
vis[i][j]=vis[i][j+1]=0;
}
if(check(i,j)) {
vis[i][j]=1;
DFS(x+1);
vis[i][j]=0;
return ;
}
}
}
}
//int _ans[5]={0,0,18,284,4862};
int main() {
int T;S
scanf("%d",&T);
bigfor(T) {
scanf("%d",&n);
memset(vis,0);
flag=0;
ans=0;
DFS(0);
printf("%d\n",ans);
}
return 0;
}