Accept: 503 Submit: 1054
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem 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

Sample Output

Hint

Here are 2 possible ways for the Hua Rong Dao 2*4.

Source

#define _CRT_SECURE_NO_DEPRECATE
#include 
 
<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<bitset>
#include<set>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
#define N_MAX 7
#define MOD 1000000007
#define INF 0x3f3f3f3f
typedef long long ll;
int n, k;
bool vis[N_MAX][N_MAX],cc;
int ans = 0;
void dfs(int x,int y) {
    if (x==n) {//搜索結束
        if (cc)ans++;//有曹操
        return;
    }
    int xx=x, yy=y+1;//下次要去的點
    if (yy == 4) {
        xx++, yy = 0;
    }
    
    if (vis[x][y])dfs(xx, yy);
    else {
        for (int cs = 0; cs < 4;cs++) {
            if (cs == 0&&!cc) {
                if (x + 1 < n&&y + 1 < 4 && !vis[x][y] && !vis[x + 1][y] && !vis[x][y + 1] && !vis[x + 1][y + 1]) {
                    cc = true;
                    vis[x][y] = vis[x + 1][y] = vis[x][y + 1] = vis[x + 1][y + 1] = true;
                    dfs(xx,yy);
                    vis[x][y] = vis[x + 1][y] = vis[x][y + 1] = vis[x + 1][y + 1] = false;
                    cc = false;
                }
            }
            if (cs == 1) {
                if (x + 1 < n && !vis[x][y] && !vis[x + 1][y]) {
                    vis[x][y] = vis[x + 1][y] = true;
                    dfs(xx,yy);
                    vis[x][y] = vis[x + 1][y] = false;
                }
            }
            if (cs == 2) {
                if (y + 1 < 4 && !vis[x][y] && !vis[x][y + 1]) {
                    vis[x][y] = vis[x][y + 1] = true;
                    dfs(xx, yy);
                    vis[x][y] = vis[x][y + 1] = false;
                }
            }
            if (cs == 3) {
                if (!vis[x][y]) {
                    vis[x][y] = true;
                    dfs(xx, yy);
                    vis[x][y] = false;
                }
            }
        }
    }
}

int main() {
    int t; cin >> t;
    while(t--){
        memset(vis, 0, sizeof(vis)); cc = 0; ans = 0;
        cin >> n;
        dfs(0, 0);
    cout << ans<<endl;
    }
    return 0;
}

foj Problem 2107 Hua Rong Dao