A bee larva living in a hexagonal cell of a large honey comb decides to creep
for a walk. In each “step” the larva may move into any of the six adjacent cells
and after n steps, it is to end up in its original cell.
Your program has to compute, for a given n, the number of different such larva walks.

Input

The first line contains an integer giving the number of test cases to follow.
Each case consists of one line containing an integer n, where 1 ≤ n ≤ 14. SAMPLE INPUT
2
2
4

Output

For each test case, output one line containing the number of walks. Under the
assumption 1 ≤ n ≤ 14, the answer will be less than 2^31. SAMPLE OUTPUT
6
90

dp

/* ***********************************************
Author :guanjun
Created Time :2016/10/5 13:32:45
File Name :spojMBEEWALK.cpp
************************************************ */
#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
int x,y;
};
struct cmp{
bool operator()(Node a,Node b){
if(a.x==b.x) return a.y> b.y;
return a.x>b.x;
}
}; bool cmp(int a,int b){
return a>b;
}
ll dp[][][];
//第i步 到达位置 j k的方案数
int dir[][]={
,,-,,,,,-,,-,-,
};
void init(){
cle(dp);
dp[][][]=;
for(int i=;i<=;i++){
for(int x=;x<=;x++){
for(int y=;y<=;y++){
for(int j=;j<;j++){
int nx=x+dir[j][];
int ny=y+dir[j][];
dp[i][x][y]+=dp[i-][nx][ny];
}
}
}
}
} int main()
{
#ifndef ONLINE_JUDGE
//freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int t,n;
init();
cin>>t;
while(t--){
cin>>n;
if(n==)puts("");
else{
cout<<dp[n][][]<<endl;
}
}
return ;
}
A bee larva living in a hexagonal cell of a large honey comb decides to creep
for a walk. In each “step” the larva may move into any of the six adjacent cells
and after n steps, it is to end up in its original cell.
Your program has to compute, for a given n, the number of different such larva walks.

Input

The first line contains an integer giving the number of test cases to follow.
Each case consists of one line containing an integer n, where 1 ≤ n ≤ 14. SAMPLE INPUT
2
2
4

Output

For each test case, output one line containing the number of walks. Under the
assumption 1 ≤ n ≤ 14, the answer will be less than 2^31. SAMPLE OUTPUT
6
90

MBEEWALK - Bee Walk的更多相关文章

  1. BZOJ1695 : [Usaco2007 Demo]Walk the Talk

    观察单词表可以发现: 对于长度为3的单词,前两个字母相同的单词不超过7个 对于长度为4的单词,前两个字母相同的单词不超过35个 于是首先$O(26*26*nm)$预处理出 s1[x][i][j]表示( ...

  2. BZOJ 1695 [Usaco2007 Demo]Walk the Talk 链表+数学

    题意:链接 方法:乱搞 解析: 出这道题的人存心报复社会. 首先这个单词表-先上网上找这个单词表- 反正总共2265个单词.然后就考虑怎么做即可了. 刚開始我没看表,找不到怎么做,最快的方法我也仅仅是 ...

  3. python os.walk()

    os.walk()返回三个参数:os.walk(dirpath,dirnames,filenames) for dirpath,dirnames,filenames in os.walk(): 返回d ...

  4. LYDSY模拟赛day1 Walk

    /* 依旧考虑新增 2^20 个点. i 只需要向 i 去掉某一位的 1 的点连边. 这样一来图的边数就被压缩到了 20 · 2^20 + 2n + m,然后 BFS 求出 1 到每个点的最短路即可. ...

  5. How Google TestsSoftware - Crawl, walk, run.

    One of the key ways Google achievesgood results with fewer testers than many companies is that we ra ...

  6. poj[3093]Margaritas On River Walk

    Description One of the more popular activities in San Antonio is to enjoy margaritas in the park alo ...

  7. os.walk()

    os.walk() 方法用于通过在目录树种游走输出在目录中的文件名,向上或者向下. walk()方法语法格式如下: os.walk(top[, topdown=True[, onerror=None[ ...

  8. bee使用

    beego虽然是一个简单的框架,但是其中用到了很多第三方的包,所以在你安装beego的过程中Go会自动安装其他关联的包. 当然第一步你需要安装Go,如何安装Go请参考我的书 安装beego go ge ...

  9. 使用bee自动生成api文档

    beego中的bee工具可以方便的自动生成api文档,基于数据库字段,自动生成golang版基于beego的crud代码,方法如下: 1.进入到gopath目录的src下执行命令: bee api a ...

随机推荐

  1. 【DVWA】【SQL Injection】SQL注入 Low Medium High Impossible

    1.初级篇 low.php 先看源码,取得的参数直接放到sql语句中执行 if( isset( $_REQUEST[ 'Submit' ] ) ) { // Get input $id = $_REQ ...

  2. dapper未将对象引用设置到对象的实例

    现象是这样的dapper在reader.Read<T>()方法时报:未将对象引用设置到对象的实例 解决:实体类里属性类型与数据库表字段类型不匹配 我用的mysql varchar(50)保 ...

  3. Java学习1_一些基础1——16.5.4

    每个java程序中都必须有一个main方法,格式为: public class ClassName { public static void main(String[] args) { program ...

  4. jmeter元件的作用域和顺序

    jmeter是一个开源的性能测试工具,它可以通过鼠标拖拽来随意改变元件之间的顺序以及元件的父子关系,那么随着它们的顺序和所在的域不同,它们在执行的时候,也会有很多不同. jmeter的test pla ...

  5. Django线上部署教程:腾讯云+Ubuntu+Django+Uwsgi(转载)

    网站名称: 向东的笔记本 本文链接: https://www.eastnotes.com/post/29 版权声明: 本博客所有文章除特别声明外,均采用 BY-NC-SA 许可协议.转载请注明出处! ...

  6. JAVA基础——toString()方法

    toString()方法返回反映这个对象的字符串 因为toString方法是Object里面已经有了的方法,而所有类都是继承Object,所以“所有对象都有这个方法”. 它通常只是为了方便输出,比如S ...

  7. Django 命令行调用模版渲染

    首先我们需要进入 Django 的 shell python manage.py shell 渲染模版中的 name from django.template import Context,Templ ...

  8. Hashing - Hard Version

    Hashing - Hard Version Given a hash table of size N, we can define a hash function . Suppose that th ...

  9. Codeforces 934D/933B - A Determined Cleanup

    传送门:http://codeforces.com/contest/934/problem/D 给定两个正整数p(p≥1).k(k>1).多项式f(x)的系数的取值集合为{0,1,2,...,k ...

  10. Apache Maven Cookbook(一)maven 使用命令创建、编译java项目

    一.创建 使用命令创建项目分几步: 1.打开命令行窗口,比如cmd,把目录切换至想要创建项目地方. 2.执行如下命令: mvn archetype:generate -DgroupId=com.zua ...