MBEEWALK - Bee Walk
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
- 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( ...
- 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 ...
- 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 ...
- 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. ...
- How Google TestsSoftware - Crawl, walk, run.
One of the key ways Google achievesgood results with fewer testers than many companies is that we ra ...
- poj[3093]Margaritas On River Walk
Description One of the more popular activities in San Antonio is to enjoy margaritas in the park alo ...
- 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[ ...
- 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 ...
- 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
- 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 ...
- 抓取猫眼电影top100的正则、bs4、pyquery、xpath实现方法
import requests import re import json import time from bs4 import BeautifulSoup from pyquery import ...
- 洛谷—— P1450 [HAOI2008]硬币购物
P1450 [HAOI2008]硬币购物 硬币购物一共有$4$种硬币.面值分别为$c1,c2,c3,c4$.某人去商店买东西,去了$tot$次.每次带$di$枚$ci$硬币,买$si$的价值的东西.请 ...
- Python,subprocess模块(补充)
1.subprocess模块,前戏 res = os.system('dir') 打印到屏幕,res为0或非0 os.popen('dir') 返回一个内存对象,相当于文件流 a = os.popen ...
- java--删除链表偶数节点
public class ListNode { int data;//当前节点的值 ListNode next = null;//是指向下一个节点的指针/引用 public ListNode(int ...
- 在vue项目中快速使用element UI
推荐使用npm安装 1.安装:npm install element-ui -S 2.整体引入: 在你项目的main.js中写入: import ElementUI from 'element-ui' ...
- ansible playbooks loop循环
在一个task中循环某个操作 1.标准循环 - name: add several users user: name: "{{ item }}" state: present gr ...
- [BZOJ1163&1339]Mafia
[Baltic2008]Mafia 题目 匪徒准备从一个车站转移毒品到另一个车站,警方准备进行布控. 对于每个车站进行布控都需要一定的代价,现在警方希望使用最小的代价控制一些车站,使得去掉这些车站后, ...
- hdu 5017 模拟退火算法
hdu 5017 http://blog.csdn.net/mypsq/article/details/39340601 #include <cstdio> #include <cs ...
- [bzoj1176]Mokia[CDQ分治]
啃了一天论文,发现CDQ分治的原理其实很简单,大概就是这样的一类分治:将左右区间按一定规律排序后分开处理,递归到底时直接计算答案,对于一个区间,按照第二关键字split成两个区间,先处理左区间,之后因 ...