Description

In how many ways can you tile a 3xn rectangle with 2x1 dominoes? 
Here is a sample tiling of a 3x12 rectangle.

Input

Input consists of several test cases followed by a line containing -1. Each test case is a line containing an integer 0 <= n <= 30.

Output

For each test case, output one integer number giving the number of possible tilings.

Sample Input

2
8
12
-1

Sample Output

3
153
2131

题解

看了别人的博客都是递推啥的,

之前自己推了半天推不出来,智商捉鸡,就放着没做了

直到我做了POJ 2411后

再回来看这题,就是把宽度设为3而已...

#include <iostream>
#include <cstdio> //EOF,NULL
#include <cstring> //memset
#include <cmath> //ceil,floor,exp,log(e),log10(10),hypot(sqrt(x^2+y^2)),cbrt(sqrt(x^2+y^2+z^2))
#include <algorithm> //fill,reverse,next_permutation,__gcd,
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
using namespace std;
#define rep(i, a, n) for (int i = a; i < n; ++i)
#define sca(x) scanf("%d", &x)
#define sca2(x, y) scanf("%d%d", &x, &y)
#define sca3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define pri(x) printf("%d\n", x)
#define pb push_back
#define mp make_pair
typedef pair<int, int> P;
typedef long long ll;
const ll inf = ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+;
const int maxn = ;
const int N = 1e4 + ;
int t, n, m;
int cnt, ans, ed;
ll dp[][<<];
int path[][];
int h, w;
void dfs(int l, int now, int pre)
{
if (l > w) {
return;
}
if (l == w) {
path[cnt][] = pre;
path[cnt++][] = now;
return;
}
dfs(l + , (now << )|, (pre << )|); // 竖放,当前行为1,上一行为0
dfs(l + , (now << )|, (pre << )); // 横放 当前行和上一行都为11
dfs(l + , (now << ), (pre << )|); //不放,上一行为1,当前行为0
} int main()
{
while (sca(h) && h != -)
{
w = ;
cnt = ;
dfs(, , );
memset(dp, , sizeof dp);
ed = ( << w) - ;
dp[][ed] = ;
for (int i = ; i < h; i++)
{
for (int j = ; j < cnt; j++)
{
dp[i + ][path[j][]] += dp[i][path[j][]];
}
}
printf("%lld\n", dp[h][ed]);
}
return ();
}

POJ 2663 Tri Tiling 【状压DP】的更多相关文章

  1. POJ 3254 Corn Fields (状压dp)

    题目链接:http://poj.org/problem?id=3254 给你n*m的菜地,其中1是可以种菜的,而菜与菜之间不能相邻.问有多少种情况. 状压dp入门题,将可以种菜的状态用一个数的二进制表 ...

  2. POJ 3254 - Corn Fields - [状压DP水题]

    题目链接:http://poj.org/problem?id=3254 Time Limit: 2000MS Memory Limit: 65536K Description Farmer John ...

  3. [ An Ac a Day ^_^ ] POJ 3254 Corn Fields 状压dp

    题意: 有一块n*m的土地 0代表不肥沃不可以放牛 1代表肥沃可以放牛 且相邻的草地不能同时放牛 问最多有多少种放牛的方法并对1e8取模 思路: 典型的状压dp 能状态压缩 能状态转移 能状态压缩的题 ...

  4. poj 3254Corn Fields (入门状压dp)

    Farmer John has purchased a lush ≤ M ≤ ; ≤ N ≤ ) square parcels. He wants to grow some yummy corn fo ...

  5. POJ 1684 Corn Fields(状压dp)

    描述 Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ ...

  6. POJ 2923 Relocation(状压DP)题解

    题意:有2辆车运货,每次同时出发,n(<10),各自装货容量c1 c2,问最少运几次运完. 思路:n比较小,打表打出所有能运的组合方式,用背包求出是否能一次运走.然后状压DP运的顺序. 代码: ...

  7. poj 2663 Tri Tiling (状压dp+多米诺骨牌问题+滚动数组反思)

    本来直接一波状压dpAC的 #include<cstdio> #include<cstring> #include<algorithm> #define REP(i ...

  8. POJ 2663 Tri Tiling

                                                                                    Tri Tiling   Time Li ...

  9. POJ 2663 Tri Tiling 矩阵快速幂 难度:3

    Tri Tiling Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7841   Accepted: 4113 Descri ...

随机推荐

  1. jmeter之regular expression extractor ,并循环调用匹配到的多个值

    jmeter之regular expression extractor 官方介绍:http://jmeter.apache.org/usermanual/regular_expressions.htm ...

  2. Spring boot +mybatis 连接mysql数据库,获取JDBC失败,服务器时区价值”Oйu±e×¼e±¼的识别或代表多个时区

    报出的错误 Cause: org.springframework.jdbc.CannotGetJdbcConnectionException: Failed to obtain JDBC Connec ...

  3. UML用例关系一览

  4. div 拖拽

    html <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF- ...

  5. selenium 对浏览器标签页进行关闭和切换

    关闭标签页 # 1.关闭浏览器全部标签页 driver.quit() # 2.关闭当前标签页(从标签页A打开新的标签页B,关闭标签页A) driver.close() 切换标签页 from selen ...

  6. BeautifulSoup库的使用

    1.简介 BeautifulSoup库也是一个HTML/XML的解析器,其使用起来很简单,但是其实解析网站用xpath和re已经足矣,这个库其实很少用到.因为其占用内存资源还是比xpath更高. '' ...

  7. kali在vbox上运行设置共享文件夹

    mount -t vboxsf VBoxShared /root/Desktop/vbox 0x00 使用共享文件夹的前提 需要自行安装增强功能:https://jingyan.baidu.com/a ...

  8. EntityFramework Core:版本不一致问题

    code first 更新数据库时候报版本不对或者未找到错误 解决方法: 在项目文件中添加以下节点: <PropertyGroup> <OutputType>Library&l ...

  9. encode和decode区别

    在python2 中是这种,编解码格式.在python3 中编码是会转换成byte类型即只显示ASCII码里的,编码会将byte转换成字符串类型.因此在py3中不需要使用,如果想要特定编码,在文件开头 ...

  10. 迷宫问题 (bfs广度优先搜索记录路径)

    问题描述: 定义一个二维数组: int maze[5][5] = { 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, ...