codeforces 9D(非原创)
1 second
64 megabytes
standard input
standard output
In one very old text file there was written Great Wisdom. This Wisdom was so Great that nobody could decipher it, even Phong — the oldest among the inhabitants of Mainframe. But still he managed to get some information from there. For example, he managed to learn that User launches games for pleasure — and then terrible Game Cubes fall down on the city, bringing death to those modules, who cannot win the game...
For sure, as guard Bob appeared in Mainframe many modules stopped fearing Game Cubes. Because Bob (as he is alive yet) has never been defeated by User, and he always meddles with Game Cubes, because he is programmed to this.
However, unpleasant situations can happen, when a Game Cube falls down on Lost Angles. Because there lives a nasty virus — Hexadecimal, who is... mmm... very strange. And she likes to play very much. So, willy-nilly, Bob has to play with her first, and then with User.
This time Hexadecimal invented the following entertainment: Bob has to leap over binary search trees with n nodes. We should remind you that a binary search tree is a binary tree, each node has a distinct key, for each node the following is true: the left sub-tree of a node contains only nodes with keys less than the node's key, the right sub-tree of a node contains only nodes with keys greater than the node's key. All the keys are different positive integer numbers from 1 to n. Each node of such a tree can have up to two children, or have no children at all (in the case when a node is a leaf).
In Hexadecimal's game all the trees are different, but the height of each is not lower than h. In this problem «height» stands for the maximum amount of nodes on the way from the root to the remotest leaf, the root node and the leaf itself included. When Bob leaps over a tree, it disappears. Bob gets the access to a Cube, when there are no trees left. He knows how many trees he will have to leap over in the worst case. And you?
The input data contains two space-separated positive integer numbers n and h (n ≤ 35, h ≤ n).
Output one number — the answer to the problem. It is guaranteed that it does not exceed 9·1018.
3 2
5
3 3
4
嗯--dp。。。。
学习的题解:http://blog.csdn.net/lg_csu/article/details/17084449
题意:求n个节点能构成多少个不同的高度不小于h的二叉树。
解题思路:从n=1考虑,只有一个,再往后都是在前面的基础上进行的。左二叉树的种类与右二叉树的种类数量相乘就是新的二叉树的数量。
附ac代码:
1 #include <iostream>
2 #include <cstdio>
3 #include <string>
4 #include <cstring>
5 #include <fstream>
6 #include <algorithm>
7 #include <cmath>
8 #include <queue>
9 #include <stack>
10 #include <vector>
11 #include <map>
12 #include <set>
13 #include <iomanip>
14 using namespace std;
15 typedef long long ll;
16 const int maxn = 44;
17 ll dp[maxn][maxn];
18 int main()
19 {
20 ios::sync_with_stdio(false);
21 ll n,h;
22 cin>>n>>h;
23 for(int i=0;i<=35;++i) dp[0][i]=1; //从0开始便于转移公式从1开始进行。
24 for(int i=1;i<=35;++i)
25 {
26 for(int j=1;j<=35;++j)
27 {
28 for(int k=0;k<i;++k)
29 {
30 dp[i][j]+=dp[k][j-1]*dp[i-k-1][j-1];
31 }
32 }
33 }
34 cout<<dp[n][n]-dp[n][h-1]<<endl;
35 return 0;
36 }
codeforces 9D(非原创)的更多相关文章
- codeforces 6E (非原创)
E. Exposition time limit per test 1.5 seconds memory limit per test 64 megabytes input standard inpu ...
- Linux下high CPU分析心得【非原创】
非原创,搬运至此以作笔记, 原地址:http://www.cnitblog.com/houcy/archive/2012/11/28/86801.html 1.用top命令查看哪个进程占用CPU高ga ...
- CSS样式命名整理(非原创)
非原创,具体出自哪里忘了,如果侵害您的利益,请联系我. CSS样式命名整理 页面结构 容器: container/wrap 整体宽度:wrapper 页头:header 内容:content 页面主体 ...
- 非原创。使用ajax加载控件
非原创.来自博客园老赵. public class ViewManager<T> where T : System.Web.UI.UserControl { private System. ...
- Codeforces 9D How many trees? 【计数类DP】
Codeforces 9D How many trees? LINK 题目大意就是给你一个n和一个h 问你有多少个n个节点高度不小于h的二叉树 n和h的范围都很小 感觉有无限可能 考虑一下一个很显然的 ...
- Java 表达式解析(非原创)
因项目需要,在网上找来一套表达式解析方法,由于原来的方法太过于零散,不利于移植,现在整理在同一文件内: 文件中包含5个内部类,源码如下: import java.util.ArrayList; imp ...
- Java Interface 是常量存放的最佳地点吗?(转帖学习,非原创)
Java Interface 是常量存放的最佳地点吗?(转帖学习,非原创) 由于java interface中声明的字段在编译时会自动加上static final的修饰符,即声明为常量.因而inter ...
- 用RD,GR,BL三个方法内代码生成一张图片(非原创,我只是完整了代码)
我公开以下图片的源代码,,是ppm格式的,,自己找到能打开的工具.. (非原创,我加工的代码,可直接执行运行输出,缩略图能看到效果) 这是原博客 http://news.cnblogs.com/n/ ...
- tp5.1 phpspreadsheet- 工具类 导入导出(整合优化,非原创,抄一抄,加了一些自己的东西,)
phpspreadsheet-工具类 导入导出(整合优化,非原创,抄一抄,加了一些自己的东西)1. composer require phpoffice/phpspreadsheet2. 看最下面的两 ...
- Vue 仿QQ左滑删除功能(非原创)
非原创,摘选来源:http://www.jb51.net/article/136221.htm. 废话不多说,相当实用,先记录. Html代码: <div class="contain ...
随机推荐
- Java基础复习2
三目运算符 语法:条件判断?表达式1:表达式2; 如果条件判断成立则获取值1否则获取值2 public class demo1{ public static void main(String[ ...
- vue3.0改变概况
一.slot API在render实现原理上的变化 二.全局API使用规范变化 三.Teleport添加 四.composition API变化 五.v-model变化
- SDNU_ACM_ICPC_2021_Winter_Practice_4th [个人赛]
传送门 D - Odd Divisor 题意: 给你一个n,问你n是否至少有一个奇数因子(这里题意没说清,没说是不是只有一个还是可以有多个!AC以后才发现是不止一个 思路: 如果这个数没有奇数因子,那 ...
- Vue中:error 'XXXXX' is not defined no-undef解决办法
Vue中:error 'XXXXX' is not defined no-undef解决办法 报错内容: × Client Compiled with some errors in 7.42s √ S ...
- CSRF Laravel Cross Site Request Forgery protection¶
Laravel 使得防止应用 遭到跨站请求伪造攻击变得简单. Laravel 自动为每一个被应用管理的有效用户会话生成一个 CSRF "令牌",该令牌用于验证授权用 户和发起请求者 ...
- 服务器端IO模型的简单介绍及实现 阻塞 / 非阻塞 VS 同步 / 异步 内核实现的拷贝效率
小结: 1.在多线程的基础上,可以考虑使用"线程池"或"连接池","线程池"旨在减少创建和销毁线程的频率,其维持一定合理数量的线程,并让空闲 ...
- 在不同情况下connect失败和ping不通的数据分析
- loj10173
炮兵阵地 司令部的将军们打算在 N×M 的网格地图上部署他们的炮兵部队.一个 N×M的地图由 N 行 M 列组成,地图的每一格可能是山地(用 H 表示),也可能是平原(用 P表示),如下图.在每一格平 ...
- VMware Workstation Pro下载
VMware Workstation Pro 下载地址:https://pan.baidu.com/s/1XXhFFh0Fx0vzvcd1A543Yg,提取码:2o19(下载得到的压缩包中含有 VMw ...
- ElasticSearch的查询(二)
一.Query String search 添加测试数据 PUT test_search { "mappings": { "test_type": { &quo ...