D. How many trees?
time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

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?

Input

The input data contains two space-separated positive integer numbers n and h (n ≤ 35, h ≤ n).

Output

Output one number — the answer to the problem. It is guaranteed that it does not exceed 9·1018.

Examples
input
3 2
output
5
input
3 3
output
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

  1. codeforces 6E (非原创)

    E. Exposition time limit per test 1.5 seconds memory limit per test 64 megabytes input standard inpu ...

  2. 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 ...

  3. 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 ...

  4. 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. ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 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/ ...

  9. 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 ...

  10. 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

  1. Angular学习资料大全和常用语法汇总(让后端程序员轻松上手)

    前言: 首先为什么要写这样的一篇文章呢?主要是因为前段时间写过一些关于Angualr的相关实战文章,有些爱学习的小伙伴对这方面比较感兴趣,但是又不知道该怎么入手(因为认识我的大多数小伙伴都是后端的同学 ...

  2. Mybatis【15】-- Mybatis一对一多表关联查询

    注:代码已托管在GitHub上,地址是:https://github.com/Damaer/Mybatis-Learning ,项目是mybatis-11-one2one,需要自取,需要配置maven ...

  3. MATLAB中load和imread的读取方式区别

    load是导入文件,一般从mat文件中,读取的是结构体imread是图像处理工具箱的库函数,处理图像比较方便,读取的是矩阵 1.之前将数组或者矩阵保存为一个mat格式的文件,在进行load命令读取时: ...

  4. UNIX DOMAIN SOCKETS IN GO unix域套接字

    Unix domain sockets in Go - Golang News https://golangnews.org/2019/02/unix-domain-sockets-in-go/ pa ...

  5. Convert a string into an ArrayBuffer

    https://github.com/mdn/dom-examples/blob/master/web-crypto/import-key/spki.js How to convert ArrayBu ...

  6. 从源码解析Nginx对 Native aio支持_运维_youbingchen的博客-CSDN博客 https://blog.csdn.net/youbingchen/article/details/51767587

    从源码解析Nginx对 Native aio支持_运维_youbingchen的博客-CSDN博客 https://blog.csdn.net/youbingchen/article/details/ ...

  7. Golang 性能优化实战

    小结: 1. 性能查看工具 pprof,trace 及压测工具 wrk 或其他压测工具的使用要比较了解. 代码逻辑层面的走读非常重要,要尽量避免无效逻辑. 对于 golang 自身库存在缺陷的,可以寻 ...

  8. (Sql Server)SQL FOR XML

    摘要:sql中的for xml语法为表转化为xml提供了很好的支持,当然使用同样的程序语言也能够达到同样的效果,但是有了for xml将使得这一切更加的方便. 主要内容: Select 的查询结果会作 ...

  9. Micro Frontends 微前端

    Micro Frontends https://martinfowler.com/articles/micro-frontends.html Integration approaches Server ...

  10. 内存屏障 WriteBarrier 垃圾回收 屏障技术

    https://baike.baidu.com/item/内存屏障 内存屏障,也称内存栅栏,内存栅障,屏障指令等, 是一类同步屏障指令,是CPU或编译器在对内存随机访问的操作中的一个同步点,使得此点之 ...