http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3602 Count the Trees Time Limit: 2 Seconds      Memory Limit: 65536 KB A binary tree is a tree data structure in which each node has at most two child nodes, usually distinguished as "left&…

Count the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1248    Accepted Submission(s): 812 Problem DescriptionAnother common social inability is known as ACM (Abnormally Compulsive Medi…

 Count the Trees  Another common social inability is known as ACM (Abnormally Compulsive Meditation). This psychological disorder is somewhat common among programmers. It can be described as the temporary (although frequent) loss of the faculty of sp…

描述 A binary tree is a tree data structure in which each node has at most two child nodes, usually distinguished as "left" and "right". A subtree of a tree T is a tree consisting of a node in T and all of its descendants in T. Two binar…

题目是说给出一个数字,然后以1到这个数为序号当做二叉树的结点,问总共有几种组成二叉树的方式.这个题就是用卡特兰数算出个数,然后因为有编号,不同的编号对应不同的方式,所以结果是卡特兰数乘这个数的阶乘种方案.因为数字比较大,所以要用高精度的方法也就是用字符数组来做,我分别写了三个函数,一个算加法,一个算乘法,最后一个打表,等打出表来最后只要判断一下输入的数是第几个,直接输出就行了,下面是我的代码,第一次写高精度的这种大数处理,可能看上去比较繁琐= = #include<iostream> #inc…

题目链接:UVa 10007 题意:统计n个节点的二叉树的个数 1个节点形成的二叉树的形状个数为:1 2个节点形成的二叉树的形状个数为:2 3个节点形成的二叉树的形状个数为:5 4个节点形成的二叉树的形状个数为:14 5个节点形成的二叉树的形状个数为:42 把n个节点对号入座有n!种情况 所以有n个节点的形成的二叉树的总数是:卡特兰数F[n]*n! 程序: import java.math.BigInteger; import java.util.Scanner; public class Ma…

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=948 卡特兰数*n! import java.math.BigInteger; import java.util.*; public class Main { public static void main(String[]args) { Scanner cin=new Scanner(Sys…

卡特兰数再乘上n的阶乘 #include<iostream> #include<cstdio> using namespace std; #define base 10000 #define len 100 void multiply(int a[],int max,int b) { ; ;i>=;i--) { array+=b*a[i]; a[i]=array%base; array/=base; } } void divide(int a[],int max,int b)…

我是在neuqvj上交的这题:http://vj.acmclub.cn/problem/viewProblem.action?id=17848 本来是挺容易的树同构题,可是节点数比较多,愣是把普通hash卡掉了(出题人麻烦您过来一下) 只能用map映射一下,给每个状态一个标号,而状态的表示是它两个儿子的标号. #include<map> #include<cstdio> #include<cstring> #include<algorithm> #defin…

The 9th Zhejiang Provincial Collegiate Programming Contest A    Taxi Fare    25.57% (166/649)     (水题)    计算两种计程车计费方式的差值, 注意四舍五入    B    Unrequited Love    14.00% (7/50)    水题) C    Count the Trees    9.87% (8/81)   (二叉树)    卡特兰数应用的问题     D    Draw S…