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…

 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…

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…

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&…

题目链接:uva 10712 - Count the Numbers 题目大意:给出n,a.b.问说在a到b之间有多少个n. 解题思路:数位dp.dp[i][j][x][y]表示第i位为j的时候.x是否前面是相等的.y是否已经出现过n.对于n=0的情况要特殊处理前导0,写的很乱.搓死. #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using nam…

转载请注明: 仰望高端玩家的小清新 http://www.cnblogs.com/luruiyuan/ 题目大意: 题目传送门:UVa 10562Undraw the Trees 给定字符拼成的树,将这些树转换成特定的括号表示的树 思路: 首先,观察样例,可以发现就是先序遍历的顺序,因此可以确定dfs 但是,还有几个地方需要考虑: 同一级的结点,在同一级的括号中 由于顺序满足先序遍历,因此不需要存储树的括号表示法,更不需要构建树,直接在遍历过程中输出即可. 空树:即输入为:# 时的树的处理,我不…

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:一个数的素因子的和假设也是素数就叫做DePrimes,统计给定区间内的DePrimes. 分析:数论.本题使用用一种素数的筛法,欧拉筛法,也加线性筛法. 这样的方法,每次删选分两种情况:1.素因子不反复:2.素因子反复: 利用这个性质,统计DePrimes,假设素因子不同就加和,否则就取对应的最小不同集合: 打表计算,做差输出就可以. 说明:线性筛法,比传统的筛法在数据量大时会快(⊙_⊙). #include <iostream> #include <cstdlib> #in…

Problem D How Many Trees? Input: standard input Output: standard output Memory Limit: 32 MB A binary search tree is a binary tree with root k such that any node v in the left subtree of k has label (v) <label (k) and any node w in the right subtree o…