http://codeforces.com/contest/568/problem/B 题意就是给一个有n个元素的集合,现在需要求有多少个A的二元关系p,使得p是对称的,是传递的,但不是自反的. 首先只用(x1, x1), (x2, x2).....这种二元对形成的传递,对称,非自反的满足条件的方法数为2^n - 1(每一对可以选择出现或者不出现,全部出现的情况是自反的,所以减掉1) 其次,由于如果存在(a, b)a!=b的二元关系对,那么a,b这两个元素一定在某一个环中(根据对称一定有(b,…

题目链接: D. Symmetric and Transitive time limit per test 1.5 seconds memory limit per test 256 megabytes input standard input output standard output Little Johnny has recently learned about set theory. Now he is studying binary relations. You've probabl…

http://codeforces.com/contest/568/problem/B 题意:题意还挺绕的,其实就是说:要你求出一个图,要求保证其中有至少一个点不连任何边,然后其他连边的点构成的每个联通块都必须构成完全连通图 思路:f[i][j]代表i个点,构成j个联通块的方案数 f[i][j]=f[i][j-1]*j(代表与其中一个联通块合并)+f[i-1][j-1](代表新开一个联通块) 然后答案是Σc[n][i]*f[i][j] #include<cstdio> #include<…

题意: 根据离散数学的内容知道,一个二元关系是一个二元有序组<x, y>的集合. 然后有一些特殊的二元关系,比如等价关系,满足三个条件: 自反性,任意的x,都有二元关系<x, x> 对称性,如果有<x, y>则有<y, x> 传递性,如果有<x, y>和<y, z>,则有<x, z> 现在要统计满足后两条,但不满足第一个条件的二元关系的个数. 题中的证明是对的: If , then (according to proper…

1.透明 https://en.wikipedia.org/wiki/Equivalence_relation In mathematics, an equivalence relation is a binary relation that is reflexive, symmetric and transitive. The relation "is equal to" is the canonical example of an equivalence relation, whe…

Part I:equals() (javadoc) must define an equivalence relation (it must be reflexive, symmetric, and transitive). In addition, it must be consistent (if the objects are not modified, then it must keep returning the same value). Furthermore, o.equals(n…

-------------------------------------------------------------- Chapter 1: Introduction to Discrete Differential Geometry: The Geometry of Plane Curves . A better approximation than the tangent is the circle of curvature. . If the curve is sufficientl…

equals() (javadoc) must define an equality relation (it must be reflexive, symmetric, and transitive). In addition, it must be consistent (if the objects are not modified, then it must keep returning the same value). Furthermore, o.equals(null) must…

Object对象是除了基础对象之外,所有的对象都需要继承的父对象,包括数组也继承了Object Object里面的关键函数罗列如下: clone();调用该函数需要实现 Cloneable,否则会抛出  CloneNotSupportedException的异常. equals();用来判断两个对象是否相等的函数,默认的是两个对象的地址比较,在使用Compare计算的时候,往往需要重写这个函数. finalize();这个函数在对象被消解的时候,例如垃圾回收的时候等,会被调用,如果对象使用了不可…

比赛链接:http://codeforces.com/contest/569 A. Music time limit per test:2 seconds memory limit per test:256 megabytes Little Lesha loves listening to music via his smartphone. But the smartphone doesn't have much memory, so Lesha listens to his favorite…

读<Effect Java中文版> 译者序 序 前言 第1章引言 1   第2章创建和销毁对象 4 第1条:考虑用静态工厂方法代替构造函数 4 第2条:使用私有构造函数强化singleton属性 8 第3条:通过私有构造函数强化不可实例化的能力 10 第4条:避免创建重复的对象 11 第5条:消除过期的对象引用 14 第6条:避免使用终结函数 17   第3章对于所有对象都通用的方法 21 第7条:在改写equals的时候请遵守通用约定 21 第8条:改写equals时总是要改写hashCod…

The theory (for the language lawyers and the mathematically inclined): equals() (javadoc) must define an equivalence relation (it must be reflexive, symmetric, and transitive). In addition, it must be consistent (if the objects are not modified, then…

题目链接:Symmetric Tree | LeetCode OJ Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not: 1 / \ 2 2 \ \ 3 3 Note: B…

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not: 1 / \ 2 2 \ \ 3 3 Note:Bonus points if you could solve it b…

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not: 1 / \ 2 2 \ \ 3 3 递归解法: /** * Definition for binary tree *…

COMPUTER ORGANIZATION AND ARCHITECTURE DESIGNING FOR PERFORMANCE NINTH EDITION As demands for perfor-mance increase and as the cost of microprocessors continues to drop, vendors have introduced systems with an SMP organization. The term SMP refers to…

Problem Link: https://oj.leetcode.com/problems/symmetric-tree/ To solve the problem, we can traverse the tree level by level. For each level, we construct an array of values of the length 2^depth, and check if this array is symmetric. The tree is sym…

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not: 1 / \ 2 2 \ \ 3 3 /** * Definition for a binary tree node.…

1 问题描述: ubuntu环境下用eclipse+maven开发Scala的时候出现错误:scalac error: bad option: '-make:transitive' on mvn package via command line 2 解决方法: (1)打开pom.xml,删除 <parameter value="-make:transitive"/> (2)添加dependance <dependency>             <gro…

创建一个函数,接受两个或多个数组,返回所给数组的 对等差分(symmetric difference) (△ or ⊕)数组. 给出两个集合 (如集合 A = {1, 2, 3} 和集合 B = {2, 3, 4}), 而数学术语 "对等差分" 的集合就是指由所有只在两个集合其中之一的元素组成的集合(A △ B = C = {1, 4}). 对于传入的额外集合 (如 D = {2, 3}), 你应该安装前面原则求前两个集合的结果与新集合的对等差分集合 (C △ D = {1, 4} △…

Symmetric Tree Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not: 1 / \ 2 2 \ \ 3 3 Note:Bonus points if you c…

https://en.wikipedia.org/wiki/Symmetric_multiprocessor_system A symmetric multiprocessor system (SMP) is a multiprocessor system with centralized shared memory called main memory (MM) operating under a single operating system with two or more homogen…

COMPUTER ORGANIZATION AND ARCHITECTURE DESIGNING FOR PERFORMANCE NINTH EDITION 17.5 CLUSTERSAn important and relatively recent development computer system design is clus-tering. Clustering is an alternative to symmetric multiprocessing as an approach…

Same Tree Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical and the nodes have the same value. 思想: 无.能遍历即可. /** * Definition for binary tree * struct T…

Dependencies declared in your project's pom.xml file often have their own dependencies. The main dependencies are called direct dependencies. And the sub dependencies on which the direct dependencies relied on are called transitive dependencies. eg:…

题目来源 https://leetcode.com/problems/symmetric-tree/ Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). 题意分析 Input:一个二叉树 Output:True or False Conditions: 判断一个二叉树是不是对称的 题目思路 一一对应判断值是否相等,注意要清楚谁对应谁. AC代码(Python)…

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not: 1 / \ 2 2 \ \ 3 3代码如下: /** * Definition for a binary tree n…

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following [1,2,2,null,3,null,3] is not: 1 / \ 2 2 \ \ 3 3 Not…

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following [1,2,2,null,3,null,3] is not: 1 / \ 2 2 \ \ 3 3 Not…

题目: Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not: 1 / \ 2 2 \ \ 3 3 Note:Bonus points if you could solve…