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…

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…

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,…

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…

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…

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…

传送门 time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and many others. Inspired by the…

Sereja and Mirroring Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description Let's assume that we are given a matrix b of size x × y, let's determine the operation of mirroring matrix b. The mirroring of…

http://codeforces.com/contest/426/problem/B B. Sereja and Mirroring time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Let's assume that we are given a matrix b of size x × y, let's determine…

B. s-palindrome 题目连接: http://www.codeforces.com/contest/691/problem/B Description Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the…

B. Problems for Round 题目连接: http://www.codeforces.com/contest/673/problem/B Description There are n problems prepared for the next Codeforces round. They are arranged in ascending order by their difficulty, and no two problems have the same difficult…

F. Magic Matrix 题目连接: http://www.codeforces.com/contest/632/problem/F Description You're given a matrix A of size n × n. Let's call the matrix with nonnegative elements magic if it is symmetric (so aij = aji), aii = 0 and aij ≤ max(aik, ajk) for all…

B. Problems for Round time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are n problems prepared for the next Codeforces round. They are arranged in ascending order by their difficulty,…

题目链接: B. Problems for Round time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are n problems prepared for the next Codeforces round. They are arranged in ascending order by their diffi…

连接在这里,->点击<- A. Bear and Game time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Bear Limak likes watching sports on TV. He is going to watch a game today. The game lasts 90 minutes and ther…

  C. Bracket Sequences Concatenation Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A bracket sequence is a string containing only characters "(" and ")". A regular…

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…

上一次我们拿学校的URP做了个小小的demo.... 其实我们还可以把每个学生的证件照爬下来做成一个证件照校花校草评比 另外也可以写一个物理实验自动选课... 但是出于多种原因,,还是绕开这些敏感话题.. 今天,我们来扒一下cf的题面! PS:本代码不是我原创 1. 必要的分析 1.1 页面的获取 一般情况CF的每一个 contest 是这样的: 对应的URL是:http://codeforces.com/contest/xxx 还有一个Complete problemset页面,它是这样的:…

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…

http://codeforces.com/contest/738/problem/D Galya is playing one-dimensional Sea Battle on a 1 × n grid. In this game a ships are placed on the grid. Each of the ships consists of bconsecutive cells. No cell can be part of two ships, however, the shi…

http://codeforces.com/contest/738/problem/C Vasya is currently at a car rental service, and he wants to reach cinema. The film he has bought a ticket for starts in t minutes. There is a straight road of length s from the service to the cinema. Let's…

http://codeforces.com/contest/738/problem/A Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters. There is…

http://codeforces.com/problemset/problem/662/A 题目大意: 给定n(n <= 500000)张卡片,每张卡片的两个面都写有数字,每个面都有0.5的概率是在正面,各个卡牌独立.求把所有卡牌来玩Nim游戏,先手必胜的概率. (⊙o⊙)-由于本人只会在word文档里写公式,所以本博客是图片格式的. Code #include <cstdio> #include <cstring> #include <algorithm> u…

http://codeforces.com/problemset/problem/274/B 题目大意: 给定你一颗树,每个点上有权值. 现在你每次取出这颗树的一颗子树(即点集和边集均是原图的子集的连通图)且这颗树中必须包含节点1 然后将这颗子树中的所有点的点权+1或-1 求把所有点权全部变为0的最小次数(n<=10^5) 题解: 因为每一次的子树中都必须有1,所以我们得知每一次变换的话1的权值都会变化 所以我们以1为根 现在,我们发现,如果一个节点的权值发生变化,那么他的父节点的权值一定发生变…