Problem Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan instructs how to deploy soldiers on the four corners of the city wall. Unfortunately, when Fei opened the pocket he found there are only four…

Secrete Master Plan Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 429    Accepted Submission(s): 244 Problem Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a…

D. Duff in Beach Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan instructs how to deploy soldiers on the four corners of the city wall. Unfortunately, w…

Problem Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed × matrix, but Fei didn't know the correct direction to hold the sheet. What a pity! Given two secrete master plans. The first one is the master's original plan. The…

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 801    Accepted Submission(s): 470 Problem Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan ins…

D. Duff in Beach Description Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan instructs how to deploy soldiers on the four corners of the city wall. Unfortunately, when Fei opened the pocket he found there are o…

题目链接:http://acm.uestc.edu.cn/#/problem/show/1215 题目大意就是问一个2*2的矩阵能否通过旋转得到另一个. 代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <algorithm> #include <set> #in…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5540 题目大意:给一个两个2*2的矩阵,第二个矩阵能不能通过旋转得到第一个矩阵 题目思路:模拟 #include <stdio.h> #include <iostream> using namespace std; ][]; ][]; ][]; void solve(int T){ ;i<=;i++) ;j<=;j++) scanf("%d",&…

Secrete Master Plan Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others) Submit Status Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan instructs how to deploy soldiers on the four…

Secrete Master Plan Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)     Master Mind KongMing gave Fei Zhang a secrete master plan stashed in a pocket. The plan instructs how to deploy soldiers on the four corners o…

题目:Sudoku 匪夷所思的方法,匪夷所思的速度!!! https://github.com/ttlast/ACM/blob/master/Dancing%20Link%20DLX/poj%203074.cpp #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; int flag; typedef long long LL; #define FF(i,A,s)…

101. 如何从其他的机器访问脚本? 答案:将包含脚本的文件夹共享出来…非常简单…你可以使用connect()在你本机运行脚本从而使得它们在其他的一些机器上执行…但是其他人无法访问这些脚本,除非你将它们共享出来或者试试别的什么方法. 103. 有什么办法可以让默认的复原系统顺序关闭对话框么? 谁能够告诉我有什么办法可以让默认的复原系统顺序关闭对话框么?这些对话框的双亲不是MainWin,而是ChildWin,我将会十分感激.默认的复原系统只能够检查那些双亲是由wMainWindow常量表示的Ma…

团队展示: 1.队名:六个核桃 2.队员学号: 王婧(201421123065).柯怡芳(201421123067组长).陈艺菡(201421123068). 钱惠(201421123071).尼玛(201421123072).林凯(201421123075) 3.拟作的团队项目描述: 博客作业查重系统:根据班级博客学生每次提交的博客作业,检查作业的重复率. 4.队员风采: a.王婧:擅长技术的:java.GUI:编程的兴趣:软件.图形界面:希望的软工角色:开发:一句话宣言:为共同远景而工作 b…

2016-2017 CT S03E06: Codeforces Trainings Season 3 Episode 6 比赛连接: http://codeforces.com/gym/101124/ 代码地址 http://git.oschina.net/qscqesze/Acm/tree/master/contest/2016-2017%20CT%20S03E06%20Codeforces%20Trainings%20Season%203%20Episode%206?dir=1&filepa…

网赛的时候看了这道题,发现就是平常的那种基础搜索题. 由于加了一个特殊条件:可以一次消耗3秒或原地停留1秒. 那就不能使用简单的队列了,需要使用优先队列才行. 题意 告诉一副地图:一个起点,一个终点,若干墙,若干监视器,剩下的是空地. 起点,终点,监视器都算空地. 监视器初始值会指定一个方向,共有四个方向. 监视器每秒顺时针转动到下个方向. 监视器视野距离为2. 在监视器的位置或在监视器面向的格子是监视区域. 普通的移动一格需要消耗1秒时间. 在监视器下移动一格需要消耗3秒时间. 如果呆在原地不…

声明 笔者最近意外的发现 笔者的个人网站http://tiankonguse.com/ 的很多文章被其它网站转载,但是转载时未声明文章来源或参考自 http://tiankonguse.com/ 网站,因此,笔者添加此条声明. 郑重声明:这篇记录<[百度之星2014~复赛 解题报告~正解]The Query on the Tree>转载自 http://tiankonguse.com/的这条记录:http://tiankonguse.com/record/record.php?id=674 前…

声明 笔者最近意外的发现 笔者的个人网站http://tiankonguse.com/ 的很多文章被其它网站转载,但是转载时未声明文章来源或参考自 http://tiankonguse.com/ 网站,因此,笔者添加此条声明. 郑重声明:这篇记录<[百度之星2014~复赛)解题报告]The Query on the Tree>转载自 http://tiankonguse.com/ 的这条记录:http://tiankonguse.com/record/record.php?id=673 前言…

声明 笔者最近意外的发现 笔者的个人网站http://tiankonguse.com/ 的很多文章被其它网站转载,但是转载时未声明文章来源或参考自 http://tiankonguse.com/ 网站,因此,笔者添加此条声明. 郑重声明:这篇记录<关于 double sort 这道题的思考>转载自 http://tiankonguse.com/ 的这条记录:http://tiankonguse.com/record/record.php?id=651 前言 前言的前言 昨天本来写好了这篇记录,…

小组:BLACK PANDA 需求&原型改进   20分 需求&原型改进: https://git.lug.ustc.edu.cn/black-panda/mblogs_plan/blob/master/plan/%E9%9C%80%E6%B1%82_%E5%8E%9F%E5%9E%8B%E6%94%B9%E8%BF%9B.pdf 软件需求规格说明书更新至V1.1: https://git.lug.ustc.edu.cn/black-panda/mblogs_plan/blob/maste…

题目描述 http://main.edu.pl/en/archive/oi/18/smi The Byteotian Waste Management Company (BWMC) has drastically raised the price of garbage collection lately. This caused some of the citizens to stop paying for collecting their garbage and start disposing…

本文主要介绍 COM 的基础知识,倾向于理论性的理解,面向初学者,浅尝辄止. 1. COM 是什么: COM 的英文全称是,Component Object Model,中文译为,组件对象模型.它官方的概念是:The Microsoft Component Object Model (COM) is a platform-independent, distributed, object-oriented system for creating binary software components…

Awesome Metaverse 关于 Metaverse 的精彩项目和信息资源列表. 由于关于 Metaverse 是什么存在许多相互竞争的想法,请随时以拉取请求.问题和评论的形式留下反馈. WebXR WebXR Explainer - 什么是 WebXR,有哪些用例? Awesome WebVR - 精选的 WebVR 软件包和资源列表 Exokit - WebXR 元浏览器.引擎.头像系统等 Exokit overview - Exokit 工具包的非官方概述 Exokit Web -…

Cable master Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21071   Accepted: 4542 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to…

Cable master Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21071   Accepted: 4542 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to…

记得前几章的组合数吧 我们学了O(n^2)的做法,加上逆元,我们又会了O(n)的做法 现在来了新问题,如果n和m很大呢, 比如求C(n, m) % p  , n<=1e18,m<=1e18,p<=1e5 看到没有,n和m这么大,但是p却很小,我们要利用这个p (数论就是这么无聊的东西,我要是让n=1e100,m=1e100,p=1e100你有本事给我算啊(°□°),还不是一样算不出来) 然后,我们著名的卢卡斯(Lucas)在人群中站了出来(｀･д･´)说:“让老子来教你这题” 卢卡斯说:…

传送门 题意 将n个数分成m个集合,\(V_i表示max(x-y),x,y∈第\)i个集合,\(求minΣV_i\) 分析 我们先对难度排序,令dp[i][j]表示前i个数分成j个集合的最小费用 转移方程为 \[dp[i][j]=min(dp[k][j-1]+(a[i]-a[k+1])^2,dp[i][j])\] 预处理dp[i][i],dp[i][1],开 long long 吐槽 开lld过,开I64d wa,用线段树可以达到\(O(n^2)\) 类似题目:Codeforces Round…

杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…

传送门 Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 28416   Accepted: 11109 Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be f…

对于深度优先算法,第一个直观的想法是只要是要求输出最短情况的详细步骤的题目基本上都要使用深度优先来解决.比较常见的题目类型比如寻路等,可以结合相关的经典算法进行分析. 常用步骤: 第一道题目:Dungeon Master  http://poj.org/problem?id=2251 Input The input consists of a number of dungeons. Each dungeon description starts with a line containing th…

IEEE/ACM International Conference on Advances in Social Network Analysis and Mining (ASONAM) 2014 Industry Track Call for Papers * Apologies if you received multiple copies of this CFP * Beijing China August 17-20, 2014Home Page: www.asonam2014.org F…