铁树银花 - 博客园

DOC常用命令（转）

摘要：常用doc命令大全 dos和windows最大的不同在于dos命令方式操作，所以使用者需要记住大量命令及其格式使用方法，dos命令分为内部命令和外部命令，内部命令是随每次启动的command.com装入并常驻内存，而外部命令是一条单独的可执行文件。在操作时要记住的是，内部命令在任何时候都可以使用，而外部命令需要保证命令文件在当前的目录中，或在autoexec.bat文件已经被加载了路径。常用的内部命令 dos的内部命令是dos操作的基础，下面就来介绍一些常用的dos内部命令。 1、dir 含义：显示指定路径上所有文件或目录的信息格式：dir [盘符：][路径][文件名] [参... 阅读全文

posted @ 2013-03-05 16:54 铁树银花阅读(278) 评论(0) 推荐(0) 编辑

C++ GUI Qt4 自学笔记

摘要：一、创建Button 1 #include <QApplication> 2 #include <QPushButton> 3 4 int main(int argc, char *argv[]) 5 { 6 QApplication app(argc, argv); 7 QPushButton *mybutton = new QPushButton("Quit"); 8 QObject::connect(mybutton, SIGNAL(clicked()), &app, SLOT(quit())); 9 mybutton->show 阅读全文

posted @ 2013-03-03 20:58 铁树银花阅读(375) 评论(0) 推荐(0) 编辑

Qt无法运行程序：Cannot find file: F:\@@@@\GUI\1\1.pro.

摘要：出现以下信息Cannot find file: F:\@@@@\GUI\1\1.pro.“@@@@”是工程文件的目录路径中的一个以中文命名的文件夹，显然，我安装的Qt不识别中文。把该文件改为英文名称，问题解决。^_^2013-03-0315:56:31 阅读全文

posted @ 2013-03-03 15:56 铁树银花阅读(503) 评论(0) 推荐(0) 编辑

Max Sum（经典DP）

摘要： Max Sum Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1003DescriptionGiven a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 阅读全文

posted @ 2013-03-03 13:29 铁树银花阅读(3608) 评论(0) 推荐(0) 编辑

最大连续子序列（经典DP）

摘要：最大连续子序列 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1231Description给定K个整数的序列{ N1, N2, ..., NK }，其任意连续子序列可表示为{ Ni, Ni+1, ..., Nj }，其中 1 <= i <= j <= K。最大连续子序列是所有连续子序列中元素和最大的一个，例如给定序列{ -2, 11, -4, 13, -5, -2 }，其最大连续子序列为{ 11, -4, 13 }，最大阅读全文

posted @ 2013-03-03 12:15 铁树银花阅读(434) 评论(0) 推荐(0) 编辑

Codeforces Round #170 (Div. 2) B. New Problem（好题）

摘要： B. New Problemtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputComing up with a new problem isn't as easy as many people think. Sometimes it is hard enough to name it. We'll consider a title original if it doesn't occur as a substring 阅读全文

posted @ 2013-03-02 21:42 铁树银花阅读(345) 评论(0) 推荐(0) 编辑

Codeforces Round #169 (Div. 2) D. Little Girl and Maximum XOR（贪心，中等）

摘要： D. Little Girl and Maximum XORtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputA little girl loves problems on bitwise operations very much. Here's one of them.You are given two integers l and r. Let's consider the values of for all pairs 阅读全文

posted @ 2013-03-02 12:16 铁树银花阅读(336) 评论(0) 推荐(0) 编辑

Codeforces Round #169 (Div. 2) C. Little Girl and Maximum Sum（线段树区间更新）

摘要： C. Little Girl and Maximum Sumtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe little girl loves the problems on array queries very much.One day she came across a rather well-known problem: you've got an array of n elements (the elements o 阅读全文

posted @ 2013-03-01 17:20 铁树银花阅读(487) 评论(0) 推荐(0) 编辑

Codeforces Round #169 (Div. 2) B. Little Girl and Game（博弈）

摘要： B. Little Girl and Gametime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe Little Girl loves problems on games very much. Here's one of them.Two players have got a string s, consisting of lowercase English letters. They play a game that is de 阅读全文

posted @ 2013-02-26 23:45 铁树银花阅读(667) 评论(0) 推荐(0) 编辑

Codeforces Round #167 (Div. 2) C. Dima and Staircase（线段树·成段更新，繁琐）

摘要： C. Dima and Staircasetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputDima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 ≤ a1 ≤ a2 ≤ ... ≤ an).Dima decided to pl 阅读全文

posted @ 2013-02-26 16:17 铁树银花阅读(462) 评论(0) 推荐(0) 编辑

微言

摘要：孔庆东《侠骨柔情顾明道》1.在当代武侠小说里，不但有武有侠，而且还有社会，有言情，既有英雄，也有美女，既有吕布，也有貂蝉。……所以武侠小说好比百货大楼，而言情小说好比连锁专卖店，哪里的顾客多，就可想而知了。2.中国古代的武侠小说，基本没有女人的地位。《水浒传》梁山108将，男人占105，女人只有3个。其中一个叫母老虎，一个叫母夜叉，都是比男人还要杀人不眨眼的部门女经理，剩下个一丈青扈三娘，是唯一具有女性美的，却被宋江嫁给了窝囊废矮脚虎王英，使得梁山之上根本没有爱情发生的可能性。3.“男女搭配，打架不累”的武侠小说诞生了。4.煮字疗饥《左氏春秋》1.多行不义必自毙。说的是齐庄公的母亲姜氏和他的弟阅读全文

posted @ 2013-02-26 15:25 铁树银花阅读(277) 评论(0) 推荐(0) 编辑

Javabeans（递推）

摘要： JavabeansTime Limit: 2 Seconds Memory Limit: 65536 KBJavabeans are delicious. Javaman likes to eat javabeans very much.Javaman has n boxes of javabeans. There are exactly i javabeans in the i-th box (i = 1, 2, 3,...n). Everyday Javaman chooses an integer x. He also chooses several boxes where the... 阅读全文

posted @ 2013-02-25 17:43 铁树银花阅读(398) 评论(0) 推荐(0) 编辑

Kagome Kagome（简单）

摘要： 3492 - Kagome Kagome时间限制:Java: 2000 ms / Others: 2000 ms内存限制: Java: 65536 KB / Others: 65536 KB问题描述Kagome kagome, kago no naka no tori wa Itsu itsu deyaru? Yoake no ban ni Tsuru to kame to subetta. Ushiro no shoumen daare? Translation: Kagome kagome, the bird in the cage, when will you come out? In 阅读全文

posted @ 2013-02-25 17:38 铁树银花阅读(935) 评论(0) 推荐(0) 编辑

Ordinal Numbers（简单）

摘要： Ordinal NumbersTime Limit: 2 Seconds Memory Limit: 65536 KBOrdinal numbers refer to a position in a series. Common ordinals include zeroth, first, second, third, fourth and so on. Ordinals are not often written in words, they are written using digits and letters. An ordinal indicator is a sign ad... 阅读全文

posted @ 2013-02-25 17:35 铁树银花阅读(1656) 评论(0) 推荐(0) 编辑

Edge（简单）

摘要： Edge 时间限制(普通/Java):1000MS/10000MS 运行内存限制:65536KByte 总提交: 1 测试通过: 0 描述For products that are wrapped in small packings it is necessary that the sheet of paper containing the directions for use is folded until its size becomes small enough. We assume that a sheet of paper is rectangular ... 阅读全文

posted @ 2013-02-25 17:33 铁树银花阅读(324) 评论(0) 推荐(0) 编辑

Conic Section（圆锥曲线基础）

摘要： Conic SectionTime Limit: 2 Seconds Memory Limit: 65536 KBThe conic sections are the nondegenerate curves generated by the intersections of a plane with one or two nappes of a cone. For a plane perpendicular to the axis of the cone, a circle is produced. For a plane that is not perpendicular to th... 阅读全文

posted @ 2013-02-25 17:31 铁树银花阅读(1503) 评论(0) 推荐(0) 编辑

敌兵布阵（线段树基础）

摘要：浙大研究生复试历年试题，欢迎练习~敌兵布阵Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23576Accepted Submission(s): 10236Problem Description C国的死对头A国这段时间正在进行军事演习，所以C国间谍头子Derek和他手下Tidy又开始忙乎了。A国在海岸线沿直线布置了N个工兵营地,Derek和Tidy的任务就是要监视这些工兵营地的活动情况。由于采取了某种先进的监测手段，所以每个工兵营地的人数C 阅读全文

posted @ 2013-02-25 17:29 铁树银花阅读(253) 评论(0) 推荐(0) 编辑

I Hate It（线段树基础）

摘要： I Hate ItTime Limit: 9000/3000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22019Accepted Submission(s): 8792Problem Description很多学校流行一种比较的习惯。老师们很喜欢询问，从某某到某某当中，分数最高的是多少。这让很多学生很反感。不管你喜不喜欢，现在需要你做的是，就是按照老师的要求，写一个程序，模拟老师的询问。当然，老师有时候需要更新某位同学的成绩。Input本题目包含多组测试，请处理到文件结束。在每个阅读全文

posted @ 2013-02-25 17:27 铁树银花阅读(243) 评论(0) 推荐(0) 编辑

线段树成段更新之延迟更新

摘要：线段树成段更新之延迟更新成段更新的重点是延迟更新，以区间[1,3]为例说明。注意，此例中“更新”为修改元素的值为a，“查询”为求区间中所有元素之和。建立二叉树如图示：每一个圆圈代表一个结点，圆内数字分别为结点标号和所对应区间，[i]表示只含一个数，只出现在叶节点中。当要更新区间[1,2]中所有元素时，对应上图即要更新结点4,5。一种方法是依次访问结点4和5并更新，但这样时间和空间开销都较大，当要修改的区间的长度较长时尤甚。于是可以采取“延迟更新”，即将更新信息储存在这段区间对应的结点处，此例中[1,2]对应的区间是结点2，给结点一个属性tag用以记录这个更新信息a，原来tag初始化为0，阅读全文

posted @ 2013-02-25 17:22 铁树银花阅读(1221) 评论(1) 推荐(1) 编辑

Just a Hook（线段树，成段更新）

摘要： Just a HookTime Limit : 4000/2000ms (Java/Other)Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 7Accepted Submission(s) : 4Problem DescriptionIn the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive me 阅读全文

posted @ 2013-02-25 17:21 铁树银花阅读(275) 评论(0) 推荐(0) 编辑

Codeforces Round #167 (Div. 2) B. Dima and Sequence（暴力）

摘要： B. Dima and Sequencetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputDima got into number sequences. Now he's got sequence a1, a2, ..., an, consisting of n positive integers. Also, Dima has got a function f(x), which can be defined with the fo 阅读全文

posted @ 2013-02-25 11:51 铁树银花阅读(249) 评论(0) 推荐(0) 编辑

Codeforces Round #167 (Div. 2) A. Dima and Friends (模拟，简单)

摘要： A. Dima and Friendstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputDima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean th 阅读全文

posted @ 2013-02-24 21:14 铁树银花阅读(354) 评论(0) 推荐(0) 编辑

Codeforces Round #168 (Div. 2)D. Zero Tree（DP，中等难度）

摘要： D. Zero Treetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputA tree is a graph with n vertices and exactly n - 1 edges; this graph should meet the following condition: there exists exactly one shortest (by number of edges) path between any pair of 阅读全文

posted @ 2013-02-24 11:46 铁树银花阅读(519) 评论(0) 推荐(0) 编辑

Codeforces Round #168 (Div. 2) C. k-Multiple Free Set（二分查找）

摘要： C. k-Multiple Free Settime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputA k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y 阅读全文

posted @ 2013-02-23 09:39 铁树银花阅读(630) 评论(0) 推荐(0) 编辑

Codeforces Round #168 (Div. 2) B. Convex Shape（暴力）

摘要： B. Convex Shapetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputConsider an n × m grid. Initially all the cells of the grid are colored white. Lenny has painted some of the cells (at least one) black. We call a painted grid convex if one can 阅读全文

posted @ 2013-02-22 15:15 铁树银花阅读(366) 评论(0) 推荐(0) 编辑

Codeforces Round #168 (Div. 2) A. Lights Out（模拟）

摘要： A. Lights Outtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputLenny is playing a game on a 3 × 3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. 阅读全文

posted @ 2013-02-22 11:04 铁树银花阅读(438) 评论(0) 推荐(0) 编辑

Codeforces Round #166 (Div. 2)D. Good Substrings（字符串散列）

摘要： D. Good Substringstime limit per test2 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputYou've got string s, consisting of small English letters. Some of the English letters are good, the rest are bad.A substring s[l...r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2...s|s 阅读全文

posted @ 2013-02-21 23:57 铁树银花阅读(398) 评论(0) 推荐(0) 编辑

BKDR Hash Function

摘要：字符串哈希函数繁多，据说在信息学竞赛中，高效而易于记忆的是BKDR Hash Function. 代码如下： 1 // BKDR Hash Function 2 int BKDRHash(char *str) 3 { 4 int seed = 131; // 31 131 1313 13131 131313 etc.. 5 int hash = 0; 6 7 while (*str) 8 { 9 hash = hash * seed + (*str++);10 }11 12 return (hash & 0x7FFFFFFF);... 阅读全文

posted @ 2013-02-21 21:05 铁树银花阅读(1439) 评论(0) 推荐(0) 编辑

开始菜单中找不到“运行”命令如何解决？（转）

摘要：原文链接：http://www.zgdnjj.com/html/33/n-3333.html开始菜单中找不到“运行”命令如何解决？在此Windows版本中，“运行”命令未显示在开始菜单上。开始菜单上的搜索框提供了很多与“运行”命令相同的功能。但是，如果需要，仍可以使用“运行”命令。甚至可以将其添加到开始菜单以方便使用。步骤如下：1.通过依次单击「开始」按钮、“控制面板”、“外观和个性化”以及“任务栏和「开始」菜单”，打开“任务栏和「开始」菜单属性”。2.单击“「开始」菜单”选项卡，然后单击“自定义”。3.在「开始」菜单选项列表中，选中“运行命令”复选框，然后单击“确定”。“运行”命令将显示在「阅读全文

posted @ 2013-02-20 22:41 铁树银花阅读(745) 评论(0) 推荐(0) 编辑

Codeforces Round #166 (Div. 2)C. Secret（构造）

摘要： C. Secrettime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), 阅读全文

posted @ 2013-02-18 13:39 铁树银花阅读(259) 评论(0) 推荐(0) 编辑

自然码双拼规则（转）

摘要：口诀：Zei Ying Qiu Nin Fen Ii W.ua.ia Ee 贼　应求　您　分　吃蛙　鸭鹅， T.ve Song Uu Lai Vui Ruan+ Mian Xie, 特约松鼠　来　追　软棉鞋， C.iao (Y) Hang P.un.vn,Geng Kao B.ou J.an D.uang.iang.此要 (外)　行　破文韵，　更靠　笔偶结案得汪洋。阅读全文

posted @ 2013-02-18 10:16 铁树银花阅读(1055) 评论(0) 推荐(0) 编辑

Codeforces Round #166 (Div. 2) B. Prime Matrix（素数筛选，简单）

摘要： B. Prime Matrixtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou've got an n × m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and i 阅读全文

posted @ 2013-02-17 21:47 铁树银花阅读(339) 评论(0) 推荐(0) 编辑

Codeforces Round #166 (Div. 2) A. Beautiful Year(水题)

摘要： A. Beautiful Yeartime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputIt seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.Now you are suggested 阅读全文

posted @ 2013-02-17 19:54 铁树银花阅读(661) 评论(0) 推荐(0) 编辑

windows如何取消开机启动项

摘要：可以利用金山卫士、360安全卫士等软件禁止系统不必要的启动项，但是有些启动项用这些软件是无法禁止的，这时就需要自己去调节系统设置。依次选择“开始”——“运行”，输入msconfig，弹出如下对话框选择“启动”选项卡，将不需要开机启动的启动项前的钩钩去掉即可。阅读全文

posted @ 2013-02-17 19:28 铁树银花阅读(247) 评论(0) 推荐(0) 编辑

如何查看IE型号

摘要：打开IE浏览器，以此选择“工具”——“关于Internet Explorer”，就能看到类似于下图的信息。阅读全文

posted @ 2013-02-17 17:40 铁树银花阅读(173) 评论(0) 推荐(0) 编辑

如何查看电脑配置

摘要： 1. 查看一般硬件配置鼠标右键点击我的电脑——属性，从弹出的“系统属性”中能看到CPU和显卡的信息。要获得更详细的信息可以点击“硬件”选项卡，再点击“设备管理器”，然后可以看到CPU、显卡、网卡乃至鼠标、键盘等的硬件信息。2.通过Dirext诊断工具查看开始——运行，输入dxdiag，按确定后显示Dirext诊断工具，通过该工具能查看电脑比较详细的硬件信息。3.查看硬盘总大小鼠标右键点击我的电脑，选择管理，在弹出的对话框中选择磁盘管理，在“磁盘0”处能查看磁盘的总大小。4.利用工具软件如鲁大师。阅读全文

posted @ 2013-02-16 10:12 铁树银花阅读(200) 评论(0) 推荐(0) 编辑

Somali Pirates（水题）

摘要： Somali PiratesTime Limit: 1 Second Memory Limit: 32768 KBIt is said that the famous Somali Pirates hate digits. So their QQ passwords never contain any digit. Given some lines of candidate passwords, you are asked to delete all the digits in the passwords and print other characters in their origi... 阅读全文

posted @ 2013-02-15 16:29 铁树银花阅读(448) 评论(0) 推荐(0) 编辑

Who is Older?（水题）

摘要： Who is Older?Time Limit: 1 Second Memory Limit: 32768 KBJavaman and cpcs are arguing who is older. Write a program to help them.InputThere are multiple test cases. The first line of input is an integer T (0 < T <= 1000) indicating the number of test cases. Then T test cases follow. The i-th li 阅读全文

posted @ 2013-02-15 16:17 铁树银花阅读(644) 评论(0) 推荐(0) 编辑

Compromise(求解最长公共子序列并输出)

摘要： CompromiseTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 5181Accepted: 2361Special JudgeDescriptionIn a few months the European Currency Union will become a reality. However, to join the club, the Maastricht criteria must be fulfilled, and this is not a trivial task for the countries (mayb 阅读全文

posted @ 2013-02-15 13:41 铁树银花阅读(266) 评论(0) 推荐(0) 编辑

warning: name lookup of `i' changed

摘要：出错现场：int main(){ int i; for(int i = 0; i < 10; i++) ……}i 这个变量名重复使用阅读全文

posted @ 2013-02-15 12:48 铁树银花阅读(1422) 评论(0) 推荐(0) 编辑

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