HDU1073:Online Judge】的更多相关文章

Problem Description Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file, then the system compare the two files. If the two f…
题目链接 http://codeforces.com/gym/101102/problem/C problem description Judge Bahosain was bored at ACM AmrahCPC 2016 as the winner of the contest had the first rank from the second hour until the end of the contest. Bahosain is studying the results of t…
Problem 1074: Hey Judge Time Limits:  1000 MS   Memory Limits:  65536 KB 64-bit interger IO format:  %lld   Java class name:  Main   Description Judge Nicole collected 7 ideas for problems of different levels, she wants to create 5 problems for the n…
转自:http://www.cnblogs.com/chouti/p/5752819.html Special Judge:当正确的输出结果不唯一的时候需要的自定义校验器 首先有个框架 #include<fstream> ifstream fin,fout,fstd ofstream fscore,freport; double Judge(){ } int main(int argc,char *argv[]) { //put something to fstreams... //Judge…
前段时间准备华为机试,正好之前看了一遍<剑指 Offer>,就在九度 Online Judge 上刷了书中的题目,使用的语言为 C++:只有3题没做,其他的都做了. 正如 Linus Torvalds 所言“Talk is cheap, show me the code!",详见托管在 Github 的源码:点击打开链接(核心部分有注释). 难度总结:有些可以不必要像书中那样保存中间结果,直接输出,也就可以不使用书中的算法:有些题目在书中题目的基础上加了更多的要求,也更难实现一点.…
UVa Online Judge 工具網站   UVa中译题uHuntAlgorithmist Lucky貓的ACM園地,Lucky貓的 ACM 中譯題目 Mirror UVa Online Judge題目中譯!非常偉大的工作,請大家要心懷敬意! Ruby兔的ACM園地 UVa Online Judge題目中譯!非常偉大的工作,請大家要心懷感激! uHunt 這個站製作了一些網頁小工具, 可以查詢自己在UVa Online Judge的解題進度.簡單題列表.世界排名等等. 另外也整理了一套題庫,…
Submissions of online judge(1021) 问题描述 An online judge is a system to test programs in programming contests. The system can compile and execute codes, and test them with pre-constructed data. Submitted code may be run with restrictions, including tim…
Problem Description Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file, then the system compare the two files. If the two f…
Big endian means the most significant byte stores first in memory. int a=0x01020304, if the cpu is big endian, data are store 01 02 03 04 in memory in increasing address. Below is a simple code to make the judgement. //execute the following cmd on th…
 Hangman Judge  In ``Hangman Judge,'' you are to write a program that judges a series of Hangman games. For each game, the answer to the puzzle is given as well as the guesses. Rules are the same as the classic game of hangman, and are given as follo…
杭州电子科技大学Online Judge 之 "确定比赛名次(ID1285)"解题报告 巧若拙(欢迎转载,但请注明出处:http://blog.csdn.net/qiaoruozhuo) Problem Description 有N个比赛队(1<=N<=500).编号依次为1,2,3,.....N进行比赛.比赛结束后.裁判委员会要将全部參赛队伍从前往后依次排名. 但如今裁判委员会不能直接获得每一个队的比赛成绩,仅仅知道每场比赛的结果.即P1赢P2,用P1.P2表示,排名时P…
Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move sequence is represented by a string. And each move is represent by a characte…
Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move sequence is represented by a string. And each move is represent by a characte…
Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move sequence is represented by a string. And each move is represent by a characte…
项目名 Piers 在线评测 项目需求 用户: 获取题库.题目的相关信息. 在线对代码进行编译.执行.保存.返回运行(编译)结果. 总体题目评测成绩查询. 用户信息服务,包括注册.登录.忘记密码.邮箱验证等功能. 管理员: 题库.题目相关信息的增删改查. 用户导入,包括文件导入.输入等方式. 用户成绩查询与修改. 题目分析. 用户抄袭作弊检测. 项目特色 安全性高,采用 Docker.Tomcat 安全策略等多重机制保护运行时安全. 拥有抄袭作弊检测的功能,方便管理员对用户的作业检查. 基于 B…
Spring 配置一些本地类,还有 HTML form 提交文件的解析器. package per.piers.onlineJudge.config; import org.springframework.context.annotation.Configuration; import org.springframework.web.servlet.support.AbstractAnnotationConfigDispatcherServletInitializer; import javax…
Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place. The move sequence is represented by a string. And each move is represent by a characte…
watch out 本文是博主的 csdn 上搬过来的,格式有点崩,看不下去的可以去 博主的 csdn上看(上面 格式会好很多,并且有些公式也用 $\LaTeX$  update 上去了) 最近有点颓废啊,写篇blog振作一下…(不过没图的数论blog真是不对我胃口) emmm …首先介绍一下这是一篇关于数论中较为重要 (主要可能经常要用到) 的一个知识分支—逆元 相信你听到数论之后可能鼠标就想往网页上的 X 键上移了,但是还是劝你看一看, 总能有点收获的吧 (反正我觉得自己讲的相对网上其他 b…
In a town, there are N people labelled from 1 to N.  There is a rumor that one of these people is secretly the town judge. If the town judge exists, then: The town judge trusts nobody. Everybody (except for the town judge) trusts the town judge. Ther…
https://leetcode.com/problems/find-the-town-judge/ In a town, there are N people labelled from 1 to N.  There is a rumor that one of these people is secretly the town judge. If the town judge exists, then: The town judge trusts nobody. Everybody (exc…
关于一些逆元知识的拓展 刚艹完一道 提高- 的黄题(曹冲养猪) ,于是又来混一波讲解了 ——承接上文扫盲篇   四.Lucas定理(求大组合数取模)   题外话 这里Lucas定理的证明需要用到很多关于组合数的定理知识,  那么关于一些组合数的知识,详情你可以看这里:Binamoto' blog. 再讲讲lucas定理这个东西(扩展lucas就不讲了,因为不大会…咳咳,然后也不怎么会用到吧) 基本公式: C(n,m) ≡ C(n/p,m/p)*C(n%p,m%p) (mod p) (也就是: C…
详解OJ(Online Judge)中PHP代码的提交方法及要点 Introduction of How to submit PHP code to Online Judge Systems  Introduction of How to commit submission in PHP to Online Judge Systems 在目前常用的在线oj中,codeforces.spoj.uva.zoj 等的题目可使用PHP实现基本算法,zoj是目前对PHP支持较好的中文OJ. PHP是一门比…
题目描述 初始位置 (0, 0) 处有一个机器人.给出它的一系列动作,判断这个机器人的移动路线是否形成一个圆圈,换言之就是判断它是否会移回到原来的位置. 移动顺序由一个字符串表示.每一个动作都是由一个字符来表示的.机器人有效的动作有 R(右),L(左),U(上)和 D(下).输出应为 true 或 false,表示机器人移动路线是否成圈. 示例 1: 输入: "UD" 输出: true 示例 2: 输入: "LL" 输出: false 思路 设置初始点的坐标为x=0…
Description Given a positive integer n, find two non-negative integers a, b such that a2 + b2 = n. Input The first line contains the number of test cases T (1 <= T <= 1000). For each test case, there is only one line with an integer n (1 <= n <…
做道题,并没有太多的技巧,关键在与对Accepted,presented error 和wa的判断,第一步如果两者完全一样,那么很定是AC了 ,否则如果去掉多余换行,空格,制表后还有不同说明是数据 不同,就wa,如果相同就是pe了........ Online Judge Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4121    …
功能 judge 模块主要从transfer中接收数据,并从HBS中获取报警策略,然后进行阈值报警判断 从HBS获取报警策略 接收transfer 上报的数据,并存储最新几个点 判断阈值,产生报警事件 判断报警事件是否写入redis 老旧报警数据的清理 配置文件 { "debug": true, "debugHost": "nil", #用于调适,在log中打印指定host的具体策略 "remain": 11, #指定内存中存…
一 安装 JDK 1.5 1 下载 到 Sun 官方网站( http://java.sun.com/j2se/1.5.0 /download.jsp )下载 j2sdk ,注意下载为 JDK 5.0 Update 22 (大约 33.1M ). 2 安装 运行下载的文件,我的安装路径按照的是默认设置,即:C:/Program Files/Java/jdk1.5.0_22 ,也可以根据自己的需要进行修改. 3 配置 在我的电脑-> 属性-> 高级-> 环境变量-> 系统变量中添加以下…
Automatic Judge Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 16    Accepted Submission(s): 11 Problem Description Welcome to HDU to take part in the second CCPC girls' competition! A new a…
UVA.12169 Disgruntled Judge ( 拓展欧几里得 ) 题意分析 给出T个数字,x1,x3--x2T-1.并且我们知道这x1,x2,x3,x4--x2T之间满足xi = (a * xi-1 + b ) MOD 10001, 求出x2,x4--x2T. 由于本题中的a和b是未知的,所以需要根据已知条件求出a和b,据说有人暴力枚举a和b然后过了. 所以我来换另一种方法. 其实我们可以枚举a,并根据x1,x3算出求出可行的b的值.如何做到呢? 首先我们已经知道 x2 = (a *…
题目链接 Problem Description Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file, then the system compare the two files. If the…