Rescue The Princess

Time Limit: 1 Sec  Memory Limit: 128 MB Submit: 412  Solved: 168 [Submit][Status][Web Board]

Description

Several days ago, a beast caught a beautiful princess and the princess was put in prison. To rescue the princess, a prince who wanted to marry  the princess set out immediately. Yet, the beast set a maze. Only if the prince find out the maze’s exit can he save the princess.
Now, here comes the problem. The maze is a dimensional plane. The beast is smart, and he hidden the princess snugly. He marked two coordinates of an equilateral triangle in the maze. The two marked coordinates are A(x1,y1) and B(x2,y2). The third coordinate C(x3,y3) is the maze’s exit. If the prince can find out the exit, he can save the princess. After the prince comes into the maze, he finds out the A(x1,y1) and B(x2,y2), but he doesn’t know where the C(x3,y3) is. The prince need your help. Can you calculate the C(x3,y3) and tell him?

Input

The first line is an integer T(1 <= T <= 100) which is the number of test cases. T test cases follow. Each test case contains two coordinates A(x1,y1) and B(x2,y2), described by four floating-point numbers x1, y1, x2, y2 ( |x1|, |y1|, |x2|, |y2| <= 1000.0). 
        Please notice that A(x1,y1) and B(x2,y2) and C(x3,y3) are in an anticlockwise direction from the equilateral triangle. And coordinates A(x1,y1) and B(x2,y2) are given by anticlockwise.
 

Output

For each test case, you should output the coordinate of C(x3,y3), the result should be rounded to 2 decimal places in a line.

Sample Input

4
-100.00 0.00 0.00 0.00
0.00 0.00 0.00 100.00
0.00 0.00 100.00 100.00
1.00 0.00 1.866 0.50

Sample Output

(-50.00,86.60)
(-86.60,50.00)
(-36.60,136.60)
(1.00,1.00)

HINT

已知一向量为(x , y) 则将它旋转θ后的坐标为(x*cosθ- y * sinθ , y*cosθ + x * sinθ)

转载:http://www.tuicool.com/articles/FnEZJb

山东省第四届acm.Rescue The Princess(数学推导)的更多相关文章

  1. 山东省第四届ACM大学生程序设计竞赛解题报告(部分)

    2013年"浪潮杯"山东省第四届ACM大学生程序设计竞赛排名:http://acm.upc.edu.cn/ranklist/ 一.第J题坑爹大水题,模拟一下就行了 J:Contes ...

  2. Alice and Bob(2013年山东省第四届ACM大学生程序设计竞赛)

    Alice and Bob Time Limit: 1000ms   Memory limit: 65536K 题目描述 Alice and Bob like playing games very m ...

  3. 2013年山东省第四届ACM大学生程序设计竞赛-最后一道大水题:Contest Print Server

    点击打开链接 2226: Contest Print Server Time Limit: 1 Sec  Memory Limit: 128 MB Submit: 53  Solved: 18 [Su ...

  4. sdut Mountain Subsequences 2013年山东省第四届ACM大学生程序设计竞赛

    Mountain Subsequences 题目描述 Coco is a beautiful ACMer girl living in a very beautiful mountain. There ...

  5. 山东省第四届ACM程序设计竞赛A题:Rescue The Princess(数学+计算几何)

    Rescue The Princess Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 412  Solved: 168[Submit][Status][ ...

  6. 山东省第四届ACM程序设计竞赛A题:Rescue The Princess

    Description Several days ago, a beast caught a beautiful princess and the princess was put in prison ...

  7. 计算几何 2013年山东省赛 A Rescue The Princess

    题目传送门 /* 已知一向量为(x , y) 则将它旋转θ后的坐标为(x*cosθ- y * sinθ , y*cosθ + x * sinθ) 应用到本题,x变为(xb - xa), y变为(yb ...

  8. 山东省第四届acm解题报告(部分)

    Rescue The PrincessCrawling in process... Crawling failed   Description Several days ago, a beast ca ...

  9. 山东省第四届ACM省赛

    排名:http://acm.sdut.edu.cn/sd2012/2013.htm 解题报告:http://www.tuicool.com/articles/FnEZJb A.Rescue The P ...

随机推荐

  1. 学习笔记——Maven实战(二)POM重构之增还是删

    重构是广大开发者再熟悉不过的技术,在Martin Fowler的<重构——改善既有代码的设计>一书中,其定义为“重构(名词):对软件内部结构的一种调整,目的是在不改变软件之可察行为前提下, ...

  2. 如何部署Iveely.Computing分布式实时计算系统

    Iveely.Computing是参考Storm的分布式实时计算系统的部分原理,用纯Java实现的轻量级.迷你型,适合于搜索引擎的实时计算系统, Iveely 搜索引擎是一款基于Iveely.Comp ...

  3. (旧)子数涵数·Flash——Flash Player的操作命令

    一.什么是Flash Player? Flash Player就是官方指定的一种FLash播发器. 用百度的话来讲,Adobe Flash Player 是一款高级客户端运行时使用的播放器.它短小精悍 ...

  4. C#中判断一个数组中是否存在某个数组值 及相关

    声明:reference:http://www.cnblogs.com/icebutterfly/archive/2010/06/22/1762738.html:http://blog.csdn.ne ...

  5. Lambda表达式和表达式树

    在C# 2.0中,通过方法组转换和匿名方法,使委托的实现得到了极大的简化.但是,匿名方法仍然有些臃肿,而且当代码中充满了匿名方法的时候,可读性可能就会受到影响.C# 3.0中出现的Lambda表达式在 ...

  6. Bootstrap3.0学习第十五轮(大屏幕介绍、页面标题、缩略图、警示框、Well)

    详情请查看 http://aehyok.com/Blog/Detail/22.html 个人网站地址:aehyok.com QQ 技术群号:206058845,验证码为:aehyok 本文文章链接:h ...

  7. VMware v12.1.1 专业版以及永久密钥

    热门虚拟机软件VMware Workstation 现已更新至v12.1.1 专业版!12.0属于大型更新,专门为Win10的安装和使用做了优化,支持DX10.4K高分辨率显示屏.OpenGL 3.3 ...

  8. 传智168期JavaEE就业班 day04-dom

    * 课程回顾: * js语法 * js的动态函数和匿名函数 * js动态函数 Function new Function(); * 匿名函数:没有名称的函数,起个名称 var add = functi ...

  9. DML语言练习,数据增删改查,复制清空表

    Connected Connected as TEST@ORCL SQL> select * from t_hq_bm; BUMBM BUMMC DIANH ---------- ------- ...

  10. iOS之单例

    今天在看多线程同步时,突然想到了单例的同步问题.自从dispatch_once出现后,我们创建单例非常简单且安全: static dispatch_once_t pred; static Single ...