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. [Json.net]Linq to Json

    引言 上篇学习了json.net的基本知识,这篇学习linq to json. 上篇文章:[Json.net]快速入门 Linq to Json Linq to Json是用来快速操作json对象的, ...

  2. 小结-Splay

    参照陈竞潇学长的模板写的BZOJ 3188: #include<cstdio> #include<cstring> #include<algorithm> #def ...

  3. nginx配置反向代理示例

    环境: nginx1:192.168.68.41 tomcat1:192.168.68.43 tomcat2:192.168.68.45 nginx安装网上很多教程,我是用yum安装的. 配置ngin ...

  4. poj1509 最小表示法

    #include<stdio.h> #include<string.h> #define maxn 10010 char s[maxn]; int getmin() { int ...

  5. poj1679 kruskal

    判断最小生成树是否唯一.kruskal时记录需要的边,然后枚举删除它们,每次删除时进行kruskal,如果值未变,表明不唯一. #include<stdio.h> #include< ...

  6. spring 第一篇(1-1):让java开发变得更简单(下)

    切面(aspects)应用 DI能够让你的软件组件间保持松耦合,而面向切面编程(AOP)能够让你捕获到在整个应用中可重用的组件功能.在软件系统中,AOP通常被定义为提升关注点分离的一个技术.系统由很多 ...

  7. BZOJ3172 后缀数组

    题意:求出一篇文章中每个单词的出现次数 对样例的解释: 原文是这样的: a aa aaa 注意每个单词后都会换行 所以a出现次数为6,aa为3 (aa中一次,aaa中两次),aaa为1 标准解法好像是 ...

  8. Threat Risk Modeling Learning

    相关学习资料 http://msdn.microsoft.com/en-us/library/aa302419(d=printer).aspx http://msdn.microsoft.com/li ...

  9. OPENSSL编程入门学习

    相关学习资料 http://bbs.pediy.com/showthread.php?t=92649 https://www.openssl.org https://www.google.com.hk ...

  10. Linux mount/unmount命令(转)

    格式:mount [-参数] [设备名称] [挂载点] 其中常用的参数有:-a 安装在/etc/fstab文件中类出的所有文件系统.-f 伪装mount,作出检查设备和目录的样子,但并不真正挂载文件系 ...