A

Rescue The Princess

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

    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 anequilateral 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?

输入

    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.

输出

    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.

示例输入

4-100.00 0.00 0.00 0.000.00 0.00 0.00 100.000.00 0.00 100.00 100.001.00 0.00 1.866 0.50

示例输出

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

提示

 

来源

2013年山东省第四届ACM大学生程序设计竞赛

示例程序

推广结论：对于任意两个不同点A和B，A绕B旋转θ角度后的坐标为：

(Δx*cosθ- Δy * sinθ+ xB, Δy*cosθ + Δx * sinθ+ yB ) 

[cpp] view plaincopyprint?

1. #include<iostream>  
2. #include<cstdio>  
3. #include<cstring>  
4. #include<cstdio>  
5. #include<math.h>  
6. using namespace std;  
7. int main()  
8. {  
9.     int T;  
10.     scanf("%d",&T);  
11.     while(T--)  
12.     {  
13.         double x1,x2,y1,y2,x3,y3,x,y;  
14.         scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);  
15.         x=x2-x1;  
16.         y=y2-y1;  
17.         x3=x/2-y*sqrt(3)/2+x1;  
18.         y3=y/2+x*sqrt(3)/2+y1;  
19.         printf("(%.2f,%.2f)\n",x3,y3);  
20.     }  
21.     return 0;  
22. }