山东省第四届ACM省赛 A SDUT 2603 Rescue The Princess（简单数学题）_sduta

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在这款游戏中，玩家扮演一位王子，为了营救被怪兽囚禁的公主，必须找到通往迷宫出口的路径。怪兽巧妙地隐藏了公主，并在迷宫中设置了一个等边三角形作为陷阱。玩家需要利用提供的两个坐标点，计算出第三个未知的坐标点，从而找到出口，成功解救公主。

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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 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?

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 0.00
0.00 0.00 0.00 100.00
0.00 0.00 100.00 100.00
1.00 0.00 1.866 0.50

示例输出

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

#include <bits/stdc++.h>

using namespace std;

int main()
{
    double x1, x2, x3, y1, y2, y3;
    int n;
    while(cin>>n)
    {
        while(n--)
        {
            cin>>x1>>y1>>x2>>y2;
            x3 = (x1+x2)/2-(y2-y1)*sqrt(3.0)/2;
            y3 = (y1+y2)/2+(x2-x1)*sqrt(3.0)/2;
            printf("(%.2lf,%.2lf)\n",x3,y3);
        }
    }
    return 0;
}