山东省第四届ACM大学生程序设计竞赛-Rescue The Princess（计算几何）

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 equilateraltriangle 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 youcalculate 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 equilateraltriangle. 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)

提示

 

来源

2013年山东省第四届ACM大学生程序设计竞赛

题目意思：
给定A、B两个点的坐标，求另外一个点C，使得ABC构成一个等边三角形，且ABC三点是逆时针顺序。

解题思路如图（From 队友大神CXF）：



注意到数学反三角函数double atan2 (double, double); 反正切(整圆值), 结果介于[-PI/2, PI/2]
区别于double atan (double); 反正切(主值), 结果介于[-PI/2, PI/2]，因为斜率为零的情况 atan2不返回0而不是错误值。

/* 
* Copyright (c) 2016, 烟台大学计算机与控制工程学院 
* All rights reserved. 
* 文件名称：cal.cpp 
* 作    者：单昕昕 
* 完成日期：2016年4月9日 
* 版 本 号：v1.0 
*/  
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
const double PI=acos(-1.0);

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double x1,y1,x2,y2;
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        double l=sqrt((y1-y2)*(y1-y2)+(x1-x2)*(x1-x2));
        double s=atan2(y2-y1,x2-x1);
        double l1=l*sin(s+(PI/3.0));
        double l2=l*cos(s+(PI/3.0));
        double x=x1+l2;
        double y=y1+l1;
        printf("(%.2lf,%.2lf)\n",x,y);
    }
    return 0;
}