山東省第四屆ACM大學生程式設計競賽-Rescue The Princess（計算幾何）

Rescue The Princess

Time Limit: 1000ms Memory limit: 65536K 有疑問？點這裡^_^

題目描述

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?

輸入

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.

輸出

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.

示例輸入

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

示例輸出

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

提示

來源

2013年山東省第四屆ACM大學生程式設計競賽

題目意思：
給定A、B兩個點的座標，求另外一個點C，使得ABC構成一個等邊三角形，且ABC三點是逆時針順序。

解題思路如圖（From 隊友大神CXF）：

注意到數學反三角函式double atan2 (double, double); 反正切(整圓值), 結果介於[-PI/2, PI/2]
區別於double atan (double); 反正切(主值), 結果介於[-PI/2, PI/2]，因為斜率為零的情況atan2不返回0而不是錯誤值。

/* 
* Copyright (c) 2016, 煙臺大學計算機與控制工程學院 
* All rights reserved. 
* 檔名稱：cal.cpp 
* 作    者：單昕昕 
* 完成日期：2016年4月9日 
* 版 本 號：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;
}