[2013山东省第四届ACM大学生程序设计竞赛]——Rescue The Princess

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)

来源

2013年山东省第四届ACM大学生程序设计竞赛

题目：http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2603

这道题就是当初我卡的那道。。很简单，就是自己*%&了。。。

当时不知道脑子咋了，愣是看了好久好久。。。

无论计算几何还是解析几何都可以做出来。

用计算几何搞得这道题。

题意： 给你等边三角形的两个点，逆时针的第三个点坐标。

用的公式：OB=(x*cosα-y*sinα，x*sinα+y*cosα)

这是逆时针的，还有顺时针的：

OA=(y*sinα - x*cosα，x*sinα+y*cosα)

OA，OB看晕了吧，当然还要配图：

然后，这道题就迎刃而解了。。。

O,还有，这道题会卡精度。。。π值最好用acos（-1）来求，

反正我用3.1415926不行（我只能背到这了，o(╯□╰)o）

/**************************************
***************************************
*        Author:Tree                  *
*From :http://blog.csdn.net/lttree    *
* Title : Rescue The Princess         *
*Source: 第四届山东省ACM比赛         *
* Hint  : 计算几何                   *
***************************************
**************************************/
#include <stdio.h>
#include <cmath>
using namespace std;
#define Pi acos(-1.0)

struct Point
{
    double x,y;
};
struct Line
{
    Point a,b;
};
// 向量l,绕l.a点逆时针转a弧度,返回转到的点。
// OB=(x*cosα-y*sinα，x*sinα+y*cosα)
Point rotate_anticw(Line l,double a)
{
    Point p;
    p.x = ( l.b.x-l.a.x ) * cos(a) - ( l.b.y-l.a.y ) * sin(a);
    p.y = ( l.b.x-l.a.x ) * sin(a) + ( l.b.y-l.a.y ) * cos(a);
    p.x += l.a.x;
    p.y += l.a.y;
    return p;
}
// 向量l,绕l.a点顺时针转a弧度,返回转到的点。
// OA=(y*sinα - x*cosα，x*sinα+y*cosα)
Point rotate_cw(Line l,double a)
{
    Point p;
    p.x = ( l.b.y-l.a.y ) * sin(a) - ( l.b.x-l.a.x ) * cos(a);
    p.y = ( l.b.x-l.a.x ) * sin(a) + ( l.b.y-l.a.y ) * cos(a);
    p.x += l.a.x;
    p.y += l.a.y;
    return p;
}
int main()
{
    int t;
    Point c;
    Line l;
    scanf("%d",&t);
    while( t-- )
    {
        scanf("%lf%lf%lf%lf",&l.a.x,&l.a.y,&l.b.x,&l.b.y);
        c = rotate_anticw(l,60*Pi/180);
        printf("(%.2lf,%.2lf)\n",c.x,c.y);
    }
    return 0;
}

[2013山东省第四届ACM大学生程序设计竞赛]——Rescue The Princess