2013山东省“浪潮杯”省赛 A.Rescue The Princess

A.Rescue The PrincessDescription

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?

Input

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.

Output

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.

Sample Input

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

Sample Output

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

最原始方法是算算算，在别人博客学到的是用旋转矩阵的，感觉厉害多了..

例如：把向量写成坐标形式并视为一个2*1列向量，再左乘一个2*2的矩阵，这样得到的新的2*1列向量就是旋转后的向量的坐标。（2*2矩阵结构如下：设旋转 角度为a, 那么左上角和右下角的元素为cos(a), 右上角的元素为-sin(a), 左下角的元素为sin(a)）。

#include<cstdio>
#include<cstring>
#include<cmath>
const double sq3=sqrt(3.0);
double m1[1][2],m3[1][2];
double m2[2][2]= {{0.5,sq3/2},{-sq3/2,0.5}};
void so()
{
    memset(m3,0,sizeof(m3));
    for(int k=0; k<2; ++k)
        for(int j=0; j<=1; ++j)
        {
            //printf("m1[0][%d]=%lf m2[%d][%d]=%lf\n",k,m1[0][k],k,j,m2[k][j]);
            m3[0][j]+=m1[0][k]*m2[k][j];
        }


}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        double x1,x2,y1,y2;
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        m1[0][0]=x2-x1;
        m1[0][1]=y2-y1;
        so();
        printf("(%.2lf,%.2lf)\n",m3[0][0]+x1,m3[0][1]+y1);
    }
    return 0;
}