http://acm.sdibt.edu.cn/JudgeOnline/problem.php?id=3230

Description

Several days ago, a beast caught a beautiful princess and the princess was put in *. 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.

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).

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)

题意：给出A与B的坐标，要求从A到B逆时针转动位置找到一个C点，使他们组成一个等边三角形

思路：

1、求已知线段的斜角：tgα=(y1-y2)/(x1-x2)

2、求已知线段的长度：L=√((y1-y2)^2+(x1-x2)^2)

3、求第三点的坐标：

x3=x2+L*cos(α+60)；y3=y2+L*sin(α+60)

而且我们可以发现，如果全部以A点为基底的话，那么求出的C点一定全部都是在A->B的逆时针上，还记得训练时本来已经推出这个公式了，但是当时脑筋很死，没有想到这一点，一直在考虑多种情况，结果反而落于下乘了，在这里我要好好反省

#include <stdio.h>

#include <math.h>

const double pi = acos(-1.0);

int main()

{

    int t;

    double x1,x2,x3,y1,y2,y3,l,at;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);

        at = atan2(y2-y1,x2-x1);

        printf("%lf\n",(y2-y1)/(x2-x1));

        printf("%lf\n",at);

        l = sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));

        x3 = x1+l*cos(at+pi/3.0);

        y3 = y1+l*sin(at+pi/3.0);

        printf("(%.2lf,%.2lf)\n",x3,y3);

    }

    return 0;

}