山东省赛A题：Rescue The Princess

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.

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)

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)

向左转 |向右转

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;
}