SDUT 2603:Rescue The Princess

SDUT 2603: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.

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 0.000.00 0.00 0.00 100.000.00 0.00 100.00 100.001.00 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)

提示

题意：按逆时针顺序给出两点坐标，求第三点坐标。使这三点构成一个等边三角形！

#include#include#define PI acos(-1.0)int main(){ int n; scanf("%d",&n); while(n--) { double x1,x2,y1,y2; scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); double t=atan2(y2-y1,x2-x1)+PI/3.0; double zz=sqrt(pow(x1-x2,2)+pow(y1-y2,2)); double x=zz*cos(t)+x1,y=zz*sin(t)+y1; printf("(%.2lf,%.2lf)\n",x,y); } return 0;}

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