rescue the princess 省赛四1


Rescue The Princess

Time Limit: 1000MS Memory limit: 65536K

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 anequilateraltriangle
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 theequilateraltriangle.
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)

提示

即将AB线段绕A点逆时针旋转60°后B点的位置

用到平面几何求解

x3=x1+L*cos(60°+angle);

y3=y1+L*sin(60°+angle);

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

60°化成弧度

const double PI=acos(-1.0);

而PI/3.0就是60°

#include <iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const double PI=acos(-1.0);
int main()
{
int t;
cin>>t;
double x1,y1,x2,y2,x3,y3,angle,l;
while(t--)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
angle=atan2(y2-y1,x2-x1);
l=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
x3=x1+l*cos(angle+PI/3.0);
y3=y1+l*sin(angle+PI/3.0);
printf("(%.2lf,%.2lf)\n",x3,y3);
printf("%.2lf",acos(PI));
}
return 0;
}


atan2(a, b) 与 ATAN(a/b)稍有不同，atan2(a,b)的取值范围介于 -pi 到 pi 之间（不包括 -pi），
而ATAN(a/b)的取值范围介于-pi/2到pi/2之间（不包括±pi/2)。