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 youcalculate 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 0.000.00 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

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)。