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 *. 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?

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 0.00

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

来源

2013年山东省第四届ACM大学生程序设计竞赛

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<stdio.h>

#include<math.h>

#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;

}

秒客网