sdut 2603 Rescue The Princess(算是解析几何吧)(山东省第四届ACM省赛A题)

题目地址:sdut 2603

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.

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

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应该是解析几何的:已知等边三角形两个顶点坐标,求第三个顶点的坐标

我转换成向量做的: 已知的两个点可以组成一个向量,然后逆时针旋转60度即得到另外一个顶点和已知顶点组成的一个向量,坐标相加即可得到等边三角形第三个顶点的坐标

向量旋转公式:

若已知向量为(x,y),则它旋转角度A后得到的向量为:(x*cosA-y*sinA,x*sinA+y*cosA) ,有了公式就显得很水了,不过还是记下来。

想请参见:http://www.cnblogs.com/woodfish1988/archive/2007/09/10/888439.html

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Code:

#include <stdio.h>
#include <stdlib.h>
#define cosA 0.5
#define sinA 0.86602540
int main()
{
    int t;
    double x1,y1,x2,y2,x,y,xx,yy;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        x = x2-x1;y = y2-y1;
        xx = x*cosA-y*sinA;yy = x*sinA+y*cosA;
        x = xx+x1;y = yy+y1;
        printf("(%.2lf,%.2lf)\n",x,y);
    }
    return 0;
}

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