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

2024-07-03 10:04:09   224人阅读

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

*************************/

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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