第四届 山东省ACM Rescue The Princess(计算几何)
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Rescue ThePrincess
Time Limit:1000ms Memory limit: 65536K 有疑问?点这里^_^
题目描述
Several daysago, a beast caught a beautiful princess and the princess was put in prison. Torescue the princess, a prince who wanted to marry the princess set outimmediately. Yet, the beast set a maze. Only if the prince find out the maze’sexit can he save the princess.
Now, here comes theproblem. The maze is a dimensional plane. The beast is smart, and he hidden theprincess snugly. He marked two coordinates of an equilateral triangle in the maze. The two markedcoordinates are A(x1,y1) and B(x2,y2).The third coordinate C(x3,y3) is the maze’s exit. If theprince can find out the exit, he can save the princess. After the prince comesinto 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 princeneed your help. Can you calculate the C(x3,y3) and tell him?
输入
The first line is aninteger T(1 <= T <= 100) which is the number of test cases. T test casesfollow. 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 fromthe equilateral triangle. Andcoordinates 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 0.000.00
0.00 0.00 0.00100.00
0.00 0.00 100.00100.00
1.00 0.00 1.8660.50
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
示例程序
推广结论:对于任意两个不同点A和B,A绕B旋转θ角度后的坐标为:
(Δx*cosθ- Δy * sinθ+ xB, Δy*cosθ + Δx * sinθ+ yB )
#include<iostream>#include<cstdio>#include<cstring>#include<cstdio>#include<math.h>using namespace std;int main(){ int T; scanf("%d",&T); while(T--) { double x1,x2,y1,y2,x3,y3,x,y; scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); x=x2-x1; y=y2-y1; x3=x/2-y*sqrt(3)/2+x1; y3=y/2+x*sqrt(3)/2+y1; printf("(%.2f,%.2f)\n",x3,y3); } return 0;}
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