sdutoj 2603 Rescue The Princess

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=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

示例程序

 

分析：

 

已知等边三角形的两个按逆时针给出的两个顶点，求第三个点。

 

官方代码：

 

 1 #include <stdio.h>

 2 #include <math.h>

 3 const double pi = acos(-1.0);  4 int main()  5 {  6     int t;  7     double x1,x2,x3,y1,y2,y3,l,at;  8     scanf("%d",&t);  9     while(t--) 10  { 11         scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); 12         at = atan2(y2-y1,x2-x1); 13         printf("%lf\n",(y2-y1)/(x2-x1)); 14         printf("%lf\n",at); 15         l = sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); 16         x3 = x1+l*cos(at+pi/3.0); 17         y3 = y1+l*sin(at+pi/3.0); 18         printf("(%.2lf,%.2lf)\n",x3,y3); 19  } 20 

21     return 0; 22 }

View Code

 

AC代码：

 

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cmath>

 4 using namespace std;

 5 #define D double

 6 #define PI acos(-1.0)

 7 D x3,y3;

 8 int d1, d2;

 9 D Length(D x1,D y1,D x2,D y2 )

10 {

11     return sqrt((y2-y1)*(y2-y1) + (x2-x1)*(x2-x1));

12 }

13 void Solve(D x1, D y1, D x2,D y2)

14 {

15     D Radian = atan((y2 - y1)/(x2 - x1));

16     D Angle = Radian * 180.0 / PI;

17     if (x1 > x2)

18         Angle += 180.0;

19     Radian += PI/3.0;

20     Angle += 60.0;

21      if (Angle <= 90 || Angle >= 270)

22          d1 = 1;

23     else

24         d1 = -1;

25      if (Angle >= 0 && Angle <= 180)

26          d2 = 1;

27     else

28         d2 = -1;

29     D dis = Length(x1,y1,x2,y2);

30     x3 = x1 + d1 * dis * fabs(cos(Radian));

31     y3 = y1 + d2 * dis * fabs(sin(Radian));

32 }

33 int main()

34 {

35     int T;

36     D x1, y1, x2, y2;

37     scanf ("%d", &T);

38     while (T --)

39     {

40         scanf ("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);

41         Solve(x1, y1, x2, y2);

42         printf ("(%.2lf,%.2lf)\n", x3, y3);

43     }

44 

45     return 0;

46 }

View Code

 

 

 

还可以推出公式的：

 

利用向量的旋转：http://www.cnblogs.com/jeff-wgc/p/4468038.html

 

AC代码：

 1 #include<stdio.h>

 2 #include<math.h>

 3 int main()

 4 {

 5     int t;

 6     scanf("%d",&t);

 7     while(t--)

 8     {

 9         double x1,x2,y1,y2,x3,y3,x,y;

10         scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);

11         x=x2-x1;

12         y=y2-y1;

13         x3=x/2-y*sqrt(3)/2+x1;

14         y3=y/2+x*sqrt(3)/2+y1;

15         printf("(%.2f,%.2f)\n",x3,y3);

16     }

17     return 0;

18 }

View Code