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
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
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 }
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 }
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 }
