sdut2603&hdu1700（向量旋转）
sdut2603
地址：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)
题意：给出两个点A,B，求出一个可以与这两点组成等边三角形的点C，要求A,B,C顺序为逆时针。
#include
#include
#include
using namespace std;
int main()
{
int t;
double x1,x2,x3,y1,y2,y3;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
x3=(x1+x2)/2-sqrt(3.0)*((y2+y1)/2-y1); //这里sqrt(3)就是A`C/BC
y3=(y1+y2)/2+sqrt(3.0)*((x2+x1)/2-x1);
printf("(%.2lf,%.2lf)\n",x3,y3);
}
} hdu1700
Points on Cycle
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1290 Accepted Submission(s): 451
Now give you a point on the cycle, you are to find out the other two points on it, to maximize the sum of the distance between each other
you may assume that the radius of the cycle will not exceed 1000.
Alway output the lower one first(with a smaller Y-coordinate value), if they have the same Y value output the one with a smaller X.
#include
#include
#include
using namespace std;
int main()
{
int t;
double x1,x2,x3,y1,y2,y3;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf",&x1,&y1);
x2=x1/2-sqrt(3.0)*y1/2; //因为三个点与圆心连线之间的角度为120°（等边三角形）
y2=y1/2+sqrt(3.0)*x1/2; //所以我们可以看为输入的这个点与圆心连线逆时针旋转两个60°
x1=x2/2-sqrt(3.0)*y2/2;
y1=y2/2+sqrt(3.0)*x2/2;
x2=x1;y2=y1;
x3=x1/2-sqrt(3.0)*y1/2;
y3=y1/2+sqrt(3.0)*x1/2;
x1=x3/2-sqrt(3.0)*y3/2;
y1=y3/2+sqrt(3.0)*x3/2;
x3=x1;y3=y1;
if(y2-y3||(fabs(y3-y2)<0.0005&&(x2-x3)>0.0005)) //关于0.0005的判定在题目中有交代
{
x1=x3;x3=x2;x2=x1;
y1=y3;y3=y2;y2=y1;
}
printf("%.3lf %.3lf %.3lf %.3lf\n",x2,y2,x3,y3);
}
}