hdu4998 Rotate【计算几何】

Noting is more interesting than rotation!

Your little sister likes to rotate things. To put it easier to analyze, your sister makes n rotations. In the i-th time, she makes everything in the plane rotate counter-clockwisely around a point ai by a radian of pi.

Now she promises that the total effect of her rotations is a single rotation around a point A by radian P (this means the sum of pi is not a multiplier of 2π).

Of course, you should be able to figure out what is A and P :).

InputThe first line contains an integer T, denoting the number of the test cases.

For each test case, the first line contains an integer n denoting the number of the rotations. Then n lines follows, each containing 3 real numbers x, y and p, which means rotating around point (x, y) counter-clockwisely by a radian of p.

We promise that the sum of all p's is differed at least 0.1 from the nearest multiplier of 2π.

T<=100. 1<=n<=10. 0<=x, y<=100. 0<=p<=2π.
OutputFor each test case, print 3 real numbers x, y, p, indicating that the overall rotation is around (x, y) counter-clockwisely by a radian of p. Note that you should print p where 0<=p<2π.

Your answer will be considered correct if and only if for x, y and p, the absolute error is no larger than 1e-5.

Sample Input

Sample Output

1.8088715944 0.1911284056 3.0000000000

划重点！

一个点绕定点旋转的坐标公式：

在平面坐标上，任意点P(x1,y1)，绕一个坐标点Q(x2,y2)旋转θ角度后,新的坐标设为(x, y)的计算公式：

x= (x1 - x2)*cos(θ) - (y1 - y2)*sin(θ) + x2 ;
y= (x1 - x2)*sin(θ) + (y1 - y2)*cos(θ) + y2 ;

这道题里最后的旋转角度就是前面所有的旋转角度加起来因为不管绕哪一个点旋转整个画面都转过了相同的角度画面上的每一个点也都转过了相同的角度

先设一个点对于每一次旋转都对这个点进行旋转就可以得到最后的时候这个点的坐标

在根据这个点最后的坐标与最开始的坐标可以反推出旋转中心

emmm感觉自己高中学的几何的公式都忘光了计算能力也大大下降解一个复杂一点的方程都解不出了啊天哪好难过

#include <iostream>

#include <algorithm>

#include <cstring>

#include <cstdio>

#include <cmath>

using namespace std;

#define PI 3.1415926

#define EPS 1.0e-5

int t, n;

struct point{

    double x, y;

}p[15];

point rot(point& p, point& ding, double theta)

{

    point c;

    c.x = (p.x - ding.x) * cos(theta) - (p.y - ding.y) * sin(theta) + ding.x;

    c.y = (p.x - ding.x) * sin(theta) + (p.y - ding.y) * cos(theta) + ding.y;

    return c;

}

int main()

{

    cin>>t;

    while(t--){

        scanf("%d",&n);

        double anangle = 0.0;

        point a, b;

        a.x = -1.0; a.y = -20.0;

        b.x = -1.0; b.y = -20.0;

        for(int i = 0; i < n; i++){

            double angle;

            scanf("%lf%lf%lf",&p[i].x, &p[i].y, &angle);

            a = rot(a, p[i], angle);

            anangle += angle;

            while(anangle >= 2 * PI){

                anangle -= 2 * PI;

            }

        }

        point ans;

        ans.y =(a.x*sin(anangle)+a.y*(1-cos(anangle))-(b.x*cos(anangle)-b.y*sin(anangle))*sin(anangle)-(b.x*sin(anangle)+b.y*cos(anangle))*(1-cos(anangle)))/(2-2*cos(anangle));

        ans.x=(a.x*(1-cos(anangle))-a.y*sin(anangle)-(b.x*cos(anangle)-b.y*sin(anangle))*(1-cos(anangle))+(b.x*sin(anangle)+b.y*cos(anangle))*sin(anangle))/(2-2*cos(anangle));

        printf("%.6f %.6f %.6f\n", ans.x, ans.y, anangle);

    }

    return 0;

}

巴特西

hdu4998 Rotate【计算几何】

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