Rotate

                                                                          Time Limit: 2000/1000 MS (Java/Others)    Memory Limit:
65536/65536 K (Java/Others)

Special Judge

Problem Description

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 :).

Input

The 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.

Output

For 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

1
3
0 0 1
1 1 1
2 2 1

Sample Output

1.8088715944 0.1911284056 3.0000000000

Source

2014 ACM/ICPC Asia Regional Anshan
Online

比赛时写这个题写了几个小时，最终也没有调出来，自己还是太弱了。

题意：一个物体每次绕着一个点旋转一个角度，旋转n次后等价于从开始状态绕一个点旋转一定角度后直接到达最终状态。求这个点的坐标和旋转角度。

分析：因为旋转次数很少，所以可以直接模拟旋转过程。选两个点作为开始状态，求出这两个点旋转后对应的坐标，然后连接旋转前和旋转后的对应点，求出两条直线的交点，然后求出旋转角度。

#include<cstdio>
#include<cmath>
using namespace std;

#define PI acos(-1.0)

struct Point {
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}
};

int n;
Point p[15];  //旋转点
double rad[15]; //旋转角度

typedef Point Vector;

Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Point A, Point B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x * p, A.y * p); }
Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); }

bool operator < (const Point& a, const Point& b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

const double eps = 1e-10;
int dcmp(double x) {
    if(fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point& b) {
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

double Dot(Vector A, Vector B) { return A.x * B.x + A.y * B.y; }  //点积
double Length(Vector A) { return sqrt(Dot(A, A)); } //求向量的模
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } //求两个向量的夹角
double Cross(Vector A, Vector B) { return A.x * B.y - A.y * B.x; } //叉乘

Vector Rotate(Vector A, double rad) {  //向量旋转
    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}

Vector Normal(Vector A) {  //求A向量的法向量
    double L = Length(A);
    return Vector(-A.y / L, A.x / L);
}

Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {  //求直线交点
    Vector u = P - Q;
    double t = Cross(w, u) / Cross(v, w);
    return P + v * t;
}

Vector Rotate_Point(Vector A) {
    for(int i = 0; i < n; i++) {
        A = p[i] + Rotate(A - p[i], rad[i]);  //转化为向量旋转
    }
    return A;
}

Vector Get_Mid_Point(Point A, Point B) {  //求中点
    return Vector((A.x + B.x) / 2, (A.y + B.y) / 2);
}

void Get_Ans() {
    Point f1[2], f2[2], mid[2], vec[2];
    f1[0].x = -1;
    f1[0].y = -1;
    f1[1].x = -10;
    f1[1].y = -50;
    for(int i = 0; i < 2; i++) {
        f2[i] = Rotate_Point(f1[i]);
        mid[i] = Get_Mid_Point(f1[i], f2[i]);
        vec[i] = Normal(f1[i] - f2[i]);
    }

    Point ans = GetLineIntersection(mid[0], vec[0], mid[1], vec[1]);
    double ansp = Angle(f1[0] - ans, f2[0] - ans);

    if(Cross(f1[0] - ans, f2[0] - ans) < 0)
        ansp = 2 * PI - ansp;
    if(dcmp(ans.x) == 0) ans.x = 0;
    if(dcmp(ans.y) == 0) ans.y = 0;

    printf("%.10lf %.10lf %.10lf\n", ans.x, ans.y, ansp);
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        for(int i = 0; i < n; i++) {
            scanf("%lf%lf%lf", &p[i].x, &p[i].y, &rad[i]);
            if(dcmp(rad[i] - 2 * PI) == 0 || dcmp(rad[i]) == 0) {
                rad[i] = 0;
                n--;
                i--;
            }
        }
        Get_Ans();
    }
    return 0;
}