HDU 4998 Rotate

Rotate

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 210    Accepted Submission(s): 110

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

130 0 11 1 12 2 1

 

Sample Output

1.8088715944 0.1911284056 3.0000000000

 

Source

2014 ACM/ICPC Asia Regional Anshan Online


题目链接 :http://acm.hdu.edu.cn/showproblem.php?pid=4998

题目大意 :某个平面绕每个点逆时针旋转一个弧度值，将该过程等效为该平面绕着一个点逆时针旋转一个弧度值，求出等效点的坐标和旋转弧度值。

题目分析 :任意取两个点通过一组旋转得到另外两个对应的点，再求对应直线的垂直平分线交于一点，该点就是所要求的点的坐标，弧度考虑可以用反三角函数求出，这里需要注意一个问题，比赛时wa到死，就是旋转的方向问题，这里可以用向量的差乘来处理

代码 :

#include <cstdio>#include <cstring>#include <cmath>#define PI 4.0 * atan(1.0)using namespace std;struct Move{       double x, y, p;}move[15];int n;double Dist(double x1, double y1, double x2, double y2)  //求两点间距离{    return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));}double calx(double x, double y)  //求逆时针旋转后的x坐标{    double x1, y1;    for(int i = 0; i < n; i++)    {        x1 = (x - move[i].x) * cos(move[i].p) - (y - move[i].y) * sin(move[i].p) + move[i].x;        y1 = (x - move[i].x) * sin(move[i].p) + (y - move[i].y) * cos(move[i].p) + move[i].y;        x = x1;        y = y1;    }    return x1;}double caly(double x, double y)  //求逆时针旋转后的y坐标{    double x1, y1;    for(int i = 0; i < n; i++)    {        x1 = (x - move[i].x) * cos(move[i].p) - (y - move[i].y) * sin(move[i].p) + move[i].x;        y1 = (x - move[i].x) * sin(move[i].p) + (y - move[i].y) * cos(move[i].p) + move[i].y;        x = x1;        y = y1;    }    return y1;}int main(){    int ca;    scanf("%d", &ca);    while(ca--)    {        scanf("%d", &n);        for(int i = 0; i < n; i++)            scanf("%lf %lf %lf", &move[i].x, &move[i].y, &move[i].p);        //任意取两个点        double x1 = 34.11111111;        double y1 = 1.4211111111;        double x3 = 42.3111111111;        double y3 = 2.32111111111;        //求出这两个点逆时针旋转后的两个点        double x2 = calx(x1, y1);        double y2 = caly(x1, y1);        double x4 = calx(x3, y3);        double y4 = caly(x3, y3);        //通过y = kx + b;联立解得两垂直平分线的交点        double x = ((y1 + y2 - y3 - y4) * (y1 - y2) * (y3 - y4) + (x4 * x4 - x3 * x3) * (y1 - y2) - (x2 * x2 - x1 * x1) * (y3 - y4))        / (2 * (x4 - x3) * (y1 - y2) - 2 * (x2 - x1) * (y3 - y4));        double y = (((x2 - x1) / (y1 - y2)) * (x - (x1 + x2) / 2) + (y1 + y2) / 2);        //圆的半径        double r = Dist(x, y, x1, y1);        //某两个对应点的距离        double dist = Dist(x1, y1, x2, y2);        //通过三角函数关系求出弧度        double angel = 2 * asin(dist / (2 * r));        //通过向量叉乘确定方向        double d1 = (x1 - x) * (y2 - y) - (x2 - x) * (y1 - y);        //方向判定        if(d1 < 0)            angel = 2 * PI - angel;         //精度特判        if(fabs(x) < 1e-10)            x = 0;        if(fabs(y) < 1e-10)            y = 0;        printf("%.10f %.10f %.10f\n", x, y, angel);    }}