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
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
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
#include
#include
#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); } }