题意：给n个点（x,y,p），从1~n，一次每次所有点绕着第 i 个点（原来的）逆时针转pi个弧度，问最后所有点的位置相当于绕哪个点旋转多少弧度，求出那点X和弧度P
解法：直接模拟旋转，每次计算新的坐标，最后选两个新的点分别和他们原来的点连一条线，两条线的中垂线的交点即为圆心，求出了圆心就可以求出转了多少弧度了。
注意判中垂线垂直x轴的情况以及n==1的情况。
最后角度要根据位置关系判下正负。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define pii acos(-1.0)
#define eps 1e-8
using namespace std;
#define N 17 typedef struct point
{
double x,y,radi;
int ind;
point(double x=,double y=):x(x),y(y){}
}Vector; struct Point
{
double x,y;
Point(double x = ,double y = ):x(x),y(y){ }
}; Vector pi[N];
Point np[N];
int n; int dcmp(double x)
{
if(fabs(x)<eps)
return ;
return x < ? -:;
}
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 dcmp(a.x-b.x)==&&dcmp(a.y-b.y)==;}
bool operator < (const point& a,const point& b){return a.x<b.x ||(a.x==b.x&&a.y<b.y);} //比较和排序可用
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;} //叉积 ，大于零说明B在A的左边,小于零说明B在A的右边
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));} //向量长度
double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));};
Vector Rotate(Vector A,double rad){return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}//rad为弧度，向量逆时针旋转rad int main()
{
int t,i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=;i<=n;i++)
{
scanf("%lf%lf%lf",&pi[i].x,&pi[i].y,&pi[i].radi),pi[i].ind = i;
np[i].x = pi[i].x;
np[i].y = pi[i].y;
}
if(n == )
{
printf("%.10f %.10f %.10f\n",pi[].x,pi[].y,pi[].radi);
continue;
}
for(i=;i<=n;i++)
{
for(j=;j<=n;j++)
{
Vector k = Vector(np[j].x-pi[i].x,np[j].y-pi[i].y);
k = Rotate(k,pi[i].radi);
np[j].x = pi[i].x + k.x;
np[j].y = pi[i].y + k.y;
}
}
Point A,B,C,D;
A = Point(pi[].x,pi[].y);
B = np[];
C = Point(pi[].x,pi[].y);
D = np[];
Point Mid1 = Point((A.x+B.x)/2.0,(A.y+B.y)/2.0);
Point Mid2 = Point((C.x+D.x)/2.0,(C.y+D.y)/2.0);
double k1,k2;
double Ix,Iy;
if(dcmp(A.y-B.y) == )
{
if(dcmp(D.x-C.x) == )
k2 = 0.0;
else
{
k2 = (D.y-C.y)/(D.x-C.x);
k2 = -1.0/k2;
}
Ix = Mid1.x;
Iy = Mid2.y + k2*(Mid1.x-Mid2.x);
}
else if(dcmp(D.y-C.y) == )
{
if(dcmp(B.x-A.x) == )
k1 = 0.0;
else
{
k1 = (B.y-A.y)/(B.x-A.x);
k1 = -1.0/k1;
}
Ix = Mid2.x;
Iy = Mid1.y + k1*(Mid2.x-Mid1.x);
}
else
{
k1 = (B.y-A.y)/(B.x-A.x);
k1 = -1.0/k1;
k2 = (D.y-C.y)/(D.x-C.x);
k2 = -1.0/k2;
double b1 = -k1*Mid1.x + Mid1.y;
double b2 = -k2*Mid2.x + Mid2.y;
Ix = (b2-b1)/(k1-k2);
Iy = k1*Ix + b1;
}
Vector ka = Vector(pi[].x-Ix,pi[].y-Iy);
Vector kb = Vector(np[].x-Ix,np[].y-Iy);
double ang = Angle(ka,kb);
double coss = Cross(ka,kb);
if(dcmp(coss-0.0) == -)
ang = 2.0*pii-ang;
printf("%.10f %.10f %.10f\n",Ix,Iy,ang);
}
return ;
}