Rotate

hdu4998:http://acm.hdu.edu.cn/showproblem.php?pid=4998

题意：给你n个点，以及绕每个点旋转的弧度。然后，问你经过这n次旋转，平面中的点总的效果是相当于哪个点旋转了多少弧度。

题解：我的第一道计算几何。可以选两个点，求出旋转之后的对应点，然后分别求出这两个点的中垂线，中垂线的交点就是要求的点，弧度就是所有弧度之和mod（2*pi）

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 const double pi=3.14159265;
 using namespace std;
 int n;
 struct Point{
   double x;
   double y;
   double s;
 }num[];
 struct LINE{
    double a,b,c;
    Point s;
 };
 Point HHH(Point o,double alpha,Point p) {//点p绕着o旋转alpha弧度
     Point tp;
     p.x-=o.x;
     p.y-=o.y;
     tp.x=p.x*cos(alpha)-p.y*sin(alpha)+o.x;
     tp.y=p.y*cos(alpha)+p.x*sin(alpha)+o.y;
     return tp;
  }
 bool CCC(LINE l1,LINE l2,Point&p){//两条直线是否相交，如果相交，则交点为p
   double d=l1.a*l2.b-l2.a*l1.b;
     if(abs(d)<1e-)    return false;
     p.x = (l2.c*l1.b-l1.c*l2.b)/d;
     p.y = (l2.a*l1.c-l1.a*l2.c)/d;
     return true;
 }
 LINE DDD(const Point &_a, const Point &_b){//求两点之间垂直平分线
         LINE ret;
         ret.s.x = (_a.x + _b.x)/;
         ret.s.y = (_a.y + _b.y)/;
         ret.a = _b.x - _a.x;
         ret.b = _b.y - _a.y;
         ret.c = (_a.y - _b.y) * ret.s.y + (_a.x - _b.x) * ret.s.x;
         return ret;
 }

 LINE EEE(Point a,Point b){//求经过两点的中垂线
     LINE ret;
     if(abs(a.x-b.x)<1e-){
         ret.a=;
         ret.b=;
         ret.c=-a.x;
         return ret;
     }
     ret.a=a.y-b.y;
     ret.b=b.x-a.x;
     ret.c=a.x*b.y-a.y*b.x;
     return ret;
 }
 int main(){
    int T;
    scanf("%d",&T);
    while(T--){
       scanf("%d",&n);
       double sum=;
       for(int i=;i<=n;i++){
          scanf("%lf%lf%lf",&num[i].x,&num[i].y,&num[i].s);
          sum+=num[i].s;
       }
       Point s1,s2;
       s1.x=2.345,s1.y=4.123;
       s2.x=11.345,s2.y=12.123;
       Point t1=s1,t2=s2;
       for(int i=;i<=n;i++){
          Point temp=HHH(num[i],num[i].s,t1);
          t1=temp;
       }
        for(int i=;i<=n;i++){
          Point temp=HHH(num[i],num[i].s,t2);
          t2=temp;
       }
       LINE one1=DDD(s1,t1);
       LINE one2=DDD(s2,t2);
       Point a;
       if(CCC(one1,one2,a)){
         if(sum>*pi)
         sum-=*pi;
         printf("%lf %lf %lf\n",a.x,a.y,sum);
       }
       else{
           LINE ttt=EEE(s1,s2);
           CCC(one1,ttt,a);
         if(sum>*pi)
          sum-=*pi;
          printf("%lf %lf %lf\n",a.x,a.y,sum);
       }
    }
 }