cf 975E

出于某种不可抗力我翻了一下以前绿名时打的div2，然后插旗：这种傻逼div2我现在不是随手ak吗？ 然后就没有然后了
给一个凸包，一开始固定两个点，每次拆下一个点(取消固定)，待稳定后再固定另外一个点。
支持查询第几个点的坐标。
首先求凸包重心，全部划分成三角形那样子。
考虑一次旋转操作，其实是将重心旋转到了固定点的正下方。
那么在知道了每个点相对于重心的向量后，这其实蛮好求的。
同样，重心每次旋转的角度也容易知道。
所以我们翻过来考虑，维护重心坐标和旋转角度，对于一个查询，我们用它的相对位置*角度+重心即可。
很好的一道题。
有点神志不清调了好久

#include <bits/stdc++.h>
#define mp make_pair
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef double db;
const db eps=1e-6;
const db pi=acos(-1);
int sign(db k){
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
int cmp(db k1,db k2){return sign(k1-k2);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内
struct point{
    db x,y;
    point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
    point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
    point operator * (db k1) const{return (point){x*k1,y*k1};}
    point operator / (db k1) const{return (point){x/k1,y/k1};}
    int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
    // 逆时针旋转
    point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
    point turn90(){return (point){-y,x};}
    bool operator < (const point k1) const{
        int a=cmp(x,k1.x);
        if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
    }
    db abs(){return sqrt(x*x+y*y);}
    db abs2(){return x*x+y*y;}
    db dis(point k1){return ((*this)-k1).abs();}
    point unit(){db w=abs(); return (point){x/w,y/w};}
    void scan(){double k1,k2; scanf("%lf%lf",&k1,&k2); x=k1; y=k2;}
    void print(){printf("%.11lf %.11lf
",x,y);}
    db getw(){return atan2(y,x);}
    point getdel(){if (sign(x)==-1||(sign(x)==0&&sign(y)==-1)) return (*this)*(-1); else return (*this);}
    int getP() const{return sign(y)==1||(sign(y)==0&&sign(x)==-1);}
};
int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
db rad(point k1,point k2){return atan2(cross(k1,k2),dot(k1,k2));}
// -pi -> pi
int compareangle (point k1,point k2){//极角排序+
    return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,k2))>0);
}
point proj(point k1,point k2,point q){ // q 到直线 k1,k2 的投影
    point k=k2-k1;return k1+k*(dot(q-k1,k)/k.abs2());
}
point reflect(point k1,point k2,point q){return proj(k1,k2,q)*2-q;}
int clockwise(point k1,point k2,point k3){// k1 k2 k3 逆时针 1 顺时针 -1 否则 0
    return sign(cross(k2-k1,k3-k1));
}
int checkLL(point k1,point k2,point k3,point k4){// 求直线 (L) 线段 (S)k1,k2 和 k3,k4 的交点
    return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))!=0;
}
point getLL(point k1,point k2,point k3,point k4){
    db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
}
int intersect(db l1,db r1,db l2,db r2){
    if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-1&&cmp(r2,l1)!=-1;
}
int checkSS(point k1,point k2,point k3,point k4){
    return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&&
           sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=0&&
           sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0;
}
db disSP(point k1,point k2,point q){
    point k3=proj(k1,k2,q);
    if (inmid(k1,k2,k3)) return q.dis(k3); else return min(q.dis(k1),q.dis(k2));
}
db disSS(point k1,point k2,point k3,point k4){
    if (checkSS(k1,k2,k3,k4)) return 0;
    else return min(min(disSP(k1,k2,k3),disSP(k1,k2,k4)),min(disSP(k3,k4,k1),disSP(k3,k4,k2)));
}
int onS(point k1,point k2,point q){return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;}
point Centroid(vector<point> v){
    point ans={0,0};db S=0;
    for(int i=1;i<v.size()-1;i++){
        db s = fabs(cross(v[i]-v[0],v[i+1]-v[0]));
        ans = ans+(v[0]+v[i]+v[i+1])*s/3;
        S+=s;
    }
    //S!=0
    return ans/S;
}
struct circle{
    point o; db r;
    void scan(){o.scan(); scanf("%lf",&r);}
    int inside(point k){return cmp(r,o.dis(k));}
};
struct line{
    // p[0]->p[1]
    point p[2];
    line(point k1,point k2){p[0]=k1; p[1]=k2;}
    point& operator [] (int k){return p[k];}
    int include(point k){return sign(cross(p[1]-p[0],k-p[0]))>0;}
    point dir(){return p[1]-p[0];}
    line push(){ // 向外 ( 左手边 ) 平移 eps
        const db eps = 1e-6;
        point delta=(p[1]-p[0]).turn90().unit()*eps;
        return {p[0]-delta,p[1]-delta};
    }
};
point getLL(line k1,line k2){return getLL(k1[0],k1[1],k2[0],k2[1]);}
int n,q,b[10005];
vector<point> p;point o;
db ang=0;
int main(){
    scanf("%d%d",&n,&q);
    p.resize(n);
    for(int i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
    o=Centroid(p);
//    o.print();
    for(int i=0;i<n;i++)p[i]=p[i]-o;
    int op,f,t,a1=0,a2=1;b[0]=b[1]=1;
    while (q--){
        scanf("%d",&op);
        if(op==1){
            scanf("%d%d",&f,&t);
            f--;t--;
            b[f]--;
            if(b[f]==0){
                int otr = (f==a1?a2:a1);
                point tmp = o+p[otr].turn(ang);//旋转点
//                tmp.print();
                db jiao = rad(o-tmp,point{0,-1.0});
//                printf("%.11f
",jiao);
//                printf("%.11f
",p[otr].abs());
                o=tmp+(point){0,-1.0}*p[otr].abs();
                ang+=jiao;
            }
//            o.print();
            b[t]++;
            if(f==a1)a1=t;else a2=t;
        }else{
            scanf("%d",&f);
            f--;
            point tmp=o+p[f].turn(ang);
            tmp.print();
        }
    }
}