Codeforce 975E

E. Hag's Khashba

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Hag is a very talented person. He has always had an artist inside him but his father forced him to study mechanical engineering.

Yesterday he spent all of his time cutting a giant piece of wood trying to make it look like a goose. Anyway, his dad found out that he was doing arts rather than studying mechanics and other boring subjects. He confronted Hag with the fact that he is a spoiled son that does not care about his future, and if he continues to do arts he will cut his 25 Lira monthly allowance.

Hag is trying to prove to his dad that the wooden piece is a project for mechanics subject. He also told his dad that the wooden piece is a strictly convex polygon with $n$ vertices.

Hag brought two pins and pinned the polygon with them in the $1$-st and $2$-nd vertices to the wall. His dad has $q$ queries to Hag of two types.

• $1$ $f$ $t$: pull a pin from the vertex $f$, wait for the wooden polygon to rotate under the gravity force (if it will rotate) and stabilize. And then put the pin in vertex $t$.
• $2$ $v$: answer what are the coordinates of the vertex $v$.

Please help Hag to answer his father's queries.

You can assume that the wood that forms the polygon has uniform density and the polygon has a positive thickness, same in all points. After every query of the 1-st type Hag's dad tries to move the polygon a bit and watches it stabilize again.

Input

The first line contains two integers $n$ and $q$ ($3\le n\le 10000$, $1\le q\le 200000$) — the number of vertices in the polygon and the number of queries.

The next $n$ lines describe the wooden polygon, the $i$-th line contains two integers $x_i$ and $y_i$ ($|x_i|,|y_i|\le {10}^8$) — the coordinates of the $i$-th vertex of the polygon. It is guaranteed that polygon is strictly convex and the vertices are given in the counter-clockwise order and all vertices are distinct.

The next $q$ lines describe the queries, one per line. Each query starts with its type $1$ or $2$. Each query of the first type continues with two integers $f$ and $t$ ($1\le f,t\le n$) — the vertex the pin is taken from, and the vertex the pin is put to and the polygon finishes rotating. It is guaranteed that the vertex $f$ contains a pin. Each query of the second type continues with a single integer $v$ ($1\le v\le n$) — the vertex the coordinates of which Hag should tell his father.

It is guaranteed that there is at least one query of the second type.

Output

The output should contain the answer to each query of second type — two numbers in a separate line. Your answer is considered correct, if its absolute or relative error does not exceed ${10}^{-4}$.

Formally, let your answer be $a$, and the jury's answer be $b$. Your answer is considered correct if $\frac{|a-b|}{max(1,|b|)}\le {10}^{-4}$

Examples

input

Copy

3 4
0 0
2 0
2 2
1 1 2
2 1
2 2
2 3

output

Copy

3.4142135624 -1.4142135624
2.0000000000 0.0000000000
0.5857864376 -1.4142135624

input

Copy

3 2
-1 1
0 0
1 1
1 1 2
2 1

output

Copy

1.0000000000 -1.0000000000

Note

In the first test note the initial and the final state of the wooden polygon.

Red Triangle is the initial state and the green one is the triangle after rotation around $(2,0)$.

In the second sample note that the polygon rotates $180$ degrees counter-clockwise or clockwise direction (it does not matter), because Hag's father makes sure that the polygon is stable and his son does not trick him.

题意：
一个n个点的凸包，一开始两个钉子钉在1号和2号点，给m种操作，第一类是将某个点上的钉子拔出，然后凸包由于重力作用绕一个点旋转，再将钉子插在另一个点上；第二类是询问某个点的坐标。

题解:

凸包的旋转依据是，最终状态是凸包的重心和旋转所绕的点在铅锤线上，这样可以求出重心旋转的角度，我们也可以理解为n个点反方向绕重心旋转了相同的角度。这样的好处是不用每转一次就更新n个点的坐标，而是只更新重心的真实坐标和总共的旋转角度，利用n个点与重心的相对位置不变就可算出n个点的真实位置了。

代码:

#include<bits/stdc++.h>
#define LL long long
#define N 10010
const double eps=1e-3;
using namespace std;
int dcmp(double x){if (fabs(x)<eps)return 0;else return x<0?-1:1;}
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y){ }
    void read(){scanf("%lf%lf",&x,&y);}
};
typedef Point Vector;
Point a[N];
int g[N];

Vector operator + (Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}
Vector operator - (Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}
Vector operator * (Vector a,double b){return Vector(a.x*b,a.y*b);}
Vector operator / (Vector a,double b){return Vector(a.x/b,a.y/b);}
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));}
double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;} //叉积
Vector Rotate(Vector a,double rad)// 向量 a 逆时针旋转 rad
{return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));}
Point Zhongxin(Point *a,int n)
{
    Point ans=Point(0,0);double ss=0;
    a[n]=a[0];
    for (int i=1;i<n-1;i++)
    {
        double s=fabs(Cross(a[i]-a[0],a[i+1]-a[0]));
        ans=ans+(a[0]+a[i]+a[i+1])*s/3;
        ss+=s;
    }
    if (dcmp(ss)==0) return Point(0,0);
    return ans/ss;
}

double GetAngle(Vector a)
{
    return atan2(a.x,a.y);
}

int main()
{
    int n,m;double ang=0; g[0]=g[1]=1; int a1=0,a2=1;
    scanf("%d%d",&n,&m);
    for (int i=0;i<n;i++) a[i].read();
    Point cc=Zhongxin(a,n);
    for (int i=0;i<n;i++) a[i]=a[i]-cc;
    for (int i=1;i<=m;i++)
    {
        int fg,j,k;
        scanf("%d%d",&fg,&j);
        if (fg&1)
        {
            scanf("%d",&k);j--;k--;
            g[j]--;
            if (!g[j])
            {
                int k=j==a1?a2:a1;
                Point t=cc+Rotate(a[k],ang);
                double r=GetAngle(t-cc);
                cc=t+Point(0,-1)*Length(a[k]);
                ang+=r;
            }
            if (j==a1) a1=k;else a2=k;
            g[k]++;
        }else
        {
            Point t=cc+Rotate(a[j-1],ang);
            double x=t.x,y=t.y;
            printf("%.8f %.8f\n",x,y);
        }
    }
}