codeforces 1017E

两个凸包判断经过旋转平移能否重合。
我一看。哇傻逼题十行秒掉。
交上去跑的飞快然后wa55。
。。。
然后这个题一共就55个点，这网友的数据竟该死的强。
看了眼数据是两个反转的平行四边形，再判下角度就好了。
怎么大家都在hash然后kmp啊。这好难啊。我根本不会这东西啊。。。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
#include <deque>
#include <cstring>

#define rep(n) for(int i=0;i<n;i++)
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);}
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};}
    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;
    }
    int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
    db abs(){return sqrt(x*x+y*y);}
    db dis(point k1){return ((*this)-k1).abs();}
    int dis2(point k1){ return (dis(k1)*dis(k1));}
    point unit(){db w=abs(); return point{x/w,y/w};}
    point turn90(){ return point{-y,x};}
    db getP()const { return sign(y)==1||(sign(y)==0&&sign(x)==-1);}
    void print(){printf("%.11lf %.11lf
",x,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));}
int compareangle(point k1,point k2){
    return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,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);
}
struct line{
    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(db eps){//向左手边平移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 parallel(line k1,line k2){ return sign(cross(k1.dir(),k2.dir()))==0;}
int sameDir(line k1,line k2){
    return parallel(k1,k2)&&sign(dot(k1.dir(),k2.dir()))==1;
}
int operator <(line k1,line k2){
    if(sameDir(k1,k2))return k2.include(k1[0]);
    return compareangle(k1.dir(),k2.dir());
}
int checkpos(line k1,line k2,line k3){ return k3.include(getLL(k1,k2));}
vector<point> convexHull(vector<point>ps){
    int n = ps.size();if(n<=1)return ps;
    sort(ps.begin(),ps.end());
    vector<point> qs(n*2);int k=0;
    for(int i=0;i<n;qs[k++]=ps[i++])
        while (k>1&&cross(qs[k-1]-qs[k-2],ps[i]-qs[k-2])<=0)--k;
    for(int i=n-2,t=k;i>=0;qs[k++]=ps[i--])
        while (k>t&&cross(qs[k-1]-qs[k-2],ps[i]-qs[k-2])<=0)--k;
    qs.resize(k-1);
    return qs;
}
vector<point> a,b;
int n,m;point o1,o2;
int main(){
    scanf("%d%d",&n,&m);
    a.resize(n);b.resize(m);
    rep(n)scanf("%lf%lf",&a[i].x,&a[i].y);
    rep(m)scanf("%lf%lf",&b[i].x,&b[i].y);
    a=convexHull(a);b=convexHull(b);
    if(a.size()!=b.size()){cout<<"NO"<<endl;return 0;}
    n=a.size();rep(n)o1=o1+a[i],o2=o2+b[i];o1=o1/n;o2=o2/n;
    rep(n)a[i]=a[i]-o1;
    rep(n)b[i]=b[i]-o2;
    sort(a.begin(),a.end(),compareangle);
    sort(b.begin(),b.end(),compareangle);
    rep(n){
        bool f=1;
        db pre=-1;
        for(int j=0;j<n&&f;j++){
            if(cmp(a[j].abs(),b[(i+j)%n].abs())!=0)f=0;
            db now = rad(a[j],b[(i+j)%n]);
            if(pre==-1)pre=fabs(now);
            else if(cmp(fabs(now),pre))f=0;
        }
        if(f){cout<<"YES"<<endl;return 0;}
    }
    cout<<"NO"<<endl;
}