Codeforces Round #432 B

Arpa and an exam about geometry

题意：a b c3个点在二维平面上构成一个三角形，问是否存在一个点，使得三角形绕点旋转一定角度后可以使得a到b，b到c的位置上

思路：易得，所求的点一定是abc三点共圆的圆心，且ab，bc所对的圆心角相等，即ab=bc，且abc不共线

AC代码：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "time.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("\n")
#define ft first
#define sd second 1
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const int N=1e3+100;

int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    ll a1,a2,b1,b2,c1,c2;
    cin>>a1>>a2>>b1>>b2>>c1>>c2;
    ll ab=(b1-a1)*(b1-a1)+(b2-a2)*(b2-a2);
    ll bc=(c1-b1)*(c1-b1)+(c2-b2)*(c2-b2);
    if(ab!=bc){
        cout<<"No";
        return 0;
    }
    else{
        ll k1,k2;
        k1=(b2-a2)*(c1-b1);
        k2=(c2-b2)*(b1-a1);
        if(k1==k2) cout<<"No";
        else cout<<"Yes";
    }
    return 0;
}