Codeforces Round#432 B. Arpa and an exam about geometry

题目如下：

B. Arpa and an exam about geometry

time limit per test2 secondsmemory limit per test256 megabytes

Arpa is taking a geometry exam. Here is the last problem of the exam.

You are given three points a, b, c.

Find a point and an angle such that if we rotate the page around the point by the angle, the new position of a is the same as the old position of b, and the new position of b is the same as the old position of c.

Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.

Input

The only line contains six integers ax, ay, bx, by, cx, cy (|ax|, |ay|, |bx|, |by|, |cx|, |cy| ≤ 109). It's guaranteed that the points are distinct.

Output

Print "Yes" if the problem has a solution, "No" otherwise.You can print each letter in any case (upper or lower).

Examples

input0 1 1 1 1 0outputYesinput1 1 0 0 1000 1000outputNo

Note

In the first sample test, rotate the page around (0.5, 0.5) by .In the second sample test, you can't find any solution.

题意&分析：
给定三个点ABC，围绕一点旋转一定的角度，使得A可以与B重合，B可以与C重合，所以只需要AB=BC并且ABC三点不共线即可。

代码如下：

#include <algorithm>#include <bitset>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <map>#include <queue>#include <set>#include <stack>#include <string>#include <cctype>#include <fstream>#define INF 0x3f3f3f3f#define TEST cout<<"stop here"<<endlusing namespace std;typedef long long ll;const ll mod = 1e9 + 7;int main(){    std::ios::sync_with_stdio(false);    std::cin.tie(0);    double ax,ay,bx,by,cx,cy;    while(cin>>ax>>ay>>bx>>by>>cx>>cy){        ll A1 = by - ay,B1 = ax - bx;        ll A2 = cy - by,B2 = bx - cx;        ll midx1 = (ax + bx)/2,midy1 = (ay + by)/2;        ll midx2 = (bx + cx)/2,midy2 = (by + cy)/2;        ll ab = ((ax-bx)*(ax-bx)+(ay-by)*(ay-by));        ll bc = ((bx-cx)*(bx-cx)+(cy-by)*(cy-by));         if(ab == bc  && A1*B2 != A2*B1)            cout<< "Yes" <<endl;        else            cout<< "No" <<endl;    }    return 0;}