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.
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.
Print "Yes" if the problem has a solution, "No" otherwise.
You can print each letter in any case (upper or lower).
0 1 1 1 1 0
Yes
1 1 0 0 1000 1000
No
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.
【题解】
大致题意：三个点a,b,c，问能否在平面上找到一点O，使得aOb角旋转一定的度数后可以与bOc重合，即a点与b点重合，b点与c点重合。
经过分析，要重合，只能是ab长等于bc长，同时三个点不在同一条直接上就可以了。
【AC代码】
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef __int64 ll;
ll x1,x2,x3,y1,y2,y3;ll check(ll a,ll b,ll c,ll d)
{return (a-c)*(a-c)+(b-d)*(b-d);
}int main()
{while(~scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&x1,&y1,&x2,&y2,&x3,&y3)){ll xx1=x2-x1;ll xx2=x3-x1;ll yy1=y2-y1;ll yy2=y3-y1;if((check(x1,y1,x2,y2)==check(x2,y2,x3,y3))&&(xx1*yy2 != xx2*yy1))printf("Yes\n");elseprintf("No\n");}return 0;
}