Arpa and an exam about geometry
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).
Example
Input
0 1 1 1 1 0
Output
Yes
Input
1 1 0 0 1000 1000
Output
No
Note
In the first sample test, rotate the page around (0.5, 0.5) by 90.
In the second sample test, you can't find any solution.
题意：旋转任意角度使得A点与B点重合，B点与C点重合。
思路：只要AB的长度等于BC的长度，而且不是平行的。
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
long long int a,b,c,d,e,f;
cin>>a>>b>>c>>d>>e>>f;
if((a-c)*(a-c)+(b-d)*(b-d)==(c-e)*(c-e)+(d-f)*(d-f)&&(a-c)*(f-d)!=(b-d)*(e-c))
printf("Yes
");
else
printf("No
");
}