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

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本文探讨了一个几何旋转问题，即是否存在一个点使得三个不同点通过绕该点旋转后能分别到达另外两点的位置。文章提供了判断方法及代码实现。

B. Arpa and an exam about geometry

[题目(http://codeforces.com/contest/851/problem/B%20%E4%BC%A0%E9%80%81%E9%97%A8)

        time limit per test2 seconds
    memory limit per test256 megabytes
            inputstandard input
           outputstandard output

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
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 .

In the second sample test, you can’t find any solution.

题意为给出三个不同的点A,B,C，是否能在平面上找到另一个点，使平面绕这个点旋转，使A旋转到B原来的位置，B旋转到C原来的位置。很明显，这三个点要满足条件，则它们必须在一个圆上（不在同一直线上的三点确定一个圆），即三点不共线，且|AB|==|BC|（此题数据较大，应该用long long）。代码如下：

#include <iostream>

using namespace std;

int main()
{
    long long ax,ay,bx,by,cx,cy;
    cin>>ax>>ay>>bx>>by>>cx>>cy;
    if(ax==bx&&bx==cx)cout<<"No\n";
    else if((cy-by)*(bx-ax)==(cx-bx)*(by-ay)) cout<<"No\n";
    else if((by-ay)*(by-ay)+(bx-ax)*(bx-ax)==(cy-by)*(cy-by)+(cx-bx)*(cx-bx))cout<<"Yes\n";
    else cout<<"No\n";
    return 0;
}