CodeForces - 851B -Arpa and an exam about geometry(计算几何)
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.
题解:若三个点可以通过旋转使a到b,b到c,那只需要两个条件,共圆和弧长相等,共圆可以转换为AB和BC不在一条直线上,即AB,BC斜率不相等,弧长相等转换为弦相等,及AB之间的距离等于BC之间的距离
代码:
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main() {
long double ax,ay,bx,by,cx,cy;
cin>>ax>>ay>>bx>>by>>cx>>cy;
if((by-ay)/(bx-ax)!=(cy-by)/(cx-bx)&&(by-ay)*(by-ay)+(bx-ax)*(bx-ax)==(cy-by)*(cy-by)+(cx-bx)*(cx-bx)) {
printf("Yes\n");
} else {
printf("No\n");
}
return 0;
}
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