Codeforces Round #432 (Div. 2） B. Arpa and an exam about geometry（数学水题）-蒲公英云

Codeforces Round #432 (Div. 2） B. Arpa and an exam about geometry（数学水题）

野性酷女 2022-06-09 11:24 298阅读 0赞

B. Arpa and an exam about geometry

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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 a**x, a**y, b**x, b**y, c**x, c**y (|a**x|, |a**y|, |b**x|, |b**y|, |c**x|, |c**y| ≤ 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：

比赛还在进行但是我已经写不动了，10分钟过a题（太水不想写题解），20分钟过b题。。c题看不懂，d题不会做，e题。。。算了吧，虽然这题很水但还是水一发博客

题意：

给你3个点的坐标，问你是否能找到一个点，以该点为中心旋转一定的角度使得a与b重合，b与c重合

思路：

直接判断只要两个线段长度相同而且三个点不在同一直线上就行了

ps：

这题之前用double型40组数据出错估计是因为double精度不够，还有就是斜率不存在的时候会出问题，把数据类型改成long long，把斜率判等写成乘法就好了

暂时ac的代码：

#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<deque>
#include<algorithm>
using namespace std;
#define INF 100861111
#define ll long long
#define eps 1e-7
#define maxn 20
int main()
{
    long long x1,y1,x2,y2,x3,y3,k1,k2,d1,d2;
    scanf("%lld%lld%lld%lld%lld%lld",&x1,&y1,&x2,&y2,&x3,&y3);
    d1=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
    d2=(x2-x3)*(x2-x3)+(y2-y3)*(y2-y3);
    if(d1!=d2)
    {
        printf("No\n");
        return 0;
    }
    if((x2-x3)*(y1-y2)==(x1-x2)*(y2-y3))
    {
        printf("No\n");
    }
    else
        printf("Yes\n");
    return 0;
}