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

野性酷女 2022-06-09 11:24 192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

Note

In the first sample test, rotate the page around (0.5, 0.5) by ![3beb169548b758ea60b306458d03196407afcd56.png][].

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;
    }

[3beb169548b758ea60b306458d03196407afcd56.png]: /images/20220609/e574adc2f921453dadaabe6a66720799.png