Codeforces Round #432 Div2 B. Arpa and an exam about geometry

最新推荐文章于 2022-07-13 02:01:45 发布

Pandauncle

最新推荐文章于 2022-07-13 02:01:45 发布

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本文介绍了一个几何旋转问题，探讨了如何判断三个给定点通过旋转能否使一点变为另一点的条件。主要内容包括直线判断、距离比较及实现代码。

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B. Arpa and an exam about geometry
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.

题目意思就是给出a,b,c三个点的坐标，然后绕着某个点旋转问是否能够a点变成b点，b点变成c点。若能输出Yes,否则输出No。
思路：判断三个点是否在一条直线上，若在一条直线上则不能。再判断a到b点的距离是否和b和c点的相等，若相等且不在一条直线上则可以。
40的数据：
0 0 1000000000 1 1000000000 -999999999
输出No.
需要注意的一点为需要用long long int 。一开始用的double精度不够过不了。
其他一些数据：

589824 196608 262144 196608 0 0
输出 Yes.
999999999 1000000000 0 0 -1000000000 -999999999
输出 Yes.

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<iostream>
using namespace std;
int main()
{
    long long int  ax,ay,bx,by,cx,cy;
    scanf("%lld%lld%lld%lld%lld%lld",&ax,&ay,&bx,&by,&cx,&cy);
    long long int  l1=((by-ay)*(by-ay)+(bx-ax)*(bx-ax));
    long long int  l2=((cy-by)*(cy-by)+(cx-bx)*(cx-bx));
    int mark=0;
//  printf("l1=%lf,l2=%lf\n",l1,l2);
//  printf("by-ay=%lf\n",by-ay);
    if((ax==bx&&ax==cx)||(ay==by&&ay==cy))
    mark=1;
    else 
    {
        double k1=(by-ay)*1.0/(bx-ax);
        double k2=(cy-by)*1.0/(cx-bx);
        if(k1==k2)
        mark=1;
    }
//  printf("%d\n",mark);
    if(!mark&&l1==l2)
    printf("Yes\n");
    else
    printf("No\n");
    return 0;
}