CF#195(div2) A:Vasily the Bear and Triangle

最新推荐文章于 2016-10-16 23:39:32 发布

原创最新推荐文章于 2016-10-16 23:39:32 发布 · 1.9k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#CF

CF 专栏收录该内容

17 篇文章

订阅专栏

针对已知直角位于原点的等腰直角三角形，本篇介绍如何确定该三角形斜边两端点坐标的方法。通过输入直角三角形内一已知点坐标，利用直线方程计算出斜边两端点坐标，确保所有矩形点位于三角形内部。

http://codeforces.com/contest/336/problem/A

Vasily the bear has a favorite rectangle, it has one vertex at point (0, 0), and the opposite vertex at point (x, y). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes.

Vasya also loves triangles, if the triangles have one vertex at point B = (0, 0). That's why today he asks you to find two points A = (x₁, y₁) and C = (x₂, y₂), such that the following conditions hold:

the coordinates of points: x₁, x₂, y₁, y₂ are integers. Besides, the following inequation holds: x₁ < x₂;
the triangle formed by point A, B and C is rectangular and isosceles ( $\text{[math]}$ is right);
all points of the favorite rectangle are located inside or on the border of triangle ABC;
the area of triangle ABC is as small as possible.

Help the bear, find the required points. It is not so hard to proof that these points are unique.

Input

The first line contains two integers x, y ( - 10⁹ ≤ x, y ≤ 10⁹, x ≠ 0, y ≠ 0).

Output

Print in the single line four integers x₁, y₁, x₂, y₂ — the coordinates of the required points.

Sample test(s)

Input

10 5

Output

0 15 15 0

Input

-10 5

Output

-15 0 0 15

Note

$\text{[math]}$

Figure to the first sample

题意：一个以x,y轴为直角边的等腰直角三角形，给出一个斜边上的点，要求斜边的两个端点，先输出x比较小的点

思路：求直线方程y=k*x+d，求出k,d,即可

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int main()
{
    double x,y,a[2],b[2],k,m;
    while(~scanf("%lf%lf",&x,&y))
    {
        if(x*y>0)//斜率为-1
        {
            k = -1;
            m = y+x;
            a[0] = -m/k;
            a[1] = 0;
            b[0] = 0;
            b[1] = m;
        }
        else//斜率为1
        {
            k = 1;
            m = y-x;
            a[0] = -m/k;
            a[1] = 0;
            b[0] = 0;
            b[1] = m;
        }
        if(a[0]<b[0])
        {
            printf("%.0lf %.0lf %.0lf %.0lf\n",a[0],a[1],b[0],b[1]);
        }
        else
        {
            printf("%.0lf %.0lf %.0lf %.0lf\n",b[0],b[1],a[0],a[1]);
        }
    }

    return 0;
}