CFボス195(div 2)A:Vasily the Bear and Triangle

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 farite rectores.parate the.ores.parectores.
Vasya also love s triangles,if the triangles have one vertex at point B?=?(0,8201).That's today he asks to find two points A?=;(x 1,820)(x 1,x1,tototototototos))))s s s=thethe.inininininintsA(x=820)(x=_)=_)=_)(x 1,x1,eb=_)(x 1,eb=_)(x 1,x1,820))(x 1,x1,x1,

the coordinans of points:x 1,x 2,y 1,y 2 are integers.Besides,the follwing inequation holds:x 1*<8201;x 2;

the triangle formed by point A,B and C is rectanglar and isoscers( is right);

all points of the favorite rectangle are located inside or on the border of triangle ABC.

the ara 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.
Input
The first line contains two integers x,?y(?-?109̵≦?x,?y≠109,?x≠
Output
Print in the single line four integers x 1,̵y1,̵x2,the coordinas of the required points.
Sample test(s)
Input

10 5

Output

0 15 15 0

Input

-10 5

Output

-15 0 0 15

ノート

Figure to the first sample
 
 
x,y軸を直角にした二等辺直角三角形は、斜辺の点を与え、斜辺の二つの端点を求めて、まずxの小さい点を出力します.
構想:直線方程式y=k*x+dを求めて、k、dを求めて、すぐできます.
 

#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)//   -1
        {
            k = -1;
            m = y+x;
            a[0] = -m/k;
            a[1] = 0;
            b[0] = 0;
            b[1] = m;
        }
        else//   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
",a[0],a[1],b[0],b[1]);
        }
        else
        {
            printf("%.0lf %.0lf %.0lf %.0lf
",b[0],b[1],a[0],a[1]);
        }
    }

    return 0;
}