[Codeforces Round #195 (Div. 2)] A. Vasily the Bear and Triangle

A. Vasily the Bear and Triangle

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ( 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

Figure to the first sample

 

题解：简单数学题。求过输入坐标(x,y)的直线y=kx+b与横轴与纵轴的交点，如图所示两交点与原点组成的三角形是等腰直角三角形。注意先输出x较小的点。利用相似三角形计算点的坐标即可。

代码：

 

#include<stdio.h>
#include<math.h>
int x,y,xi,yi;

int
main()
{
    scanf("%d%d",&x,&y);
    xi=abs(x)+abs(y);
    if(x>0&&y>0)
    printf("%d %d %d %d\n",0,xi,xi,0);
    if(x<0&&y>0)
    printf("%d %d %d %d\n",-xi,0,0,xi);
    if(x<0&&y<0)
    printf("%d %d %d %d\n",-xi,0,0,-xi);
    if(x>0&&y<0)
    printf("%d %d %d %d\n",0,-xi,xi,0);
    
    return 0;
}