Codeforces Round #195 (Div. 2) / 336A Vasily the Bear and Triangle(模拟&数学)

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本文介绍了一个CodeForces上的编程题目“Vasily the Bear and Triangle”的解法。该题目要求找到两个整数坐标点A(x1, y1)和C(x2, y2)，使得形成的三角形ABC为直角等腰三角形，并且包含给定矩形的所有点，同时面积尽可能小。

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

简单模拟。

完整代码：

/*30ms,0KB*/

#include<cstdio>

int main()
{
	int x, y;
	scanf("%d%d", &x, &y);
	if (x > 0 && y > 0)
		printf("0 %d %d 0", x + y, x + y);
	else if (x > 0 && y < 0)
		printf("0 %d %d 0", y - x, x - y);
	else if (x < 0 && y > 0)
		printf("%d 0 0 %d", x - y, y - x);
	else
		printf("%d 0 0 %d", x + y, x + y);
}