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

此博客介绍了如何通过给定的矩形坐标,找到一个最小的等腰直角三角形,该三角形包含矩形内所有点,并且其面积最小。通过数学方法计算三角形的两个顶点坐标。

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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: x1x2y1y2 are integers. Besides, the following inequation holds: x1 < x2;
  • the triangle formed by point AB 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:

 

 1 #include<stdio.h>
 2 #include<math.h>
 3 int x,y,xi,yi;
 4 
 5 int
 6 main()
 7 {
 8     scanf("%d%d",&x,&y);
 9     xi=abs(x)+abs(y);
10     if(x>0&&y>0)
11     printf("%d %d %d %d\n",0,xi,xi,0);
12     if(x<0&&y>0)
13     printf("%d %d %d %d\n",-xi,0,0,xi);
14     if(x<0&&y<0)
15     printf("%d %d %d %d\n",-xi,0,0,-xi);
16     if(x>0&&y<0)
17     printf("%d %d %d %d\n",0,-xi,xi,0);
18     
19     return 0;
20 }

 

 

 

转载于:https://www.cnblogs.com/sxiszero/p/3637061.html

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