CF 336A(Vasily the Bear and Triangle-推公式)

本文探讨了如何在给定一个矩形的情况下，找到两个点来构成一个等腰直角三角形，使得该三角形包含矩形的所有点且面积尽可能小。通过输入矩形的长和宽，输出满足条件的三角形顶点坐标。

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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

Figure to the first sample

大水题。。但是我没看到等腰(isosceles)

不说了。。说多了都是泪。。。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
ll a,b,t;
int main()
{
//	freopen("tri.in","r",stdin);
	cin>>a>>b;
	int b1=a<0?-1:1,b2=b<0?-1:1;
	t=abs(a)+abs(b);
	//(t,0),(0,t)
	int x1=t*b1,y1=0,x2=0,y2=t*b2;
	if (x1>x2) swap(x1,x2),swap(y1,y2);
	cout<<x1<<' '<<y1<<' '<<x2<<' '<<y2<<endl;
	
	return 0;
}