ZOJ 3488 Conic Section

最新推荐文章于 2019-05-02 14:42:46 发布

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最新推荐文章于 2019-05-02 14:42:46 发布

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本文介绍了一种通过输入六个实数参数来判断二次曲线类型的算法。该算法首先判断曲线是否为椭圆或圆，接着判断是否为双曲线，并提供了完整的代码实现。

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The conic sections are the nondegenerate curves generated by the intersections of a plane with one or two nappes of a cone. For a plane perpendicular to the axis of the cone, a circle is produced. For a plane that is not perpendicular to the axis and that intersects only a single nappe, the curve produced is either an ellipse or a parabola. The curve produced by a plane intersecting both nappes is a hyperbola.

conic section	equation
circle	`x²+y²=a²`
ellipse	`x²/a²+y²/b²=1`
parabola	`y²=4ax`
hyperbola	`x²/a²-y²/b²=1`

Input

There are multiple test cases. The first line of input is an integer T ≈ 10000 indicating the number of test cases.

Each test case consists of a line containing 6 real numbers a, b, c, d, e, f. The absolute value of any number never exceeds 10000. It's guaranteed that a²+c²>0, b=0, the conic section exists and it is non-degenerate.

Output

For each test case, output the type of conic section ax²+bxy+cy²+dx+ey+f=0. See sample for more details.

Sample Input

5
1 0 1 0 0 -1
1 0 2 0 0 -1
0 0 1 1 0 0
1 0 -1 0 0 1
2 0 2 4 4 0

Sample Output

circle
ellipse
parabola
hyperbola
circle

References

Weisstein, Eric W. "Conic Section." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/ConicSection.html

简单题，先判断是否椭圆或圆，再判断是否双曲线即可

#include<map>
#include<cmath>    
#include<queue> 
#include<string>
#include<vector>
#include<cstdio>    
#include<cstring>  
#include<iostream>
#include<algorithm>    
using namespace std;
#define ms(x,y) memset(x,y,sizeof(x))    
#define rep(i,j,k) for(int i=j;i<=k;i++)    
#define per(i,j,k) for(int i=j;i>=k;i--)    
#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])    
#define inone(x) scanf("%d",&x)    
#define intwo(x,y) scanf("%d%d",&x,&y)    
#define inthr(x,y,z) scanf("%lf%lf%lf",&x,&y,&z)    
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
int T, n;
double a, b, c, d, e, f;

int main()
{
	for (inone(T); T--;)
	{
		inthr(a, b, c); inthr(d, e, f);
		if (a * c > 0) puts(a == c ? "circle" : "ellipse");
		else puts(a*c ? "hyperbola" : "parabola");
 	}
	return 0;
}