190 - Circle Through Three Points

[Mohammad Mahmudur Rahman]

My AC program gives the follwoing output for your input -

Code: Select all

(x - 3.000)^2 + (y + 2.000)^2 = 5.000^2
x^2 + y^2 - 6.000x + 4.000y - 12.000 = 0

(x - 3.921)^2 + (y - 2.447)^2 = 5.409^2
x^2 + y^2 - 7.842x - 4.895y - 7.895 = 0

(x - 2.500)^2 + (y - 0.000)^2 = 2.500^2
x^2 + y^2 - 5.000x - 0.000y - 0.000 = 0

Note the (y - 0.000) in the 3rd test case.

[tan_Yui]

My AC code prints same output as yours.
I think, the bugs of your code is the part of division.
In some parts, your code is in danger of the division by zero.
Parameters; ma, mb, h; are faced with this problem.

Best regards.

[tan_Yui]

I think both +0.000 and -0.000 will be accepted.

[xintactox]

Thanks you for uor reply

I fixed my code to generate answer like ur code do, but still getting WA...

Here's my new code

Code: Select all

#include <iostream>
#include <stdio.h>
#include <math.h>

using namespace std;

class point
{
	public:
		double x, y;
		
};

int main()
{
	point p1, p2, p3;
	double ma, mb, h, k, r, c, d, e;
	while(cin>>p1.x)
	{
		cin>>p1.y>>p2.x>>p2.y>>p3.x>>p3.y;
		ma = (p2.y - p1.y)/(p2.x - p1.x);
		mb = (p3.y - p2.y)/(p3.x - p2.x);
		h = (ma*mb*(p1.y - p3.y) + mb*(p1.x + p2.x) - ma*(p2.x + p3.x))/(2.0*(mb - ma));		
		k = (-1.0/(ma+0.000001))*(h - ((p1.x + p2.x))/2) + (p1.y + p2.y)/2.0;					
		r = sqrt((h - p1.x)*(h - p1.x)+(k - p1.y)*(k - p1.y));
								
		//cout<<"c = "<<c<<endl;
		//cout<<"d = "<<d<<endl;
				
		if(h >= 0.001 || h == 0)
			printf("(x - %.3f)^2 + (y ", fabs(h));
		else
			printf("(x + %.3f)^2 + (y ", fabs((-1)*h));
			
		if(k >= 0.001 || k == 0)
			printf("- %.3f)^2 = %.3f^2", fabs(k), fabs(r));
		else 
			printf("+ %.3f)^2 = %.3f^2", fabs((-1)*k),fabs(r));		
		
		c = 2.0*h;
		d = 2.0*k;
		e = (h*h) + (k*k) - (r*r);
		
		
		if(c >= 0.001 || c == 0)
			printf("\nx^2 + y^2 - %.3fx ", fabs(c));
		else
			printf("\nx^2 + y^2 + %.3fx ", fabs(c));;	
		
		if(d >= 0.001 || d == 0)
			printf("- %.3fy ", fabs(d));
		else
			printf("+ %.3fy ", fabs(d));
		
		if(e > 0.00000001)
			printf("+ %.3f = 0\n", fabs(e));
		else
			printf("- %.3f = 0\n", fabs(e));
		
		printf("\n");		
	}
	return 0;
}

This problem is driving me mad for about three days now...

Someone help me!

Thanks!

[xintactox]

Thanks!

Any clue on how avoiding division by zero in this case?
I tried to add very small vlaues in the denominators of the fractions (like 0.000000001 or something) but it changes the output...

[tan_Yui]

Thanks!

Any clue on how avoiding division by zero in this case?
I tried to add very small vlaues in the denominators of the fractions (like 0.000000001 or something) but it changes the output...

Try the following case.
In current your code, if (p2.x - p1.x == 0.000) or (p3.x - p2.x == 0.000) or (mb - ma == 0.000), it'll be occured.

My code tries to separate some cases by the parameters (e.g. when p2.x - p1.x == 0.000, calculate in some function).
, then no need to use tiny values for avoiding precision error.

Code: Select all

 6.6  2.5 -4.0 -4.0  6.6 -8.0
 3.7 -4.1  3.7 -1.5  1.0  1.0
 9.0 -2.0  1.0  1.0 -5.0 -2.0
 5.4  6.3  0.0  6.3 -7.8 -0.8
 3.1  4.1  5.9  2.6  5.3  5.8
10.0 11.0 12.0 13.0  0.0  0.0

My output is :

Code: Select all

(x - 2.526)^2 + (y + 2.750)^2 = 6.645^2
x^2 + y^2 - 5.053x + 5.500y - 30.211 = 0

(x + 0.011)^2 + (y + 2.800)^2 = 3.932^2
x^2 + y^2 + 0.022x + 5.600y - 7.622 = 0

(x - 2.000)^2 + (y + 8.500)^2 = 9.552^2
x^2 + y^2 - 4.000x + 17.000y - 15.000 = 0

(x - 2.700)^2 + (y + 4.501)^2 = 11.133^2
x^2 + y^2 - 5.400x + 9.001y - 96.399 = 0

(x - 4.883)^2 + (y - 4.066)^2 = 1.784^2
x^2 + y^2 - 9.767x - 8.131y + 37.195 = 0

(x - 142.500)^2 + (y + 119.500)^2 = 185.974^2
x^2 + y^2 - 285.000x + 239.000y - 0.000 = 0

Best regards.

[Mohammad Mahmudur Rahman]

Try this input -

Code: Select all

0.0 0.0 1.1 0.0 0.0 1.1

output should be -

Code: Select all

(x - 0.550)^2 + (y - 0.550)^2 = 0.778^2
x^2 + y^2 - 1.100x - 1.100y - 0.000 = 0

[Mohammad Mahmudur Rahman]

I think both +0.000 and -0.000 will be accepted.

Note that this program doesn't have a special corrector program. So, how both can get AC? However, I modified my code to print +0.000 & that got AC as well. So, apparantly there's no such input.

[tan_Yui]

Note that this program doesn't have a special corrector program. So, how both can get AC? However, I modified my code to print +0.000 & that got AC as well. So, apparantly there's no such input.

I checked both case by changing my code like you did, then thought simply both can get AC.
I also believe there's no such input.

Thank you.

[=viki=]

here is my code
plz check why PE

Code: Select all

#include<stdio.h>
#include<math.h>

int main(void)
{
	float x[3],y[3];
	float mid1X,mid2X,mid1Y,mid2Y,m1,m2,c1,c2,temp;
	float h,k,r,c,d,e;
	while(scanf("%f%f%f%f%f%f",&x[0],&y[0],&x[1],&y[1],&x[2],&y[2])==6)
	{
/*	while(!feof(in))
	{	for(l=0;l<3;l++)
			fscanf(in,"%f%f",&x[l],&y[l]); */
		mid1X=(x[0]+x[1])/2;
		mid1Y=(y[0]+y[1])/2;
		m1=-(x[0]-x[1])/(y[0]-y[1]);
		mid2X=(x[1]+x[2])/2;
		mid2Y=(y[1]+y[2])/2;
		m2=-(x[1]-x[2])/(y[1]-y[2]);
		c1=mid1Y-(m1*mid1X);
		c2=mid2Y-(m2*mid2X);
		h=(c2-c1)/(m1-m2);
		k=(m1*h)+c1;
		temp=pow((x[0]-h),2)+pow((y[0]-k),2);
		r=sqrt(temp);
		e=pow(h,2)+pow(k,2)-temp;
		d=-2*k;
		c=-2*h;
		if(h<0)
			printf("(x + %.3f)^2 +  ",-h);
		else
			printf("(x - %.3f)^2 + ",h);
		if(k<0)
			printf("(y + %.3f)^2 = ",-k);
		else
			printf("(y - %.3f)^2 = ",k);
		printf("%.3f^2\n",r);
		printf("x^2 + y^2 ");
		if(c<0)
			printf("- %.3fx ",-c);
		else if(c>0)
			printf("+ %.3fx ",c);
		else
			printf("- 0.000x ");
		if(d<0)
			printf("- %.3fy ",-d);
		else if(d>0)
			printf("+ %.3fy ",d);
		else
			printf("- 0.000y ");
		if(e<0)
			printf("- %.3f = 0\n",-e);
		else
			printf("+ %.3f = 0\n",e);
		printf("\n");
	}
	return 0;
}

i gave up looking for any presentation error...
plz help me with this.. where is the error..

[=viki=]

im getting PE on this code

Code: Select all

#include<stdio.h>
#include<math.h>

int main(void)
{
	float x[3],y[3];
	float mid1X,mid2X,mid1Y,mid2Y,m1,m2,c1,c2,temp;
	float h,k,r,c,d,e;
	while(scanf("%f%f%f%f%f%f",&x[0],&y[0],&x[1],&y[1],&x[2],&y[2])==6)
	{
/*	while(!feof(in))
	{	for(l=0;l<3;l++)
			fscanf(in,"%f%f",&x[l],&y[l]); */
		mid1X=(x[0]+x[1])/2;
		mid1Y=(y[0]+y[1])/2;
		m1=-(x[0]-x[1])/(y[0]-y[1]);
		mid2X=(x[1]+x[2])/2;
		mid2Y=(y[1]+y[2])/2;
		m2=-(x[1]-x[2])/(y[1]-y[2]);
		c1=mid1Y-(m1*mid1X);
		c2=mid2Y-(m2*mid2X);
		h=(c2-c1)/(m1-m2);
		k=(m1*h)+c1;
		temp=pow((x[0]-h),2)+pow((y[0]-k),2);
		r=sqrt(temp);
		e=pow(h,2)+pow(k,2)-temp;
		d=-2*k;
		c=-2*h;
		if(h<0)
			printf("(x + %.3f)^2 +  ",-h);
		else
			printf("(x - %.3f)^2 + ",h);
		if(k<0)
			printf("(y + %.3f)^2 = ",-k);
		else
			printf("(y - %.3f)^2 = ",k);
		printf("%.3f^2\n",r);
		printf("x^2 + y^2 ");
		if(c<0)
			printf("- %.3fx ",-c);
		else if(c>0)
			printf("+ %.3fx ",c);
		else
			printf("- 0.000x ");
		if(d<0)
			printf("- %.3fy ",-d);
		else if(d>0)
			printf("+ %.3fy ",d);
		else
			printf("- 0.000y ");
		if(e<0)
			printf("- %.3f = 0\n",-e);
		else
			printf("+ %.3f = 0\n",e);
		printf("\n");
	}
	return 0;
}

im printing -0.000 if theres any...
tried bothe +0.000 nad -0.000 still getting PE....plz help..

[tobby]

Code: Select all

		if(h<0)
			printf("(x + %.3f)^2 +  ",-h);
		else
			printf("(x - %.3f)^2 + ",h);

The format of this part is incorrect. I wonder why you didn't see this.

[tobby]

Please give up the habit of making cross posting. Your question has been answered in another thread.

[Psyco]

Thank you Source!

I inserted some parts, and I got AC!!

[sluga]

I didn't explicitly take care of printing +0.000 or -0.000, and got P.E.
I solved one problem here on ACM in which you have to judge some outputs. It stated that if al numerals are correct, but some letters are wrong, this is considered P.E. I believe this is the case here.