190 - Circle Through Three Points

[artless]

#include<stdio.h>
#include<math.h>
int main()
{
double x1,x2,x3,y1,y2,y3,a1,a2,a3,b1,b2,b3,c1,c2,c3,d1,d2,d3,D,g,f,c,G,F,r;
while(scanf("%lf %lf %lf %lf %lf %lf",&x1,&y1,&x2,&y2,&x3,&y3)==6)
{a1=2*x1;
a2=2*x2;
a3=2*x3;
b1=2*y1;
b2=2*y2;
b3=2*y3;
c1=c2=c3=1.0;
d1=(x1*x1)+(y1*y1);
d2=(x2*x2)+(y2*y2);
d3=(x3*x3)+(y3*y3);
// slove of 3 equations having 3 variable a*g+b*f+c=0
D=(a1*b2*c3+b1*a3*c2+c1*a2*b3)-(a1*c2*b3+b1*a2*c3+c1*b2*a3);
g=((b1*c3*d2+c1*b2*d3+d1*c2*b3)-(b1*c2*d3+c1*b3*d2+d1*b2*c3))/D;
f=((a1*c2*d3+c1*a3*d2+d1*a2*c3)-(a1*c3*d2+c1*a2*d3+d1*c2*a3))/D;
c=((a1*b3*d2+b1*a2*d3+d1*b2*a3)-(a1*b2*d3+b1*a3*d2+d1*a2*b3))/D;
G=2.0*g;
F=2.0*f;
r=sqrt(g*g+f*f-c);

printf("(x ");if(g<0) printf("- %.3lf)^2 + (y ",-g);
else if(g>=0)printf("+ %.3lf)^2 + (y ",g);
if(f<0) printf("- %.3lf)^2 = %.3lf^2\n",-f,r);
else if(f>=0) printf("+ %.3lf)^2 = %.3lf^2\n",f,r);
printf("x^2 + y^2 ");
if(G<0)printf("- %.3lfx ",-G);
else if(G>=0)printf("+ %.3lfx ",G);
if(F<0)printf("- %.3lfy ",-F);
else if(F>=0)printf("+ %.3lfy ",F);
if(c<0)printf("- %.3lf = 0\n",-c);
else if(c>=0)printf("+ %.3lf =0\n",c);
}
return 0;
}

[artless]

#include<stdio.h>
#include<math.h>
int main()
{
double x1,x2,x3,y1,y2,y3,a1,a2,a3,b1,b2,b3,c1,c2,c3,d1,d2,d3,D,g,f,c,G,F,r;
while(scanf("%lf %lf %lf %lf %lf %lf",&x1,&y1,&x2,&y2,&x3,&y3)==6)
{a1=2.0*x1; a2=2.0*x2; a3=2.0*x3;
b1=2.0*y1; b2=2.0*y2; b3=2.0*y3;
c1=c2=c3=1.0;
d1=(x1*x1)+(y1*y1); d2=(x2*x2)+(y2*y2); d3=(x3*x3)+(y3*y3);

D=(a1*b2*c3+b1*a3*c2+c1*a2*b3)-(a1*c2*b3+b1*a2*c3+c1*b2*a3);
g=((b1*c3*d2+c1*b2*d3+d1*c2*b3)-(b1*c2*d3+c1*b3*d2+d1*b2*c3))/D;
f=((a1*c2*d3+c1*a3*d2+d1*a2*c3)-(a1*c3*d2+c1*a2*d3+d1*c2*a3))/D;
c=((a1*b3*d2+b1*a2*d3+d1*b2*a3)-(a1*b2*d3+b1*a3*d2+d1*a2*b3))/D;
G=2.0*g;
F=2.0*f;
r=sqrt(g*g+f*f-c);

printf("(x ");if(g<0) printf("- %.3lf)^2 + (y ",-g);
else if(g>=0)printf("+ %.3lf)^2 + (y ",g);
if(f<0) printf("- %.3lf)^2 = %.3lf^2\n",-f,r);
else if(f>=0) printf("+ %.3lf)^2 = %.3lf^2\n",f,r);
printf("x^2 + y^2 ");
if(G<0)printf("- %.3lfx ",-G);
else if(G>=0)printf("+ %.3lfx ",G);
if(F<0)printf("- %.3lfy ",-F);
else if(F>=0)printf("+ %.3lfy ",F);
if(c<0)printf("- %.3lf = 0\n",-c);
else if(c>=0)printf("+ %.3lf = 0\n",c);
}
return 0;
}

[Neto_o]

I'm not reading your code since it's not written with the needed spaces and indentation, I can give you some inputs/outputs from my Accepted code, so you can test:

Input:

Code: Select all

-34 23 -18 16 20 15
0 0 0 12 12 0
0 12 0 0 12 0
-143 200 -143 -100 200 200
0 0 0.0134 0.0325 0.0888 0.087787
0 0 0.00134 0.00325 0.0888 0.0087787
-42350 -800 0 0 4235 -800

Output:

Code: Select all

(x - 2.840)^2 + (y - 85.420)^2 = 72.481^2
x^2 + y^2 - 5.680x - 170.840y + 2051.200 = 0

(x - 6.000)^2 + (y - 6.000)^2 = 8.485^2
x^2 + y^2 - 12.000x - 12.000y + 0.000 = 0

(x - 6.000)^2 + (y - 6.000)^2 = 8.485^2
x^2 + y^2 - 12.000x - 12.000y + 0.000 = 0

(x - 28.500)^2 + (y - 50.000)^2 = 227.843^2
x^2 + y^2 - 57.000x - 100.000y - 48600.000 = 0

(x - 0.116)^2 + (y + 0.029)^2 = 0.120^2
x^2 + y^2 - 0.233x + 0.058y + 0.000 = 0

(x - 0.047)^2 + (y + 0.017)^2 = 0.050^2
x^2 + y^2 - 0.093x + 0.035y + 0.000 = 0

(x + 19057.500)^2 + (y + 112495.156)^2 = 114097.978^2
x^2 + y^2 + 38115.000x + 224990.312y + 0.000 = 0

[artless]

thank you... i got AC

[pecijackson]

i did't see anything wrong bout your codes sir.try to rewrite the codes. if it's doesn't work. i'll re-write the codes for you

[S.H.Bouwhuis]

Here are some sample inputs and outputs for my accepted submission:

Input

Code: Select all

7.0 -5.0 -1.0 1.0 0.0 -6.0
1.0 7.0 8.0 6.0 7.0 -2.0
5 5 4 7 16 5
56 40 66 76 31 8
44 39 26 23 37 38
18 82 29 41 33 15
0 49 19 56 98 3
7 0 11 0 9 2
0 9 0 15 3 12
0.0 0.0 1.0 0.0 0.0 2.0
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
0.0 0.0 1.1 0.0 0.0 1.1

Output

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 - 10.500)^2 + (y - 9.000)^2 = 6.801^2
x^2 + y^2 - 21.000x - 18.000y + 145.000 = 0

(x + 33.686)^2 + (y - 84.302)^2 = 100.031^2
x^2 + y^2 + 67.372x - 168.603y - 1764.717 = 0

(x - 42.830)^2 + (y - 22.191)^2 = 16.849^2
x^2 + y^2 - 85.660x - 44.383y + 2042.957 = 0

(x + 279.295)^2 + (y + 19.738)^2 = 314.221^2
x^2 + y^2 + 558.590x + 39.475y - 20339.607 = 0

(x - 32.340)^2 + (y + 9.494)^2 = 66.838^2
x^2 + y^2 - 64.679x + 18.987y - 3331.372 = 0

(x - 9.000)^2 + y^2 = 2.000^2
x^2 + y^2 - 18.000x + 77.000 = 0

x^2 + (y - 12.000)^2 = 3.000^2
x^2 + y^2 - 24.000y + 135.000 = 0

(x - 0.500)^2 + (y - 1.000)^2 = 1.118^2
x^2 + y^2 - 1.000x - 2.000y = 0

(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

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

[emranbuet2002]

Code: Select all

Removed after accepted!

[brianfry713]

Print a single blank line after each equation pair including the last one.

[emranbuet2002]

Thanks!

[laituanksa245]

I have tried all the test cases posted in this forum, but i still got WA.
Help me please. I feel so frustrated

Code: Select all

Code Removed
AC :P

[brianfry713]

Check the sample I/O, you're printing extra spaces.

[laituanksa245]

Thank you so much.
You have helped me AC so many problems

[alimbubt]

Code: Select all

This is very easy problem...Just follow the process...

(1)try to make three equation with three coordinates.
(2)then try to find the value of unknown three variables with the process of determinant.
(3)be careful what the statement says "Plus, minus, and equal signs must be separated from the adjacent characters by a single space on each side."

[mahade hasan]

ACCCC

[?????]

what is my problem in this prob i cant find help please

#include<stdio.h>
#include<math.h>
#include<string.h>
void p(double x,double y,double temp,double ax,double ay)
{
if(x==0)
{
printf("x^2 + ");
}
else if(x<0)
{
printf("(x + %0.3lf)^2 + ",(x*(-1)));
}
else if(x>0)
{
printf("(x - %0.3lf)^2 + ",x);
}
if(y==0)
{
printf("y^2 = %0.3lf^2\n",temp);
}
else if(y<0)
{
printf("(y + %0.3lf)^2 = %0.3lf^2\n",(y*(-1)),temp);
}
else if(y>0)
{
printf("(y - %0.3lf)^2 = %0.3lf^2\n",y,temp);
}
temp=pow(x,2)+pow(y,2)-pow(temp,2);
printf("x^2 + y^2 ");
if(x<0)
{
printf("+ %0.3lfx ",(x*(-2)));
}
else if(x>0)
{
printf("- %0.3lfx ",(x*2));
}
if(y<0)
{
printf("+ %0.3lfy ",(y*(-2)));
}
else if(y>0)
{
printf("- %0.3lfy ",(y*2));
}

if(temp<=0)
{
printf("- %0.3lf ",(temp*(-1)));
}
else if(temp>0)
{
printf("+ %0.3lf ",temp);
}
printf("= 0\n");
return ;
}
void britto(double ax,double ay,double bx,double by,double cx,double cy)
{
double temp1,temp2,temp3,temp4,temp,x,y,ax1,ay1,bx1,by1,cx1,cy1;
ax1=pow(ax,2);
ay1=pow(ay,2);
bx1=pow(bx,2);
by1=pow(by,2);
cx1=pow(cx,2);
cy1=pow(cy,2);
temp1=(((bx1+by1-cx1-cy1)*(ay-by)*(-1))/2);
temp2=((ax1+ay1-bx1-by1)*(by-cy))/2;
temp3=((ax1+ay1-bx1-by1)*(bx-cx)*(-1))/2;
temp4=((bx1+by1-cx1-cy1)*(ax-bx))/2;
temp=((ax-bx)*(by-cy)) -((ay-by)*(bx-cx));
x=(temp1+temp2)/(temp);
y=(temp3+temp4)/(temp);
temp1=pow((x-ax),2);
temp2=pow((y-ay),2);
temp=sqrt((temp1+temp2));
p(x,y,temp,ax,ay);
return ;
}
int main()
{
double ax,ay,bx,by,cx,cy;
for( ;(scanf("%lf%lf%lf%lf%lf%lf",&ax,&ay,&bx,&by,&cx,&cy))==6;)
{
britto(ax,ay,bx,by,cx,cy);
}

return 0;
}