/home/toolbox/public_html/solutions/4/438/i1.c
1 /* @JUDGE_ID: 823125 Prob 438 C "Use of equation of the circle to compute circumference." */
2 /**************
3 Solution to Problem 438
4 by: Theodore A. Irra
5 date: 02/09/2016
6 **************/
7 #include <stdio.h>
8 #include <math.h> /* for sqrt function */
9 #define PI 3.141592653589793 /* Since input data is accurate to only 1 decimal */
10 /* place and circumference is to be rounded off to 2 decimal places pi should */
11 /* be 3.142 but this value used to match sample output. Algebraic manipulation */
12 /* of equation (x-a)^2+(y-b)^2=r^2 with substitution of the point coordinates */
13 /* yields system of 3 linear equations, one of which is a linear combination of */
14 /* the other two, which by Gauss elimination and back substitution allows computation */
15 /* of radius and circumference. Two pecial cases to avoid division by zero are handled */
16 double r, /* radius */
17 ctrX,ctrY, /* x-coordinate and y-coordinate of center */
18 circ; /* circumference */
19 double X1,Y1,X2,Y2,X3,Y3; /* input point coordinates for P1, P2, P3 */
20 int main()
21 {
22 /* EOFin std input is(<ctrl+d> in UNIX, <ctrl+z> in DOS) */
23 while(scanf("%lf %lf %lf %lf %lf %lf",&X1,&Y1,&X2,&Y2,&X3,&Y3)!=EOF)
24 {
25 ctrX=((((X2*X2)-(X3*X3)+(Y2*Y2)-(Y3*Y3))*((2*Y1)-(2*Y2)))-(((X1*X1)-(X2*X2)+(Y1*Y1)-(Y2*Y2))*((2*Y2)-(2*Y3))))/
26 ((((2*Y1)-(2*Y2))*((2*X2)-(2*X3)))-(((2*X1)-(2*X2))*((2*Y2)-(2*Y3))));
27 if (!(Y2==Y3)) /* special case to avoid division by zero */
28 ctrY=(((X2*X2)-(X3*X3)+(Y2*Y2)-(Y3*Y3))/((2*Y2)-(2*Y3)))-((((2*X2)-(2*X3))*ctrX)/((2*Y2)-(2*Y3)));
29 if (!(Y1==Y2)) /* special case to avoid division by zero */
30 ctrY=(((X1*X1)-(X2*X2)+(Y1*Y1)-(Y2*Y2))/((2*Y1)-(2*Y2)))-((((2*X1)-(2*X2))*ctrX)/((2*Y1)-(2*Y2)));
31 r=sqrt(((X1-ctrX)*(X1-ctrX))+((Y1-ctrY)*(Y1-ctrY)));
32 circ=2*PI*r;
33 printf("%.2lfx\n",circ);
34 }
35 return(0);
36 }
37