题目的链接如下： http://acm.hdu.edu.cn/showproblem.php?pid=1374
The Circumference of the Circle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 920 Accepted Submission(s): 685
Problem Description To calculate the circumference of a circle seems to be an easy task - provided you know its diameter. But what if you don’t?
You are given the cartesian coordinates of three non-collinear points in the plane. Your job is to calculate the circumference of the unique circle that intersects all three points.
Input The input file will contain one or more test cases. Each test case consists of one line containing six real numbers x1,y1, x2,y2,x3,y3, representing the coordinates of the three points. The diameter of the circle determined by the three points will never exceed a million. Input is terminated by end of file.
Output For each test case, print one line containing one real number telling the circumference of the circle determined by the three points. The circumference is to be printed accurately rounded to two decimals. The value of pi is approximately 3.141592653589793.
Sample Input
0.0 -0.5 0.5 0.0 0.0 0.5 0.0 0.0 0.0 1.0 1.0 1.0 5.0 5.0 5.0 7.0 4.0 6.0 0.0 0.0 -1.0 7.0 7.0 7.0 50.0 50.0 50.0 70.0 40.0 60.0 0.0 0.0 10.0 0.0 20.0 1.0 0.0 -500000.0 500000.0 0.0 0.0 500000.0
Sample Output
3.14 4.44 6.28 31.42 62.83 632.24 3141592.65
完全的模板，求不共线的三点的外接圆。
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #define eps 1e-8 #define PI 3.141592653589793 struct Tpoint { double x,y; Tpoint operator-(Tpoint&a){ Tpoint p1; p1.x=x-a.x; p1.y=y-a.y; return p1; } }; double dist(Tpoint p1,Tpoint p2) // 返回两点之间欧氏距离 { return( sqrt( (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y) ) ); } Tpoint q;//圆心 double r, l; void cocircle(Tpoint p1,Tpoint p2,Tpoint p3) { double x12=p2.x-p1.x; double y12=p2.y-p1.y; double x13=p3.x-p1.x; double y13=p3.y-p1.y; double z2=x12*(p1.x+p2.x)+y12*(p1.y+p2.y); double z3=x13*(p1.x+p3.x)+y13*(p1.y+p3.y); double d=2.0*(x12*(p3.y-p2.y)-y12*(p3.x-p2.x)); q.x=(y13*z2-y12*z3)/d; q.y=(x12*z3-x13*z2)/d; r=dist(p1,q); l = 2 * PI * r; return ; } using namespace std; int main() { Tpoint ans; Tpoint a[10]; while(~scanf("%lf%lf%lf%lf%lf%lf",&a[1].x,&a[1].y,&a[2].x,&a[2].y,&a[3].x,&a[3].y)){ cocircle(a[1], a[2], a[3]); printf("%.2f\n", l); } return 0; }