TOJ 1132 ZOJ 1090 The Circumference of the Circle

The Circumference of the Circle

时间限制(普通/Java):1000MS/10000MS     运行内存限制:65536KByte

描述

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.

输入

The input 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.

输出

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.

样例输入

0.0 -0.5 0.5 0.0 0.0 0.50.0 0.0 0.0 1.0 1.0 1.05.0 5.0 5.0 7.0 4.0 6.00.0 0.0 -1.0 7.0 7.0 7.050.0 50.0 50.0 70.0 40.0 60.00.0 0.0 10.0 0.0 20.0 1.00.0 -500000.0 500000.0 0.0 0.0 500000.0

样例输出

3.144.446.2831.4262.83632.243141592.65

主要是让自己记得一个数学公式

#include<stdio.h>#include<math.h>#define PI acos(-1)int main(){double x1,y1,x2,y2,x3,y3,a,b,c,d,s,l,x,y;while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3)!=EOF){a=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));b=sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));c=sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));l=(a+b+c)/2;s=sqrt(l*(l-a)*(l-b)*(l-c));d=s/c*2;//d是第三条边上的高 x=a*b/d/2;//对于一个三角形的外接圆  半径等于 两条边的乘积除以第三条边上的高再除以2 y=2*PI*x;printf("%.2f\n",y);}return 0;}