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 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
#include <iostream>
#include<iomanip>
#include <cmath>
using namespace std;
#define PI 3.141592653589793
int main()
{
double x1,y1,x2,y2,x3,y3;
while(cin>>x1>>y1>>x2>>y2>>x3>>y3){
double l,a1,b1,a2,b2,k1,k2,a,b;
a1=x1/+x2/;
a2=x1/+x3/;
b1=y1/+y2/;
b2=y1/+y3/;
if(y1!=y2&&y3!=y1){
k1=(x1-x2)/(y2-y1);
k2=(x1-x3)/(y3-y1);
a=(k1*a1-k2*a2+b2-b1)/(k1-k2);
b=k1*(a-a1)+b1;
}
else if(y1==y2){
k2=(x1-x3)/(y3-y1);
a=(x1+x2)/;
b=k2*(a-a2)+b2;
}
else {
k1=(x1-x2)/(y2-y1);
a=(x1+x3)/;
b=k1*(a-a1)+b1;
}
l=*PI*sqrt((a-x1)*(a-x1)+(b-y1)*(b-y1));
cout.precision();
cout.setf(ios::fixed);
cout<<l<<endl;
}
//system("pause");
return ;
}

F - The Circumference of the Circle的更多相关文章

  1. poj 1090:The Circumference of the Circle(计算几何,求三角形外心)

    The Circumference of the Circle Time Limit: 2 Seconds      Memory Limit: 65536 KB To calculate the c ...

  2. ZOJ Problem Set - 1090——The Circumference of the Circle

      ZOJ Problem Set - 1090 The Circumference of the Circle Time Limit: 2 Seconds      Memory Limit: 65 ...

  3. ZOJ 1090 The Circumference of the Circle

    原题链接 题目大意:已知三角形的三个顶点坐标,求其外接圆的周长. 解法:刚看到这道题时,马上拿出草稿纸画图,想推导出重心坐标,然后求出半径,再求周长.可是这个过程太复杂了,写到一半就没有兴致了,还是求 ...

  4. POJ2242 The Circumference of the Circle(几何)

    题目链接. 题目大意: 给定三个点,即一个任意三角形,求外接圆的周长. 分析: 外接圆的半径可以通过公式求得(2*r = a/sinA = b/sinB = c/sinC),然后直接求周长. 注意: ...

  5. 【POJ2242】The Circumference of the Circle(初等几何)

    已知三点坐标,算圆面积. 使用初等几何知识,根据海伦公式s = sqrt(p(p - a)(p - b)(p - c)) 和 外接圆直径 d = a * b * c / (2s) 来直接计算. #in ...

  6. POJ 2242 The Circumference of the Circle

    做题回顾:用到海伦公式,还有注意数据类型,最好统一 p=(a+b+c)/2; s=sqrt(p*(p-a)*(p-b)*(p-c));//三角形面积,海伦公式 r=a*b*c/(4*s);//这是外接 ...

  7. H - Ones

    Description Given any integer 0 <= n <= 10000 not divisible by 2 or 5, some multiple of n is a ...

  8. [Swift]LeetCode478. 在圆内随机生成点 | Generate Random Point in a Circle

    Given the radius and x-y positions of the center of a circle, write a function randPoint which gener ...

  9. [LeetCode] Generate Random Point in a Circle 生成圆中的随机点

    Given the radius and x-y positions of the center of a circle, write a function randPoint which gener ...

随机推荐

  1. uva 508 Morse Mismatches

    Samuel F. B. Morse is best known for the coding scheme that carries his name. Morse code is still us ...

  2. (原)VS2013在Release情况下使用vector有时候会崩溃的一个可能原因

    转载请注明出处: http://www.cnblogs.com/darkknightzh/p/5016352.html 参考网址: http://www.cnblogs.com/BryZ/archiv ...

  3. [Leetcode] Container With Most Water ( C++)

    题目: Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, a ...

  4. JavaScript this 局部变量全局变量 作用域 作用域链 闭包

    从阮老师博客的一道测试题说起: 代码段一: var name = "The Window"; var object = { name : "My Object" ...

  5. C# 对类中的保护成员进行写操作(邀请大家拍砖)

    假如我有一个类库 Lib,提供一个类 ClassA 对外服务,ClassA 中有若干只读属性 PropA, PropB 等等,外部调用者无法对 ClassA 中的 PropA 和 PropB 进行写操 ...

  6. PHP 获取客户端IP

    function get_ip() { static $realIP; if (isset($_SERVER)){ if (isset($_SERVER["HTTP_X_FORWARDED_ ...

  7. yii2中的url美化

    在yii2中,如果想实现类似于post/1,post/update/1之类的效果,官方文档已经有明确的说明 但是如果想把所有的controller都实现,这里采用yii1的方法 'rules' =&g ...

  8. 【Xamarin挖墙脚系列:关闭 OS X El Capitan 中 SIP 安全设置功能】

    比如需要修改内核配置文件: com.apple.Boot.plist 那么我们需要解锁权限. 禁止SIP模式,那么就可以修改此文件了. 在 OS X El Capitan 中有一个跟安全相关的模式叫 ...

  9. 【转】Android下编译jni库的二种方法(含示例) -- 不错

    原文网址:http://blog.sina.com.cn/s/blog_3e3fcadd01011384.html 总结如下:两种方法是:1)使用Android源码中的Make系统2)使用NDK(从N ...

  10. codecomb 2085【肥得更高】

    题目背景 自2009年以来,A.B站的历史就已经步入了农业变革的黎明期. 在两站的娱乐及音乐区,金坷垃制造业早已得到长足的发展,甚至有些地方还出现了坷垃翻唱的萌芽. 新兴肥料人开始走上历史的舞台. 他 ...