Codeforces Beta Round #1 C.Ancient Berland Circus

C. Ancient Berland Circus
time limit per test2 seconds
memory limit per test64 megabytes
inputstandard input
outputstandard output
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.

In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.

Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.

You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.

Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.

Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.

Examples
input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
output
1.00000000

思路: 这题就是精度坑得一逼

ac = acos(1 - a*a/2/r/r);
bc = acos(1 - b*b/2/r/r);
cc = 2*pi - ac - bc;

不能写成

ac = acos(1 - a*a/2/r/r);
bc = acos(1 - b*b/2/r/r);
cc = acos(1 - c*c/2/r/r);

不然会对角度极小时会造成精度丢失。

而且如果用ac = acos(sqrt(r*r-a*a/4)/r);

算完角度再来*2也是不行，精度也会严重丢失。gcd可能还会形成死循环。。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

double eps = 1e-2;

double dist(double x1,double y1,double x2,double y2){
    return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
}

double gcd(double a,double b){

    return b<eps?a:gcd(b,fmod(a,b));
}
const double pi = 3.1415926535898;

int main(){
    
    double x1,x2,x3,y1,y2,y3,a,b,c,r,s,p,ac,bc,cc;
    scanf("%lf%lf",&x1,&y1);
    scanf("%lf%lf",&x2,&y2);
    scanf("%lf%lf",&x3,&y3);
    
    a = dist(x1,y1,x2,y2);
    b = dist(x2,y2,x3,y3);
    c = dist(x1,y1,x3,y3);
    p = (a+b+c)/2;
    s = sqrt(p*(p-a)*(p-b)*(p-c));
    r = a*b*c/4/s;
    ac = acos(1 - a*a/2/r/r);
    bc = acos(1 - b*b/2/r/r);
    cc = 2*pi - ac - bc;
    double unit=0;
    unit = gcd(ac,bc);
    unit = gcd(unit,cc);
    
    printf("%.6lf
",pi/unit*r*r*sin(unit));
    
    return 0;
}