Ancient Berland Circus
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Appoint description:
System Crawler (2015-01-06)
Description
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.
Sample Input
Input
0.000000 0.000000 1.000000 1.000000 0.000000 1.000000
Output
1.00000000
题目大意：给出三个点，求出以这三个点为定点的最小正多边形。
求最小正多边形，边数越多，面积越大，所以要是求得的多边形的边尽量的小。
由三个点组成的三角形，可以确定一个外接圆，那么正多边形的所有的定点应该都在圆上，求出三边对应的圆心角，找出圆心角的最大公约数，也就得到了多边形的最小的边数。
防止钝角的情况，边长最长的对应的圆心角 = 2*PI - 其他两个圆心角。
#include <cstdio>
#include <cstring>
#include <math.h>
#include <algorithm>
using namespace std ;
#define PI acos(-1)
#define eqs 0.01
double gcd(double a,double b)
{
return a < eqs ? b : gcd(fmod(b,a),a);
}
int main()
{
double x1 , y1 , x2 , y2 , x3 , y3 ;
double a , b , c , p , s , r , k ;
double A , B , C ;
scanf("%lf %lf %lf %lf %lf %lf", &x1, &y1, &x2, &y2, &x3, &y3) ;
a = sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) ) ;
b = sqrt( (x2-x3)*(x2-x3) + (y2-y3)*(y2-y3) ) ;
c = sqrt( (x1-x3)*(x1-x3) + (y1-y3)*(y1-y3) ) ;
p = ( a + b + c ) / 2.0 ;
s = sqrt( p * (p-a) * (p-b) * (p-c) ) ;
r = a * b * c / ( 4 * s ) ;
if( a > c )
{
k = a ; a = c ; c = k ;
}
if( b > c )
{
k = b ; b = c ; c = k ;
}
A = 2 * asin(a/(2*r)) ;
B = 2 * asin(b/(2*r)) ;
C = 2 * PI - A - B ;
//printf("%lf %lf %lf\n", A, B, C) ;
p = gcd(A,B);
p = gcd(p,C) ;
//printf("%lf %lf\n", r, p) ;
printf("%.6lf\n", (PI*r*r*sin(p))/p ) ;
return 0;
}