AC日记——codeforces Ancient Berland Circus 1c

1C - Ancient Berland Circus

思路：

　　求出三角形外接圆；

　　然后找出三角形三条边在小数意义下的最大公约数;

　　然后n=pi*2/fgcd;

　　求出面积即可；

代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define INF (1e9)
#define eps (1e-4)
#define pi (3.1415926535)

struct point
{
    double x,y;
};
struct point ai[];

struct line {
    double k,b;

    line *another;

    void updata(struct point a,struct point bb)
    {
        another=new line;
        if(a.x==bb.x) k=INF,b=a.x;
        else
        {
            if(a.y==bb.y) k=,b=a.y;
            else
            {
                k=(bb.y-a.y)/(bb.x-a.x);
                b=a.y-k*a.x;
            }
        }
        if(k==)
        {
            another->k=INF,another->b=(a.x+bb.x)/2.0;
        }
        else if(k==INF)
        {
            another->k=,another->b=(a.y+bb.y)/2.0;
        }
        else
        {
            another->k=-1.0/k;
            struct point temp;
            temp.x=(a.x+bb.x)/2.0;
            temp.y=(a.y+bb.y)/2.0;
            another->b=temp.y-another->k*temp.x;
        }
    }
};
struct line bi[],ci[];

struct point getnode(line a,line b)
{
    struct point res;
    res.x=(b.b-a.b)/(a.k-b.k);
    res.y=a.k*res.x+a.b;
    return res;
}

double dist(struct point a,struct point b)
{
    double aa=(a.x-b.x)*(a.x-b.x);
    double bb=(a.y-b.y)*(a.y-b.y);
    aa=sqrt(aa+bb);
    return aa;
}

double fgcd(double x, double y) {
    while (fabs(x)>eps&&fabs(y)>eps)
    {
        if (x > y) x-=floor(x/y)*y;
        else y-=floor(y/x)*x;
    }
    return x + y;
}

int main()
{
    for(int i=;i<=;i++) cin>>ai[i].x>>ai[i].y;
    bi[].updata(ai[],ai[]);
    bi[].updata(ai[],ai[]);
    bi[].updata(ai[],ai[]);
    for(int i=;i<=;i++) ci[i]=*bi[i].another;
    struct point o=getnode(ci[],ci[]);
    double R=dist(o,ai[]);
    double vi[];
    vi[]=dist(ai[],ai[]);
    vi[]=dist(ai[],ai[]);
    vi[]=dist(ai[],ai[]);
    sort(vi+,vi+);
    double an[];
    an[]=2.0*asin(vi[]/(2.0*R));
    an[]=2.0*asin(vi[]/(2.0*R));
    an[]=2.0*pi-an[]-an[];
    sort(an+,an+);
    double ans1=fgcd(an[],an[]);
    double ans2=fgcd(an[],an[]);
    double ans=fgcd(max(ans1,ans2),min(ans1,ans2));
    double ans_=(2.0*pi-ans)/2.0;
    printf("%.9lf",R*R*sin(ans)*pi/ans);
    return ;
}