三个顶点可以组成什么正多边形——几何hdu 4195

http://acm.hdu.edu.cn/showproblem.php?pid=4195

组成多边形的边最小为多少

1。先找三角形的垂心（注意线段垂直的情况）

2。以垂心画圆YY,这个三角形可以把这个圆分成多少弧度（注意钝角三角形的第三边所对应的边的弧度为 2*PI的补数）

3。接着枚举最小弧度的 n分之几，是否可以把其它弧度近似整除（如1801可以近似整除600）

 

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#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;

double dis(double x,double y,double x1,double y1)
{
double dd=sqrt((x-x1)*(x-x1)+(y-y1)*(y-y1));

return dd;
}

int ha[100009];
int add;

double min(double a,double b)
{
if(a<b)return a;
else return b;
}

int gcd(int m,int n)
{
int t;
if(m<n){t=n,n=m,m=t;}
while(n!=0)
    {
        t=m%n;
        m=n;
        n=t;
    }

return m;
}

int main()
{
double x1,y1,x2,y2,x3,y3,k1,k2,xx1,yy1,xx2,yy2,xx,yy,r;
char ss[199];
while(scanf("%s",ss))
    {
if(strcmp(ss,"END")==0)return 0;

        sscanf(ss,"%lf",&x1);
        scanf("%lf",&y1);
        scanf("%lf",&x2);
        scanf("%lf",&y2);
        scanf("%lf",&x3);
        scanf("%lf",&y3);

double k1,k2,k3;
double x,y;
if(y2==y1||y3==y2)
        {
if(y1==y2&&y2!=y3)
            {xx=(x1+x2)/2;
            k2=(x2-x3)/(y3-y2); 
            xx2=(x2+x3)/2;
            yy2=(y2+y3)/2;
            yy=k2*(xx-xx2)+yy2;
            }
if(y1!=y2&&y2==y3)
            {
                xx=(x2+x3)/2;
                k1=(x1-x2)/(y2-y1);
                xx1=(x1+x2)/2;
                yy1=(y1+y2)/2;
                yy=k1*(xx-xx1)+yy1;
            }
        }
else
        {
            k1=(x1-x2)/(y2-y1);
            k2=(x2-x3)/(y3-y2);
            xx1=(x1+x2)/2;
            yy1=(y1+y2)/2;
            xx2=(x2+x3)/2;
            yy2=(y2+y3)/2;
            xx=(xx1*k1-xx2*k2+yy2-yy1)/(k1-k2);
            yy=k1*(xx-xx1)+yy1;

        }

        r=sqrt((x1-xx)*(x1-xx)+(y1-yy)*(y1-yy));

double l1,l2,l3; 
        l1=dis(x1,y1,x2,y2);
        l2=dis(x2,y2,x3,y3);
        l3=dis(x3,y3,x1,y1);

double th1,th2,th3;

double temp=(2*r*r-l1*l1)*1.0/2/r/r;
if(temp>1.0)temp=1.0;
if(temp<-1.0)temp=-1.0;
        th1=acos(temp);

        temp=(2*r*r-l2*l2)*1.0/2/r/r;
if(temp>1.0)temp=1.0;
if(temp<-1.0)temp=-1.0;
        th2=acos(temp);

        temp=(2*r*r-l3*l3)*1.0/2/r/r;
if(temp>1.0)temp=1.0;
if(temp<-1.0)temp=-1.0;
        th3=acos(temp);
if(th1<th2)
            temp=th1,th1=th2,th2=temp;
if(th1<th3)
            temp=th1,th1=th3,th3=temp;

if(th1-(th2+th3)<0.00000001)
            th1=2*acos(-1.0)-th2-th3;

int i,j,k;

int max=1,shi1;
double xiao=min(min(th1,th2),th3);

        add=0;
int num;
while(1)
        {
            add++;
double temp=xiao*1.0/add;
if(fabs(th1/temp-(int)(th1/temp+0.0001))>0.0001)continue;
if(fabs(th2/temp-(int)(th2/temp+0.0001))>0.0001)continue;
if(fabs(th3/temp-(int)(th3/temp+0.0001))>0.0001)continue;
            num=(th1+0.0001)/temp+(th2+0.0001)/temp+(th3+0.0001)/temp;
break;
        }

        printf("%d\n",num);

    }
return 0;
}