题解去看它

 

精度真是卡的我醉生梦死，w(ﾟДﾟ)w    Ｏ(≧口≦)Ｏ

bzoj改成long double 就过了

洛谷仍处于

输出x.99999，答案输出x+1.00000

输出-0.00000，答案输出0.00000

救命啊~~~~(>_<)~~~~

 

来自大佬的建议：输出double时用%f

 

#include
     
      #include
      
       #include
       
         #define N 50001 using namespace std; const long double eps=1e-10; int dcmp(long double x){    if(fabs(x)
        
         1 && Cross(c[m-1]-c[m-2],p[i]-c[m-2])<=0) m--;        c[m++]=p[i];    }    int k=m;    for(int i=n-2;i>=0;--i)    {        while(m>k && Cross(c[m-1]-c[m-2],p[i]-c[m-2])<=0) m--;        c[m++]=p[i];    }    m--;    return m;} long double getdis(Point A,Point B){    return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));} void RotatingCaliper(Point *c,int m){    long double AnsArea=1e20,AnsPeri=1e20;    int q=1,l=0,r=0;    long double d,h,w,rw;    for(int p=0;p
         
          fabs(Cross(c[p]-c[p+1],c[q]-c[p+1]))) q=(q+1)%m;        while(dcmp(Dot(c[p+1]-c[p],c[r+1]-c[r]))>0)         r=(r+1)%m;        if(!l) l=q;        while(dcmp(Dot(c[p+1]-c[p],c[l+1]-c[l]))<0)         l=(l+1)%m;        d=Length(c[p+1]-c[p]);        h=fabs(Area2(c[p],c[p+1],c[q]))/d;        w=Dot(c[p+1]-c[p],c[r]-c[l])/d;        rw=Dot(c[r]-c[p],c[p+1]-c[p])/d;        if(w*h