masterpiece002_9_MARTIE

#include <fstream>
#include <cmath>
#define Dmax 505550
using namespace std;
int n,X1,Y1,X2,Y2,vf,Max,a,b,r,st,dr,mij,poz;
float T[Dmax],s[Dmax];

int main()
{
    ifstream f("masterpiece002.in");
    ofstream g("masterpiece002.out");
    while(f>>X1>>Y1>>X2>>Y2)
    {
        a=abs(X2-X1),b=abs(Y2-Y1);
        if(b!=0)
            do
            {
                r=a%b;
                a=b;
                b=r;
            }
            while(r); /// in final a va fi cel mai mare divizor comun
        else
            if(a==0) /// pentru a prinde cazul particular in care ambele valori sunt 0
                a=-1;
        if(a==1||(Y1==Y2&&abs(X2-X1)==1)||(X1==X2&&abs(Y2-Y1)==1)||(X2==X1&&Y2==Y1&&abs(Y1-X1)==1)) /// Daca e segment simplu,
            T[n++]=sqrt((X2-X1)*(X2-X1)+(Y2-Y1)*(Y2-Y1)),g<<"n= "<<n<<' '<<X1<<' '<<Y1<<' '<<X2<<' '<<Y2<<' '<<endl; /// calculam distanta si o memoram
    }
    f.close();
    for(int i=1;i<=n;++i)
    {
        int st=1,dr=vf,mij,poz=0; /// Cautare binara
        while(st<=dr)
        {
            mij=(st+dr)/2;
            if(s[mij]<T[i])
                st=mij+1;
            else
                if(s[mij]>T[i])
                    dr=mij-1,poz=mij;
                else
                {
                    poz=mij;
                    break;
                }
        }
        if(poz==0)
            ++vf,s[vf]=T[i],poz=vf;
        else
            s[poz]=T[i];
        if(poz>Max)
            Max=poz;
    }
    ///ofstream g("masterpiece002.out");
    g<<Max<<'\n';
    for(int i=1;i<=n;++i)
        g<<T[i]<<' ';
    g<<endl;
    g.close();
    return 0;
}