[CodeForces-1036E] Covered Points 暴力 GCD 求交点

　　题意：

　　　　在二维平面上给出n条不共线的线段，问这些线段总共覆盖到了多少个整数点

　　

　　解法：

　　　　　　用GCD可求得一条线段覆盖了多少整数点，然后暴力枚举线段，求交点，对于相应的

　　　　整数交点，结果-1即可

　　

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<cmath>
 #include<vector>
 #include<queue>
 #include<set>
 #include<map>
 using namespace std;
 #define eps 1e-6
 #define For(i,a,b) for(int i=a;i<=b;i++)
 #define Fore(i,a,b) for(int i=a;i>=b;i--)
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define mkp make_pair
 #define pb push_back
 #define sz size()
 #define met(a,b) memset(a,b,sizeof(a))
 #define iossy ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
 #define fr freopen
 #define pi acos(-1.0)
 #define Vector Point
 typedef pair<int,int> pii;
 const long long linf=1LL<<;
 const int iinf=<<;
 const double dinf=1e17;
 const int Mod=1e9+;
 typedef long long ll;
 typedef long double ld;
 const int maxn=;
 int n;
 struct Point{
     ll x,y;
     int id;
     Point(ll x=,ll y=):x(x),y(y) {}
     Point operator - (const Point &a)const { return Point(x-a.x,y-a.y);}
     bool operator == (const Point &a)const { return x==a.x && y==a.y; }
 };
 ll Cross(Vector a,Vector b){
     return a.x*b.y-a.y*b.x;
 }
 ll Dot(Vector a,Vector b) {
     return a.x*b.x+a.y*b.y;
 }
 bool onsg(Point p,Point a1,Point a2){
     return Cross(a1-p,a2-p)== && Dot(a1-p,a2-p)<;
 }
 void ck(ll &c){
     if(c>) c=;
     else if(c<) c=-;
 }
 int Ins(Point a1,Point a2,Point b1,Point b2){
     if(a1==b1 || a1==b2 || a2==b1 || a2==b2) return ;
     if(onsg(a1,b1,b2) || onsg(a2,b1,b2) || onsg(b1,a1,a2) || onsg(b2,a1,a2)) return ;
     ll c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1);
     ll c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
     ck(c1);ck(c2);ck(c3);ck(c4);
     return c1*c2< && c3*c4<;
 }
 set<pair<ll,ll> >c;
 void chk(Point p,Vector v,Point q,Vector w){
     Vector u=p-q;
     ll v1=Cross(w,u),v2=Cross(v,w);
     if(abs(v1*v.x)%v2!= || abs(v1*v.y)%v2!=) return ;
     ll xx,yy;
     xx=p.x+v.x*v1/v2;yy=p.y+v.y*v1/v2;
     c.insert(mkp(xx,yy));
 }
 struct segm{
     Point p1,p2;
 };
 segm ss[maxn];
 Point p1,p2;
 void solve(){
     iossy;
     cin>>n;
     int ans=;
     For(i,,n){
         cin>>p1.x>>p1.y>>p2.x>>p2.y;
         ss[i].p1=p1;ss[i].p2=p2;
         ans+=__gcd(abs(ss[i].p2.x-ss[i].p1.x),abs(ss[i].p2.y-ss[i].p1.y))+;
     }
     For(i,,n){
         c.clear();
         For(j,i+,n){
             int ct=Ins(ss[i].p1,ss[i].p2,ss[j].p1,ss[j].p2);
             if(ct) chk(ss[i].p1,ss[i].p2-ss[i].p1,ss[j].p1,ss[j].p2-ss[j].p1);
         }
         ans-=c.sz;
     }
     //cout<<ans-c.sz<<endl;
     cout<<ans<<endl;
 }
 int main(){
     int t=;
    // For(i,1,t) printf("Case #%d: ",i);
     solve();
     return ;
 }