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

 

　　

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<iostream>
  5 #include<cmath>
  6 #include<vector>
  7 #include<queue>
  8 #include<set>
  9 #include<map>
 10 using namespace std;
 11 #define eps 1e-6
 12 #define For(i,a,b) for(int i=a;i<=b;i++)
 13 #define Fore(i,a,b) for(int i=a;i>=b;i--)
 14 #define lson l,mid,rt<<1
 15 #define rson mid+1,r,rt<<1|1
 16 #define mkp make_pair
 17 #define pb push_back
 18 #define sz size()
 19 #define met(a,b) memset(a,b,sizeof(a))
 20 #define iossy ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
 21 #define fr freopen
 22 #define pi acos(-1.0)
 23 #define Vector Point
 24 typedef pair<int,int> pii;
 25 const long long linf=1LL<<62;
 26 const int iinf=1<<30;
 27 const double dinf=1e17;
 28 const int Mod=1e9+9;
 29 typedef long long ll;
 30 typedef long double ld;
 31 const int maxn=1000005;
 32 int n;
 33 struct Point{
 34     ll x,y;
 35     int id;
 36     Point(ll x=0,ll y=0):x(x),y(y) {}
 37     Point operator - (const Point &a)const { return Point(x-a.x,y-a.y);}
 38     bool operator == (const Point &a)const { return x==a.x && y==a.y; }
 39 };
 40 ll Cross(Vector a,Vector b){
 41     return a.x*b.y-a.y*b.x;
 42 }
 43 ll Dot(Vector a,Vector b) {
 44     return a.x*b.x+a.y*b.y;
 45 }
 46 bool onsg(Point p,Point a1,Point a2){
 47     return Cross(a1-p,a2-p)==0 && Dot(a1-p,a2-p)<0;
 48 }
 49 void ck(ll &c){
 50     if(c>0) c=1;
 51     else if(c<0) c=-1;
 52 }
 53 int Ins(Point a1,Point a2,Point b1,Point b2){
 54     if(a1==b1 || a1==b2 || a2==b1 || a2==b2) return 1;
 55     if(onsg(a1,b1,b2) || onsg(a2,b1,b2) || onsg(b1,a1,a2) || onsg(b2,a1,a2)) return 1;
 56     ll c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1);
 57     ll c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
 58     ck(c1);ck(c2);ck(c3);ck(c4);
 59     return c1*c2<0 && c3*c4<0;
 60 }
 61 set<pair<ll,ll> >c;
 62 void chk(Point p,Vector v,Point q,Vector w){
 63     Vector u=p-q;
 64     ll v1=Cross(w,u),v2=Cross(v,w);
 65     if(abs(v1*v.x)%v2!=0 || abs(v1*v.y)%v2!=0) return ;
 66     ll xx,yy;
 67     xx=p.x+v.x*v1/v2;yy=p.y+v.y*v1/v2;
 68     c.insert(mkp(xx,yy));
 69 }
 70 struct segm{
 71     Point p1,p2;
 72 };
 73 segm ss[maxn];
 74 Point p1,p2;
 75 void solve(){
 76     iossy;
 77     cin>>n;
 78     int ans=0;
 79     For(i,1,n){
 80         cin>>p1.x>>p1.y>>p2.x>>p2.y;
 81         ss[i].p1=p1;ss[i].p2=p2;
 82         ans+=__gcd(abs(ss[i].p2.x-ss[i].p1.x),abs(ss[i].p2.y-ss[i].p1.y))+1;
 83     }
 84     For(i,1,n){
 85         c.clear();
 86         For(j,i+1,n){
 87             int ct=Ins(ss[i].p1,ss[i].p2,ss[j].p1,ss[j].p2);
 88             if(ct) chk(ss[i].p1,ss[i].p2-ss[i].p1,ss[j].p1,ss[j].p2-ss[j].p1);
 89         }
 90         ans-=c.sz;
 91     }
 92     //cout<<ans-c.sz<<endl;
 93     cout<<ans<<endl;
 94 }
 95 int main(){
 96     int t=1;
 97    // For(i,1,t) printf("Case #%d: ",i);
 98     solve();
 99     return 0;
100 }
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