E. Covered Points （线段上的整点数）

题目链接：https://codeforces.com/contest/1036/problem/E

思路：学会了一个在线段上的整数点等于 GCD（x1 - x2, y1 - y2） +  1,然后去重线段相交的重复整点。

AC代码：

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const double eps = 1e-;
 const int maxn = 1e3 + ;
 int sgn(double x)
 {
     if(fabs(x) < eps) return ;
     else return x <  ? - : ;
 }
 struct Point{
     double x, y;
     Point(){}
     Point(double _x, double _y){
         x = _x, y = _y;
     }
     void input(){
         scanf("%lf%lf", &x, &y);
     }
     bool operator == (Point b) const{
         return sgn(x - b.x) ==  && sgn(y - b.y) == ;
     }
     bool operator < (Point b)const{
         return sgn(x - b.x) ==  ? sgn(y - b.y < ) : x < b.x;
     }
     Point operator - (const Point &b)const{
         return Point(x - b.x, y - b.y);
     }
     double operator ^(const Point &b){
         return x * b.y - y * b.x;
     }
     double operator *(const Point &b){
         return x * b.x + y * b.y;
     }
 };
 struct Line{
     Point s, e;
     Line(){}
     Line(Point _s, Point _e){s = _s, e = _e;}
     bool operator == (Line v){
         return (s == v.s) && (e == v.e);
     }
     void input(){
         s.input();
         e.input();
     }
     int segcrossing(Line v)
     {
         int d1 = sgn((e - s)^(v.s - s));
         int d2 = sgn((e - s)^(v.e - s));
         int d3 = sgn((v.e - v.s)^(s - v.s));
         int d4 = sgn((v.e - v.s)^(e - v.s));
         if( (d1^d2) == - && (d3^d4) == - )return ;
         return (d1 ==  && sgn((v.s - s)*(v.s - e)) <= ) ||
         (d2 ==  && sgn((v.e - s)*(v.e - e)) <= ) ||
         (d3 ==  && sgn((s - v.s)*(s - v.e))<=) ||
         (d4 ==  && sgn((e - v.s)*(e - v.e))<=);
     }
     Point crosspoint(Line v){
         double a1 = (v.e - v.s)^(s - v.s);
         double a2 = (v.e - v.s)^(e - v.s);
         return Point((s.x*a2 - e.x*a1)/(a2 - a1),(s.y*a2 - e.y*a1)/(a2 - a1));
     }
 };
 Line l[maxn];
 int main()
 {
     std::ios::sync_with_stdio(false);
     int n;
     cin >> n;
     ll ans = ;
     int x1, x2, y1, y2;
     for(int i = ;i < n;i++){
         cin >> x1 >> y1 >> x2 >> y2;
         l[i] = Line(Point((double)x1, (double)y1), Point((double)x2, (double)y2));
         ans += __gcd(abs(x1 - x2), abs(y1 - y2)) + ;
     }
     for(int i = ;i < n;i++)
     {
         set< pair<int, int> >st;
         for(int j = i + ;j < n;j++)
         {
             if(l[i].segcrossing(l[j])){
                 Point v = l[i].crosspoint(l[j]);
                 if((int)v.x == v.x && (int)v.y == v.y)
                     st.insert({v.x,v.y});
             }
         }
         ans -= st.size();
     }
     cout << ans << endl;
     return ;
 }