Codeforces 1036E. Covered Points

又一次写起了几何。。。。

特殊处理在于有可能出现多条线段交于一点的情况，每次考虑时，对每条线段与其他所有线段的交点存在一个set里，对每一个set，每次删除set.size()即可

重点在于判断两条线段的交点是否是一个整数点，需要特殊考虑，平行和y=kx+b关系式不能成立的情况

我的代码中没有判断除数为0的情况（因为不会报错，我也就没写23333，但不影响结果

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<vector>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<fstream>
#include<cstdlib>
#include<ctime>
#include<list>
#include<climits>
#include<bitset>
#include<random>
#include <ctime>
#include <cassert>
#include <complex>
#include <cstring>
#include <chrono>
using namespace std;
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("input.in", "r", stdin);freopen("output.in", "w", stdout);
#define left asfdasdasdfasdfsdfasfsdfasfdas1
#define tan asfdasdasdfasdfasfdfasfsdfasfdas
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef long long ll;
typedef unsigned int un;
const int desll[][]={{,},{,-},{,},{-,}};
const int mod=1e9+;
const int maxn=5e5+;
const int maxm=1e5+;
const double eps=1e-;
int n,k,m;
int ar[maxn][];
int xl[],yl[];
set<pair<int,int>>se;
bool is_interge(double x){
    return fabs(x-round(x))<=eps;
}

bool equal(double a,double b){
    return fabs(a-b)<=eps;
}
void getKB(int x,double& k,double& b)
{
    k=(double)(ar[x][]-ar[x][])/(ar[x][]-ar[x][]);
    b=(double)ar[x][]-k*ar[x][];
}
void func(int i,int j)
{
    if(ar[i][]==ar[i][]&&ar[j][]==ar[j][])return ;
    if(ar[i][]==ar[i][]&&ar[j][]==ar[j][])return ;
    if(equal((double)(ar[i][]-ar[i][])/(ar[i][]-ar[i][]),
            (double)(ar[j][]-ar[j][])/(ar[j][]-ar[j][])))return ;

    double k1,k2,b1,b2;
    getKB(i,k1,b1);
    getKB(j,k2,b2);
    double x,y;
    if(ar[i][]==ar[i][]){
        x=ar[i][];
        y=k2*x+b2;
    }
    else if(ar[j][]==ar[j][]){
        x=ar[j][];
        y=k1*x+b1;
    }
    else{
        x=(b2-b1)/(k1-k2);
        y=k1*x+b1;
    }
    if(!is_interge(x) || !is_interge(y))return ;
    int xx=round(x),yy=round(y);
    if(xx>=min(ar[i][],ar[i][])&&xx<=max(ar[i][],ar[i][])&&
            yy>=min(ar[i][],ar[i][])&&yy<=max(ar[i][],ar[i][])&&
            xx>=min(ar[j][],ar[j][])&&xx<=max(ar[j][],ar[j][])&&
            yy>=min(ar[j][],ar[j][])&&yy<=max(ar[j][],ar[j][]))
        se.insert(make_pair(xx,yy));
}

int main()
{
    scanf("%d",&n);
    for(int i=;i<n;i++){
        for(int j=;j<;j++)scanf("%d",&ar[i][j]);
    }
    ll ans=;
    for(int i=;i<n;i++){
        int xx=abs(ar[i][]-ar[i][]);
        int yy=abs(ar[i][]-ar[i][]);
        ans += __gcd(xx,yy)+;
    }
    for(int i=;i<n;i++){
        se.clear();
        for(int j=i+;j<n;j++){
            func(i,j);
        }
        ans-=se.size();
    }
    printf("%I64d\n",ans);

    return ;
}