Expert.
Put your skills to the test.

Problem summary

Count the distinct inclinations of nondegenerate segments. Inclination is measured from the left endpoint, or upward for a vertical segment, so reversing endpoints does not change the slope.

Constraints

0 < n < 100; coordinates are positive integers less than 100; endpoints differ.

Worked example

Official sample; other valid outputs may exist.

6
3 1 6 2
3 2 6 3
1 2 1 3
2 1 2 3
1 4 4 5
6 4 5 5

3

Solution analysis

1. Represent a segment by its integer direction (dx,dy). Flip signs if dx<0, or if dx=0 and dy<0, to choose the left-to-right or upward direction.
2. Divide both components by gcd(|dx|,|dy|), producing one canonical primitive vector per slope. Insert those vectors into a set and output its size.
3. The sample has normalized directions (3,1), (0,1), and (1,−1). Parallel segments share a direction regardless of their lengths.

Why it works

Two segments have equal inclination exactly when their oriented direction vectors are positive scalar multiples. Dividing by the greatest common divisor gives identical primitive integer vectors exactly in this case. The set therefore has one element for every distinct inclination.

Watch out. Do not use floating-point angles. All vertical segments have the same positive inclination.

An independently written complete program with standard headers.

// Key idea: Two segments have equal inclination exactly when their oriented direction vectors are
// positive scalar multiples. Dividing by the greatest common divisor gives identical primitive integer
// vectors exactly in this case.
#include<iostream>
#include<set>
#include<utility>
#include<numeric>
#include<cstdlib>
using namespace std;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin>>n;
    set<pair<int,int>>slopes;
    for(int i=0;i<n;++i){
        int x1,y1,x2,y2;
        cin>>x1>>y1>>x2>>y2;
        int dx=x2-x1,dy=y2-y1;
        if(dx<0||(dx==0&&dy<0)){
            dx=-dx;
            dy=-dy;
        }
        int divisor=gcd(abs(dx),abs(dy));
        slopes.insert({dx/divisor,dy/divisor});
    }
    cout<<slopes.size()<<'\n';
}