2 solutions

The first method is to exceed the memory limit.Recommend using the second 
method.

1st method:

#include <bits/stdc++.h>
using namespace std;
using P = pair<long long, long long>;
P operator+(P a, P b){
    return {a.first + b.first, a.second + b.second};
}
P operator-(P a, P b){
    return {a.first - b.first, a.second - b.second};
}
vector<pair<P, int>> all_subsets(const vector<P> &dirs){
	vector<pair<P, int>> v{{}};
	for(const P &d : dirs){
		v.resize(2 * v.size());
		for(int i = 0; i < v.size() / 2; i++){
			v[i + v.size() / 2] = {v[i].first + d, v[i].second + 1};
		}
	}
	return v;
}
struct hsh{
	size_t operator()(const P &p) const {
		return p.first * 239 + p.second;
	}
};
int main(){
	int N;
	cin >> N;
	P goal;
	cin >> goal.first >> goal.second;
	vector<P> dirs(N);
	for(auto &d : dirs){
		cin >> d.first >> d.second;
	}
	vector<pair<P, int>> a =
		all_subsets(vector<P>(begin(dirs), begin(dirs) + N / 2));
	vector<pair<P, int>> b =
		all_subsets(vector<P>(begin(dirs) + N / 2, end(dirs)));
	unordered_map<P, map<int, int>, hsh> first_half;
	for(const auto &[offset, num] : a){
		++first_half[offset][num];
	}
	vector<long long> ans(N + 1);
	for(const auto &[offset, num] : b){
		auto it = first_half.find(goal - offset);
		if(it != end(first_half)){
			for(const auto &[num2, ways] : it->second){
				ans[num + num2] += ways;
			}
		}
	}
	for (int i = 1; i <= N; i++){
		cout << ans[i] << "\n";
	}
}

2nd method:

#include <bits/stdc++.h>
using namespace std;
using P = pair<long long, long long>;
P operator+(P a, P b){
    return {a.first + b.first, a.second + b.second};
}
P operator-(P a, P b){
    return {a.first - b.first, a.second - b.second};
}
vector<pair<P, int>> all_subsets(const vector<P> &dirs){
	vector<pair<P, int>> v{{}};
	for(const P &d : dirs){
		v.resize(2 * v.size());
		for(int i = 0; i < v.size() / 2; i++){
			v[i + v.size() / 2] = {v[i].first + d, v[i].second + 1};
		}
	}
	sort(v.begin(), v.end());
	return v;
}

int main(){
	int N;
	cin >> N;
	P goal;
	cin >> goal.first >> goal.second;
	vector<P> dirs(N);
	for(auto &d : dirs){
		cin >> d.first >> d.second;
	}
	vector<pair<P, int>> a =
		all_subsets(vector<P>(begin(dirs), begin(dirs) + N / 2));
	vector<pair<P, int>> b =
		all_subsets(vector<P>(begin(dirs) + N / 2, end(dirs)));
	reverse(b.begin(), b.end());
	vector<long long> ans(N + 1);
	vector<int> with_num;
	P rest_prev{1e18, 1e18};
	int ib = 0;
	for(const auto &[offset, num] : a){
		const P rest = goal - offset;
		if(rest != rest_prev){
			rest_prev = rest;
			with_num = vector<int>(N - N / 2 + 1);
			for(; ib < b.size() && b.at(ib).first > rest; ++ib);
			for(; ib < b.size() && b.at(ib).first == rest; ++ib){
				++with_num.at(b.at(ib).second);
			}
		}
		for(int i = 0; i < with_num.size(); i++){
			ans[i + num] += with_num[i];
		}
	}
	for(int i = 1; i <= N; i++){
		cout << ans[i] << "\n";
	}
}

The following code is not the solution to this preblem.

#include <iostream>
using namespace std;
int main(){
    cout << "subscribe~";
    return 0;
}