(Analysis by Nick Wu)

The first observation we make in this problem is that the order in which operations happen in does not matter, it only matters which operations we do. This means there are $2^N$ different sets of operations to consider. To solve the subtask, we can manually consider each set.

Below is my Python 3 code that solves the subtask where $N \le 20$.

def solve():
n = int(input())
wantx, wanty = (int(z) for z in input().split())
cand = [(0, 0, 0)] # [(x, y, number of instructions used)]
for _ in range(n):
x, y = (int(z) for z in input().split())
now = len(cand)
for i in range(now):
cand.append((cand[i][0] + x, cand[i][1] + y, cand[i][2] + 1))
ret = [0] * n
for x, y, z in cand:
if (x, y) == (wantx, wanty):
ret[z-1] += 1
for x in ret:
print(x)

solve()


We note that the original bounds on the problem have $N \le 40$, which is only twice as big. This suggests cutting the given operations into two halves of roughly equal sizes and running the above algorithm on both halves. If we iterate over one of the given lists, we know exactly how far we need to move from the other half of the instructions.

This technique is sometimes known as meet-in-the-middle. To motivate this, imagine that half of the instructions are actually instructions to move the goal location in some direction. Note that in this variant of the problem, even though we are effectively brute-forcing all possible ways to move the robot and all possible ways to move the goal, we can quickly check when the goal and robot are in the same place. The time complexity for each of the solutions below is $O(N\cdot 2^{N/2})$.

Benjamin Qi's C++ code:

#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";
}
}


Alternatively, a hashmap can be used instead of sorting, though the constant factor is somewhat worse:

#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";
}
}


Danny Mittal's Java code:

import java.io.BufferedReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.StringTokenizer;

public class RobotInstructionsSort {

public static void main(String[] args) throws IOException {
long xGoal = Long.parseLong(tokenizer.nextToken());
long yGoal = Long.parseLong(tokenizer.nextToken());
long[] instructionsX = new long[n];
long[] instructionsY = new long[n];
for (int j = 0; j < n; j++) {
instructionsX[j] = Long.parseLong(tokenizer.nextToken());
instructionsY[j] = Long.parseLong(tokenizer.nextToken());
}
InstructionChoice[] choicesLeft = new InstructionChoice[1 << (n / 2)];
long x = 0;
long y = 0;
for (int j = 0; j < n / 2; j++) {
if ((mask & (1 << j)) != 0) {
x += instructionsX[j];
y += instructionsY[j];
}
}
}
Arrays.sort(choicesLeft);
InstructionChoice[] choicesRight = new InstructionChoice[1 << ((n + 1) / 2)];
long x = 0;
long y = 0;
for (int j = n / 2; j < n; j++) {
if ((mask & (1 << (j - (n / 2)))) != 0) {
x += instructionsX[j];
y += instructionsY[j];
}
}
}
Arrays.sort(choicesRight);
long[] answer = new long[n + 1];
long[] amounts = new long[(n / 2) + 1];
int j = 0;
int k = 0;
for (InstructionChoice choice : choicesRight) {
while (j < choicesLeft.length && choicesLeft[j].compareTo(choice) <= 0) {
amounts[choicesLeft[j].instructions]++;
j++;
}
while (k < choicesLeft.length && choicesLeft[k].compareTo(choice) < 0) {
amounts[choicesLeft[k].instructions]--;
k++;
}
for (int instructions = 0; instructions <= n / 2; instructions++) {
}
}
StringBuilder out = new StringBuilder();
for (int instructionsUsed = 1; instructionsUsed <= n; instructionsUsed++) {
}
System.out.print(out);
}

static class InstructionChoice implements Comparable<InstructionChoice> {
final long x;
final long y;
final int instructions;

InstructionChoice(long x, long y, int instructions) {
this.x = x;
this.y = y;
this.instructions = instructions;
}

@Override
public int compareTo(InstructionChoice other) {
if (y != other.y) {
return Long.compare(y, other.y);
} else {
return Long.compare(x, other.x);
}
}
}
}