#include "hack.h"
#include <vector>
#include <numeric>
#include <algorithm>
#include <random>
#include <cmath>
// Declare the interactor provided by the judge
namespace {
// Helper to find all divisors of val that are strictly greater than limit.
// Since we constrain input values to ~10^10, val is small enough to factorize quickly.
std::vector<long long> get_divisors(long long val, long long limit) {
std::vector<long long> divs;
for (long long i = 1; i * i <= val; ++i) {
if (val % i == 0) {
if (i > limit) divs.push_back(i);
if (val / i > limit && val / i != i) divs.push_back(val / i);
}
}
std::sort(divs.begin(), divs.end());
return divs;
}
// Check if a candidate n is the correct answer
// If n is correct, 1 and n+1 will collide (both map to bucket 1)
bool check_n(long long n) {
if (n <= 0) return false;
std::vector<long long> q = {1, n + 1};
return collisions(q) > 0;
}
}
int hack() {
// Phase 1: Check for small n efficiently
// If n <= 2500, a set of size 2501 {1..2501} guarantees a collision.
const int SMALL_LIMIT = 2500;
std::vector<long long> small_q(SMALL_LIMIT + 1);
std::iota(small_q.begin(), small_q.end(), 1);
if (collisions(small_q) > 0) {
// Binary search to find the smallest k such that {1..k} has a collision.
// The first collision occurs when k = n + 1.
int low = 2, high = SMALL_LIMIT + 1;
int ans = high;
while (low <= high) {
int mid = low + (high - low) / 2;
std::vector<long long> q(mid);
std::iota(q.begin(), q.end(), 1);
if (collisions(q) > 0) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans - 1;
}
// Phase 2: Solve for large n using Birthday Paradox
// Use a fixed seed for consistency, or random_device if allowed.
std::mt19937_64 rng(1337);
// CRITICAL OPTIMIZATION:
// We restrict the random range to [1, 2*10^10].
// This is > 10^9 (so collisions occur naturally based on n)
// but small enough that |u - v| can be factorized instantly without TLE.
std::uniform_int_distribution<long long> dist(1, 20000000000LL);
// K = 50,000 fits the budget (approx 100k-110k total cost) and gives high collision prob.
int K = 50000;
// We loop until success. Even if one attempt fails, the retry will likely succeed.
while (true) {
std::vector<long long> current_set;
current_set.reserve(K);
// Generate distinct random numbers
while (current_set.size() < K) {
current_set.push_back(dist(rng));
}
std::sort(current_set.begin(), current_set.end());
current_set.erase(std::unique(current_set.begin(), current_set.end()), current_set.end());
// Refill if duplicates were removed (extremely rare)
while (current_set.size() < K) {
current_set.push_back(dist(rng));
std::sort(current_set.begin(), current_set.end());
current_set.erase(std::unique(current_set.begin(), current_set.end()), current_set.end());
}
if (collisions(current_set) > 0) {
// Collision found. Use Randomized Divide & Conquer to find the pair.
while (current_set.size() > 2) {
int sz = current_set.size();
int mid = sz / 2;
std::vector<long long> left_part(current_set.begin(), current_set.begin() + mid);
std::vector<long long> right_part(current_set.begin() + mid, current_set.end());
if (collisions(left_part) > 0) {
current_set = left_part;
} else if (collisions(right_part) > 0) {
current_set = right_part;
} else {
// Collision is between Left and Right. Shuffle and retry split.
std::shuffle(current_set.begin(), current_set.end(), rng);
}
}
// Calculate difference D = |u - v|. We know n | D.
long long diff = std::abs(current_set[0] - current_set[1]);
// Check divisors of D to find n.
std::vector<long long> cands = get_divisors(diff, SMALL_LIMIT);
for (long long cand : cands) {
if (check_n(cand)) {
return (int)cand;
}
}
}
// If no collision in this random set, loop continues and tries a new set.
// This prevents WA by ensuring we eventually find n.
}
return -1; // Should not be reached
}
