제출 #1311809

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1311809		Boycl07	Hack (APIO25_hack)	C++20	46 / 100	508 ms	2456 KiB

hack

#include <vector>
#include <numeric>
#include <algorithm>
#include <random>
#include <cmath>
#include <set>

// Declare the interactor provided by the judge
long long collisions(std::vector<long long> x);

namespace {
    // Helper to find all divisors of val that are strictly greater than limit
    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 correct (Cost: 2)
    bool check_n(long long n) {
        if (n <= 0) return false;
        std::vector<long long> q = {1, n + 1};
        return collisions(q) > 0;
    }

    // Isolate a colliding pair from a set known to have collisions
    // Returns the hidden N
    int solve_collision(std::vector<long long>& set_with_collision, long long limit_n, std::mt19937_64& rng) {
        std::vector<long long> current_set = set_with_collision;
        
        while (true) {
            // Optimization: If set is small, brute force all pairs to save queries
            if (current_set.size() <= 20) { 
                for (size_t i = 0; i < current_set.size(); ++i) {
                    for (size_t j = i + 1; j < current_set.size(); ++j) {
                        long long diff = std::abs(current_set[i] - current_set[j]);
                        if (diff == 0) continue;
                        
                        // We check divisors. If limit_n is set, we can skip very small divisors 
                        // if we are sure they were ruled out, but checking > 1 is always safe.
                        std::vector<long long> cands = get_divisors(diff, 1);
                        for (long long cand : cands) {
                            if (check_n(cand)) return (int)cand;
                        }
                    }
                }
                // If we reach here, the collision was 'lost' (unlikely) or logic error.
                // Break to allow caller to handle or retry (though for deterministic Phase 1 this shouldn't happen)
                break; 
            }

            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 {
                // Crossing collision. Shuffle and retry split.
                std::shuffle(current_set.begin(), current_set.end(), rng);
            }
        }
        return -1;
    }
}

int hack() {
    std::mt19937_64 rng(1337); 

    // --- Phase 1: Deterministic Check for Subtasks 1 & 2 (N <= 1,000,000) ---
    // Construct Difference Set S = {1..B} U {B, 2B..}
    // With B = 1000, we cover differences up to 1,000,000.
    // Size ~ 2000. Cost to query: 2000. Cost to isolate: ~4000.
    // Total Phase 1 Cost: ~6000 (Very Cheap).
    {
        const long long B = 1000; 
        std::vector<long long> diff_set;
        diff_set.reserve(2 * B);
        for (int i = 1; i <= B; ++i) diff_set.push_back(i);
        for (int i = 1; i <= B; ++i) diff_set.push_back(i * B);
        
        std::sort(diff_set.begin(), diff_set.end());
        diff_set.erase(std::unique(diff_set.begin(), diff_set.end()), diff_set.end());

        if (collisions(diff_set) > 0) {
            // Collision found! N <= 1,000,000 (roughly).
            // Do NOT use Binary Search (too expensive). Use D&C Isolation.
            int res = solve_collision(diff_set, 1, rng);
            if (res != -1) return res;
        }
    }

    // --- Phase 2: Randomized Birthday Attack for Subtask 3 (N <= 10^9) ---
    // Remaining Budget: ~104,000.
    // We use K = 26,000.
    // Cost per attempt: ~26k (query) + ~26k (iso) = ~52k.
    // We can afford ~2 clean attempts within budget.
    // For N=10^9, P(collision) with 26k is ~50%.
    // Even if we go over budget slightly on retries, we get high partial score.
    
    int K = 26000;
    std::uniform_int_distribution<long long> dist(1, 20000000000LL);

    while (true) {
        std::vector<long long> current_set;
        current_set.reserve(K);
        std::set<long long> distinct_check;
        
        while (current_set.size() < K) {
            long long val = dist(rng);
            if (distinct_check.insert(val).second) {
                current_set.push_back(val);
            }
        }

        if (collisions(current_set) > 0) {
            int res = solve_collision(current_set, 1, rng);
            if (res != -1) return res;
        }
    }
    
    return -1;
}

• Subtask 0: 0 / 0
• Subtask 1: 8 / 8
• Subtask 2: 17 / 17
• Subtask 3: 20.9909 / 75