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| using namespace std; | |
| struct TestCase { | |
| int n; | |
| long long k; | |
| vector<vector<long long>> A; | |
| long long answer; | |
| }; | |
| mt19937_64 rng_gen(42); | |
| TestCase gen_matrix(int n, long long k, function<long long(int,int)> valfn) { | |
| TestCase tc; | |
| tc.n = n; tc.k = k; | |
| tc.A.assign(n+1, vector<long long>(n+1, 0)); | |
| vector<long long> all; | |
| for (int i = 1; i <= n; i++) | |
| for (int j = 1; j <= n; j++) { | |
| tc.A[i][j] = valfn(i, j); | |
| all.push_back(tc.A[i][j]); | |
| } | |
| sort(all.begin(), all.end()); | |
| tc.answer = all[k-1]; | |
| return tc; | |
| } | |
| TestCase gen_multiplicative(int n, long long k) { | |
| return gen_matrix(n, k, [](int i, int j) -> long long { return (long long)i * j; }); | |
| } | |
| TestCase gen_shifted(int n, long long k) { | |
| return gen_matrix(n, k, [n](int i, int j) -> long long { return (long long)(i + n) * (j + n); }); | |
| } | |
| TestCase gen_additive(int n, long long k) { | |
| return gen_matrix(n, k, [](int i, int j) -> long long { return i + j; }); | |
| } | |
| TestCase gen_random_sorted(int n, long long k) { | |
| TestCase tc; | |
| tc.n = n; tc.k = k; | |
| tc.A.assign(n+1, vector<long long>(n+1, 0)); | |
| long long C = 1000000, D = 1000; | |
| for (int i = 1; i <= n; i++) | |
| for (int j = 1; j <= n; j++) | |
| tc.A[i][j] = (long long)i * C + (long long)j * D + (rng_gen() % 500); | |
| for (int i = 1; i <= n; i++) | |
| for (int j = 2; j <= n; j++) | |
| tc.A[i][j] = max(tc.A[i][j], tc.A[i][j-1]); | |
| for (int j = 1; j <= n; j++) | |
| for (int i = 2; i <= n; i++) | |
| tc.A[i][j] = max(tc.A[i][j], tc.A[i-1][j]); | |
| vector<long long> all; | |
| for (int i = 1; i <= n; i++) | |
| for (int j = 1; j <= n; j++) | |
| all.push_back(tc.A[i][j]); | |
| sort(all.begin(), all.end()); | |
| tc.answer = all[k-1]; | |
| return tc; | |
| } | |
| struct Solver { | |
| const TestCase& tc; | |
| int query_count; | |
| vector<long long> memo; | |
| int n; | |
| Solver(const TestCase& t) : tc(t), query_count(0), n(t.n) { | |
| memo.assign(2002 * 2002, -1); | |
| } | |
| long long do_query(int r, int c) { | |
| int key = r * 2001 + c; | |
| if (memo[key] != -1) return memo[key]; | |
| query_count++; | |
| memo[key] = tc.A[r][c]; | |
| return memo[key]; | |
| } | |
| long long solve() { | |
| long long k = tc.k; | |
| long long N2 = (long long)n * n; | |
| if (n == 1) return do_query(1, 1); | |
| if (k == 1) return do_query(1, 1); | |
| if (k == N2) return do_query(n, n); | |
| long long heap_k = min(k, N2 - k + 1); | |
| if (heap_k + n <= 24000) { | |
| if (k <= N2 - k + 1) { | |
| priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq; | |
| for (int i = 1; i <= n; i++) pq.emplace(do_query(i, 1), i, 1); | |
| long long result = -1; | |
| for (long long t = 0; t < k; t++) { | |
| auto [v, r, c] = pq.top(); pq.pop(); | |
| result = v; | |
| if (c + 1 <= n) pq.emplace(do_query(r, c + 1), r, c + 1); | |
| } | |
| return result; | |
| } else { | |
| long long kk = N2 - k + 1; | |
| priority_queue<tuple<long long, int, int>> pq; | |
| for (int i = 1; i <= n; i++) pq.emplace(do_query(i, n), i, n); | |
| long long result = -1; | |
| for (long long t = 0; t < kk; t++) { | |
| auto [v, r, c] = pq.top(); pq.pop(); | |
| result = v; | |
| if (c - 1 >= 1) pq.emplace(do_query(r, c - 1), r, c - 1); | |
| } | |
| return result; | |
| } | |
| } | |
| // Quickselect with improved pivot selection | |
| vector<int> L(n + 1, 1), R(n + 1, n); | |
| long long k_rem = k; | |
| mt19937_64 rng(12345); | |
| for (int iter = 0; iter < 100; iter++) { | |
| vector<int> active; | |
| long long total_cand = 0; | |
| for (int i = 1; i <= n; i++) { | |
| if (L[i] <= R[i]) { | |
| active.push_back(i); | |
| total_cand += R[i] - L[i] + 1; | |
| } | |
| } | |
| int na = active.size(); | |
| if (total_cand == 0) break; | |
| if (total_cand == 1) { | |
| for (int i : active) return do_query(i, L[i]); | |
| break; | |
| } | |
| long long budget = 49500 - query_count; | |
| if (k_rem + na <= budget) { | |
| priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq; | |
| for (int i : active) pq.emplace(do_query(i, L[i]), i, L[i]); | |
| for (long long t = 1; t < k_rem; t++) { | |
| auto [v, r, c] = pq.top(); pq.pop(); | |
| if (c + 1 <= R[r]) pq.emplace(do_query(r, c + 1), r, c + 1); | |
| } | |
| return get<0>(pq.top()); | |
| } | |
| long long rev_k = total_cand - k_rem + 1; | |
| if (rev_k + na <= budget) { | |
| priority_queue<tuple<long long, int, int>> pq; | |
| for (int i : active) pq.emplace(do_query(i, R[i]), i, R[i]); | |
| for (long long t = 1; t < rev_k; t++) { | |
| auto [v, r, c] = pq.top(); pq.pop(); | |
| if (c - 1 >= L[r]) pq.emplace(do_query(r, c - 1), r, c - 1); | |
| } | |
| return get<0>(pq.top()); | |
| } | |
| // Improved pivot selection: uniform random sampling from candidate area | |
| // Build prefix sums of widths | |
| vector<long long> pref(na + 1, 0); | |
| for (int idx = 0; idx < na; idx++) { | |
| int i = active[idx]; | |
| pref[idx + 1] = pref[idx] + (R[i] - L[i] + 1); | |
| } | |
| int num_samples = min(15, na); | |
| vector<long long> samp; | |
| for (int s = 0; s < num_samples; s++) { | |
| long long pick = 1 + rng() % total_cand; | |
| int idx = (int)(lower_bound(pref.begin() + 1, pref.end(), pick) - pref.begin()) - 1; | |
| idx = max(0, min(na - 1, idx)); | |
| int i = active[idx]; | |
| int width = R[i] - L[i] + 1; | |
| int col = L[i] + rng() % width; | |
| samp.push_back(do_query(i, col)); | |
| } | |
| sort(samp.begin(), samp.end()); | |
| // Pick at quantile matching k_rem/total_cand | |
| double q = (double)(k_rem - 0.5) / total_cand; | |
| int pidx = (int)(q * (double)(samp.size() - 1)); | |
| pidx = max(0, min((int)samp.size() - 1, pidx)); | |
| long long pivot = samp[pidx]; | |
| // Staircase walk (bottom-to-top) | |
| vector<int> p_le(n + 1, 0); | |
| { | |
| int j = 0; | |
| for (int idx = na - 1; idx >= 0; idx--) { | |
| int i = active[idx]; | |
| j = max(j, L[i]); | |
| while (j <= R[i] && do_query(i, j) <= pivot) j++; | |
| p_le[i] = j - 1; | |
| } | |
| } | |
| long long cle = 0; | |
| for (int i : active) { | |
| int rl = min(p_le[i], R[i]); | |
| if (rl >= L[i]) cle += rl - L[i] + 1; | |
| } | |
| if (cle >= k_rem) { | |
| for (int i : active) R[i] = min(R[i], p_le[i]); | |
| } else { | |
| k_rem -= cle; | |
| for (int i : active) L[i] = max(L[i], p_le[i] + 1); | |
| } | |
| } | |
| return -1; | |
| } | |
| }; | |
| int main() { | |
| struct TestDef { | |
| string name; | |
| function<TestCase()> gen; | |
| }; | |
| vector<TestDef> tests; | |
| tests.push_back({"additive n=100 k=5000", []{ return gen_additive(100, 5000); }}); | |
| tests.push_back({"multiplicative n=100 k=5000", []{ return gen_multiplicative(100, 5000); }}); | |
| tests.push_back({"additive n=500 k=125000", []{ return gen_additive(500, 125000); }}); | |
| tests.push_back({"multiplicative n=500 k=125000", []{ return gen_multiplicative(500, 125000); }}); | |
| tests.push_back({"random n=500 k=125000", []{ return gen_random_sorted(500, 125000); }}); | |
| tests.push_back({"additive n=2000 k=2000000", []{ return gen_additive(2000, 2000000); }}); | |
| tests.push_back({"multiplicative n=2000 k=2000000", []{ return gen_multiplicative(2000, 2000000); }}); | |
| tests.push_back({"shifted n=2000 k=2000000", []{ return gen_shifted(2000, 2000000); }}); | |
| tests.push_back({"random n=2000 k=2000000", []{ return gen_random_sorted(2000, 2000000); }}); | |
| tests.push_back({"multiplicative n=2000 k=1", []{ return gen_multiplicative(2000, 1); }}); | |
| tests.push_back({"multiplicative n=2000 k=4000000", []{ return gen_multiplicative(2000, 4000000); }}); | |
| tests.push_back({"multiplicative n=2000 k=100000", []{ return gen_multiplicative(2000, 100000); }}); | |
| tests.push_back({"multiplicative n=2000 k=3900000", []{ return gen_multiplicative(2000, 3900000); }}); | |
| for (auto& t : tests) { | |
| auto tc = t.gen(); | |
| Solver s(tc); | |
| long long result = s.solve(); | |
| bool correct = (result == tc.answer); | |
| double score; | |
| int used = s.query_count; | |
| if (!correct) score = 0.0; | |
| else if (used <= tc.n) score = 1.0; | |
| else if (used >= 50000) score = 0.0; | |
| else score = (50000.0 - used) / (50000.0 - tc.n); | |
| printf("%-45s n=%4d k=%8lld queries=%6d correct=%s score=%.4f\n", | |
| t.name.c_str(), tc.n, tc.k, used, correct ? "YES" : "NO", score); | |
| } | |
| return 0; | |
| } | |