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1fd0050 | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 | #include <bits/stdc++.h>
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];
}
// count_leq with staircase walk, returns count and boundary positions
long long count_leq(long long x, vector<int>& pos) {
pos.assign(n + 1, 0);
long long cnt = 0;
int j = n;
for (int i = 1; i <= n; i++) {
while (j >= 1 && do_query(i, j) > x) j--;
pos[i] = j;
cnt += j;
if (j == 0) break;
}
return cnt;
}
long long count_lt(long long x, vector<int>& pos) {
pos.assign(n + 1, 0);
long long cnt = 0;
int j = n;
for (int i = 1; i <= n; i++) {
while (j >= 1 && do_query(i, j) >= x) j--;
pos[i] = j;
cnt += j;
if (j == 0) break;
}
return cnt;
}
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;
}
}
// Strategy: query a strategic row, then use binary search on values from that row
// Row sqrt(k) should contain values near the k-th smallest
// Actually: query row n/2. The median row has n values spanning roughly
// from a[n/2][1] to a[n/2][n]. Binary search within this row to find approximate pivot.
// Better strategy: query the anti-diagonal elements to get initial value range,
// then do value-based binary search with staircase count_leq
// Phase 1: Get bounds.
// Row ceil(k/n) column n is an upper bound for the k-th element.
int rBound = min(n, (int)((k + n - 1) / n));
long long hi = do_query(rBound, n);
long long lo = do_query(1, 1);
// Verify: count_leq(hi) >= k
vector<int> posHi;
long long cntHi = count_leq(hi, posHi);
if (cntHi < k) {
hi = do_query(n, n);
cntHi = count_leq(hi, posHi);
}
vector<int> posLo;
long long cntLo = 0;
// posLo is all zeros initially
posLo.assign(n + 1, 0);
// Phase 2: Narrow via value-based binary search
// Each count_leq costs O(n) but caches queries
// Use the row-merge approach: query a strategic row and binary search within it
// Better: use the "fractional cascading" idea.
// Query row at index ceil(sqrt(n)) and use values from that row as candidates.
// Actually, let's use a simpler approach:
// Binary search: find min value v such that count_leq(v) >= k
// But we can't binary search on long long values directly (range too large)
// Instead, generate candidate values by querying strategic positions
// Strategy: Query entire row at index ~sqrt(k/n) * something
// Actually the simplest approach that works:
// Use the quickselect approach but with the staircase from TOP (i ascending, j descending)
// and use value from the staircase boundary as next pivot
// Let me implement: quickselect with staircase walk, and use boundary values
// as pivot candidates for next iteration
for (int iter = 0; iter < 50; iter++) {
long long candTotal = cntHi - cntLo;
if (candTotal <= 0) return hi;
long long needSmall = k - cntLo;
long long needLarge = cntHi - k + 1;
// Count non-empty segments
int nonempty = 0;
for (int i = 1; i <= n; i++) {
if (posHi[i] > posLo[i]) nonempty++;
}
long long budget = 49500 - query_count;
if (min(needSmall, needLarge) + nonempty + 10 <= budget) {
// Enumerate
if (needSmall <= needLarge) {
priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq;
for (int i = 1; i <= n; i++) {
int L = posLo[i] + 1, R = posHi[i];
if (L >= 1 && L <= n && L <= R) pq.emplace(do_query(i, L), i, L);
}
long long result = 0;
for (long long t = 0; t < needSmall; t++) {
auto [v, r, c] = pq.top(); pq.pop();
result = v;
if (c + 1 <= posHi[r]) pq.emplace(do_query(r, c + 1), r, c + 1);
}
return result;
} else {
priority_queue<tuple<long long, int, int>> pq;
for (int i = 1; i <= n; i++) {
int L = posLo[i] + 1, R = posHi[i];
if (R >= 1 && R <= n && L <= R) pq.emplace(do_query(i, R), i, R);
}
long long result = 0;
for (long long t = 0; t < needLarge; t++) {
auto [v, r, c] = pq.top(); pq.pop();
result = v;
if (c - 1 >= posLo[r] + 1) pq.emplace(do_query(r, c - 1), r, c - 1);
}
return result;
}
}
if (budget < 2 * n + 200) return hi; // fallback
// Choose pivot: query a strategic position
// Use the median row's value at the fractional position
// The "median row" in terms of width-weighted is more complex
// Simple approach: pick the row with the most candidates and query at target_frac
// Build prefix sums for position-uniform sampling
vector<long long> pref(n + 1, 0);
for (int i = 1; i <= n; i++) {
int len = posHi[i] - posLo[i];
pref[i] = pref[i - 1] + max(0, len);
}
candTotal = pref[n];
if (candTotal <= 0) return hi;
// Sample multiple pivot candidates from the boundary values
// After a staircase walk for count_leq(v), the cells a[i][posHi[i]] are all <= hi
// and cells a[i][posHi[i]+1] (if exists) are > hi.
// The boundary values a[i][posHi[i]] and a[i][posLo[i]+1] are in the cache.
// Let's use the midpoint of each segment as pivot candidate.
// Actually, let me try: query the midpoint of each active segment in value terms.
// For row i, the values range from a[i, posLo[i]+1] to a[i, posHi[i]].
// The midpoint (in position) is (posLo[i]+1+posHi[i])/2.
// Sample from several rows, pick weighted quantile.
double target_q = (double)needSmall / (double)candTotal;
// Sample from ALL active rows at the target_q position
vector<pair<long long, long long>> vw; // (value, weight)
for (int i = 1; i <= n; i++) {
int L = posLo[i] + 1, R = posHi[i];
if (L > R || L < 1 || R > n) continue;
int width = R - L + 1;
int col = L + (int)(target_q * max(0, width - 1));
col = max(L, min(R, col));
vw.push_back({do_query(i, col), width});
}
// Weighted quantile
sort(vw.begin(), vw.end());
long long target_w = (long long)(target_q * candTotal);
long long cum = 0;
long long pivot = vw.back().first;
for (auto& [v, w] : vw) {
cum += w;
if (cum >= target_w) {
pivot = v;
break;
}
}
// Staircase walk for pivot
vector<int> posPivot;
long long cntPivot = count_leq(pivot, posPivot);
if (cntPivot >= k) {
if (cntPivot == cntHi && pivot >= hi) {
// Pivot didn't help - try count_lt
vector<int> posLess;
long long cntLess = count_lt(pivot, posLess);
if (cntLess < k) return pivot; // answer is pivot
hi = pivot - 1; // This is a hack for integer values
posHi = posLess;
cntHi = cntLess;
} else {
hi = pivot;
posHi = posPivot;
cntHi = cntPivot;
}
} else {
cntLo = cntPivot;
posLo = posPivot;
lo = pivot;
}
}
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=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;
}
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