File size: 13,376 Bytes
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 | #include <bits/stdc++.h>
using namespace std;
static const long long P = 998244353;
// Program structure: (n-1) POP instructions + 1 HALT at position n-1
// POP instruction i: POP (i+1) GOTO (i+1) PUSH push_val[i] GOTO push_goto[i]
// pop matches value (i+1), goes to next instruction
// push_val[i] in {1, ..., maxVal}, push_goto[i] in {0, ..., n-1}
// HALT at n-1: HALT PUSH halt_push GOTO halt_goto
//
// Stack values are 0 (empty), 1..maxVal
// State space: n * (maxVal + 1)
// Returns cost mod P, or -1 if cycle
long long evaluate(int n, const int* push_val, const int* push_goto, int halt_push, int halt_goto, int maxVal) {
int nstates = n * (maxVal + 1);
vector<int> dest(nstates, -2);
vector<long long> cost(nstates, 0);
vector<int8_t> state(nstates, 0); // 0=unvisited, 1=in-progress, 2=done
// Stack-based DFS to avoid recursion overhead
struct Frame {
int i, x;
int phase; // 0: initial, 1: after first sub-call, 2: after second sub-call
int j; // intermediate instruction
long long u; // intermediate cost
};
vector<Frame> stack;
stack.reserve(nstates);
auto idx = [&](int i, int x) { return i * (maxVal + 1) + x; };
auto push_frame = [&](int i, int x) -> bool {
int id = idx(i, x);
if (state[id] == 2) return true; // already done
if (state[id] == 1) return false; // cycle!
state[id] = 1;
stack.push_back({i, x, 0, 0, 0});
return true;
};
// We need to evaluate (0, 0)
if (!push_frame(0, 0)) return -1;
while (!stack.empty()) {
Frame& f = stack.back();
int id = idx(f.i, f.x);
if (f.i < n - 1) { // POP
int pop_v = f.i + 1;
if (f.x == pop_v) {
dest[id] = f.i + 1; // pop_goto = next instruction
cost[id] = 1;
state[id] = 2;
stack.pop_back();
continue;
}
// Not matched: push push_val[i], goto push_goto[i]
if (f.phase == 0) {
// Need solve(push_goto[i], push_val[i])
int sub_i = push_goto[f.i];
int sub_x = push_val[f.i];
int sub_id = idx(sub_i, sub_x);
if (state[sub_id] == 2) {
f.j = dest[sub_id];
f.u = cost[sub_id];
f.phase = 1;
// fall through to phase 1
} else if (state[sub_id] == 1) {
return -1; // cycle
} else {
state[sub_id] = 1;
f.phase = 1;
stack.push_back({sub_i, sub_x, 0, 0, 0});
continue;
}
}
if (f.phase == 1) {
int sub_i = push_goto[f.i];
int sub_x = push_val[f.i];
int sub_id = idx(sub_i, sub_x);
if (state[sub_id] != 2) return -1; // shouldn't happen
f.j = dest[sub_id];
f.u = cost[sub_id];
// Now need solve(j, x)
if (f.j == -1) {
// j=-1 means halted already, can't process x... this shouldn't happen
// unless the sub-program halted, which means x is still on stack
// Actually if j=-1, the program halted inside the sub-call.
// This means the whole program halted, cost = u + 1.
// But the checker would give dest=-1 meaning halt.
// Actually looking at the checker: dest=-1 only from HALT with x=0.
// After that, solve(j, x) with j=-1 would be out of bounds.
// This means if sub-call halts (returns dest=-1), then solving (dest=-1, x) is invalid.
// Wait, in the checker, dest is used as an instruction index, and -1 means halt.
// But then solve(j, x) with j=-1 would be called... but there's no instruction -1.
// Let me re-read the checker.
// Actually solve returns (k, v) where k is the next instruction or -1 (halt).
// Then the caller does solve(k, x). If k=-1, that's solve(-1, x) which is out of bounds.
// This means the program MUST be designed so that after the sub-call,
// the returned instruction is valid (not -1).
// So if a HALT with empty stack is reached during a sub-call,
// the returned dest is -1, and then solve(-1, x) would crash.
// The checker uses dp[-1][x] which is out of bounds...
// Actually in the checker, vectors are indexed from 0 to n-1.
// dp[-1] would be undefined behavior. But practically, this means
// programs where a HALT is reached with empty stack during a sub-call
// would cause UB in the checker, so they're invalid.
// CONCLUSION: in valid programs, HALT with empty stack should only be reached
// at the very end (solve(0,0)'s final halt).
return -1; // invalid program
}
int sub_id2 = idx(f.j, f.x);
if (state[sub_id2] == 2) {
dest[id] = dest[sub_id2];
cost[id] = (f.u + cost[sub_id2] + 1) % P;
state[id] = 2;
stack.pop_back();
continue;
} else if (state[sub_id2] == 1) {
return -1; // cycle
} else {
state[sub_id2] = 1;
f.phase = 2;
stack.push_back({f.j, f.x, 0, 0, 0});
continue;
}
}
if (f.phase == 2) {
int sub_id2 = idx(f.j, f.x);
if (state[sub_id2] != 2) return -1;
dest[id] = dest[sub_id2];
cost[id] = (f.u + cost[sub_id2] + 1) % P;
state[id] = 2;
stack.pop_back();
continue;
}
} else { // HALT (i = n-1)
if (f.x == 0) {
dest[id] = -1;
cost[id] = 1;
state[id] = 2;
stack.pop_back();
continue;
}
// push halt_push, goto halt_goto
if (f.phase == 0) {
int sub_i = halt_goto;
int sub_x = halt_push;
int sub_id = idx(sub_i, sub_x);
if (state[sub_id] == 2) {
f.j = dest[sub_id];
f.u = cost[sub_id];
f.phase = 1;
} else if (state[sub_id] == 1) {
return -1;
} else {
state[sub_id] = 1;
f.phase = 1;
stack.push_back({sub_i, sub_x, 0, 0, 0});
continue;
}
}
if (f.phase == 1) {
int sub_i = halt_goto;
int sub_x = halt_push;
int sub_id = idx(sub_i, sub_x);
if (state[sub_id] != 2) return -1;
f.j = dest[sub_id];
f.u = cost[sub_id];
if (f.j == -1) return -1; // invalid
int sub_id2 = idx(f.j, f.x);
if (state[sub_id2] == 2) {
dest[id] = dest[sub_id2];
cost[id] = (f.u + cost[sub_id2] + 1) % P;
state[id] = 2;
stack.pop_back();
continue;
} else if (state[sub_id2] == 1) {
return -1;
} else {
state[sub_id2] = 1;
f.phase = 2;
stack.push_back({f.j, f.x, 0, 0, 0});
continue;
}
}
if (f.phase == 2) {
int sub_id2 = idx(f.j, f.x);
if (state[sub_id2] != 2) return -1;
dest[id] = dest[sub_id2];
cost[id] = (f.u + cost[sub_id2] + 1) % P;
state[id] = 2;
stack.pop_back();
continue;
}
}
}
int id0 = idx(0, 0);
if (state[id0] != 2) return -1;
return cost[id0];
}
int main() {
long long targets[2] = {150994941LL, 150994939LL};
// Search over programs with n instructions
// Using push values 1..V where V is small
// Each POP i has: pop_val=i+1, pop_goto=i+1, push_val in 1..V, push_goto in 0..n-1
// HALT has: push_val in 1..V, push_goto in 0..n-2
for (int n = 5; n <= 20; n++) {
int d = n - 1;
int V = min(d, 5); // max push value
cerr << "n=" << n << " V=" << V << endl;
mt19937 rng(42 + n * 999979);
bool found[2] = {false, false};
auto startTime = chrono::steady_clock::now();
int timeLimit = 10000; // 10s per n
int push_val[25], push_goto[25];
int halt_push, halt_goto;
int best_push_val[2][25], best_push_goto[2][25], best_halt_push[2], best_halt_goto[2];
long long attempts = 0;
while (!(found[0] && found[1])) {
auto now = chrono::steady_clock::now();
auto ms = chrono::duration_cast<chrono::milliseconds>(now - startTime).count();
if (ms > timeLimit) break;
// Random initialization
for (int i = 0; i < d; i++) {
push_val[i] = 1 + rng() % V;
push_goto[i] = rng() % n;
}
halt_push = 1 + rng() % V;
halt_goto = rng() % d; // point to a POP instruction
long long cost = evaluate(n, push_val, push_goto, halt_push, halt_goto, V);
if (cost < 0) continue;
// Hill climbing
for (int iter = 0; iter < 20000; iter++) {
for (int ti = 0; ti < 2; ti++) {
if (!found[ti] && cost == targets[ti]) {
found[ti] = true;
memcpy(best_push_val[ti], push_val, sizeof(int) * d);
memcpy(best_push_goto[ti], push_goto, sizeof(int) * d);
best_halt_push[ti] = halt_push;
best_halt_goto[ti] = halt_goto;
cerr << "FOUND n=" << n << " target" << (ti+1) << endl;
}
}
if (found[0] && found[1]) break;
// Mutate
int save_val, save_goto, save_idx;
int mut_type = rng() % 3; // 0: push_val, 1: push_goto, 2: halt params
if (mut_type == 0) {
save_idx = rng() % d;
save_val = push_val[save_idx];
push_val[save_idx] = 1 + rng() % V;
} else if (mut_type == 1) {
save_idx = rng() % d;
save_goto = push_goto[save_idx];
push_goto[save_idx] = rng() % n;
} else {
save_val = halt_push;
save_goto = halt_goto;
halt_push = 1 + rng() % V;
halt_goto = rng() % d;
}
long long new_cost = evaluate(n, push_val, push_goto, halt_push, halt_goto, V);
bool accept = false;
if (new_cost >= 0) {
// Check if closer to any unfound target
long long bestOld = P, bestNew = P;
for (int ti = 0; ti < 2; ti++) {
if (!found[ti]) {
bestOld = min(bestOld, min((cost - targets[ti] + P) % P, (targets[ti] - cost + P) % P));
bestNew = min(bestNew, min((new_cost - targets[ti] + P) % P, (targets[ti] - new_cost + P) % P));
}
}
if (bestNew <= bestOld) accept = true;
}
if (accept) {
cost = new_cost;
} else {
// Revert
if (mut_type == 0) push_val[save_idx] = save_val;
else if (mut_type == 1) push_goto[save_idx] = save_goto;
else { halt_push = save_val; halt_goto = save_goto; }
}
attempts++;
}
}
auto ms = chrono::duration_cast<chrono::milliseconds>(chrono::steady_clock::now() - startTime).count();
cerr << " attempts=" << attempts << " time=" << ms << "ms found=" << found[0] << "," << found[1] << endl;
if (found[0] && found[1]) {
for (int ti = 0; ti < 2; ti++) {
cout << "Target " << (ti+1) << " (n=" << n << "):" << endl;
cout << n << endl;
for (int i = 0; i < d; i++) {
cout << "POP " << (i+1) << " GOTO " << (i+2)
<< " PUSH " << best_push_val[ti][i] << " GOTO " << (best_push_goto[ti][i]+1) << endl;
}
cout << "HALT PUSH " << best_halt_push[ti] << " GOTO " << (best_halt_goto[ti]+1) << endl;
}
break;
}
}
return 0;
}
|