| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 1066514 | myst6 | Ancient Machine 2 (JOI23_ancient2) | C++17 | 23 ms | 980 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "ancient2.h"
#include <bits/stdc++.h>
using namespace std;
const int MAX_BITS = 1000;
const int BOUND = 51;
#define bs bitset<MAX_BITS>
vector<bs> MIS(const vector<bs>& vectors) {
vector<bs> basis; // Store the basis vectors here
vector<bs> mis;
for (const auto& vec : vectors) {
bs v = vec; // Copy the vector
// Try to eliminate the current vector with the basis vectors
for (const auto& b : basis) {
if (v.none()) break; // If the vector is already zero, skip
if (v.test(b._Find_first())) { // If leading bit matches
v ^= b; // Subtract the basis vector (add in GF(2))
}
}
// If the vector is not zero after elimination, it's linearly independent
if (v.any()) {
basis.push_back(v);
mis.push_back(vec); // Add the original vector to the mis
}
}
return mis;
}
void gaussian_elimination(vector<bs> &A, bs &S, bs &b, int N) {
int row = 0;
for (int col = 0; col < N; ++col) {
// Find pivot row
int pivot = -1;
for (int i = row; i < N; ++i) {
if (A[i][col]) {
pivot = i;
break;
}
}
if (pivot == -1) continue; // No pivot found, move to the next column
// Swap pivot row with the current row
swap(A[row], A[pivot]);
// swap(b[row], b[pivot]);
int tmp = b[row];
b[row] = b[pivot];
b[pivot] = tmp;
// Eliminate below
for (int i = row + 1; i < N; ++i) {
if (A[i][col]) {
A[i] ^= A[row];
// b[i] ^= b[row];
if (b[row]) b[i] = !b[i];
}
}
++row;
}
// Backward substitution
S.reset();
for (int i = row - 1; i >= 0; --i) {
if (A[i].any()) {
int leading_one = A[i]._Find_first();
S[leading_one] = b[i];
for (int j = leading_one + 1; j < N; ++j) {
if (A[i][j]) {
// S[leading_one] ^= S[j];
if (S[j]) S[leading_one] = !S[leading_one];
}
}
}
}
}
// info[m][r] => index === r (mod m)
string solve(int N, vector<array<int,3>> &info) {
// Clear old stuff
vector<bs> A(N);
bs S, b;
// Populate the matrix A and vector b with parity information
int equation_index = 0;
for (auto [i, j, parity] : info) {
for (int k = i; k < N; k += j) {
A[equation_index][k] = 1;
}
// cerr << i << "," << j << "," << parity << "\n";
b[equation_index] = parity /* Your parity information for p_{i,j} */;
equation_index++;
if (equation_index == N) break;
}
// Perform Gaussian elimination
gaussian_elimination(A, S, b, N);
// Output the reconstructed binary string
string s = S.to_string();
reverse(s.begin(), s.end());
return s.substr(0, N);
}
vector<int> P = {
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61,
67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149,
151, 157, 163
};
string Solve1(int N) {
vector<array<int,3>> info;
function<void(int,int)> add = [&](int i, int j) -> void {
int m = 2 * j;
vector<int> a(m), b(m);
// nodes r=0..j-1 is even parity modulo r
// nodes r=j..2j-1 is odd parity modulo r
for (int r=0; r<j; r++) {
// both transition to next node in chain
a[r] = (r + 1) % j;
b[r] = (r + 1) % j;
a[j + r] = j + ((r + 1) % j);
b[j + r] = j + ((r + 1) % j);
}
// if one, transition to opposite node in chain
b[i] = j + ((i + 1) % j);
b[i + j] = (i + 1) % j;
info.push_back({i, j, Query(m, a, b) / j});
};
for (int j : P) {
if (info.size() == N) break;
if (j > BOUND) continue;
// while (j * p <= BOUND) j *= p;
for (int i=0; i<j-1; i++) {
if (info.size() == N) break;
add(i, j);
}
}
for (int p : P) {
if (info.size() == N) break;
for (int q : P) {
if (info.size() == N) break;
if (p >= q) continue;
int j = p * q;
if (j > BOUND) continue;
for (int i=0; i<j-q; i++) {
if (info.size() == N) break;
if (i > q) {
add(i, j);
}
}
}
}
// cout << info.size() << "\n";
assert(info.size() >= N);
// cout << solve(N, info) << "\n";
return solve(N, info);
}
string Solve(int N) {
vector<array<int,3>> info;
function<void(int,int)> add = [&](int i, int j) -> void {
int m = 2 * j;
vector<int> a(m), b(m);
// nodes r=0..j-1 is even parity modulo r
// nodes r=j..2j-1 is odd parity modulo r
for (int r=0; r<j; r++) {
// both transition to next node in chain
a[r] = (r + 1) % j;
b[r] = (r + 1) % j;
a[j + r] = j + ((r + 1) % j);
b[j + r] = j + ((r + 1) % j);
}
// if one, transition to opposite node in chain
b[i] = j + ((i + 1) % j);
b[i + j] = (i + 1) % j;
info.push_back({i, j, Query(m, a, b) / j});
};
vector<bs> all;
vector<array<int,2>> ij;
for (int j=1; j<=BOUND; j++) {
for (int i=0; i<j; i++) {
bs here;
for (int k=i; k<N; k+=j) {
here[k] = 1;
}
all.push_back(here);
ij.push_back({i, j});
}
}
int A = all.size();
vector<bs> mis = MIS(all);
int M = mis.size();
// cout << M << "\n";
// for (bs &HELP : mis) {
// cout << HELP.to_string() << "\n";
// }
// random_shuffle(mis.begin(), mis.end());
for (bs &HELP : mis) {
if (info.size() == N) break;
bool found = false;
for (int i=0; i<A; i++) {
if (HELP == all[i]) {
add(ij[i][0], ij[i][1]);
found = true;
}
}
assert(found);
}
// cout << solve(N, info) << "\n";
return solve(N, info);
}
Compilation message (stderr)
| # | Verdict | Execution time | Memory | Grader output |
|---|---|---|---|---|
| Fetching results... | ||||