| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 572800 | vovamr | Longest beautiful sequence (IZhO17_subsequence) | C++17 | 0 ms | 0 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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("-O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,abm,mxx,popcnt,tune=native")
#define fi first
#define se second
#define ll long long
#define ld long double
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define mpp make_pair
#define ve vector
using namespace std;
using namespace __gnu_pbds;
template<class T> using oset = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
const ll inf = 1e18; const int iinf = 1e9;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T> inline bool chmin(T& a, T b) { return (a > b ? a = b, 1 : 0); }
template <typename T> inline bool chmax(T& a, T b) { return (a < b ? a = b, 1 : 0); }
#define cnt(x) __builtin_popcount((x))
const int N = 10;
const int MX = 1e6 + 10;
int ek[MX];
int dp[1 << N][1 << N][N << 1];
inline void solve() {
int n;
cin >> n;
ve<int> a(n), b(n);
for (auto &i : a) cin >> i;
for (auto &i : b) cin >> i;
for (int i = 0; i < n; ++i) {
int x = a[i];
int les10 = x & ((1 << 10) - 1);
int big10 = x >> 10;
chmax(ek[i], 1);
for (int p = 0; p < (1 << 10); ++p) {
int hui = cnt(p & x);
if (hui > b[i] || b[i] - hui > 10) continue;
chmax(ek[i], dp[p][big10][b[i] - hui] + 1);
}
for (int p = 0; p < (1 << 10); ++p) {
int hui = cnt(p & big10);
chmax(dp[les10][p][hui], ek[i]);
}
}
ve<int> ans;
int mx = max_element(ek, ek + n) - ek;
int pos = mx, i = mx - 1;
ans.pb(pos);
while (~i) {
if (cnt(a[i] & a[pos]) == b[pos] && ek[pos] == ek[i] + 1) {
ans.pb(i);
pos = i;
--i;
}
else --i;
}
reverse(all(ans));
cout << sz(ans) << '\n';
for (auto &i : ans) cout << i + 1 << " ";
}
signed main() {
ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int q = 1; // cin >> q;
while (q--) solve();
cerr << fixed << setprecision(3) << "Time execution: " << (double)clock() / CLOCKS_PER_SEC << endl;
}