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제출 #58099

#제출 시각아이디문제언어결과실행 시간메모리
58099BenqLongest beautiful sequence (IZhO17_subsequence)C++14
100 / 100
5516 ms132020 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/tree_policy.hpp> #include <ext/pb_ds/assoc_container.hpp> using namespace std; using namespace __gnu_pbds; typedef long long ll; typedef long double ld; typedef complex<ld> cd; typedef pair<int, int> pi; typedef pair<ll,ll> pl; typedef pair<ld,ld> pd; typedef vector<int> vi; typedef vector<ld> vd; typedef vector<ll> vl; typedef vector<pi> vpi; typedef vector<pl> vpl; typedef vector<cd> vcd; template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; #define FOR(i, a, b) for (int i=a; i<(b); i++) #define F0R(i, a) for (int i=0; i<(a); i++) #define FORd(i,a,b) for (int i = (b)-1; i >= a; i--) #define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--) #define sz(x) (int)(x).size() #define mp make_pair #define pb push_back #define f first #define s second #define lb lower_bound #define ub upper_bound #define all(x) x.begin(), x.end() const int MOD = 1000000007; const ll INF = 1e18; const int MX = 100001; int n, a[MX], k[MX], pre[MX], bp[1024]; pi ans; pi bes[1<<10][1<<10][11]; pi query(int x, int y) { pi ret = {0,0}; int X = x&((1<<10)-1); F0R(i,1<<10) { int t = y-bp[(x>>10)&i]; if (t >= 0 && t <= 10) ret = max(ret,bes[i][X][t]); } return ret; } void tri(pi& x, pi y) { x = max(x,y); } void ins(int x, pi y) { F0R(i,1<<10) tri(bes[x>>10][i][bp[i&x]],y); } int main() { ios_base::sync_with_stdio(0); cin.tie(0); F0R(i,1024) bp[i] = __builtin_popcount(i); cin >> n; FOR(i,1,n+1) cin >> a[i]; FOR(i,1,n+1) cin >> k[i]; FOR(i,1,n+1) { pi x = query(a[i],k[i]); pre[i] = x.s; x.f ++; x.s = i; ans = max(ans,x); ins(a[i],x); } cout << ans.f << "\n"; vi v = {ans.s}; while (v.back()) v.pb(pre[v.back()]); v.pop_back(); reverse(all(v)); for (int i: v) cout << i << " "; } /* Look for: * the exact constraints (multiple sets are too slow for n=10^6 :( ) * special cases (n=1?) * overflow (ll vs int?) * array bounds */
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