| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 722254 | urosk | Longest beautiful sequence (IZhO17_subsequence) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#define here cerr<<"===========================================\n"
#define dbg(x) cerr<<#x<<": "<<x<<endl;
#include "bits/stdc++.h"
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
#define ld double
#define ll long long
#define llinf 100000000000000000LL // 10^17
#define pb push_back
#define popb pop_back
#define fi first
#define sc second
#define endl '\n'
#define pll pair<ll,ll>
#define pld pair<ld,ld>
#define all(a) a.begin(),a.end()
#define ceri(a,l,r) {cerr<<#a<<": ";for(ll i_ = l;i_<=r;i_++) cerr<<a[i_]<< " ";cerr<<endl;}
#define cer(a) {cerr<<#a<<": ";for(ll x_ : a) cerr<<x_<< " ";cerr<<endl;}
#define daj_mi_malo_vremena ios_base::sync_with_stdio(false);cerr.tie(0);cout.tie(0);cin.tie(0);
using namespace std;
//using namespace __gnu_pbds;
/*
ll add(ll x,ll y){
x+=y;
if(x<0){
x%=mod;
x+=mod;
}else{
if(x>=mod) x%=mod;
}
return x;
}
ll mul(ll a,ll b){
ll ans = (a*b)%mod;
if(ans<0) ans+=mod;
return ans;
}
ll po(ll x,ll y){
if(y==0) return 1LL;
ll ans = po(x,y/2);
ans = mul(ans,ans);
if(y&1) ans = mul(ans,x);
return ans;
}
ll inv(ll x){return po(x,mod-2);}
*/
/*
typedef tree<int,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef tree<int,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll rnd(ll l,ll r){
return uniform_int_distribution<ll>(l,r)(rng);
}
*/
#define maxn 100005
#define maxx 1025
ll n;
ll a[maxn];
ll k[maxn];
pll dp[maxx][maxx][11];
ll mask = (1<<10)-1;
ll bc[maxx][maxx];
ll last[maxn];
ll ans = 1;
ll ansid = 0;
void tc(){
for(ll i = 0;i<=mask;i++) for(ll j = 0;j<=mask;j++) bc[i][j] = __builtin_popcount(i&j);
cin >> n;
for(ll i = 1;i<=n;i++) cin >> a[i];
for(ll i = 1;i<=n;i++) cin >> k[i];
for(ll i = 1;i<=n;i++){
ll x = a[i];
ll rx = (x>>10)
ll lx = x%(1<<10);
ll cur = 1;
ll curid = i;
for(ll l = 0;l<=mask;l++){
if(k[i]-bc[l][lx]>0||k[i]-bc[l][lx]>10) continue;
if(dp[l][rx][k[i]-bc[l][lx]].fi+1>cur){
cur = dp[l][rx][k[i]-bc[l][lx]].fi+1;
last[i] = dp[l][rx][k[i]-bc[l][lx]].sc;
}
}
for(ll r = 0;r<=mask;r++){
if(dp[lx][r][bc[rx][r]].fi<cur){
dp[lx][r][bc[rx][r]].fi = cur;
dp[lx][r][bc[rx][r]].sc = i;
}
}
if(cur>ans){
ansid = i;
ans = cur;
}
}
cout<<ans<<endl;
vector<ll> v;
while(ansid){
v.pb(ansid);
ansid = last[ansid];
}
reverse(all(v));
for(ll x : v) cout<<x<< " ";
cout<<endl;
}
int main(){
daj_mi_malo_vremena
int t; t = 1;
while(t--){
tc();
}
return 0;
}
/**
4
1 2 3 4
10 0 1 0
2
8 9
20 0
5
5 3 5 3 5
10 1 20 1 20
**/