이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#pragma GCC optimize("unroll-loops,Ofast,O3")
#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define spc << " " <<
#define endl "\n"
#define all(x) x.begin(), x.end()
//#define int long long
#define ii pair<long long,int>
#define vi vector<int>
#define vii vector<ii>
#define st first
#define nd second
#define mid (l+r)/2
#define inf 1e15
#define MOD 1000000007
#define MX 1505
using namespace std;
ii dp[1<<10][1<<10][10];
void solve(){
int n; cin >> n;
int a[n], k[n];
for(int i=0; i<n; i++) cin >> a[i];
for(int i=0; i<n; i++) cin >> k[i];
for(int i=0; i<1<<10; i++) for(int j=0; j<1<<10; j++) for(int k=0; k<10; k++) dp[i][j][k]={0, -1};
int ans=0, bestie=0;
int pre[n];
memset(pre, -1, sizeof(pre));
for(int i=0; i<n; i++){
ii cur={a[i]>>10, a[i]&((1<<10)-1)};
int res=1;
for(int j=0; j<1<<10; j++){
if(k[i]-__builtin_popcount(cur.st&j)>=10) continue;
if(dp[j][cur.nd][k[i]-__builtin_popcount(cur.st&j)].st+1>res){
res = dp[j][cur.nd][k[i]-__builtin_popcount(cur.st&j)].st+1;
pre[i] = dp[j][cur.nd][k[i]-__builtin_popcount(cur.st&j)].nd;
}
}
if(res>ans){
ans=res;
bestie=i;
}
for(int j=0; j<1<<10; j++){
if(res > dp[cur.st][j][__builtin_popcount(cur.nd&j)].st){
dp[cur.st][j][__builtin_popcount(cur.nd&j)].st = res;
dp[cur.st][j][__builtin_popcount(cur.nd&j)].nd = i;
}
}
}
cout << ans << endl;
vi wow;
while(bestie>-1){
wow.pb(bestie);
bestie=pre[bestie];
}
for(int i=wow.size()-1; i>=0; i--) cout << wow[i]+1 << " ";
}
signed main(){
ios_base::sync_with_stdio(false);cin.tie(0);
#ifdef Local
freopen("in","r",stdin);
freopen("out","w",stdout);
#endif
/*freopen(".in","r",stdin);
freopen(".out","w",stdout);*/
int t=1;
//cin >> t;
while(t--) solve();
}
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