Submission #158049

#TimeUsernameProblemLanguageResultExecution timeMemory
158049hentai_loverLongest beautiful sequence (IZhO17_subsequence)C++14
23 / 100
62 ms1656 KiB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define lft(x) x * 2
#define rgt(x) x * 2 + 1

#define tm hui_pizda
#define ft first
#define sd second
#define pb push_back
#define pf push_front
#define sz size()
#define cnt continue
#define m_p make_pair
#define fr(i, l, r) for(int i = l; i <= r; ++ i)
#define rf(i, r, l) for(int i = r; i >= l; -- i)
#define all(x) x.begin(), x.end()

using namespace __gnu_pbds;
using namespace std;

template <typename T>
using _set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

typedef int ll;
typedef long double ld;

typedef pair <ll, ll> pll;
typedef pair <int, int> pii;
typedef vector <int> vi;
typedef vector <ll> vl;
typedef vector <pii> vpi;
typedef vector <pll> vpl;

mt19937_64 rnd(time(NULL));

const ll N = 1e6 + 10;
const ll mtrxN = 10;
//const ll oo = 1e18 + 10;
const ll oo = 1e9 + 10;
const ll B = 500;
const ll mod = 1e9 + 7;

struct mtrx{
    ll m[mtrxN][mtrxN] = {};
};
mtrx mtrx_mult(mtrx a, mtrx b){
    mtrx c;
    fr(i, 0, mtrxN - 1){
        fr(j, 0, mtrxN - 1){
            ll sum = 0;
            fr(x, 0, mtrxN - 1){
                sum += a.m[i][x] * b.m[x][j];
                sum %= mod;
            }
            c.m[i][j] = sum;
        }
    }
    return c;
}
mtrx mtrx_pow(mtrx a, ll n){
    mtrx res;
    fr(i, 0, mtrxN - 1)fr(j, 0, mtrxN - 1)res.m[i][j] = a.m[i][j];
    n --;
    while(n){
        if(n&1)res = mtrx_mult(res, a);
        a = mtrx_mult(a, a);
        n >>= 1;
    }
    return res;
}
ll _pow(ll a, ll n){
    ll r = 1;
    while(n){
        if(n&1)r = r * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return r;
}
ll div(ll x, ll y, ll md){
    return x * _pow(y, md - 2) % md;
}

ll a[N], k[N], pt[N];


ll n;

int main(){
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    fr(i, 1, n)cin >> a[i];
    fr(i, 1, n)cin >> k[i];

    if(n <= 5000){
        vector <ll> dp(n + 10);
        fr(i, 1, n)dp[i] = 1;
        fr(i, 1, n)
            fr(j, 1, i - 1)
                if(__builtin_popcount(a[i] & a[j]) == k[i])
                    if(dp[i] < dp[j] + 1)dp[i] = dp[j] + 1, pt[i] = j;


        ll p = 0;
        fr(i, 1, n)
            if(dp[p] < dp[i])p = i;

        vl ans;
        while(p != 0){
            ans.pb(p);
            p = pt[p];
        }
        reverse(ans.begin(), ans.end());
        cout << ans.sz << "\n";
        for(auto i : ans)cout << i << ' ';
        cout << endl;
    }
    return 0;
}
/*
*/
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