제출 #1091725

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1091725		ULTRABIG7	Longest beautiful sequence (IZhO17_subsequence)	C++14	23 / 100	359 ms	262144 KiB

subsequence

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using db = long double;

#define ff first
#define ss second
#define pb push_back
#define all(x) begin(x),end(x)

#define FOR(i,a,b) for(int i = (a);i<(b);i++)
#define F0R(i,a) FOR(i,0,a)
#define rep(a) F0R(_,a)
#define each(a,x) for(auto& a:x)
#define sz(x) (int)size(x)
#define vi vector<int>
const int MOD = 1e9+7;
const db PI = acos((db)-1);


int main(){
    cin.tie(0)->sync_with_stdio(0);
    int n; cin>>n;

    vi A(n+1), K(n+1);
    int mx = -1;
    FOR(i,1,n+1){
        cin>>A[i];
        mx = max(mx,A[i]);
    }

    FOR(i,1,n+1){
        cin>>K[i];
    }

    if(n<=5000){
        vi dp(n+1,1), memo(n+1,-1);


        FOR(i,1,n+1){
            FOR(j,1,i){
                if(__builtin_popcount(A[i]&A[j]) == K[i] && dp[j]+1>dp[i]){
                    dp[i] = dp[j] + 1;
                    memo[i] = j;
                }
            }
        }

        int ans = 0, id = -1;
        FOR(i,1,n+1){
            if(dp[i]>ans){
                id = i;
                ans = dp[i];
            }
        }

        cout<<ans<<"\n";
        vi sol;
        while(id != -1){
            sol.pb(id);
            id = memo[id];
        }

        reverse(all(sol));

        each(x,sol){
            cout<<x<<" ";
        }
    }else if(mx < 256){
        vi dp(mx+1,-1),memo(mx+1,-1),ant(n+1,-1);

        FOR(i,1,n+1){
            dp[A[i]] = 1;
            if(memo[A[i]] == -1) memo[A[i]] = i;
        }

        // dp(x) é a maior subsequência que termina no valor x

        FOR(i,1,n+1){
            FOR(j,0,mx+1)if(dp[j] != -1 && __builtin_popcount(A[i] & j) == K[i]){
                if(dp[j]+1 > dp[A[i]]){
                    dp[A[i]] = dp[j]+1;
                    memo[A[i]] = i;
                    ant[i] = memo[j]; 
                }
            }
        }

        int ans = 0, val = -1;
        FOR(i,0,mx+1){
            if(ans<dp[i]){
                val = i;
                ans = dp[i];
            }
        }

        cout<<ans<<"\n";
        val = memo[val];
        vi sol;
        while(val != -1){
            sol.pb(val);
            val = ant[val];
        }

        reverse(all(sol));

        each(x,sol){
            cout<<x<<" ";
        }
    }
    
    return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 16 / 16
• Subtask 3: 0 / 17
• Subtask 4: 0 / 60