제출 #298642

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
298642		caoash	Longest beautiful sequence (IZhO17_subsequence)	C++14	40 / 100	115 ms	4608 KiB

subsequence

#include <bits/stdc++.h> 
using namespace std;

using ll = long long;

using vi = vector<int>;
using vl = vector<ll>;
#define pb push_back
#define rsz resize
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()

using pi = pair<int,int>;
#define f first
#define s second
#define mp make_pair

const int MX = 200005;
const int MOD = (int) (1e9 + 7);

namespace output {
    void pr(int x) { cout << x; }
    void pr(long x) { cout << x; }
    void pr(ll x) { cout << x; }
    void pr(unsigned x) { cout << x; }
    void pr(unsigned long x) { cout << x; }
    void pr(unsigned long long x) { cout << x; }
    void pr(float x) { cout << x; }
    void pr(double x) { cout << x; }
    void pr(long double x) { cout << x; }
    void pr(char x) { cout << x; }
    void pr(const char* x) { cout << x; }
    void pr(const string& x) { cout << x; }
    void pr(bool x) { pr(x ? "true" : "false"); }
    
    template<class T1, class T2> void pr(const pair<T1,T2>& x);
    template<class T> void pr(const T& x);
    
    template<class T, class... Ts> void pr(const T& t, const Ts&... ts) { 
        pr(t); pr(ts...); 
    }
    template<class T1, class T2> void pr(const pair<T1,T2>& x) { 
        pr("{",x.f,", ",x.s,"}"); 
    }
    template<class T> void pr(const T& x) { 
        pr("{"); // const iterator needed for vector<bool>
        bool fst = 1; for (const auto& a: x) pr(!fst?", ":"",a), fst = 0; 
        pr("}");
    }
    
    void ps() { pr("\n"); } // print w/ spaces
    template<class T, class... Ts> void ps(const T& t, const Ts&... ts) { 
        pr(t); if (sizeof...(ts)) pr(" "); ps(ts...); 
    }
    
    void pc() { cout << "]" << endl; } // debug w/ commas
    template<class T, class... Ts> void pc(const T& t, const Ts&... ts) { 
        pr(t); if (sizeof...(ts)) pr(", "); pc(ts...); 
    }
    #define dbg(x...) pr("[",#x,"] = ["), pc(x);
}
 
#ifdef LOCAL
using namespace output;
#endif

int n; 

void solve1() {
    vi a(n), k(n), fr(n), dp(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> k[i];
    }
    dp[0] = 1;
    for (int i = 1; i < n; i++) {
        dp[i] = 1;
        for (int j = 0; j < i; j++) {
            if (__builtin_popcount(a[i] & a[j]) == k[i]) {
                if (dp[j] + 1 > dp[i]) {
                    dp[i] = dp[j] + 1;
                    fr[i] = j;
                }
            }
        }
    }
    pi fin = mp(INT_MIN, -1);
    for (int i = 0; i < n; i++) {
        if (dp[i] > fin.f) {
            fin = mp(dp[i], i);
        }
    }
    cout << fin.f << '\n';
    int curr = fin.s;
    vi ret;
    while (dp[curr] != 1) {
        ret.pb(curr);
        curr = fr[curr];
    }
    ret.pb(curr);
    for (int i = sz(ret) - 1; i >= 0; i--) {
        cout << ret[i] + 1 << " ";
    }
    cout << '\n';
}

void solve2() {
    vi a(n), k(n), fr(n), dp(n);
    pi best[(1 << 8)];
    memset(best, 0, sizeof(best));
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> k[i];
    }
    dp[0] = 1;
    best[a[0]] = mp(1, 0);
    for (int i = 1; i < n; i++) {
        dp[i] = 1;
        for (int j = 0; j < (1 << 8); j++) {
            if (__builtin_popcount(a[i] & j) == k[i]) {
                if (best[j].f + 1 > dp[i]) {
                    dp[i] = best[j].f + 1;
                    fr[i] = best[j].s;
                }
            }
        }
        best[a[i]] = max(best[a[i]], mp(dp[i], i));
    }
    pi fin = mp(0, 0);
    for (int i = 0; i < n; i++) {
        if (dp[i] > fin.f) {
            fin = mp(dp[i], i);
        }
    }
    cout << fin.f << '\n';
    int curr = fin.s;
    vi ret;
    while (dp[curr] != 1) {
        ret.pb(curr);
        curr = fr[curr];
    }
    ret.pb(curr);
    for (int i = sz(ret) - 1; i >= 0; i--) {
        cout << ret[i] + 1 << " ";
    }
    cout << '\n';
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    if (n <= 5000) {
        solve1();
    }
    else {
        solve2();
    }
}

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