Submission #1200037

#	Time	Username	Problem	Language	Result	Execution time	Memory
1200037		countless	Sequence (APIO23_sequence)	C++20	0 / 100	2096 ms	31260 KiB

sequence

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

typedef long long ll;
typedef long double ld;

const ll MOD = 998244353;
const ll INF = 1e18;
const ld EPS = 1e-12;

#define endl "\n"
#define sp <<" "<<
#define REP(i, a, b) for(ll i = a; i < b; i++)
#define dbg(x) cout << #x << " = " << x << endl
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fast_io() ios_base::sync_with_stdio(false); cin.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

template <typename Key, typename Value>
using hash_map = unordered_map<Key, Value, custom_hash>;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// uniform_int_distribution<int>(a, b)(rng);
// shuffle(all(a), rng);

int sequence(int N, vector<int> A) {
    // brute force
    int ans = 0;
    int med1, med2;
    REP(i, 0, N) {
        vector<int> temp;
        hash_map<int, int> cnt;

        REP(j, i, N) {
          
            temp.push_back(A[j]);
            cnt[A[j]]++;
            
            if (A[i] != A[j]) continue;

            int sz = j - i + 1;
            if (sz % 2 == 0) {
                nth_element(temp.begin(), temp.begin() + sz / 2, temp.end()); 
                nth_element(temp.begin(), temp.begin() + (sz - 1) / 2, temp.end());
                med1 = temp[(sz - 1) / 2];
                med2 = temp[sz / 2];

                ans = max(ans, cnt[med1]);
                ans = max(ans, cnt[med2]);
            } else {
                nth_element(temp.begin(), temp.begin() + sz / 2, temp.end()); 
                med2 = temp[sz / 2];
                
                ans = max(ans, cnt[med2]);
            }
        }
    }

    return ans;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 17
• Subtask 3: 0 / 7
• Subtask 4: 0 / 12
• Subtask 5: 0 / 13
• Subtask 6: 0 / 22
• Subtask 7: 0 / 18