제출 #1242918

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1242918		jerzyk	Broken Device 2 (JOI22_device2)	C++20	0 / 100	51 ms	2376 KiB

Anna

#include <bits/stdc++.h>
#include "Anna.h"

using namespace std;
#define pb push_back
#define st first
#define nd second
typedef long long ll;
typedef long double ld;
const ll I = 1'000'000'000'000'000'000LL;
const int II = 2'000'000'000;
const ll M = 1'000'000'007LL;
const int N = 1'000'007;
const int K = 60;

int Declare()
{
    return K * 3;
}

pair<vector<int>, vector<int>> Anna(ll A)
{
    vector<int> v1, v2;
    for(int i = 0; i < K * 3; ++i)
        v1.pb((i % 2));
    for(ll i = 0; i < K; ++i)
        for(int r = 0; r < 3; ++r)
            v2.pb(((1LL<<i) & A) > 0LL);
    return pair{v1, v2};
}

Bruno

#include <bits/stdc++.h>
#include "Bruno.h"

using namespace std;
#define pb push_back
#define st first
#define nd second
typedef long long ll;
typedef long double ld;
const ll I = 1'000'000'000'000'000'000LL;
const int II = 2'000'000'000;
const ll M = 1'000'000'007LL;
const int N = 1'000'007;
int cnt = 0;

ll Bruno(vector<int> u)
{
    vector<pair<int, int>> v, v2;
    int il = 0;
    for(int i = 0; i < (int)u.size(); ++i)
    {
        ++il;
        if(i == (int)u.size() - 1 || u[i] != u[i + 1])
        {
            v.pb(pair{il, u[i]});
            il = 0;
        }
    }
    int lst = 1, n = (int)v.size();
    for(int i = 0; i < n; ++i)
    {
        // cout << "C: " << i << " " << v[i].st << " " << v[i].nd << "\n";
        if((int)v2.size() != 0 && v2.back().nd == v[i].nd)
        {
            v[i].st += v2.back().st;
            v2.pop_back();
        }

        if(i < n - 1 && v[i + 1].st == 1)
        {
            if(lst != v[i].nd)
                --v[i].st;
            lst = v[i].nd;
        }else if(v[i].st % 3 == 1)
        {
            --v[i].st;
            lst = v[i].nd;
        }
        if(v[i].st != 0)
            v2.pb(v[i]);
    }
    ll ans = 0LL;
    ll cnt = 0LL;
    for(int i = 0; i < (int)v2.size(); ++i)
    {
        // cout << "F: " << v2[i].st << " " << v2[i].nd << "\n";
        for(int l = 0; l < (int)v2[i].st; l += 3)
        {
            ans += (1LL<<cnt) * (ll)v2[i].nd;
            ++cnt;
        }
    }
    return ans;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 5
• Subtask 3: 0 / 3
• Subtask 4: 0 / 12
• Subtask 5: 0 / 15
• Subtask 6: 0 / 60