제출 #827574

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
827574		GrindMachine	죄수들의 도전 (IOI22_prison)	C++17	41 / 100	16 ms	1504 KiB

prison

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define ff first
#define ss second
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define conts continue
#define sz(a) a.size()

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &x, T y){
  x = min(x,y);
}

template<typename T>
void amax(T &x, T y){
  x = max(x,y);
}

#define debug(x) cout << #x << " = "; cout << x; cout << endl

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = 1e9 + 5;
const ll inf2 = (ll)1e18 + 5;

#include "prison.h"

std::vector<std::vector<int>> devise_strategy(int n) {
    int LOG = 0;
    while(1<<(LOG+1) <= n) LOG++;
    LOG++;

    vector<vector<int>> a(LOG*3-2,vector<int>(n+1));
    rep(i,LOG*3-2){
        int f = 1 << (LOG-1-(i/3));
        if(i%3 == 0){
            a[i][0] = 0;
            rep1(j,n){
                int b = 0;
                if(j&f) b = 1;
                if(f > 1){
                    a[i][j] = i+1+b;
                }
                else{
                    if(b == 0){
                        a[i][j] = -1;
                    }
                    else{
                        a[i][j] = -2;
                    }
                }
            }
        }
        else{
            a[i][0] = 1;
            int b1 = (i%3)-1;
            rep1(j,n){
                int b2 = 0;
                if(j&f) b2 = 1;
                if(b1 < b2){
                    a[i][j] = -1;
                }
                else if(b2 < b1){
                    a[i][j] = -2;
                }
                else{
                    a[i][j] = min(LOG*3-1,(i/3+1)*3);
                }
            }
        }
    }

    return a;
}

• Subtask 1: 5 / 5
• Subtask 2: 5 / 5
• Subtask 3: 31 / 90