Submission #728557

#	Time	Username	Problem	Language	Result	Execution time	Memory
728557		Nelt	Prisoner Challenge (IOI22_prison)	C++17	0 / 100	3 ms	2004 KiB

prison

#include "prison.h"

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,fma")

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

/* DEFINES */
#define S second
#define F first
#define ll long long
#define ull unsigned long long
#define ld long double
#define npos ULLONG_MAX
#define INF LLONG_MAX
#define vv(a) vector<a>
#define pp(a, b) pair<a, b>
#define pq(a) priority_queue<a>
#define qq(a) queue<a>
#define ss(a) set<a>
#define mm(a, b) map<a, b>
#define ump(a, b) unordered_map<a, b>
#define sync                      \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
#define elif else if
#define endl "\n"
#define allc(a) begin(a), end(a)
#define all(a) a, a + sizeof(a) / sizeof(a[0])
#define pb push_back
#define logi(a) __lg(a)
#define sqrt(a) sqrtl(a)
#define mpr make_pair
#define ins insert
using namespace std;
using namespace __gnu_pbds;
using namespace __cxx11;
typedef char chr;
typedef basic_string<chr> str;
template <typename T, typename key = less<T>>
using ordered_set = tree<T, null_type, key, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vv(vv(int)) devise_strategy(int n)
{
    ll B = sqrt(n);
    vv(vv(int)) ans;
    ans.pb(vv(int)(n + 1));
    ans.back()[0] = 0;
    for (ll i = 1; i <= n; i++)
        ans.back()[i] = i / B + B + 1;
    for (ll i = 1; i <= B; i++)
    {
        ans.pb(vv(int)(n + 1));
        ans.back()[0] = 0;
        for (ll j = 1; j <= n; j++)
            if (j % B < i - 1)
                ans.back()[j] = -1;
            else
                ans.back()[j] = -2;
    }
    for (ll i = B + 1; i <= B + n / B + 1; i++)
    {
        ans.pb(vv(int)(n + 1));
        ans.back()[0] = 1;
        for (ll j = 1; j < (i - (B + 1)) * B; j++)
            ans.back()[j] = -2;
        for (ll j = (i - (B + 1)) * B; j < (i - (B + 1) + 1) * B; j++)
            ans.back()[j] = j % B + 1;
        for (ll j = (i - (B + 1) + 1) * B; j <= n; j++)
            ans.back()[j] = -1;
    }
    return ans;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 5
• Subtask 3: 0 / 90