Submission #917218

#	Time	Username	Problem	Language	Result	Execution time	Memory
917218		GrindMachine	Super Dango Maker (JOI22_dango3)	C++17	100 / 100	2575 ms	1004 KiB

dango3

#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 sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#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 &a, T b) {
    a = min(a,b);
}

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

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

read the solution a long time ago, remember the key idea of the solution

*/

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

#include "dango3.h"

void Solve(int n, int m) {
    vector<vector<int>> group(m+5);
    vector<bool> there(n*m+5);

    auto ok = [&](vector<int> v){
        fill(all(there),0);
        trav(x,v) there[x] = 1;

        vector<int> v_compl;
        rep1(i,n*m){
            if(!there[i]){
                v_compl.pb(i);
            }
        }

        int res = Query(v_compl);

        if(res == m-2) return false;
        else return true;
    };

    rep1(i,n*m){
        int l = 1, r = m;
        int first = -1;

        while(l <= r){
            int mid = (l+r) >> 1;
            group[mid].pb(i);
            if(ok(group[mid])){
                first = mid;
                r = mid-1;
            }
            else{
                l = mid+1;
            }

            group[mid].pop_back();    
        }

        assert(first != -1);
        group[first].pb(i);
    }

    rep1(i,m){
        Answer(group[i]);
    }
}

• Subtask 1: 2 / 2
• Subtask 2: 5 / 5
• Subtask 3: 15 / 15
• Subtask 4: 78 / 78