Submission #1073285

#	Time	Username	Problem	Language	Result	Execution time	Memory
1073285		GrindMachine	Rarest Insects (IOI22_insects)	C++17	0 / 100	35 ms	600 KiB

insects

#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
 
/*
 
refs:
https://youtu.be/mm5Nv81P5u8?t=19948
edi

first, figure out the #of distinct guys in the array

b.s on answer
let's say we want to check if ans < mid
go over all indices and put each value into the machine
if val > mid, then pop the recently added value from the machine
at the end, count the #of guys that were added to the machine
if cnt == mid*unique, then each guy appears at least mid #of times, so return false
otherwise, cnt < mid*unique, so there is at least 1 guy that appears < mid #of times, so return true
gets around 50 points

key idea for 100 points:
try to save operations between successive calls of b.s

cnt == mid*unique:
we increase the left bound of the b.s
there are only good guys in the machine
because we increase the left bound, these good guys will remain forever
no need to remove them and add them again (put them forever in the machine)

cnt < mid*unique:
we decrease the right bound of the b.s
if a guy is bad in this iteration, he would be bad in the successive iteration too
so ignore the bad guys
also, remember to remove the good guys in this iteration from the machine

if case 1 is true, mid*unique guys removed from consideration
if case 2 is true, at least active-mid*unique guys removed from consideration
mid*unique splits the active set into almost 2 equal halves
so at each stage, around n/2 guys are removed from consideration

so we get n+n/2+n/4+... = 2n queries
n queries for finding the #of unique guys
so around 3n queries in total
refer https://codeforces.com/blog/entry/105835?#comment-942719 for more details

further optimizations for full score:
set l and r bounds of the b.s optimally (l = 1, r = (n/unique)-1, ans = n/unique) (ans is at most n/unique)
stop adding guys to the machine once the limit (cnt == mid*unique) is reached
randomize the order in which the guys are added to the machine

*/
 
const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
 
#include "insects.h"

int min_cardinality(int n) {
    int d = 0;

    {
        vector<int> v;
        rep(i,n){
            move_inside(i);
            if(press_button() > 1){
                move_outside(i);
            }       
            else{
                v.pb(i);
                d++;
            }  
        }

        trav(i,v){
            move_outside(i);
        }
    }

    vector<bool> state(n,1); // 0 = ignored, 1 = active
    int lo = 1, hi = n/d;
    int ans = -1;
    int cnt = 0;

    auto ok = [&](int mid){
        vector<int> stay,leave;
        rep(i,n){
            if(!state[i]) conts;
            move_inside(i);
            cnt++; 
            if(press_button() > mid){
                move_outside(i);
                cnt--;
                leave.pb(i);
            }
            else{
                stay.pb(i);
            }
        }

        bool ok = (cnt == mid*d);
        
        if(ok){
            trav(i,stay){
                state[i] = 0;
            }
        }
        else{
            trav(i,leave){
                move_outside(i);
                state[i] = 0;
            }
            cnt = 0;
        }

        return ok;
    };

    while(lo <= hi){
        int mid = (lo+hi) >> 1;
        if(ok(mid)){
            ans = mid;
            lo = mid+1;
        }
        else{
            hi = mid-1;
        }
    }

    return ans;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 15
• Subtask 3: 0 / 75