Submission #448406

#	Time	Username	Problem	Language	Result	Execution time	Memory
448406		dxz05	Counting Mushrooms (IOI20_mushrooms)	C++14	82.18 / 100	11 ms	332 KiB

mushrooms

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

using namespace std;

int f(int n, int k){
    return k + (n + k - 2) / k - 2;
}

int count_mushrooms(int n) {
    if (n <= 227){
        int ans = 1;
        for (int i = 1; i < n; i++){
            ans += 1 - use_machine({0, i});
        }
        return ans;
    }

    vector<int> vec;
    int k = 85;

    vector<int> A = {0}, B;

    if (n >= 2000) {
        for (int i = 1; i <= 2; i++) {
            if (use_machine({0, i})) B.push_back(i);
            else
                A.push_back(i);
        }

        for (int i = 3; i < min(n, 2 * k - 1); i += 2) {
            if (i + 1 == n) {
                if (use_machine({0, i})) B.push_back(i);
                else
                    A.push_back(i);
                break;
            }

            if (A.size() >= 2) {
                vec = {i, A[0], i + 1, A[1]};
                int res = use_machine(vec);
                if (res % 2) B.push_back(i);
                else
                    A.push_back(i);
                if (res >= 2) B.push_back(i + 1);
                else
                    A.push_back(i + 1);
            } else {
                vec = {i, B[0], i + 1, B[1]};
                int res = use_machine(vec);
                if (res % 2) A.push_back(i);
                else
                    B.push_back(i);
                if (res >= 2) A.push_back(i + 1);
                else
                    B.push_back(i + 1);
            }
        }
    }

    int beg = 2 * k - 1;

    int ans = A.size();
    for (int i = beg; i < n;){
        vec.clear();
        k = max(A.size(), B.size());
        bool ok = A.size() >= B.size();

        for (int j = i; j < min(n, i + k); j++){
            vec.push_back(j);
            vec.push_back(ok ? A[j - i] : B[j - i]);
        }

        int res = use_machine(vec);
        if (ok) ans += vec.size() / 2 - (res + 1) / 2; else
            ans += (res + 1) / 2;

        if (res % 2){
            if (ok) B.push_back(i); else
                A.push_back(i);
        }

        i += k;
    }

	return ans;
}

• Subtask 1: 82.1818 / 100