#include "mushrooms.h"
#include <bits/stdc++.h>
using namespace std;
//a contains the a and b, but easy to swap
vector<int> a[2], ve;
//res = answer,
int res, x, c, k, r;
void update() {
if (a[0].size() < a[1].size()) {
res = r - res;
swap(a[0], a[1]);
c ^= 1;
}
}
int count_mushrooms(int n) {
//if n is small, we dont need to go through all this trouble, just check the all directly
if (n <= 227) {
for (int i = 1; i < n; i++) {
res += 1 - use_machine({0, i});
}
return res + 1;
}
a[0] = {0};
//clever way to not FALLUNTERSCHEID them
for (int i = 1; i < 3; i++) {
int x = use_machine({0, i});
a[x].push_back(i);
}
update();
//get at least 5 items
x = use_machine({3, a[0][0], 4, a[0][1]});
a[x&1].push_back(3);
a[x/2&1].push_back(4);
update();
//in the first phase we collect as much explicit using this nice method
//using 2 queries, we can determine 5 items
//we find first explicit, then we know if the others are the same or dif
//if same-> we know what, else we know they are different
//in the last case we have limited number of possiblities, so just check all of them
//with this nice trick
for (int i = 5; max(a[0].size(), a[1].size()) < 100;) {
x = use_machine({i, a[0][0], i+1, a[0][1], i+2, a[0][2]});
a[x&1].push_back(i);
if (!(x/2 % 2)) {
a[x/4].push_back(i+1);
a[x/4].push_back(i+2);
i+=3;
} else if (a[1].size() < 2) {
x = use_machine({a[0][0], i+1});
a[x].push_back(i+1);
a[x^1].push_back(i+2);
i+=3;
} else {
x = use_machine({i+4, a[0][0], i+3, a[0][1], i+2, a[0][2], a[1][0], i+1, a[1][1]})-1;
a[x&1].push_back(i+4);
a[x/2&1].push_back(i+3);
a[x/4].push_back(i+2);
a[x/4^1].push_back(i+1);
i += 5;
}
}
update();
//now phase 2
res = a[1].size();
for (int i = a[0].size() + a[1].size(); i < n; i = r) {
ve.clear();//(int)
r = min(i+(int)a[0].size(), n);
for (int j = i; j < r; j++) {
ve.push_back(j);
ve.push_back(a[0][j-i]);
}
x = use_machine(ve);
res += (x+1)/2;
a[x&1].push_back(i);
update();
}
return (c ? res : n - res);
}
