# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
661467 | evenvalue | Counting Mushrooms (IOI20_mushrooms) | C++17 | 1 ms | 464 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "mushrooms.h"
#include <bits/stdc++.h>
using namespace std;
template<typename T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template<typename T>
using max_heap = priority_queue<T, vector<T>, less<T>>;
using int64 = long long;
using ld = long double;
constexpr int kInf = 1e9 + 10;
constexpr int64 kInf64 = 1e15 + 10;
constexpr int kMod = 1e9 + 7;
constexpr int kConst = 4;
int count_mushrooms(int n) {
/*
* Let us manually query for the first (n / 4)
* Out of this at least one set will be having (n / 8) elements.
* Since everything in this set is distinct, and of the same value....
* We can query the rest of the (n / 4) elements using the (n / 8) distinct values.
* This can be optimised, but we'll think about this later. :)
*/
int used = 0;
auto get_next = [&](const vector<int> &s, const int l) {
const int x = s.size();
vector<int> next = {l};
used = 1;
for (int i = l + 1, j = 0; i < n and j < s.size(); i++, j++) {
next.push_back(s[j]);
next.push_back(i);
used++;
}
if (used == 1) next.push_back(s[0]);
return next;
};
vector<int> zero = {0};
vector<int> ones = {};
for (int i = 1; i <= n / kConst; i++) {
(use_machine({0, i}) == 0 ? zero : ones).push_back(i);
}
int ans = zero.size();
for (int i = n / kConst + 1; i < n; i += used) {
if (zero.size() > ones.size()) {
const int x = use_machine(get_next(zero, i));
ans += used - x;
} else {
assert(false);
const int x = use_machine(get_next(ones, i));
ans += x;
}
}
return ans;
}
Compilation message (stderr)
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