#pragma GCC optimize("O3,unroll-all-loops,inline")
#pragma GCC target("avx2")
#include "lang.h"
#include "grader.h"
#include <bits/extc++.h>
using namespace std;
using namespace __gnu_pbds;
using ll = long long;
using pint = pair<int, int>;
int ghost = 0;
double magic(double x) {
return exp(abs(x)) - 1;
}
template <typename T, class Hash, int MAX>
struct Data {
int tot = 0;
unordered_map<T, int, Hash> m;
set<pair<int, T>> inv;
double operator&(const Data<T, Hash, MAX> &o) const {
double ret = 0;
for (auto [k, v]: m) {
ret += magic((double) v / tot - (o.m.find(k) != m.end() ? (double) o.m.find(k)->second / o.tot : 0));
}
for (auto [k, v]: o.m) {
if (m.find(k) == m.end()) ret += magic((double) v / o.tot);
}
return ret;
}
void operator+=(const Data<T, Hash, MAX> &o) {
for (auto &[k, v]: o.m) {
inv.erase({m[k], k});
inv.emplace(m[k] += v, k);
}
tot += o.tot;
while (inv.size() > MAX) {
tot -= inv.begin()->first;
m.erase(inv.begin()->second);
inv.erase(inv.begin());
}
}};
struct hash1 {
int operator()(const int x) const {
return ((long long)x) * 327821 % 1000000007;
}
};
struct hash2 {
ll operator() (const pint &x) const { return ((x.first) * 37) ^ ((x.second) * 73); }
};
struct hash3 {
ll operator()(const array<int, 3> &x) const { return (x[0] * 37) ^ (x[1] * 73) ^ (x[2] * 409); }
};
struct Profile {
Data<int, hash1, 50> sing;
Data<pint, hash2, 300> doub;
Data<array<int, 3>, hash3, 50> trip;
Profile() {}
Profile(int *arr) {
for (int i = 0; i < 100; ++i) {
sing.m[arr[i]]++;
if (i) doub.m[{arr[i - 1], arr[i]}]++;
if (i >=2) trip.m[{arr[i - 2], arr[i - 1], arr[i]}];
}
for (auto [k, v]: sing.m) sing.inv.emplace(v, k);
for (auto [k, v]: doub.m) doub.inv.emplace(v, k);
for (auto [k, v]: trip.m) trip.inv.emplace(v, k);
sing.tot = 100;
doub.tot = 99;
trip.tot = 98;
}
double operator&(const Profile &o) const {
return (sing & o.sing) + (ghost > 0 ? (doub & o.doub) + (trip & o.trip) : 0);
}
void operator+=(const Profile &o) {
sing += o.sing;
doub += o.doub;
trip += o.trip;
}
} profile[56];
void excerpt(int *E) {
Profile p(E);
pair<double, int> mn{1e18, 0};
for (int i = 0; i < 56; ++i) {
mn = min(mn, {p & profile[i], i});
}
int ans = language(mn.second);
profile[ans] += p;
ghost++;
}
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Correct |
8542 ms |
3412 KB |
Output is correct |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Partially correct |
8666 ms |
3372 KB |
Output is partially correct - 84.95% |
