Submission #1146621

#	Time	Username	Problem	Language	Result	Execution time	Memory
1146621		mnbvcxz123	Simfonija (COCI19_simfonija)	C++20	110 / 110	35 ms	2376 KiB

simfonija

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define fi first
#define se second
#define ll long long
#define ld long double
#define sz(x) ((int)(x).size())
#define all(x) 	(x).begin(), (x).end()
#define pb push_back
#define mpp make_pair
#define ve vector
using namespace std;
using namespace __gnu_pbds;
template<class T> using oset = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
const ll inf = 1e18; const int iinf = 1e9;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T> inline bool chmin(T& a, T b) { return (a > b ? a = b, 1 : 0); }
template <typename T> inline bool chmax(T& a, T b) { return (a < b ? a = b, 1 : 0); }

inline void solve() {
	int n, k;
	cin >> n >> k;

	ve<int> a(n), b(n), p(n);
	for (auto &i : a) cin >> i;
	for (auto &i : b) cin >> i;
	for (int i = 0; i < n; ++i) p[i] = b[i] - a[i];

	sort(all(p));

	ve<ll> pf(n);
	for (int i = 0; i < n; ++i) {
		pf[i] = p[i];
		if (i) pf[i] += pf[i - 1];
	}

	auto cnt = [&](int l, int r) { return max(0, r - l + 1); };
	auto sum = [&](int l, int r) { return (l > r ? 0ll : pf[r] - (!l ? 0 : pf[l - 1])); };

	auto get = [&](int l, int r) {
		if (l >= r) return 0ll;

		int s = r - l + 1;
		int x = p[l + s / 2];

		int pos = upper_bound(all(p), x) - p.begin();
		
		ll res = 0;
		res += cnt(l, pos - 1) * 1ll * x - sum(l, pos - 1);
		res += sum(pos, r) - cnt(pos, r) * 1ll * x;

		return res;
	};

	ll ans = inf;
	for (int p = 0; p <= k; ++p) {
		chmin(ans, get( p, n - (k - p) - 1 ));
	}

	cout << ans;
}

signed main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int q = 1; // cin >> q;
	while (q--) solve();
	cerr << fixed << setprecision(3) << "Time execution: " << (double)clock() / CLOCKS_PER_SEC << endl;
}

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 11 / 11
• Subtask 5: 11 / 11
• Subtask 6: 11 / 11
• Subtask 7: 11 / 11
• Subtask 8: 11 / 11
• Subtask 9: 11 / 11
• Subtask 10: 11 / 11