제출 #1308379

#제출 시각아이디문제언어결과실행 시간메모리
1308379thieunguyenhuyStove (JOI18_stove)C++20
100 / 100
18 ms1624 KiB
#include <bits/stdc++.h>
using namespace std;

#define POPCOUNT(n) (__builtin_popcountll((n)))
#define CLZ(n) (__builtin_clzll((n)))
#define CTZ(n) (__builtin_ctzll((n)))
#define LOG(n) (63 - __builtin_clzll((n)))
#define BIT(n, i) (((n) >> (i)) & 1ll)
#define MASK(i) (1ll << (i))
#define FLIP(n, i) ((n) ^ (1ll << (i)))
#define ON(n, i) ((n) | MASK(i))
#define OFF(n, i) ((n) & ~MASK(i))

#define Int __int128
#define fi first
#define se second

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef pair<long long, int> pli;
typedef pair<int, long long> pil;
typedef vector<pair<int, int>> vii;
typedef vector<pair<long long, long long>> vll;
typedef vector<pair<long long, int>> vli;
typedef vector<pair<int, long long>> vil;

template <class T1, class T2> bool maximize(T1 &x, T2 y) {
	if (x < y) {
		x = y;
		return true;
	}
	return false;
}
template <class T1, class T2> bool minimize(T1 &x, T2 y) {
	if (x > y) {
		x = y;
		return true;
	}
	return false;
}

template <class T> void remove_duplicate(vector<T> &ve) {
	sort (ve.begin(), ve.end());
	ve.resize(unique(ve.begin(), ve.end()) - ve.begin());
}

mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
long long random(long long l, long long r) {
	return uniform_int_distribution<long long>(l, r)(rng);
}
template <class T> T random(T r) {
	return rng() % r;
}

const int N = 1e6 + 5;
const int MOD = 1e9 + 7;
const int inf = 1e9;
const long long INF = 1e18;

int n, k;
int t[N];

signed main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

	cin >> n >> k;

	for (int i = 1; i <= n; ++i) cin >> t[i];

	sort (t + 1, t + 1 + n);
	
	priority_queue<int, vector<int>, greater<int>> pq;
	for (int i = 2; i <= n; ++i) pq.emplace(t[i] - t[i - 1]);

	int ans = 0;
	while (pq.size() > k - 1) {
		ans += pq.top(); pq.pop();
	}

	cout << ans + k;
	return 0;
}
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