Submission #38982

#	Time	Username	Problem	Language	Result	Execution time	Memory
38982		RockyB	K개의 묶음 (IZhO14_blocks)	C++14	18 / 100	33 ms	45928 KiB

blocks

/// In The Name Of God

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,sse3,sse4,popcnt,abm,mmx")

#include <bits/stdc++.h>

#define f first
#define s second

#define pb push_back
#define pp pop_back
#define mp make_pair

#define sz(x) (int)x.size()
#define sqr(x) ((x) * 1ll * (x))
#define all(x) x.begin(), x.end()

#define Kazakhstan ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);

#define nl '\n'
#define ioi exit(0);

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

const int N = (int)1e5 + 7, inf = (int)1e9 + 7, mod = (int)1e9 + 7;
const ll linf = (ll)1e18 + 7;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1}, dy[] = {0, 1, 0, -1, 1, -1, 1, -1};

using namespace std;

int n, k;
int a[N];
int dp[111][N];
int main() {
	#ifdef IOI2018
		freopen ("in.txt", "r", stdin);
	#endif
	Kazakhstan
	cin >> n >> k;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	memset(dp, 0x3f3f, sizeof(dp));
	dp[0][0] = 0;
	for (int o = 1; o <= k; o++) {
		for (int i = 1; i <= n; i++) {
			int mx = a[i];
			for (int j = i; j >= max(1, i - 50); j--) {
				mx = max(mx, a[j]);
				dp[o][i] = min(dp[o][i], dp[o - 1][j - 1] + mx);
			}
		}
	}
	cout << dp[k][n];
	ioi
}

• Subtask 1: 0 / 14
• Subtask 2: 18 / 18
• Subtask 3: 0 / 21
• Subtask 4: 0 / 47