이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#include<bits/extc++.h>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define pb push_back
using namespace std;
using namespace __gnu_pbds;
using ll = long long;
using vi = vector<ll>;
using vvi = vector<vector<ll>>;
const ll mod = 998244353;
using oset = tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MK = 220, MN = 1e5 + 3;
ll n, a[MN], pref[MN], dp[MK][MN], p[MK][MN];
ll f(ll l, ll r, ll i) {
ll a = pref[i] - pref[l - 1];
ll b = pref[r] - pref[i];
return a * b;
}
void bt(int a, int b) {
if (b > n)
exit(-1);
cout << b << " ";
if (a > 1)
bt(a - 1, p[a][b]);
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
ll k;
cin >> n >> k;
for (int i = 1; i <= n; i++)
cin >> a[i], pref[i] = pref[i - 1] + a[i];
ll ans = 0, g = n - 1;
for (int a = 1; a <= k; a++) {
int j =1;
for (int i = 1; i < n; i++) {
while(j+1 <= i && (dp[a - 1][j - 1] + f(j, n, i)) <= (dp[a - 1][j] + f(j+1, n, i)))
j++;
dp[a][i] = dp[a - 1][j - 1] + f(j, n, i);
p[a][i] = j - 1;
ans = max(ans, dp[a][i]);
if (ans == dp[a][i])
g = i;
}
}
cout << ans << '\n';
bt(k, g);
}
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