#include <bits/stdc++.h>
using namespace std;
#define int long long
int inf = 5e18;
int n,k,a[100005];
int dp[205][100005];
int f[100005];
int ant[205][100005];
struct dreapta
{
int offset, panta;
int eval(int x)
{
return offset + panta * x;
}
};
deque<pair<dreapta,int>> dq;
long double inters(dreapta d1, dreapta d2)
{
long double numarator = d2.offset - d1.offset,numitor = d1.panta - d2.panta;
return numarator / numitor;
}
void update(dreapta d,int pos)
{
dq.push_back({d,pos});
while (dq.size() >= 2)
{
dreapta d1 = dq.back().first;
int pos1 = dq.back().second;
dq.pop_back();
dreapta d2 = dq.back().first;
if (d1.panta != d2.panta)
{
dq.push_back({d1,pos1});
break;
}
else
{
dq.pop_back();
dq.push_back({d1,pos1});
}
}
while (dq.size() >= 3)
{
dreapta d1 = dq.back().first;
int pos1 = dq.back().second;
dq.pop_back();
dreapta d2 = dq.back().first;
int pos2 = dq.back().second;
dq.pop_back();
dreapta d3 = dq.back().first;
int pos3 = dq.back().second;
dq.pop_back();
if (inters(d1,d2) >= inters(d2,d3))
{
dq.push_back({d3,pos3});
dq.push_back({d1,pos1});
}
else
{
dq.push_back({d3,pos3});
dq.push_back({d2,pos2});
dq.push_back({d1,pos1});
break;
}
}
}
int query(int x)
{
if (dq.empty())
return -inf;
while (dq.size() >= 2)
{
dreapta d1 = dq.front().first;
int pos1 = dq.front().second;
dq.pop_front();
dreapta d2 = dq.front().first;
if (d1.eval(x) > d2.eval(x))
{
dq.push_front({d1,pos1});
break;
}
}
return dq.front().first.eval(x);
}
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 >> a[i];
f[n] = a[n];
for (int i = n - 1; i >= 1; i--)
f[i] = f[i + 1] + a[i];
dp[0][1] = 0;
for (int i = 2; i <= n; i++)
dp[0][i] = -inf;
for (int j = 1; j <= k; j++)
{
dq.clear();
for (int i = 1; i <= n; i++)
{
dp[j][i] = query(f[i]) - f[i] * f[i];
if (!dq.empty())
ant[j][i] = dq.front().second;
dreapta aux;
aux.offset = dp[j - 1][i];
aux.panta = f[i];
update(aux,i);
}
}
int ans = 0;
for (int i = k + 1; i <= n; i++)
ans = max(ans,dp[k][i]);
cout << ans << '\n';
int cine;
for (int i = k + 1; i <= n; i++)
if (dp[k][i] == ans)
cine = i;
vector<int> lol;
for (int i = k; i >= 1; i--)
{
lol.push_back(cine - 1);
cine = ant[i][cine];
}
reverse(lol.begin(),lol.end());
for (auto it : lol)
cout << it << ' ';
return 0;
}
/*
7 3
4 1 3 4 0 2 3
*/
/**
Nu conteaza ordinea operatiilor, so dp[j][i] = facand j splituri pe locuri pana inainte de i (ultimul pe i), care e suma maxima pe care o pot obtine
dp[j][i] = max(dp[j - 1][q < i] + f[i] * (f[q] - f[i]))
f[i] = suma pe sufix de la i incolo
deci am -f[i] * f[i] mereu, vreau maximul din dp[j - 1][q < i] + f[i] * f[q]
Asta e evident CHT, cu update: dreapta dp[j - 1][i] cu panta f[i] si query in f[i]
Query-urile sunt sortate descrescator si update-urile sunt tot descrescator dupa panta
**/
