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#include <iostream>
using namespace std;
#define maxn 100000
#define maxk 200
#define ll long long
int best[maxn][maxk];
//arr is prefix sums
int main()
{
ll dp[maxn], prv[maxn], arr[maxn];
//prv is not quite necessary
int n, k;
cin >> n >> k;
k++;
//array endgth n, split into k+1 intervals, each time points = product of sums of two sides
//if decided how intervals will be picked: then how to arrange which to select?
//answer for a set of intervals = sum ab where a, b are distinct in it
//true by induction on set's size
//= ((sum a)^2 - sum a^2)/2 where first sum is known (sum of array) and second is split up
//divide up the array optimally to minimize sum (segment sum)^2? dp with prefix sums
for (int i = 0; i < n; i++)
{
cin >> arr[i];
if (i) arr[i] += arr[i-1];
//1 interval
dp[i] = arr[i]*arr[i];
best[i][0] = -1;
//cut at -1 means this is final interval
}
int stack[maxn];
ll cut[maxn+1];
cut[0] = 0;
//stack[0] is best from cut[0] to cut[1]-1, ...
int end, start;
for (int t = 1; t < k; t++)
{
//t+1 intervals to make 0..i-1; if t = 0 then just interval sum ^ 2
//else: dp[i] = min_x<i arr[i]^2 - 2*arr[x]*arr[i] + arr[x]^2 + prv[x]
//with x >= t-1 (i >= t to make it well defined)
//do all of dp[?] in O(n)? need to insert another x or query best x for given i
//monotonic stack: record pair<int,ll>: best value of x for arr[i] <= that ll
for (int i = 0; i < n; i++) prv[i] = dp[i];
end = 0;
start = 0;
for (int i = t; i < n; i++)
{
//x = i-1: insert it into the monotonic stack
ll x = 0;
bool special = false;
while (start != end)
{
//see when it beats the last function, then truncate
//definitely should work at some value of arr[i]: prv is increasing not too fast
//just assume it ought to work
//solve for smallest int x:
//arr[stack[end-1]]*arr[stack[end-1]]+prv[stack[end-1]]-2*arr[stack[end-1]]*x >
//arr[i-1]*arr[i-1]+prv[i-1]-2*arr[i-1]*x
//or tmp2x > tmp1 or x = ceil of (tmp1+1)/tmp2
//just take floor and +1 if needed
if (arr[i-1] == arr[stack[end-1]])
{
//impossible to beat previous function, so done
special = true;
break;
}
ll tmp1 = arr[i-1]*arr[i-1]+prv[i-1]-arr[stack[end-1]]*arr[stack[end-1]]-prv[stack[end-1]],
tmp2 = 2*arr[i-1]-2*arr[stack[end-1]];
x = (tmp1+1)/tmp2;
if ((tmp1+1)%tmp2) x++;
//if at some point it is optimal, it will continue to be optimal later
if (x > cut[end-1])
{
//previous one is sometimes optimal so cannot discard
break;
}
end--;
}
if (!special)
{
stack[end] = i-1;
cut[end] = x;
end++;
cut[end] = 1e18;
}
//the last curve is always optimal, but previous one is up to x-1 only
//now find which curve is best for arr[i]: pop from front of the stack
while (start != end)
{
if (cut[start+1] > arr[i]) break;
start++;
}
best[i][t] = stack[start];
dp[i] = (arr[i]-arr[stack[start]])*(arr[i]-arr[stack[start]])+prv[stack[start]];
}
}
//now dp[n-1] is the answer
cout << (arr[n-1]*arr[n-1]-dp[n-1])/2 << '\n';
int cur = n-1;
while (k)
{
cur = best[cur][--k];
if (cur != -1) cout << cur+1 << ' ';
}
}
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