| # | Time | Username | Problem | Language | Result | Execution time | Memory | 
|---|---|---|---|---|---|---|---|
| 369943 | EIMONIM | Split the sequence (APIO14_sequence) | C++14 | 0 ms | 0 KiB | 
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#define pb push_back
#define all(x) x.begin(),x.end()
using namespace std;
#define int int64_t
#define ii pair<int,int>
#define x first
#define y second
#define vi vector<int>
#define vvi vector<vi>
#define vii vector<ii>
#define vvii vector<vii>
#define vb vector<bool>
#define vvb vector<vb>
#define loop(i,s,e) for(int i=(s);i<(e);i++)
#define loopr(i,s,e) for(int i=(e)-1;i>=(s);i--)
#define chkmax(a,b) a = max(a,b)
#define chkmin(a,b) a=min(a,b)
const int INF = 1e18, MOD = 1e9 + 7;
/*mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
auto unidist = uniform_int_distribution<int>(0, MAXN);*/
const int N = 2e5;
typedef int64_t ll;
typedef pair<ll, ll> pll;
int n, k, a[N];
ll pre[N], dp[N], cnt[N];
struct CHT {
  struct Line {
    ll m, b, c;
    ll operator()(ll x) { return m * x + b; }
  } dq[N * 2];
  int l, r;
  void init() {
    dq[0] = {0, 0, 0};
    l = 0, r = 1;
  }
  bool better_ins(Line& L, Line& L1, Line& L2) {
    ll b1 = (L.b - L2.b) * (L2.m - L1.m), b2 = (L2.m - L.m) * (L1.b - L2.b);
    return b1 < b2 || (b1 == b2 && L.c <= L1.c);
  }
  bool better_qry(Line& L1, Line& L2, ll x) {
    ll b1 = L1(x), b2 = L2(x);
    return b1 > b2 || (b1 == b2 && L1.c >= L2.c);
  }
  void insert(Line L) {
    while(r - l >= 2 && better_ins(L, dq[r - 1], dq[r - 2])) --r;
    dq[r++] = L;
  }
  pll query(ll x) {
    while(r - l >= 2 && better_qry(dq[l], dq[l + 1], x)) ++l;
    return {dq[l](x), dq[l].c};
  }
} cht;
pll calc(ll x) {
  dp[0] = cnt[0] = 0;
  cht.init();
  for(int i = 1 ; i <= n ; ++i) {
    pll qry = cht.query(pre[i]);
    dp[i] = pre[i] * pre[i] + x + qry.F;
    cnt[i] = qry.S + 1;
    cht.insert({-2LL * pre[i], dp[i] + pre[i] * pre[i], cnt[i]});
  }
  return {cnt[n], dp[n]};
}
void init() {
  cin >> n >> k; k++;
  for(int i = 1 ; i <= n ; ++i) cin >> a[i], pre[i] = pre[i - 1] + a[i];
}
ll solve() {
  ll l = 0, r = pre[n] * pre[n], ans = -1;
  while(l <= r) {
    ll mid = l + (r - l) / 2;
    pll res = calc(mid);
    if(res.F > k) l = mid + 1;
    else r = mid - 1, ans = res.S - k * mid;
  }
  return ans;
}
int32_t main(){
    init();
    cout << int(pre[n]*pre[n] - solve())/2 << endl;
    Orig::n = n, Orig::k = k-1;
    loop(i,0,n) Orig::a[i+1] = a[i+1];
    Orig::calc();
    return 0;
}
/*
color a
cls
g++ seq.cpp -o a & a
7 3
4 1 3 4 0 2 3
for /l %x in (1,1,16) do a < out\big%x.in > out\big%x.out
*/
