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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define watch(x) cout<<(#x)<<"="<<(x)<<'\n'
#define mset(d,val) memset(d,val,sizeof(d))
#define setp(x) cout<<fixed<<setprecision(x)
#define forn(i,a,b) for(int i=(a);i<(b);i++)
#define fore(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
#define F first
#define S second
#define pqueue priority_queue
#define fbo find_by_order
#define ook order_of_key
typedef long long ll;
typedef pair<ll,ll> ii;
typedef vector<ll> vi;
typedef vector<ii> vii;
typedef long double ld;
template<typename T>
using pbds = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
void amin(ll &a, ll b){ a=min(a,b); }
void amax(ll &a, ll b){ a=max(a,b); }
void YES(){cout<<"YES\n";} void NO(){cout<<"NO\n";}
void SD(int t=0){ cout<<"PASSED "<<t<<endl; }
const ll INF = ll(1e18);
const int MOD = 998244353;
const bool DEBUG = 0;
const int MAXN = 300005;
int n,K;
ll a[MAXN];
ll p[MAXN];
ll dp[MAXN], cnt[MAXN];
ii pre[MAXN]; //best dp[j-1]+sum(j..i) for j<=i
ll sum(int l, int r){ return p[r]-(l?p[l-1]:0LL); }
int main()
{
ios_base::sync_with_stdio(0); cin.tie(0);
cin>>n>>K;
fore(i,1,n)
{
cin>>a[i];
p[i]=p[i-1]+a[i];
}
dp[0]=0;
cnt[0]=0;
pre[0]={-INF,0};
ll ans=0;
for(ll L=0,R=1e16;L<=R;)
{
ll mid=(L+R)>>1;
fore(i,1,n)
{
ii tmp=pre[i-1];
tmp.F+=a[i];
pre[i]=max(tmp, {(i>1?dp[i-1]:0LL)+a[i]-mid, cnt[i-1]+1});
if(pre[i].F>=dp[i-1])
{
dp[i]=pre[i].F;
cnt[i]=pre[i].S;
}
else
{
dp[i]=dp[i-1];
cnt[i]=cnt[i-1];
}
}
if(cnt[n]>=K)
{
ans=dp[n]+min((ll)K,cnt[n])*mid;
L=mid+1;
}
else R=mid-1;
if(0)
{
cout<<"mid="<<mid<<'\n';
cout<<"a: "; fore(i,1,n) cout<<a[i]<<" "; cout<<'\n';
cout<<"dp: "; fore(i,1,n) cout<<dp[i]<<" "; cout<<'\n';
cout<<"cnt: "; fore(i,1,n) cout<<cnt[i]<<" "; cout<<'\n';
cout<<"pre: "; fore(i,1,n) cout<<"("<<pre[i].F<<","<<pre[i].S<<") "; cout<<'\n';
cout<<"ans: "<<dp[n]+cnt[n]*mid<<"\n\n";
}
}
cout<<ans<<'\n';
return 0;
}
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