Submission #337679

#	Time	Username	Problem	Language	Result	Execution time	Memory
337679		ogibogi2004	수열 (APIO14_sequence)	C++14	60 / 100	2060 ms	93308 KiB

sequence

#include<bits/stdc++.h>
using namespace std;
#define ld long double
#define ll long long
const ld INF=2e9;
const ll MAXN=1e5+6;
const ll MAXK=202;
ll dp[MAXN][2],a[MAXN];
int last[MAXN][MAXK];
struct line
{
	ld a,b;
	ll pos;
	ld intersect(line const& other)
	{
		if(other.a==a)return -INF;
		else return (ld)(other.b-b)/(a-other.a);
	}
	ld gety(ld x)
	{
		return (ld)a*x+b;
	}
};
struct ConvexHull
{
	deque<line>dq;
	bool yoda(line l)
	{
		ld int1=l.intersect(dq.back());
		ld int2=dq.back().intersect(dq[dq.size()-2]);
		ld int3=l.intersect(dq[dq.size()-2]);
		if(int1==-INF)
		{
			if(l.b>=dq.back().b)return 1;
			else return 0;
		}
		if(int3>int2)return 1;
		else return 0;
	}
	void add(line l)
	{
		while(dq.size()>1&&yoda(l))
		{
			dq.pop_back();
		}
		if(dq.empty())dq.push_back(l);
		else if(!(l.a==dq.back().a&&l.b>dq.back().b))
		{
			dq.push_back(l);
		}
	}
	pair<ld,ld> query(ld x)
	{
		ll l=0,r=dq.size()-1,mid;
		while(l<r)
		{
			mid=(l+r)/2;
			if(dq[mid].gety(x)<dq[mid+1].gety(x))
			{
				l=mid+1;
			}
			else r=mid;
		}
		//cout<<x<<" "<<dq[l].a<<" "<<dq[l].b<<endl;
		return {dq[l].gety(x),dq[l].pos};
	}
}ch[2];
ll n,k;
ll pref[MAXN];
int main()
{
	cin>>n>>k;
	for(ll i=1;i<=n;i++)
	{
		cin>>a[i];
		pref[i]=pref[i-1]+a[i];
	}
	ch[0].add({0,0,0});
	for(int j=1;j<=k+1;j++)
	{
		ch[j%2].dq.clear();
		ch[j%2].add({0,-INF,-1});
		for(int i=1;i<=n;i++)
		{
			tie(dp[i][j%2],last[i][j])=ch[1-j%2].query(-pref[i]);
		}
		for(int i=1;i<=n;i++)
		{
			ch[j%2].add({(ld)-pref[i],(ld)dp[i][j%2]-pref[i]*pref[i],i});				
			/*for(auto xd:ch[j%2].dq)cout<<xd.a<<","<<xd.b<<" ";
			cout<<endl;*/
		}
	}
	cout<<dp[n][1-k%2]<<endl;
	if(dp[n][1-k%2]==0)
	{
		for(int j=1;j<=k;j++)
		{
			cout<<j<<" ";
		}
		cout<<endl;
		return 0;
	}
	vector<ll>v;
	ll n1=n;
	for(ll i=k;i>=1;i--)
	{
		v.push_back(last[n1][i+1]);
		n1=last[n1][i+1];
	}
	for(auto xd:v)cout<<xd<<" ";
	cout<<endl;
return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 0 / 11
• Subtask 4: 17 / 17
• Subtask 5: 21 / 21
• Subtask 6: 0 / 29