Submission #180287

#	Time	Username	Problem	Language	Result	Execution time	Memory
180287		MvC	Hacker (BOI15_hac)	C++11	100 / 100	314 ms	8952 KiB

hac

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#define rc(x) return cout<<x<<endl,0
#define pb push_back
#define mkp make_pair
#define in insert
#define er erase
#define fd find
#define fr first
#define sc second
typedef long long ll;
typedef long double ld;
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll llinf=(1LL<<62);
const int inf=(1<<30);
const int nmax=5e5+50;
const int mod=1e9+7;
using namespace std;
int n,i,st[4*nmax],a[nmax],cur,tot,rs,j;
int tr(int x)
{
	x+=n;
	x%=n;
	if(!x)x=n;
	return x;
}
void upd(int x,int l,int r,int p,int v)
{
	if(l==r)
	{
		st[x]=v;
		return;
	}
	int mid=(l+r)/2;
	if(p<=mid)upd(2*x,l,mid,p,v);
	else upd(2*x+1,mid+1,r,p,v);
	st[x]=max(st[2*x],st[2*x+1]);
}
int qry(int x,int l,int r,int tl,int tr)
{
	if(tl>tr || tl>n || tr>n || tl<1 || tr<1)return 0;
	if(l>tr || tl>r)return 0;
	if(tl<=l && r<=tr)return st[x];
	int mid=(l+r)/2;
	return max(qry(2*x,l,mid,tl,tr),qry(2*x+1,mid+1,r,tl,tr));
}
int main()
{
	//freopen("sol.in","r",stdin);
	//freopen("sol.out","w",stdout);
	//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	ios_base::sync_with_stdio(false);cin.tie(0);cerr.tie(0);cout.tie(0);
	cin>>n;
	for(i=1;i<=n;i++)cin>>a[i],tot+=a[i];
	for(i=1;i<=n/2;i++)cur+=a[i];
	for(j=1;j<=n;j++,i++)
	{
		i=tr(i);
		upd(1,1,n,j,cur);
		cur-=a[tr(i-n/2)];
		cur+=a[tr(i)];
	}
	for(i=1;i<=n;i++)
	{
		//cout<<i<<" "<<1<<" "<<i-n/2<<" "<<i+1<<" "<<tr(i-n/2)<<endl;
		cur=qry(1,1,n,1,i-n/2);
		if(i+1<=tr(i-n/2))cur=max(cur,qry(1,1,n,i+1,tr(i-n/2)));
		else if(i+1>tr(i-n/2))cur=max(cur,max(qry(1,1,n,i+1,n),qry(1,1,n,1,tr(i-n/2))));
		rs=max(rs,tot-cur);
	}                      
	cout<<rs<<endl;                                                                                                                                                                                                                                                                                       
	return 0;
}

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 40 / 40