Submission #1152551

#	Time	Username	Problem	Language	Result	Execution time	Memory
1152551		KluydQ	Split the sequence (APIO14_sequence)	C++20	0 / 100	33 ms	82756 KiB

sequence

#include <bits/stdc++.h>
//#include "grader.h"
 
#define respagold ios_base::sync_with_stdio(0), cin.tie(0);
#define int long long
#define ll long long
#define int2 __int128_t
#define FOR( i, x, n, d ) for( int i = x; i <= n; i += d )
#define FORR( i, x, n, d ) for( int i = x; i >= n; i -= d )
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x.size())
#define pb push_back
#define ins insert
#define lb lower_bound
#define ub upper_bound
#define pii pair <int, int>
#define mkp make_pair
 
using namespace std;
 
const int N = 1e5 + 123;
const int inf = 1e18;
 
int a[N], b[N], n, m, k, z, w, x, y, mm[N], cc[N];
int dp[N], par[N][201];

mt19937 rng( chrono::steady_clock::now().time_since_epoch().count());
 
int rand( int l, int r )
{
    uniform_int_distribution <int> uid( l, r );
    return uid( rng );
}
struct line
{
	int m, c, id;
	line( int m, int c, int id ): m(m), c(c), id(id) {}
	line(): m(0), c(-inf), id(0) {}
	int get( int x ) { return m * x + c; }
};
line t[33 * N], sv[N], sv1[N];
int L[33 * N], R[33 * N], id[4 * N], timer = 1;
 
void add( line nw, int v = 1, int l = 0, int r = 1e9 + 1 )
{
	int md = ( l + r ) >> 1;
	bool lef = ( nw.get(l) > t[v].get(l) ), mid = ( nw.get(md) > t[v].get(md) );
	
	if( mid ) swap( nw, t[v] );
	
	if( l + 1 == r ) return;
	
	if( lef == mid ) 
	{
		if( !R[v] ) R[v] = ++ timer;
		add( nw, R[v], md, r ); 
	}
	else
	{
		if( !L[v] ) L[v] = ++ timer;
		add( nw, L[v], l, md ); 
	}
}
pii get( int pos, int v = 1, int l = 0, int r = 1e9 + 1 )
{ 
	int md = ( l + r ) >> 1;
	pii res = { t[v].get(pos), t[v].id };
	
	if( l + 1 == r ) return res;
	if( pos < md && L[v] ) res = max( res, get( pos, L[v], l, md ));
	if( pos >= md && R[v] ) res = max( res, get( pos, R[v], md, r ));
	return res;
}
void solve()
{
	cin >> n >> k;
	
	FOR( i, 1, n, 1 ) cin >> a[i], b[i] = b[i - 1] + a[i];
	sv1[0] = line( 0, 0, 0 );
		
	FOR( bl, 1, k, 1 )
	{
		add( sv1[bl - 1] );
		
		FOR( i, bl, n, 1 )
		{
			// for i find k that ( dp[k] + sum(k + 1, i) * sum(i + 1, n) ) is max
			
			// dp[i] = dp[k] + ( b[i] - b[k] ) * ( b[n] - b[i] )
			// dp[i] = dp[k] + b[i] * b[n] - b[i] * b[i] - b[k] * b[n] + b[k] * b[i];
			// dp[i] = ( dp[k] - b[k] * b[n] ) + ( b[i] * b[n] - b[i] * b[i] ) + ( b[k] * b[i] )
			
			pii j = get(b[i]);
			dp[i] = j.F + b[i] * b[n] - b[i] * b[i];
			sv[i] = line( b[i], dp[i] - b[i] * b[n], i );
			par[i][bl] = j.S;
			
			add( sv1[i] );
		}
		FOR( i, 1, n, 1 ) sv1[i] = sv[i];
		FOR( i, 1, timer, 1 ) t[i] = line(), L[i] = R[i] = 0;
		timer = 1;
	}
	int ans = 0, id = 0;
	FOR( i, 1, n, 1 ) ans = max( ans, dp[i] );
	
	FOR( i, 1, n, 1 )
	{
		if( ans == dp[i] ) id = i;
	}
	vector <int> v;
	FORR( blo, k, 1, 1 ) v.pb( id ), id = par[id][blo];
	reverse( all( v ));
	cout << ans << '\n';
	FOR( i, 0, sz(v) - 1, 1 ) cout << v[i] << ' ';
}
signed main() 
{
//	freopen("connect.in", "r", stdin);
//	freopen("connect.out", "w", stdout);
	    
    respagold
    
    int test = 1;
    
    if( !test ) cin >> test;
	
	while( test -- )
	{
		solve();
	}
}
// solved by KluydQ

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