Submission #1002908

#	Time	Username	Problem	Language	Result	Execution time	Memory
1002908		Lalic	Cigle (COI21_cigle)	C++17	48 / 100	1033 ms	357716 KiB

cigle

#include <bits/stdc++.h>
#pragma GCC optimize "Ofast"
using namespace std;

#define fi first
#define se second
#define pb push_back
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()
#define mp make_pair
#define int long long

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef complex<double> cd;

const int MAXN = 5005;
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);

int tree[MAXN][4*MAXN], lazy[MAXN][4*MAXN];

void lazyP(int id, int node, int l, int r){
	if(!lazy[id][node]) return;
	
	tree[id][node]+=lazy[id][node];
	if(l!=r){
		lazy[id][node*2]+=lazy[id][node];
		lazy[id][node*2+1]+=lazy[id][node];
	}
	lazy[id][node]=0;
}

void upd(int id, int node, int l, int r, int p, int q, int val){
	lazyP(id, node, l, r);
	
	if(r<p || l>q) return;
	if(l>=p && r<=q){
		lazy[id][node]+=val;
		lazyP(id, node, l, r);
		return;
	}
	
	int mid=(l+r)>>1;
	upd(id, node*2, l, mid, p, q, val);
	upd(id, node*2+1, mid+1, r, p, q, val);
	
	tree[id][node]=max(tree[id][node*2], tree[id][node*2+1]);
}

int query(int id, int node, int l, int r, int p, int q){
	lazyP(id, node, l, r);
	
	if(r<p || l>q) return 0;
	if(l>=p && r<=q) return tree[id][node];
	
	int mid=(l+r)>>1;
	return max(query(id, node*2, l, mid, p, q), query(id, node*2+1, mid+1, r, p, q));
}

void solve(){
	int n; cin >> n;
	
	vector<int> arr(n+1);
	for(int i=1;i<=n;i++) cin >> arr[i];
	
	for(int i=3;i<=n;i++){
		int sum=arr[i], c2=i-2, tot=arr[i-1];
		
		for(int c1=i-1;c1>=0;c1--){
			upd(i, 1, 0, n, c1, c1, query(c1, 1, 0, n, 0, n));
			
			while(c2>0 && tot<sum){
				tot+=arr[c2];
				c2--;
			}
			
			if(tot==sum && c2>0) upd(c1, 1, 0, n, 0, c2-1, 1);
			
			sum+=arr[c1];
			tot-=arr[c1];
		}
	}
	
	cout << query(n, 1, 0, n, 0, n) << "\n";
}

int32_t main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL);
	
	int tt=1;
	//~ cin >> tt;
	while(tt--) solve();
	return 0;
}

• Subtask 1: 9 / 9
• Subtask 2: 11 / 11
• Subtask 3: 13 / 13
• Subtask 4: 15 / 15
• Subtask 5: 0 / 52