Submission #1166173

#	Time	Username	Problem	Language	Result	Execution time	Memory
1166173		SSSM	Bigger segments (IZhO19_segments)	C++20	100 / 100	174 ms	164908 KiB

segments

#include <bits/stdc++.h>

/*
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target ("avx2")
*/


using namespace std;

/*
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds;
template<class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
*/

#define F first
#define S second
#define pb push_back
#define FIO freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
#define md(a) ((a%mod+mod)%mod)
#define all(a) a.begin(), a.end()
#define MP make_pair
#define lc (id<<1)
#define rc (lc|1)
#define mid (l+r)/2
#define SZ(a) (ll)(a.size())
#define kill(a) cout << a << "\n", exit(0)
typedef pair<int,int> pii;
typedef pair<long long ,long long> pll;
typedef long long ll;
typedef long double ld;
typedef vector<vector<ll>> matrix;
mt19937_64  rng(chrono::steady_clock::now().time_since_epoch().count());

ll const maxn=5e6+10, mod=998244353, INF=1e18+10, LOG=21, sq=200;

ll poww(ll a, ll b, ll mod) {
 if (b == 0) return 1;
 return 1 * poww(1 * a * a % mod, b / 2, mod) * ((b % 2 == 1) ? a : 1) % mod;
}

ll n, ps[maxn], seg[maxn<<2], dp[maxn];

void Set(ll p, ll x, ll l=1, ll r=n+1, ll id=1)
{
	if(l==r-1){
		seg[id]=x;
		return;
	}
	if(p<mid) Set(p, x, l, mid, lc);
	else Set(p, x, mid, r, rc);
	seg[id]=min(seg[lc], seg[rc]);
}

ll Get(ll x, ll l=1, ll r=n+1, ll id=1)
{
	if(l==r-1)  return l;
	if(seg[rc]<=x) return Get(x, mid, r, rc);
	else return Get(x, l, mid, lc); 
}

int main(){    
    
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	
	cin>>n;
	for(ll i=1;i<=n;i++)
	{
		ll a;
		cin>>a;
		ps[i]=ps[i-1]+a;
	}

	fill(seg, seg+maxn*4, INF);

	for(ll i=1;i<=n;i++)
	{
		ll k=0;
		if(seg[1]>ps[i]) k=0;
		else k=Get(ps[i]);
		dp[i]=dp[k]+1;
		Set(i, 2*ps[i]-ps[k]);
	//	cout<<i<<" "<<k<<" "<<dp[k]<<" "<<dp[i]<<"\n";
	}
	cout<<dp[n]<<"\n";
}

• Subtask 1: 13 / 13
• Subtask 2: 14 / 14
• Subtask 3: 10 / 10
• Subtask 4: 36 / 36
• Subtask 5: 27 / 27