Submission #497572

#	Time	Username	Problem	Language	Result	Execution time	Memory
497572		armashka	Bigger segments (IZhO19_segments)	C++17	100 / 100	71 ms	16000 KiB

segments

#include <bits/stdc++.h>
 
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#define fast ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define all(v) v.begin(),v.end()
#define pb push_back
#define sz size()
#define ft first
#define sd second
 
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef unsigned long long ull;
 
const int N = 1e6 + 5;
const ll M = 1e8;
const ll inf = 1e18;
const ll mod = 1e9;
const double Pi = acos(-1); 
 
ll binpow(ll x, ll ti) { ll res = 1;while (ti){if(ti & 1)res *= x;x *= x;ti >>= 1; x %= mod; res %= mod;} return res;}
ll binmul(ll x, ll ti) { ll res = 0;while (ti){if(ti & 1)res += x;x += x;ti >>= 1; x %= mod; res %= mod;} return res;}
ll nok(ll a, ll b) { return (a*b)/__gcd(abs(a),abs(b)) * (a*b > 0 ? 1 : -1); }
bool odd(ll n) { return (n % 2 == 1); }
bool even(ll n) { return (n % 2 == 0); }            
 
ll n, a[N], pr[N];
pll dp[N];
 
const void solve(/*Armashka*/) {
	cin >> n;
	for (int i = 1; i <= n; ++ i) {
		cin >> a[i];
		pr[i] = pr[i - 1] + a[i];
		dp[i] = {1, 0};
	}
	for (int i = 1; i <= n; ++ i) {
		ll l = i + 1, r = n, j = -1;
		dp[i] = max(dp[i], dp[i - 1]);
		while (l <= r) {
			ll mid = (l + r) / 2;                                                        
			if (pr[mid] >= 2 * pr[i] - pr[dp[i].sd]) r = mid - 1, j = mid;
			else l = mid + 1;
		}
		if (j != -1)
			dp[j] = max(dp[j], {dp[i].ft + 1, i});
		//cout << i << " " << j << "\n";
	}	
	cout << dp[n].ft << "\n";
}
            
signed main() {
    fast;
    //freopen("divide.in", "r", stdin);
    //freopen("divide.out", "w", stdout);
    int tt = 1;
    //cin >> tt;
    while (tt --) {                                                
        solve();
    }
}
 
/*
5 4 4
1 2
3 1
3 4
5 3
4 5 2 3
2 1 3 1
1 3 5
2 3 4 5
2 1 3 1
*/

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