이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//я так много думал, что опять попал
#include <bits/stdc++.h>
#define all(x) x.begin(),x.end()
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define f first
#define s second
#define left(v) v + v
#define right(v) v + v + 1
#define ub upper_bound
#define lb lower_bound
#define pll pair<ll,ll>
//17 SEVENTEEN
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
typedef long long ll;
const long double Pi = acos(-1.0);
const ll dx[] = {0,0,1,-1};
const ll dy[] = {1,-1,0,0};
const ll N = (ll) 1e6 + 17;
const ll M = (ll) 5e3 + 69;
const ll inf = (ll) 1e14 + 3;
const ll mod = (ll) 1e9 + 7;
ll sq(ll x) { return x * x; }
ll zxc = 1, a[N], p[N];
pll dp[N];
void solve() {
    ll n;
    cin >> n;
    for(ll i = 1; i <= n; i++) {
        cin >> a[i];
        p[i] = p[i - 1] + a[i];
        dp[i] = {1, 0};
    }
    for(ll i = 1; i <= n; i++) {
        dp[i] = max(dp[i], dp[i - 1]);
        ll l = i + 1, r = n + 1;
        while(l < r) {
            ll m = (l + r) / 2;
            if(p[m] >= 2 * p[i] - p[dp[i].s]) r = m;
            else l = m + 1;
        }
        if(l <= n) dp[l] = max(dp[l], {dp[i].f + 1, i});
    }
    cout << dp[n].f;
}
int main(/*Уверенно*/) {
ios_base::sync_with_stdio(0);
    cin.tie(0);
/*
	freopen(".in", "r", stdin);
	freopen(".out", "w", stdout);
*/
//    cin >> zxc;
    while(zxc--) {
        solve();
    }
  	return 0;
}
// さよならさ いかなくちゃ
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