이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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); }
struct dinamika {
ll cnt, sum;
} dp[N];
ll n, a[N], pr[N], t[N * 4];
void upd(ll pos, ll val, ll v = 1, ll tl = 1, ll tr = n) {
if (tl == tr) {
t[v] = val;
return;
} else {
ll tm = (tl + tr) / 2;
if (pos <= tm) upd(pos, val, v + v, tl, tm);
else upd(pos, val, v + v + 1, tm + 1, tr);
t[v] = max(t[v + v], t[v + v + 1]);
}
}
pll get(ll l, ll r, ll v = 1, ll tl = 1, ll tr = n) {
if (r < tl || tr < l) return {0, 0};
if (tl == tr) return {t[v], tl};
ll tm = (tl + tr) / 2;
pll ans = {0, 0};
if (t[v + v] > t[v + v + 1]) {
ans = get(l, r, v + v, tl, tm);
if (ans.ft == 0) ans = get(l, r, v + v + 1, tm + 1, tr);
} else {
ans = get(l, r, v + v + 1, tm + 1, tr);
if (ans.ft == 0) ans = get(l, r, v + v, tl, tm);
}
return ans;
}
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, pr[i]};
upd(i, dp[i].cnt);
}
for (int i = 1; i <= n; ++ i) {
ll l = 1, r = i, j = 1;
while (l <= r) {
ll mid = (l + r) / 2;
if (pr[i] - pr[mid - 1] >= dp[mid - 1].sum) l = mid + 1, j = mid;
else r = mid - 1;
}
//cout << j << " ";
pll ans = get(1, j - 1);
dp[i] = {ans.ft + 1, pr[i] - pr[ans.sd]};
//cout << dp[i].cnt << " ";
upd(i, dp[i].cnt);
}
cout << dp[n].cnt << "\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
*/
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