Submission #503115

#	Time	Username	Problem	Language	Result	Execution time	Memory
503115		stanislavpolyn	Bigger segments (IZhO19_segments)	C++17	27 / 100	1567 ms	35660 KiB

segments

#include <bits/stdc++.h>

#define fr(i, a, b) for(int i = (a); i <= (b); ++i)
#define rf(i, a, b) for(int i = (a); i >= (b); --i)
#define fe(x, y) for(auto& x : y)

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple

#define all(x) (x).begin(), (x).end()
#define pw(x) (1LL << (x))
#define sz(x) (int)(x).size()

using namespace std;

mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<typename T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define fbo find_by_order
#define ook order_of_key

template<typename T>
bool umn(T& a, T b) { return (a > b ? (a = b, 1) : 0); }
template<typename T>
bool umx(T& a, T b) { return (a < b ? (a = b, 1) : 0); }

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename T>
using ve = vector<T>;

const int N = 3e3 + 5;

int n;
int a[N];
ll pref[N];
int dp[N][N];

int brute() {
    fr(i, 0, n) {
        fr(j, 0, n) {
            dp[i][j] = -1e9;
        }
    }

    fr(i, 1, n) {
        dp[i][1] = 1;

        fr(j, 1, i) {
            fr(nxt, i + 1, n) {
                if(pref[nxt] - pref[i] >= pref[i] - pref[j - 1]) {
                    umx(dp[nxt][i + 1], dp[i][j] + 1);
                }
            }
        }
    }

    int mx = 0;
    fr(i, 1, n) {
        fr(j, 1, n) {
            umx(mx, dp[i][j]);
        }
    }
    return mx;
}

int main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
#else
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
#endif

    cin >> n;

    fr(i, 1, n) {
        cin >> a[i];
        pref[i] = a[i] + pref[i - 1];
    }

    cout << brute() << "\n";

    return 0;
}

• Subtask 1: 13 / 13
• Subtask 2: 14 / 14
• Subtask 3: 0 / 10
• Subtask 4: 0 / 36
• Subtask 5: 0 / 27