Submission #531852

#	Time	Username	Problem	Language	Result	Execution time	Memory
531852		abc864197532	Bigger segments (IZhO19_segments)	C++17	100 / 100	90 ms	12996 KiB

segments

#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define pii pair <int, int>
#define X first
#define Y second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
void abc() {cout << endl;}
template <typename T, typename ...U> void abc(T a, U ...b) {
    cout << a << ' ', abc(b...);
}
template <typename T> void printv(T l, T r) {
    for (; l != r; ++l) cout << *l << " \n"[l + 1 == r];
}
template <typename A, typename B> istream& operator >> (istream& o, pair<A, B> &a) {
    return o >> a.X >> a.Y;
}
template <typename A, typename B> ostream& operator << (ostream& o, pair<A, B> a) {
    return o << '(' << a.X << ", " << a.Y << ')';
}
template <typename T> ostream& operator << (ostream& o, vector<T> a) {
    bool is = false;
    if (a.empty()) return o << "{}";
    for (T i : a) {o << (is ? ' ' : '{'), is = true, o << i;}
    return o << '}';
}
template <typename T> struct vv : vector <vector <T>> {
    vv(int n, int m, T v) : vector <vector <T>> (n, vector <T>(m, v)) {}
    vv() {}
};
template <typename T> struct vvv : vector <vv <T>> {
    vvv(int n, int m, int k, T v) : vector <vv <T>> (n, vv <T>(m, k, v)) {}
    vvv() {}
};
#ifdef Doludu
#define test(args...) abc("[" + string(#args) + "]", args)
#define owo freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout); 
#else
#define test(args...) void(0)
#define owo ios::sync_with_stdio(false); cin.tie(0)
#endif
const int mod = 1e9 + 7, N = 1001, logN = 20, K = 111, C = 7;

int main () {
    owo;
    int n;
    cin >> n;
    vector <lli> pre(n + 1);
    vector <int> dp(n + 1, 0), p(n + 1, 0);
    for (int i = 0, x; i < n; ++i) {
        cin >> x;
        pre[i + 1] = pre[i] + x;
    }
    for (int i = 1; i <= n; ++i) {
        p[i] = max(p[i], p[i - 1]);
        dp[i] = dp[p[i]] + 1;
        int x = lower_bound(all(pre), pre[i] * 2 - pre[p[i]]) - pre.begin();
        p[x] = max(p[x], i);
    }
    cout << dp[n] << '\n';
}

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