제출 #440175

#제출 시각아이디문제언어결과실행 시간메모리
440175CodeChamp_SSHacker (BOI15_hac)C++17
20 / 100
457 ms37704 KiB
#include<bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace __gnu_pbds; using namespace std; #define ff first #define ss second #define pb push_back #define eb emplace_back #define mp make_pair #define lb lower_bound #define ub upper_bound #define setbits(x) __builtin_popcountll(x) #define zrobits(x) __builtin_ctzll(x) #define sz(v) (int)v.size() #define ps(y) cout << fixed << setprecision(y) #define ms(arr, v) memset(arr, v, sizeof(arr)) #define all(v) v.begin(), v.end() #define rall(v) v.rbegin(), v.rend() #define trav(x, v) for(auto &x: v) #define w(t) int t; cin >> t; while(t--) #define rep0(i, n) for(int i = 0; i < n; i++) #define rrep0(i, n) for(int i = n - 1; i >= 0; i--) #define rep1(i, n) for(int i = 1; i <= n; i++) #define rrep1(i, n) for(int i = n; i > 0; i--) #define inp(arr, n) rep0(i, n) cin >> arr[i]; #define rep(i, a, b) for(int i = a; i <= b; i++) #define rrep(i, a, b) for(int i = a; i >= b; i--) typedef long long ll; typedef unsigned long long ull; typedef long double ld; typedef pair<ll, ll> pii; typedef vector<ll> vi; typedef vector<vi> vvi; typedef vector<pii> vp; typedef vector<bool> vb; typedef vector<string> vs; typedef map<ll, ll> mii; typedef map<char, ll> mci; typedef priority_queue<ll> pq_mx; typedef priority_queue<ll, vi, greater<>> pq_mn; typedef tree<ll, null_type, less<>, rb_tree_tag, tree_order_statistics_node_update> pbds; /* * find_by_order(i) -> returns an iterator to the element at ith position (0 based) * order_of_key(i) -> returns the position of element i (0 based) */ const int N = 1e6 + 5; const int mod = 1e9 + 7; //const int mod = 998244353; const ll inf = 1e18; mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); void fio() { ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0); } ll n, a[N], pre[N], sub[N]; int main() { fio(); cin >> n; inp(a, n) rep(x, n, 2 * n - 1) a[x] = a[x - n]; rep1(x, 2 * n) pre[x] = pre[x - 1] + a[x - 1]; trav(x, sub) x = 1e9; ll k = (n + 1) / 2; multiset<ll> st; rep0(x, 2 * n) { if (x + k - 1 < 2 * n) st.insert(pre[x + k] - pre[x]); if (x > k) st.erase(st.find(pre[x] - pre[x - k])); sub[x % n] = min(sub[x % n], *st.begin()); } ll res = 0; rep0(x, n) res = max(res, sub[x]); cout << res; return 0; }
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