Submission #440032

#TimeUsernameProblemLanguageResultExecution timeMemory
440032CodeChamp_SSHacker (BOI15_hac)C++17
0 / 100
9 ms15948 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], fw[N], bk[N]; int main() { fio(); cin >> n; inp(a, n) rep(x, n, 2 * n - 1) a[x] = a[x - n]; ms(fw, -1); ms(bk, -1); ll sum = 0; ll k = (n + 1) / 2; rep0(x, k) sum += a[x]; fw[0] = sum; rep(x, k, 2 * n - 1) { if (fw[(x % n - k + 1 + n) % n] != -1) break; sum -= a[x - k], sum += a[x]; fw[(x % n - k + 1 + n) % n] = sum; } sum = 0; for (int x = 2 * n - 1, i = 0; i < k; x--, i++) sum += a[x]; bk[n - 1] = sum; rrep0(x, 2 * n - k) { if (bk[(x % n + k - 1) % n] != -1) break; sum -= a[x + k], sum += a[x]; bk[(x % n + k - 1) % n] = sum; } ll res = 0; rep0(x, n) res = max(res, min(fw[x], bk[x])); cout << res; return 0; }
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