| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 145628 | dolphingarlic | Hacker (BOI15_hac) | C++14 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
// Basically we want Bytesar to start on the computer with
// the maximum of the minimums of semicircle sums that contain
// that computer, since the operator is able to force him to do that
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1000050;
int a[N];
ll sum[N], p[N];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i], a[i + n] = a[i];
for (int i = 1; i <= n * 2; i++) sum[i] = sum[i - 1] + a[i]; // Prefix sums
int sz = n + 1 >> 1;
// Sums of semicircles
for (int i = 1; i <= n; i++) p[i] = sum[i + sz - 1] - sum[i - 1];
for (int i = n + 1; i <= n * 2; i++) p[i] = p[i - n];
// Monotonic queue - Easily get min in range
deque<int> dq;
for (int i = n - sz + 2; i <= n; i++) {
while (dq.size() && p[dq.back()] >= p[i]) dq.pop_back();
dq.push_back(i);
}
int ans = 0;
for (int i = n + 1; i <= 2 * n; i++) {
while (dq.size() && p[dq.back()] >= p[i]) dq.pop_back();
dq.push_back(i);
// Make sure we only consider semicircle sums that contain
// the current computer i
while (dq.front() < i - sz + 1) dq.pop_front();
ans = max(ans, p[dq.front()]);
}
cout << ans << '\n';
return 0;
}