题解 | #小美走公路#

维护好前缀和b，那么 l 到 r 的要么正着走花费 b[r - 1] - b[l - 1] ，要么反着走花费 b[n] - b[r - 1] + b[l - 1]

记得保证 l<r

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e6 + 5;
int __t = 1, n, q, l, r, a[N], b[N];
void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        b[i] = b[i - 1] + a[i];
    }
    cin >> l >> r;
    if (l > r)
        swap(l, r);
    cout << min(b[r - 1] - b[l - 1], b[n] - b[r - 1] + b[l - 1]) << '\n';
    return;
}
int32_t main() {
#ifdef ONLINE_JUDGE
    ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    // cin >> __t;
    while (__t--)
        solve();
    return 0;
}