Day \(2^2+3^2+4^2\)。
HNOI2016 序列的加强版。我去年怎么这么菜啊，虽然现在也是就是了。
\[\sum\limits_{[l,r]\in [L,R]}\left(\max\limits_{i\in [l,r]}a_i\right)\left(\max\limits_{i\in [l,r]}b_i\right) \]考虑离线，对右端点 \(r\) 扫描线，对每个左端点 \(l\) 维护 \(S_l=\left(\max\limits_{i\in [l,r]}a_i\right)\left(\max\limits_{i\in [l,r]}b_i\right)\)。然后查询就变成了求历史版本和，由于需要维护最大值，建立关于 \(a_i,b_i\) 的单调栈，题目就变成了：
- 令 \(a_i\gets a_i+c,i\in [l,r]\)。
- 令 \(b_i\gets b_i+c,i\in [l,r]\)。
- 令 \(c_i\gets c_i+a_ib_i,i\in [l,r]\)。
- 查询 \(\sum \limits_{i\in [l,r]}c_i\) 的值。
类似上面那题，没有历史版本和的问题时容易的。维护 \(a_i,b_i\) 的区间和、区间加法标记以及 \(a_ib_i\) 的区间和就行了。由于要维护历史版本和，再维护 \(a_i,b_i\) 的区间历史版本和、区间历史加法标记和以及 \(a_ib_i\) 的历史版本和、\(a_i,b_i\) 标记乘积的历史版本和即可。
// Problem: #3899. 「NOIP2022」比赛
// Contest: LibreOJ
// URL: https://loj.ac/p/3899
// Memory Limit: 512 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define mt make_tuple
#define mp make_pair
#define fi first
#define se second
// #define FILE
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> pi;
typedef tuple<int, int, int> tu;
bool Mbe;
const int N = 3e5 + 300;
int n, m, a[N], b[N], tp[2], st[N][2];
ull ans[N];
vector<pi> q[N];
struct node {
ull sa, sb, sab, ta, tb, len;
ull hsa, hsb, hsab, hta, htb, htab, ti;
node () { sa = sb = sab = ta = tb = len = hsa = hsb = hsab = hta = htb = htab = ti = 0; }
} tr[N << 2];
node tag (ull _ta, ull _tb, ull _ti) { node res; res.ta = _ta, res.tb = _tb, res.ti = _ti; return res; }
#define ls x << 1
#define rs x << 1 | 1
#define mid ((l + r) >> 1)
void ptg(node &x, node y) {
x.hsab += x.sab * y.ti + x.sb * y.hta + x.sa * y.htb + x.len * y.htab;
x.hsa += x.sa * y.ti + x.len * y.hta, x.hsb += x.sb * y.ti + x.len * y.htb;
x.htab += x.ta * x.tb * y.ti + x.ta * y.htb + x.tb * y.hta + y.htab;
x.hta += x.ta * y.ti + y.hta, x.htb += x.tb * y.ti + y.htb;
x.sab += y.ta * x.sb + y.tb * x.sa + x.len * y.ta * y.tb;
x.sa += x.len * y.ta, x.sb += x.len * y.tb, x.ta += y.ta, x.tb += y.tb, x.ti += y.ti;
}
void pup(int x) {
tr[x].sab = tr[ls].sab + tr[rs].sab;
tr[x].sa = tr[ls].sa + tr[rs].sa;
tr[x].sb = tr[ls].sb + tr[rs].sb;
tr[x].hsa = tr[ls].hsa + tr[rs].hsa;
tr[x].hsb = tr[ls].hsb + tr[rs].hsb;
tr[x].hsab = tr[ls].hsab + tr[rs].hsab;
}
void pdn(int x) {
ptg(tr[ls], tr[x]), ptg(tr[rs], tr[x]);
tr[x].ta = tr[x].tb = tr[x].ti = tr[x].hta = tr[x].htb = tr[x].htab = 0;
}
void build(int l, int r, int x) {
tr[x].len = r - l + 1;
if (l == r) return;
build(l, mid, ls), build(mid + 1, r, rs);
}
void upd(int l, int r, int s, int t, int op, int c, int x) {
if (s <= l && r <= t) return ptg(tr[x], tag((ull)(op ? 0 : c), (ull)(!op ? 0 : c), 0)), void();
pdn(x);
if (s <= mid) upd(l, mid, s, t, op, c, ls);
if (t > mid) upd(mid + 1, r, s, t, op, c, rs);
pup(x);
}
ull qry(int l, int r, int s, int t, int x) {
if (s <= l && r <= t) return tr[x].hsab;
pdn(x); ull res = 0;
if (s <= mid) res += qry(l, mid, s, t, ls);
if (t > mid) res += qry(mid + 1, r, s, t, rs);
return res;
}
void solve() {
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = 1; i <= n; i++) cin >> b[i];
cin >> m;
for (int i = 1, l, r; i <= m; i++)
cin >> l >> r, q[r].pb(mp(l, i));
build(1, n, 1);
for (int i = 1; i <= n; i++) {
while (tp[0] && a[st[tp[0]][0]] < a[i]) upd(1, n, st[tp[0] - 1][0] + 1, st[tp[0]][0], 0, a[i] - a[st[tp[0]][0]], 1), tp[0]--;
while (tp[1] && b[st[tp[1]][1]] < b[i]) upd(1, n, st[tp[1] - 1][1] + 1, st[tp[1]][1], 1, b[i] - b[st[tp[1]][1]], 1), tp[1]--;
st[++tp[0]][0] = st[++tp[1]][1] = i, upd(1, n, i, i, 0, a[i], 1), upd(1, n, i, i, 1, b[i], 1);
ptg(tr[1], tag(0, 0, 1));
for (pi j : q[i]) ans[j.se] = qry(1, n, j.fi, i, 1);
}
for (int i = 1; i <= m; i++)
cout << ans[i] << '\n';
}
bool Med;
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cerr << (&Mbe - &Med) / 1048576.0 << " MB\n";
#ifdef FILE
freopen("match.in", "r", stdin);
freopen("match.out", "w", stdout);
#endif
int T = 1;
cin >> T, T = 1;
while (T--) solve();
cerr << (int)(1e3 * clock() / CLOCKS_PER_SEC) << " ms\n";
return 0;
}