CodeForces 1928F Digital Patterns
为什么我场上被卡常了。
转化题意，将 \(a, b\) 差分，答案为在 \(a, b\) 选出相同长度的不含 \(0\) 的子段方案数。
设 \(a\) 选出长度为 \(i\) 的不含 \(0\) 的子段方案数为 \(x_i\)，\(b\) 选出长度为 \(i\) 的不含 \(0\) 的子段方案数为 \(y_i\)。答案为 \(\sum x_i y_i\)。考虑线段树维护这个东西。
一个 \(a\) 中长度为 \(l\) 的极长子段会给 \(\forall k \in [1, l]\) 的 \(x_k\) 贡献 \(k - l + 1\)。\(b\) 同理。\(k + 1\) 和 \(-l\) 是独立的，先考虑 \(-l\) 这部分。也就是要支持 \(x_i \gets x_i + di\) 或 \(y_i \gets y_i + di\)。
只考虑修改 \(x\)，\(y\) 是对称的。拆式子：\(\sum (x_i + di) y_i = \sum x_i y_i + d \sum i y_i\)。再维护 \(\sum i x_i\) 和 \(\sum i y_i\) 就可以完成 \(\sum x_i y_i\) 的修改。\(\sum i x_i\) 的修改是容易的，考虑 \(\sum i (x_i + i) = \sum i x_i + \sum i^2\)，后者显然可以 \(O(1)\) 求的。
还要支持 \(x_i \gets x_i + d\) 或者 \(y_i \gets y_i + d\)。继续拆式子：\(\sum (x_i + d) y_i = \sum x_i y_i + d \sum x_i\)。再维护 \(\sum x_i, \sum y_i\) 即可。
于是线段树每个结点维护 \(5\) 个值：\(\sum x_i y_i, \sum x_i, \sum y_i, \sum i x_i, \sum i y_i\) 和 \(4\) 种懒标记即可。
差分后区间加变成了单点修改。考虑对 \(a, b\) 分别开个 set 维护非 \(0\) 极长连续段。显然单点修改只会加入或删除 \(O(1)\) 个段，讨论下就行了。
时间复杂度 \(O((n + q) \log n)\)，但是线段树的常数巨大，实现精细点就能过了。
真的有人会想看 7KB+ 的代码吗？
// Problem: F. Digital Patterns
// Contest: Codeforces - Codeforces Round 924 (Div. 2)
// URL: https://codeforces.com/contest/1928/problem/F
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb insert_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline int read() {
char c = getchar();
int x = 0;
bool f = 0;
for (; !isdigit(c); c = getchar()) f |= (c == '-');
for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return f ? -x : x;
}
const int maxn = 300100;
ll n, m, q, a[maxn], b[maxn];
struct node {
ll sxy, sx, sy, six, siy;
};
inline node operator + (const node &a, const node &b) {
node res;
res.sxy = a.sxy + b.sxy;
res.sx = a.sx + b.sx;
res.sy = a.sy + b.sy;
res.six = a.six + b.six;
res.siy = a.siy + b.siy;
return res;
}
inline ll calc1(ll l, ll r) {
return (l + r) * (r - l + 1) / 2;
}
inline ll calc2(ll n) {
return n * (n + 1) * (n * 2 + 1) / 6;
}
inline ll calc2(ll l, ll r) {
return calc2(r) - calc2(l - 1);
}
namespace SGT {
node a[maxn << 2];
ll ta1[maxn << 2], ta2[maxn << 2], tb1[maxn << 2], tb2[maxn << 2];
inline void pushup(int x) {
a[x] = a[x << 1] + a[x << 1 | 1];
}
inline void pushta1(int x, int l, int r, ll y) {
node &u = a[x];
u.sxy += y * u.siy;
u.sx += y * calc1(l, r);
u.six += y * calc2(l, r);
ta1[x] += y;
}
inline void pushta2(int x, int l, int r, ll y) {
node &u = a[x];
u.sxy += y * u.sy;
u.sx += y * (r - l + 1);
u.six += y * calc1(l, r);
ta2[x] += y;
}
inline void pushtb1(int x, int l, int r, ll y) {
node &u = a[x];
u.sxy += y * u.six;
u.sy += y * calc1(l, r);
u.siy += y * calc2(l, r);
tb1[x] += y;
}
inline void pushtb2(int x, int l, int r, ll y) {
node &u = a[x];
u.sxy += y * u.sx;
u.sy += y * (r - l + 1);
u.siy += y * calc1(l, r);
tb2[x] += y;
}
inline void pushdown(int x, int l, int r) {
int mid = (l + r) >> 1;
if (ta1[x]) {
pushta1(x << 1, l, mid, ta1[x]);
pushta1(x << 1 | 1, mid + 1, r, ta1[x]);
ta1[x] = 0;
}
if (ta2[x]) {
pushta2(x << 1, l, mid, ta2[x]);
pushta2(x << 1 | 1, mid + 1, r, ta2[x]);
ta2[x] = 0;
}
if (tb1[x]) {
pushtb1(x << 1, l, mid, tb1[x]);
pushtb1(x << 1 | 1, mid + 1, r, tb1[x]);
tb1[x] = 0;
}
if (tb2[x]) {
pushtb2(x << 1, l, mid, tb2[x]);
pushtb2(x << 1 | 1, mid + 1, r, tb2[x]);
tb2[x] = 0;
}
}
void upda1(int rt, int l, int r, int ql, int qr, ll x) {
if (ql <= l && r <= qr) {
pushta1(rt, l, r, x);
return;
}
pushdown(rt, l, r);
int mid = (l + r) >> 1;
if (ql <= mid) {
upda1(rt << 1, l, mid, ql, qr, x);
}
if (qr > mid) {
upda1(rt << 1 | 1, mid + 1, r, ql, qr, x);
}
pushup(rt);
}
void upda2(int rt, int l, int r, int ql, int qr, ll x) {
if (ql <= l && r <= qr) {
pushta2(rt, l, r, x);
return;
}
pushdown(rt, l, r);
int mid = (l + r) >> 1;
if (ql <= mid) {
upda2(rt << 1, l, mid, ql, qr, x);
}
if (qr > mid) {
upda2(rt << 1 | 1, mid + 1, r, ql, qr, x);
}
pushup(rt);
}
void updb1(int rt, int l, int r, int ql, int qr, ll x) {
if (ql <= l && r <= qr) {
pushtb1(rt, l, r, x);
return;
}
pushdown(rt, l, r);
int mid = (l + r) >> 1;
if (ql <= mid) {
updb1(rt << 1, l, mid, ql, qr, x);
}
if (qr > mid) {
updb1(rt << 1 | 1, mid + 1, r, ql, qr, x);
}
pushup(rt);
}
void updb2(int rt, int l, int r, int ql, int qr, ll x) {
if (ql <= l && r <= qr) {
pushtb2(rt, l, r, x);
return;
}
pushdown(rt, l, r);
int mid = (l + r) >> 1;
if (ql <= mid) {
updb2(rt << 1, l, mid, ql, qr, x);
}
if (qr > mid) {
updb2(rt << 1 | 1, mid + 1, r, ql, qr, x);
}
pushup(rt);
}
}
inline void updx(int x, int y) {
SGT::upda1(1, 1, n, 1, x, -y);
SGT::upda2(1, 1, n, 1, x, y * (x + 1));
}
inline void updy(int x, int y) {
SGT::updb1(1, 1, n, 1, x, -y);
SGT::updb2(1, 1, n, 1, x, y * (x + 1));
}
set<int> S, T;
int f[maxn], g[maxn];
inline void addx(int x) {
auto i = S.lower_bound(x);
if (i != S.end() && *i == x + 1) {
updx(f[*i] - *i + 1, -1);
int l = x, r = f[*i];
if (i != S.begin() && f[*prev(i)] == x - 1) {
int p = *prev(i);
l = p;
updx(f[p] - p + 1, -1);
S.erase(prev(i));
}
S.erase(i);
S.insert(l);
f[l] = r;
updx(r - l + 1, 1);
} else if (i != S.begin() && f[*prev(i)] == x - 1) {
int p = *prev(i);
S.erase(prev(i));
int l = p, r = x;
updx(f[p] - p + 1, -1);
f[l] = r;
S.insert(l);
updx(r - l + 1, 1);
} else {
updx(1, 1);
S.insert(x);
f[x] = x;
}
}
inline void delx(int x) {
auto i = --S.upper_bound(x);
int p = *i;
int q = f[p];
S.erase(i);
updx(q - p + 1, -1);
if (p < x) {
updx(x - p, 1);
S.insert(p);
f[p] = x - 1;
}
if (x < q) {
updx(q - x, 1);
S.insert(x + 1);
f[x + 1] = q;
}
}
inline void addy(int x) {
auto i = T.lower_bound(x);
if (i != T.end() && *i == x + 1) {
updy(g[*i] - *i + 1, -1);
int l = x, r = g[*i];
if (i != T.begin() && g[*prev(i)] == x - 1) {
int p = *prev(i);
l = p;
updy(g[p] - p + 1, -1);
T.erase(prev(i));
}
T.erase(i);
T.insert(l);
g[l] = r;
updy(r - l + 1, 1);
} else if (i != T.begin() && g[*prev(i)] == x - 1) {
int p = *prev(i);
T.erase(prev(i));
int l = p, r = x;
updy(g[p] - p + 1, -1);
g[l] = r;
T.insert(l);
updy(r - l + 1, 1);
} else {
updy(1, 1);
T.insert(x);
g[x] = x;
}
}
inline void dely(int x) {
auto i = --T.upper_bound(x);
int p = *i;
int q = g[p];
T.erase(i);
updy(q - p + 1, -1);
if (p < x) {
updy(x - p, 1);
T.insert(p);
g[p] = x - 1;
}
if (x < q) {
updy(q - x, 1);
T.insert(x + 1);
g[x + 1] = q;
}
}
void solve() {
n = read();
m = read();
q = read();
for (int i = 1; i <= n; ++i) {
a[i] = read();
}
for (int i = 1; i <= m; ++i) {
b[i] = read();
}
for (int i = 1; i < n; ++i) {
a[i] -= a[i + 1];
}
for (int i = 1; i < m; ++i) {
b[i] -= b[i + 1];
}
ll nn = max(n, m) - 1, tn = n, tm = m;
n = nn;
for (int i = 1, j = 1; i < tn; i = (++j)) {
if (!a[i]) {
continue;
}
while (j + 1 < tn && a[j + 1]) {
++j;
}
S.insert(i);
f[i] = j;
updx(j - i + 1, 1);
}
for (int i = 1, j = 1; i < tm; i = (++j)) {
if (!b[i]) {
continue;
}
while (j + 1 < tm && b[j + 1]) {
++j;
}
T.insert(i);
g[i] = j;
updy(j - i + 1, 1);
}
printf("%lld\n", tn * tm + SGT::a[1].sxy);
while (q--) {
ll op, l, r, x;
op = read();
l = read();
r = read();
x = read();
--l;
if (op == 1) {
if (l) {
ll p = a[l];
a[l] -= x;
if (!p && a[l]) {
addx(l);
} else if (p && !a[l]) {
delx(l);
}
}
if (r < tn) {
ll p = a[r];
a[r] += x;
if (!p && a[r]) {
addx(r);
} else if (p && !a[r]) {
delx(r);
}
}
} else {
if (l) {
ll p = b[l];
b[l] -= x;
if (!p && b[l]) {
addy(l);
} else if (p && !b[l]) {
dely(l);
}
}
if (r < tm) {
ll p = b[r];
b[r] += x;
if (!p && b[r]) {
addy(r);
} else if (p && !b[r]) {
dely(r);
}
}
}
printf("%lld\n", tn * tm + SGT::a[1].sxy);
}
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}
