Description
定义由 42 42 42 为初始数，朝后依次拼接 4 , 2 4,2 4,2 的数为 好数 .
给定序列 a = ( a 1 , a 2 , ⋯ , a n ) a=(a_1,a_2,\cdots,a_n) a=(a1,a2,⋯,an)，有 m m m 个操作分四种：
- add ( l , r , k ) \operatorname{add}(l,r,k) add(l,r,k)：对每个 i ∈ [ l , r ] i\in[l,r] i∈[l,r] 执行 a i ← a i + k a_i\gets a_i+k ai←ai+k.
- mul ( l , r , k ) \operatorname{mul}(l,r,k) mul(l,r,k)：对每个 i ∈ [ l , r ] i\in[l,r] i∈[l,r] 执行 a i ← a i × k a_i\gets a_i\times k ai←ai×k.
- assign ( l , r , k ) \operatorname{assign}(l,r,k) assign(l,r,k)：对每个 i ∈ [ l , r ] i\in[l,r] i∈[l,r] 执行 a i ← k a_i\gets k ai←k.
- query ( l , r ) \operatorname{query}(l,r) query(l,r)：求 a l ∼ a r a_l\sim a_r al∼ar 中好数个数.
Limitations
1 ≤ n , m , k ≤ 5 × 1 0 5 1\le n,m,k\le 5\times 10^5 1≤n,m,k≤5×105
1 ≤ l ≤ r ≤ n 1\le l\le r\le n 1≤l≤r≤n
保证任意时刻 1 ≤ a i ≤ 5 × 1 0 5 1\le a_i\le5\times 10^5 1≤ai≤5×105.
0.7 s , 64 MB \textcolor{red}{0.7\text s},64\text{MB} 0.7s,64MB
Solution
设 f ( x ) f(x) f(x) 表示第一个不小于 x x x 的好数与 x x x 的差.
但 f ( x ) f(x) f(x) 的图像不连续，不好维护，但 V = 5 × 1 0 5 V=5\times 10^5 V=5×105 时，好数只有 5 5 5 个.
由于要区间操作，考虑用 势能线段树 ，每个节点维护区间内 f ( a i ) f(a_i) f(ai) 的最小值 mix \textit{mix} mix，以及满足 f ( a i ) = mix f(a_i)=\textit{mix} f(ai)=mix 的 a i a_i ai 个数 cnt \textit{cnt} cnt.
对于 add \operatorname{add} add 操作，结合函数图像，若当前节点的 mix ≤ k \textit{mix}\le k mix≤k，则直接打加标记，否则递归子树继续修改.
对于 mul \operatorname{mul} mul 操作，直接暴力做，每个数至多被乘 log V \log V logV 次.
对于 assign \operatorname{assign} assign 操作，直接打标记，但它会让前两个操作的复杂度假掉，不过很好解决，维护区间最大最小值，当两者相等时，直接打标记覆盖成 max + ( 或 × ) k \textit{max} +(\text{或} \,\times)\, k max+(或×)k 即可.
对于 query \operatorname{query} query，若 mix = 0 \textit{mix}=0 mix=0，则答案为 cnt \textit{cnt} cnt，否则为 0 0 0.
时间复杂度 O ( m log n log V ) O(m\log n\log V) O(mlognlogV)，注意忽略掉 k = 1 k=1 k=1 的 mul \operatorname{mul} mul 操作，以及卡常.
Code
4.69 KB , 201 ms , 34.85 MB (in total, C++20 with O2) 4.69\text{KB},201\text{ms},34.85\text{MB}\;\texttt{(in total, C++20 with O2)} 4.69KB,201ms,34.85MB(in total, C++20 with O2)
cpp
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
using ui64 = unsigned long long;
using i128 = __int128;
using ui128 = unsigned __int128;
using f4 = float;
using f8 = double;
using f16 = long double;
template<class T>
bool chmax(T &a, const T &b){
if(a < b){ a = b; return true; }
return false;
}
template<class T>
bool chmin(T &a, const T &b){
if(a > b){ a = b; return true; }
return false;
}
// #define LOCAL
#ifdef LOCAL
#define getchar_unlocked getchar
#define putchar_unlocked putchar
#endif
template<class T>
inline T read() {
T x = 0, f = 1;
char ch = getchar_unlocked();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar_unlocked();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar_unlocked();
}
return x * f;
}
template<class T>
inline void write(T x) {
if (x < 0) {
putchar_unlocked('-');
x = -x;
}
if (x > 9) write(x / 10);
putchar_unlocked(x % 10 + '0');
return;
}
constexpr int V = 5e5;
namespace seg_tree {
inline int ls(int u) { return 2 * u + 1; }
inline int rs(int u) { return 2 * u + 2; }
struct Node {
int l, r;
int max, min, mix, cnt, add = 0, cov = 0;
};
array<int, V + 1> nxt;
void init() {
int cur = V + 1;
nxt.fill(-1);
nxt[42] = nxt[424] = nxt[4242] = nxt[42424] = nxt[424242] = 0;
for (int i = V; i >= 0; i--) {
if (nxt[i] == 0) cur = i;
else nxt[i] = cur - i;
}
}
struct SegTree {
vector<Node> tr;
inline SegTree() {}
inline SegTree(const vector<int>& a) {
const int n = a.size();
tr.resize(n << 1);
build(0, 0, n - 1, a);
}
inline void pushup(int u, int mid) {
tr[u].max = max(tr[ls(mid)].max, tr[rs(mid)].max);
tr[u].min = min(tr[ls(mid)].min, tr[rs(mid)].min);
tr[u].mix = min(tr[ls(mid)].mix, tr[rs(mid)].mix);
tr[u].cnt = ((tr[u].mix == tr[ls(mid)].mix) * tr[ls(mid)].cnt
+ (tr[u].mix == tr[rs(mid)].mix) * tr[rs(mid)].cnt);
}
inline void applyC(int u, int x) {
tr[u].max = tr[u].min = x;
tr[u].mix = nxt[x];
tr[u].cnt = tr[u].r - tr[u].l + 1;
tr[u].cov = x;
tr[u].add = 0;
}
inline void applyA(int u, int x) {
tr[u].mix -= x;
tr[u].add += x;
tr[u].min += x;
tr[u].max += x;
}
inline void pushdown(int u, int mid) {
if (tr[u].cov) {
applyC(ls(mid), tr[u].cov);
applyC(rs(mid), tr[u].cov);
tr[u].cov = 0;
}
if (tr[u].add) {
applyA(ls(mid), tr[u].add);
applyA(rs(mid), tr[u].add);
tr[u].add = 0;
}
}
void build(int u, int l, int r, const vector<int>& a) {
tr[u].l = l, tr[u].r = r;
if (l == r) {
tr[u].max = tr[u].min = a[l];
tr[u].mix = nxt[a[l]];
tr[u].cnt = 1;
return;
}
const int mid = (l + r) >> 1;
build(ls(mid), l, mid, a);
build(rs(mid), mid + 1, r, a);
pushup(u, mid);
}
void add(int u, int l, int r, int x) {
if (l <= tr[u].l && tr[u].r <= r) {
if (tr[u].mix > x) return applyA(u, x);
if (tr[u].min == tr[u].max) return applyC(u, tr[u].min + x);
}
const int mid = (tr[u].l + tr[u].r) >> 1;
pushdown(u, mid);
if (l <= mid) add(ls(mid), l, r, x);
if (r > mid) add(rs(mid), l, r, x);
pushup(u, mid);
}
void mul(int u, int l, int r, int x) {
if (x == 1) return;
if (l <= tr[u].l && tr[u].r <= r && tr[u].max == tr[u].min)
return applyC(u, tr[u].min * x);
const int mid = (tr[u].l + tr[u].r) >> 1;
pushdown(u, mid);
if (l <= mid) mul(ls(mid), l, r, x);
if (r > mid) mul(rs(mid), l, r, x);
pushup(u, mid);
}
void assign(int u, int l, int r, int x) {
if (l <= tr[u].l && tr[u].r <= r) return applyC(u, x);
const int mid = (tr[u].l + tr[u].r) >> 1;
pushdown(u, mid);
if (l <= mid) assign(ls(mid), l, r, x);
if (r > mid) assign(rs(mid), l, r, x);
pushup(u, mid);
}
int query(int u, int l, int r) {
if (l <= tr[u].l && tr[u].r <= r) return (tr[u].mix == 0) * tr[u].cnt;
const int mid = (tr[u].l + tr[u].r) >> 1;
int res = 0;
pushdown(u, mid);
if (l <= mid) res += query(ls(mid), l, r);
if (r > mid) res += query(rs(mid), l, r);
return res;
}
};
}
using seg_tree::init;
using seg_tree::SegTree;
signed main() {
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
const int n = read<int>(), m = read<int>();
vector<int> a(n);
for (int i = 0; i < n; i++) a[i] = read<int>();
init();
SegTree sgt(a);
for (int i = 0, op, l, r, v; i < m; i++) {
op = read<int>(), l = read<int>(), r = read<int>(), l--, r--;
if (op == 4) {
write(sgt.query(0, l, r));
putchar_unlocked('\n');
}
else {
v = read<int>();
if (op == 1) sgt.add(0, l, r, v);
if (op == 2) sgt.mul(0, l, r, v);
if (op == 3) sgt.assign(0, l, r, v);
}
}
return 0;
}