Description
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 个操作分三种：
- 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.
- square ( l , r ) \operatorname{square}(l,r) square(l,r)：对每个 i ∈ [ l , r ] i\in[l,r] i∈[l,r] 执行 a i ← ⌊ a i ⌋ a_i\gets \lfloor \sqrt {a_i} \rfloor ai←⌊ai ⌋.
- query ( l , r ) \operatorname{query}(l,r) query(l,r)：求 ∑ i = l r a i \sum_{i=l}^r a_i ∑i=lrai.
Limitations
1 ≤ n , m ≤ 1 0 5 1\le n,m\le 10^5 1≤n,m≤105
1 ≤ l ≤ r ≤ n 1\le l\le r\le n 1≤l≤r≤n
1 ≤ a i , k ≤ 1 0 5 1\le a_i,k\le 10^5 1≤ai,k≤105
Solution
考虑 sqrt \operatorname{sqrt} sqrt 操作怎么做.
显然当区间极差为 0 0 0 时，直接对整个区间开方.
但是当极差为 1 1 1 时，可以被开方直到极差为 1 1 1 后加回去的方法卡成 O ( n m ) O(nm) O(nm).
这个也很好解决，如果当前区间与开方后的区间极差均为 1 1 1，也直接进行开方.
Code
3.3 KB , 637 ms , 11.57 MB (in total, C++20) 3.3\text{KB},637\text{ms},11.57\text{MB}\;\texttt{(in total, C++20)} 3.3KB,637ms,11.57MB(in total, C++20)
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;
}
namespace seg_tree {
struct Node {
int l, r;
i64 max, min, sum, tag;
};
inline int ls(int u) { return 2 * u + 1; }
inline int rs(int u) { return 2 * u + 2; }
struct SegTree {
vector<Node> tr;
inline SegTree() {}
inline SegTree(const vector<i64>& 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].sum = tr[ls(mid)].sum + tr[rs(mid)].sum;
tr[u].max = max(tr[ls(mid)].max, tr[rs(mid)].max);
tr[u].min = min(tr[ls(mid)].min, tr[rs(mid)].min);
}
inline void apply(int u, i64 tag) {
tr[u].sum += tag * (tr[u].r - tr[u].l + 1);
tr[u].min += tag;
tr[u].max += tag;
tr[u].tag += tag;
}
inline void pushdown(int u, int mid) {
if (tr[u].tag) {
apply(ls(mid), tr[u].tag);
apply(rs(mid), tr[u].tag);
tr[u].tag = 0;
}
}
void build(int u, int l, int r, const vector<i64>& a) {
tr[u].l = l, tr[u].r = r;
if (l == r) return (void)(tr[u].sum = tr[u].min = tr[u].max = a[l]);
const int mid = (l + r) >> 1;
build(ls(mid), l, mid, a);
build(rs(mid), mid + 1, r, a);
pushup(u, mid);
}
void sqrt(int u, int l, int r) {
if (tr[u].max == 1) return;
if (l <= tr[u].l && tr[u].r <= r) {
i64 tmin = std::sqrt(tr[u].min), tmax = std::sqrt(tr[u].max);
if (tr[u].min == tr[u].max ||
(tmin + 1 == tmax && tr[u].min + 1 == tr[u].max)) {
return apply(u, tmax - tr[u].max);
}
}
const int mid = (tr[u].l + tr[u].r) >> 1;
pushdown(u, mid);
if (l <= mid) sqrt(ls(mid), l, r);
if (r > mid) sqrt(rs(mid), l, r);
pushup(u, mid);
}
void add(int u, int l, int r, i64 k) {
if (l <= tr[u].l && tr[u].r <= r) return apply(u, k);
const int mid = (tr[u].l + tr[u].r) >> 1;
pushdown(u, mid);
if (l <= mid) add(ls(mid), l, r, k);
if (r > mid) add(rs(mid), l, r, k);
pushup(u, mid);
}
i64 query(int u, int l, int r) {
if (l <= tr[u].l && tr[u].r <= r) return tr[u].sum;
const int mid = (tr[u].l + tr[u].r) >> 1;
i64 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;
}
inline void range_sqrt(int l, int r) { sqrt(0, l, r); }
inline void range_add(int l, int r, i64 v) { add(0, l, r, v); }
inline i64 range_sum(int l, int r) { return query(0, l, r); }
};
}
using seg_tree::SegTree;
signed main() {
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
int n, m; scanf("%d %d", &n, &m);
vector<i64> a(n);
for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
SegTree sgt(a);
for (int i = 0, op, l, r, v; i < m; i++) {
scanf("%d %d %d", &op, &l, &r), l--, r--;
if (op == 1) {
scanf("%d", &v);
sgt.range_add(l, r, v);
}
else if (op == 2) sgt.range_sqrt(l, r);
else printf("%lld\n", sgt.range_sum(l, r));
}
return 0;
}