Description

给定序列 $a=(a_1,a_2,\cdots ,a_n)$，有 $m$ 个操作分三种：

• $\mathrm{add}(l,r,k)$：对每个 $i\in [l,r]$ 执行 $a_i\leftarrow a_i+k$.
• $\mathrm{square}(l,r)$：对每个 $i\in [l,r]$ 执行 $a_i\leftarrow \lfloor \sqrt{a_i}\rfloor$.
• $\mathrm{query}(l,r)$：求 ${\sum }_{i=l}^r{a}_i$.

Limitations

$1\le n,m\le 10^5$
$1\le l\le r\le n$
$1\le a_i,k\le 10^5$

Solution

考虑 $\mathrm{sqrt}$ 操作怎么做.
显然当区间极差为 $0$ 时，直接对整个区间开方.
但是当极差为 $1$ 时，可以被开方直到极差为 $1$ 后加回去的方法卡成 $O(nm)$.
这个也很好解决，如果当前区间与开方后的区间极差均为 $1$，也直接进行开方.

Code

$3.3\text{KB},637\text{ms},11.57\text{MB}\text{(in total, C++20)}$

#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;
}