P10045 [CCPC 2023 北京市赛] 线段树

Description

给定序列 $a=(a_1,a_2,\cdots ,a_n)$，有 $m$ 个操作分两种：

• $\mathrm{add}(l,r,x)$：对每个 $i\in [l,r]$ 执行 $a_i\leftarrow a_i+x$.
• $\mathrm{query}(l,r)$：求 $(\prod _{i=l}^r{a}_i)\mathrm{mod}{2}^{20}$.

Limitations

$1\le n,m\le 2\times 10^5$
$1\le l\le r\le n$
$1\le a<2^{20}$，且为奇数.
$0\le x<2^{20}$，且为偶数.
$4\text{s},1\text{GB}$

Solution

使用线段树，每个点维护多项式 $\prod _{i=l}^r(a_i+x)$，不难发现只用维护 $x_0\sim x_{19}$ 项.
不妨设展开后 $x^i$ 项系数为 $f_i$.
考虑区间加 $k$，由于 $k$ 为偶数，可以枚举贡献，得到：
$f_i^{\prime }=\sum _{i=0}^{19}(k^j\times f_{i+j}\times (\genfrac{}{}{0ex}{}{i+j}{j}))$.
时间复杂度 $O(c^{2}n\log n)$，其中 $c=20$ 为项数.
用 & 代替 % 可显著加快速度.

Code

$3.47\text{KB},3.71\text{s},46.03\text{MB}\text{(in total, C++20 with O2)}$

#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;
}

constexpr int mask = 1048575, L = 20;

namespace poly {
	array<array<int, L + 1>, L + 1> C;
	
	struct Poly {
		vector<int> p;
		inline Poly() {}
		inline Poly(int val) : p(2) { p[0] = val; p[1] = 1; }
		inline void resize(int n) { p.resize(n); }
		inline int size() const { return p.size(); }
		inline int& operator[](int i) { return p[i]; }
		inline int operator[](int i) const { return p[i]; }
	};
	
	Poly operator*(const Poly& a, const Poly& b) {
		const int n = a.size(), m = b.size();
		Poly c; c.resize(min(n + m - 1, L));
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m && i + j < L; j++)
				c[i + j] = (c[i + j] + 1LL * a[i] * b[j]) & mask;
		return c;
	}
	
	Poly operator+(const Poly& a, int k) {
		const int l = a.size();
		Poly r; r.resize(l);
		for (int i = 0; i < l; i++) {
			for (int j = 0, pw = 1; i + j < l; j++) {
				r[i] = (((1LL * pw * a[i + j]) & mask) * C[i + j][j] + r[i]) & mask;
				pw = (1LL * pw * k) & mask;
			}	
		}
		return r;
	}
	
	inline void init() {
		for (int i = 0; i <= L; i++) {
			C[i][0] = 1;
			for (int j = 1; j <= i; j++)
				C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) & mask;
		}
	}
}

namespace seg_tree {
	using poly::Poly;
	struct Node {
		int l, r, tag;
		Poly poly;
	};
	
	inline int ls(int u) { return 2 * u + 1; }
	inline int rs(int u) { return 2 * u + 2; }
	
	struct SegTree {
		vector<Node> tr;
		SegTree() {}
		SegTree(const vector<int>& a) {
			const int n = a.size();
			tr.resize(n << 2);
		    build(0, 0, n - 1, a);
		}
		
		inline void pushup(int u) {
			tr[u].poly = tr[ls(u)].poly * tr[rs(u)].poly;
		}
		
		inline void apply(int u, int k) {
			tr[u].tag = (tr[u].tag + k) & mask;
			tr[u].poly = tr[u].poly + k;
		}
		
		inline void pushdown(int u) {
			if (!tr[u].tag) return;
			apply(ls(u), tr[u].tag), apply(rs(u), tr[u].tag);
			tr[u].tag = 0;
		}
		
		inline 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].poly = a[l];
				return;
			}
			const int mid = (l + r) >> 1;
			build(ls(u), l, mid, a), build(rs(u), mid + 1, r, a);
			pushup(u);
		}
		
		inline void add(int u, int l, int r, int k) {
			if (l <= tr[u].l && tr[u].r <= r) return apply(u, k);
			pushdown(u);
			const int mid = (tr[u].l + tr[u].r) >> 1;
			if (l <= mid) add(ls(u), l, r, k);
			if (r > mid) add(rs(u), l, r, k);
			pushup(u);
		}
		
		inline int query(int u, int l, int r) {
			if (l <= tr[u].l && tr[u].r <= r) return tr[u].poly[0];
			pushdown(u);
			const int mid = (tr[u].l + tr[u].r) >> 1;
			int res = 1;
			if (l <= mid) res = (res * query(ls(u), l, r)) & mask;
			if (r > mid) res = (res * query(rs(u), l, r)) & mask;
			return res;
		}
	};
}

using poly::init;
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<int> a(n);
	for (int i = 0; i < n; i++) scanf("%d", &a[i]);
	
	init();
	SegTree sgt(a);
	for (int i = 0, op, l, r, x; i < m; i++) {
		scanf("%d %d %d", &op, &l, &r), l--, r--;
		if (op == 1) {
			scanf("%d", &x);
			sgt.add(0, l, r, x);
		}
		else printf("%d\n", sgt.query(0, l, r));
	}
	return 0;
}