P10045 [CCPC 2023 北京市赛] 线段树 Solution

原创已于 2025-05-20 21:03:28 修改 · 1k 阅读

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#算法 #数据结构 #c++ #线段树

于 2025-03-20 22:09:16 首次发布

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Description

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

$add⁡(l,r,x)\operatorname{add}(l,r,x)$ ：对每个 $i∈[l,r]i\in[l,r]$ 执行 $ai←ai+xa_i\gets a_i+x$ .
$query⁡(l,r)\operatorname{query}(l,r)$ ：求 $220(\prod\limits_{i=l}^r a_i) \bmod 2^{20}$ .

Limitations

$\le n,m \le 2\times 10^5$
$\le l \le r \le n$
$\le a < 2^{20}$ ，且为奇数.
$\le x < 2^{20}$ ，且为偶数.
$4s,1GB4\text{s},1\text{GB}$

Solution

使用线段树，每个点维护多项式 $∏i=lr(ai+x)\prod\limits_{i=l}^r(a_i+x)$ ，不妨设 $fi=[xi]∏i=lr(ai+x)f_i=[x^i]\prod\limits_{i=l}^r(a_i+x)$ .
考虑区间加 $k$ ，由于 $k$ 为偶数，可以用二项式定理，得到：
$fi′=∑i=0∑j=0(kj×fi+j×(i+jj))f^{\prime}_i=\sum\limits_{i=0}\sum\limits_{j=0}(k^j\times f_{i+j}\times \binom{i+j}{j})$ .
不难发现只用维护 $f0∼f19f_0 \sim f_{19}$ .
因此时间复杂度 $O(c^2n\log n)$ ，空间复杂度 $O (c n)$ ，其中 $c = 20$ 为项数.
要卡常，用 & 1048575 代替 % 1048576 可显著加快速度.

Code

$O2)4.39\text{KB},3.62\text{s},30.73\text{MB}\;\texttt{(in total, C++20 with O2)}$
fastio 删了.

#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 fastio {}
using fastio::read;
using fastio::write;

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 << 1);
		    build(0, 0, n - 1, a);
		}
		
		inline void pushup(int u, int mid) {
			tr[u].poly = tr[ls(mid)].poly * tr[rs(mid)].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, int mid) {
			if (!tr[u].tag) return;
			apply(ls(mid), tr[u].tag), apply(rs(mid), 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(mid), l, mid, a), build(rs(mid), mid + 1, r, a);
			pushup(u, mid);
		}
		
		inline void add(int u, int l, int r, int 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);
		}
		
		inline int query(int u, int l, int r) {
			if (l <= tr[u].l && tr[u].r <= r) return tr[u].poly[0];
			const int mid = (tr[u].l + tr[u].r) >> 1;
			int res = 1;
			pushdown(u, mid);
			if (l <= mid) res = (res * query(ls(mid), l, r)) & mask;
			if (r > mid) res = (res * query(rs(mid), l, r)) & mask;
			return res;
		}
		
		inline void range_add(int l, int r, int x) { add(0, l, r, x); }
		inline int range_prod(int l, int r) { return query(0, l, r); }
	};
}

using poly::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; i < m; i++) {
		int op = read<int>(), l = read<int>(), r = read<int>();
		l--, r--;
		if (op == 1) sgt.range_add(l, r, read<int>());
		else write(sgt.range_prod(l, r)), putchar_unlocked('\n');
	}
	return 0;
}