P6327 区间加区间 sin 和 Solution

Description

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

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

Limitations

$1\le n,m,a_i,k\le 10^5$
$1\le l\le r\le n$
$1\text{s},125\text{MB}$

Solution

首先考虑转化，使用泰勒展开，然后维护 $k$ 次方和，但精度显然会炸.
事实上，有公式：

• $\sin (x+y)=\sin x\cos y+\cos x\sin y$.
• $\cos (x+y)=\cos x\cos y-\sin x\sin y$.

于是可用线段树维护区间 $\sin$ 和与 $\cos$ 和.
区间加考虑打 $\text{tag}$，pushdown 时用上面公式即可做到 $O(1)$.

Code

$2.64\text{KB},4.59\text{s},13.45\text{MB}\text{(in total, C++20 with O2)}$

#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;
}

namespace seg_tree {
	struct Node {
		int l, r;
		i64 tag = 0;
		f8 sin = 0., cos = 0.;
	};
	
	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);
		}
		
		void apply(int u, i64 c, f8 sc, f8 cc) {
			f8 si = tr[u].sin, ci = tr[u].cos;
			tr[u].sin = si * cc + ci * sc;
			tr[u].cos = cc * ci - si * sc;
			tr[u].tag += c;
		}
		
		void pushup(int u, int mid) {
			tr[u].sin = tr[ls(mid)].sin + tr[rs(mid)].sin;
			tr[u].cos = tr[ls(mid)].cos + tr[rs(mid)].cos;
		}
		
		void pushdown(int u, int mid) {
			if (!tr[u].tag) return;
			i64 x = tr[u].tag;
			f8 sx = sin(x), cx = cos(x);
			apply(ls(mid), x, sx, cx);
			apply(rs(mid), x, sx, cx);
			tr[u].tag = 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].sin = sin(a[l]);
				tr[u].cos = cos(a[l]);
				return;
			}
			
			int mid = (l + r) >> 1;
			build(ls(mid), l, mid, a);
			build(rs(mid), mid + 1, r, a);
			pushup(u, mid);
		}
		
		void modify(int u, int l, int r, int c) {
			if (l <= tr[u].l && tr[u].r <= r) {
				apply(u, c, sin(c), cos(c));
				return;
			}
			int mid = (tr[u].l + tr[u].r) >> 1;
			pushdown(u, mid);
			if (l <= mid) modify(ls(mid), l, r, c);
			if (mid < r) modify(rs(mid), l, r, c);
			pushup(u, mid);
		}
		
		f8 query(int u, int l, int r) {
			if (l <= tr[u].l && tr[u].r <= r) return tr[u].sin;
			f8 ans = 0;
			int mid = (tr[u].l + tr[u].r) >> 1;
			pushdown(u, mid);
			if (l <= mid) ans += query(ls(mid), l, r);
			if (mid < r) ans += query(rs(mid), l, r);
			return ans; 
		}
		
		void range_add(int l, int r, int v) { modify(0, l, r, v); }
		f8 range_sine(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; cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];
	SegTree sgt(a);
	int m; cin >> m;
	for (int i = 0, op, l, r, c; i < m; i++) {
		cin >> op >> l >> r, l--, r--;
		if (op == 1) {
			cin >> c;
			sgt.range_add(l, r, c);
		}
		else printf("%.1lf\n", sgt.range_sine(l, r));
	}
	
	return 0;
}