Luogu6327 区间加区间sin和

博主： tony102
发布时间：2020 年 11 月 28 日
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分类：题解

[Link](https://www.luogu.com.cn/problem/P6327)

## 前置知识

这个题还是要先学会和角公式(~~有的什么毒瘤复数的做法太无聊了~~)

$$
\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)
$$

$$
\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)
$$

## Sol

有了上面的公式,你应该就会做这个题了.

加入现在要给 $[l,r]$ 内的所有角加上一个 $\alpha$ 角

那么

$$
\sum_{i=l}^{r} \sin(a_i + \alpha) = \sum_{i=l}^{r} \sin(a_i) \times \cos(\alpha) + \sum_{i= l}^{r} \cos(a_i) \times \sin(\alpha)
$$

那么需要维护区间 $\sin$ 和 $\cos$ 的和

$$
\sum_{i=l}^{r} \cos(a_i + \alpha) = \sum_{i=l}^{r} \cos(a_i) \times \cos(\alpha) - \sum_{i= l}^{r} \sin(a_i) \times \sin(\alpha)
$$

也许到这里你就会了.

还要注意的地方是,每次更新 $\sin$ 和 $\cos$ 的时候要先记录一下没更新前的值,再更新.如果你这么做的话

```tree[lson(p)].sine = cos(tree[p].tag) * tree[lson(p)].sine + sin(tree[p].tag) * tree[lson(p)].cosine;```

```		tree[lson(p)].cosine = tree[lson(p)].cosine * cos(tree[p].tag) - tree[lson(p)].sine * sin(tree[p].tag);```

就会跟我一样~~sb~~

还有要开`long long`

## Code

```cpp
#include <bits/stdc++.h>
using namespace std;

#define int long long

const int SIZE = 3e5 + 5;

int n, q;
int a[SIZE];

struct node {
	int l, r, tag;
	double cosine, sine;
} tree[SIZE << 2];

namespace ae86 {
	const int bufl = 1 << 15;
	char buf[bufl], *s = buf, *t = buf;
	inline int fetch() {
		if (s == t) { t = (s = buf) + fread(buf, 1, bufl, stdin); if (s == t) return EOF; }
		return *s++;
	}
	inline int read() {
		int a = 0, b = 1, c = fetch();
		while (!isdigit(c))b ^= c == '-', c = fetch();
		while (isdigit(c)) a = a * 10 + c - 48, c = fetch();
		return b ? a : -a;
	}
}
using ae86::read;

#define lson(p) p << 1
#define rson(p) p << 1 | 1

inline void pushUp(int p)
{
	tree[p].cosine = tree[lson(p)].cosine + tree[rson(p)].cosine;
	tree[p].sine = tree[lson(p)].sine + tree[rson(p)].sine;
}

inline void pushDown(int p)
{
	if (tree[p].tag != 0.00) {
		double lsine = tree[lson(p)].sine, rsine = tree[rson(p)].sine;
		double lcos = tree[lson(p)].cosine, rcos = tree[rson(p)].cosine;
		tree[lson(p)].sine = cos(tree[p].tag) * lsine + sin(tree[p].tag) * lcos;
		tree[lson(p)].cosine = lcos * cos(tree[p].tag) - lsine * sin(tree[p].tag);
		tree[rson(p)].sine = cos(tree[p].tag) * rsine + sin(tree[p].tag) * rcos;
		tree[rson(p)].cosine = rcos * cos(tree[p].tag) - rsine * sin(tree[p].tag);
		tree[lson(p)].tag += tree[p].tag, tree[rson(p)].tag += tree[p].tag, tree[p].tag = 0;
	}
}

inline void build(int p, int l, int r)
{
	tree[p].l = l, tree[p].r = r;
	if (l == r) {
		tree[p].cosine = cos(a[l]), tree[p].sine = sin(a[l]);
		return;
	}
	int mid = (l + r) >> 1;
	build(lson(p), l, mid); build(rson(p), mid + 1, r);
	pushUp(p);
}

inline void modify(int p, int l, int r, int v)
{
	if (l <= tree[p].l && tree[p].r <= r) {
		double si = tree[p].sine, co = tree[p].cosine;
		tree[p].sine = cos(v) * si + sin(v) * co;
		tree[p].cosine = co * cos(v) - si * sin(v);
		tree[p].tag += v;
		return;
	}
	pushDown(p);
	int mid = (tree[p].l + tree[p].r) >> 1;
	if (l <= mid) modify(lson(p), l, r, v); 
	if (r > mid) modify(rson(p), l, r, v);
	pushUp(p);
}

inline double query(int p, int l, int r)
{
	if (l <= tree[p].l && tree[p].r <= r) {
		return tree[p].sine;
	}
	pushDown(p);
	int mid = (tree[p].l + tree[p].r) >> 1; double ans = 0.00;
	if (l <= mid) ans += query(lson(p), l, r);
	if (r > mid) ans += query(rson(p), l, r);
	return ans;
}

signed main()
{
	n = read();
	for (int i = 1; i <= n; ++ i) {
		a[i] = read();
	}
	build(1, 1, n);
	q = read();
	int opt, l, r, v;
	while (q --) {
		opt = read();
		if (opt == 1) {
			l = read(), r = read(), v = read();
			modify(1, l, r, v);
		}
		if (opt == 2) {    
			l = read(), r = read();
			printf("%.1lf\n", query(1, l, r));
		}
	}
	return 0;
}
```

最后修改：2021 年 09 月 04 日