阳间数据结构杂题选讲

\[\text{orz lxl sto} \]

Codechef DGCD (Weaker) / AcWing 246

给定一个长度为 \(n\) 的数列 \(A = (a_1, a_2, \dots, a_n)\)，支持两种操作：

• C L R d：将 \(a_L, a_{L+1}, \dots, a_R\) 都加上 \(d\)。

• Q L R：查询 \(\gcd(a_L, a_{L+1}, \dots, a_R)\)。

\(1 \leq n \leq 500000\)。

根据辗转相减法，有：

\[\gcd(x,y)=\gcd(x,y-x) \]

拓展一下，就有：

\[\gcd(p_1, p_2, \dots, p_m) = \gcd(p_1, p_2 - p_1, p_3 - p_2, \dots, p_m - p_{m-1}) \]

如果加上 \(d\)，影响是：

\[\gcd(p_1 + d, p_2 + d, \dots, p_m + d) = \gcd(p_1 + d, p_2 - p_1, p_3 - p_2, \dots, p_m - p_{m-1}) \]

维护两棵线段树即可。（因为单纯的区间加线段树会爆 long long，所以我们采取区间 min 的线段树，反正也只需要查询单点值）

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
inline int read () {
	int x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	return !f ? x : -x;
} 
const int N = 5e5 + 5;
namespace ori {
	int mn[N << 2], inc[N << 2], seq[N];
	inline void push_up (int u) {mn[u] = min (mn[u << 1], mn[u << 1 | 1]);}
	inline void push_down (int u) {
		if (!inc[u]) return ;
		inc[u << 1] += inc[u], inc[u << 1 | 1] += inc[u];
		mn[u << 1] += inc[u], mn[u << 1 | 1] += inc[u];
		inc[u] = 0;
	}
	inline void build (int u, int l, int r) {
		inc[u] = 0;
		if (l == r) {mn[u] = seq[l]; return ;}
		int mid = l + r >> 1;
		build (u << 1, l, mid), build (u << 1 | 1, mid + 1, r);
		push_up (u);
	}
	inline void modify (int u, int l, int r, int x, int y, int v) {
		if (x <= l && r <= y) {mn[u] += v, inc[u] += v; return ;}
		push_down (u);
		int mid = l + r >> 1;
		if (x <= mid) modify (u << 1, l, mid, x, y, v);
		if (y > mid) modify (u << 1 | 1, mid + 1, r, x, y, v);
		push_up (u); 
	}
	inline int query (int u, int l, int r, int x, int y) {
		if (x <= l && r <= y) return mn[u];
		push_down (u);
		int mid = l + r >> 1, ans = 9223372036854775807ll;
		if (x <= mid) ans = min (ans, query (u << 1, l, mid, x, y));
		if (y > mid) ans = min (ans, query (u << 1 | 1, mid + 1, r, x, y));
		return ans;
	}
}
inline int gcd (int a, int b) {
	if (!a || !b) return a + b;
	else {
		int tag = ((a < 0) && (b < 0));
		a = abs (a), b = abs (b);
		if (tag) tag = -1; else tag = 1;
		return tag * gcd (b, a % b);
	}
}
namespace diff {
	int val[N << 2], inc[N << 2], seq[N];
	inline void push_up (int u) {val[u] = gcd (val[u << 1], val[u << 1 | 1]);}
	inline void push_down (int u) {
		if (inc[u] == 9223372036854775807ll) return ;
		inc[u << 1] = inc[u << 1 | 1] = val[u << 1] = val[u << 1 | 1] = inc[u];
		inc[u] = 9223372036854775807ll;
	}
	inline void build (int u, int l, int r) {
		inc[u] = 9223372036854775807ll;
		if (l == r) {val[u] = seq[l]; return ;}
		int mid = l + r >> 1;
		build (u << 1, l, mid), build (u << 1 | 1, mid + 1, r);
		push_up (u);
	}
	inline void update (int u, int l, int r, int x, int y, int v) {
		if (x <= l && r <= y) {val[u] = inc[u] = v; return ;}
		push_down (u);
		int mid = l + r >> 1;
		if (x <= mid) update (u << 1, l, mid, x, y, v);
		if (y > mid) update (u << 1 | 1, mid + 1, r, x, y, v);
		push_up (u); 
	}
	inline int query (int u, int l, int r, int x, int y) {
		if (x <= l && r <= y) return val[u];
		int mid = l + r >> 1, ans = 0;
		if (x <= mid) ans = gcd (ans, query (u << 1, l, mid, x, y));
		if (y > mid) ans = gcd (ans, query (u << 1 | 1, mid + 1, r, x, y));
		return ans;
	}
}
int n, m;
signed main () {
	n = read (), m = read ();
	ori :: seq[0] = 0;
	for (int i = 1;i <= n; ++ i) ori :: seq[i] = read ();
	for (int i = 1;i <= n; ++ i) diff :: seq[i] = ori :: seq[i] - ori :: seq[i - 1];
	ori :: build (1, 1, n), diff :: build (1, 1, n);
	while (m --) {
		char op[2];
		scanf ("%s", op + 1);
		if (op[1] == 'C') {
			int l = read (), r = read (), d = read ();
			ori :: modify (1, 1, n, l, r, d);
			int val1 = ori :: query (1, 1, n, l, l);
			if (l > 1) val1 -= ori :: query (1, 1, n, l - 1, l - 1);
			diff :: update (1, 1, n, l, l, val1);
			if (r + 1 <= n) diff :: update (1, 1, n, r + 1, r + 1, ori :: query (1, 1, n, r + 1, r + 1) - ori :: query (1, 1, n, r, r));
		}
		else {
			int l = read (), r = read ();
			int res = ori :: query (1, 1, n, l, l);
			if (l + 1 <= r) res = gcd (res, diff :: query (1, 1, n, l + 1, r));
			printf ("%lld\n", res);
		}
	}
	return 0;
}

P6327 区间加区间sin和

给定一个长度为 \(n\) 的数列 \(A = (a_1, a_2, \dots, a_n)\)，支持两种操作：

• 1 l r v：将 \(a_l, a_{l+1}, \dots, a_r\) 都加上 \(v\)。

• 2 l r：计算 \(\displaystyle \sum_{i = l}^{r} \sin (a_i)\)。

\(1 \leq n \leq 2 \times 10^5\)。

首先需要知道两个三角函数的公式：

\[\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]

\[\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

那么我们对于每一个线段树上的区间 \([l,r]\)，都维护两个值，\(\displaystyle \sum_{i = l}^{r}\sin(a_i)\) 和 \(\displaystyle \sum_{i = l}^{r}\cos(a_i)\)。

如果现在给 \(a[l,r]\) 加上 \(d\)，会这样变：

\[\sum_{i=l}^{r}\sin(a_i + d) = \sum_{i = l}^{r}\sin a_i \cos d+\sum_{i=l}^{r}\cos a_i \sin d = \cos d \sum_{i=l}^{r}\sin(a_i) + \sin d\sum_{i=l}^{r}\cos(a_i) \]

\[\sum_{i=l}^{r}\cos(a_i + d) = \sum_{i=l}^{r}\cos a_i \cos d - \sum_{i=l}^{r} \sin a_i \sin d = \cos d\sum_{i=l}^{r}\cos(a_i) - \sin d\sum_{i=l}^{r}\sin(a_i) \]

剩下的就交给线段树就行。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
//#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
inline int read () {
	int x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	return !f ? x : -x;
}
const int N = 2e5 + 5;
int n, m, a[N];
double _cos[N << 2], _sin[N << 2], inc[N << 2];
inline void push_up (int u) {
	_sin[u] = _sin[u << 1] + _sin[u << 1 | 1];
	_cos[u] = _cos[u << 1] + _cos[u << 1 | 1];
}
inline void push_down (int u) {
	inc[u << 1] += inc[u];
	inc[u << 1 | 1] += inc[u];
	double sins = _sin[u << 1], coss = _cos[u << 1];
	_sin[u << 1] = cos (inc[u]) * sins + sin (inc[u]) * coss;
	_cos[u << 1] = cos (inc[u]) * coss - sin (inc[u]) * sins;
	sins = _sin[u << 1 | 1], coss = _cos[u << 1 | 1];
	_sin[u << 1 | 1] = cos (inc[u]) * sins + sin (inc[u]) * coss;
	_cos[u << 1 | 1] = cos (inc[u]) * coss - sin (inc[u]) * sins;
	inc[u] = 0;
}
inline void build (int u, int l, int r) {
	if (l == r) {
		_sin[u] = sin (a[l]);
		_cos[u] = cos (a[l]);
		return ;
	}
	int mid = l + r >> 1;
	build (u << 1, l, mid), build (u << 1 | 1, mid + 1, r);
	push_up (u);
}
inline void modify (int u, int l, int r, int x, int y, int v) {
	if (x <= l && r <= y) {
		inc[u] += v;
		double sins = _sin[u], coss = _cos[u];
		_sin[u] = cos (v) * sins + sin (v) * coss;
		_cos[u] = cos (v) * coss - sin (v) * sins;
		return ;
	}
	push_down (u);
	int mid = l + r >> 1;
	if (x <= mid) modify (u << 1, l, mid, x, y, v);
	if (y > mid) modify (u << 1 | 1, mid + 1, r, x, y, v);
	push_up (u); 
}
inline double query (int u, int l, int r, int x, int y) {
	if (x <= l && r <= y) return _sin[u];
	push_down (u);
	int mid = l + r >> 1;
	double res = 0.0;
	if (x <= mid) res += query (u << 1, l, mid, x, y);
	if (y > mid) res += query (u << 1 | 1, mid + 1, r, x, y);
	return res;
}
signed main () {
	n = read ();
	for (int i = 1;i <= n; ++ i) a[i] = read ();
	build (1, 1, n);
	m = read ();
	while (m --) {
		int op = read ();
		if (op == 1) {
			int l = read (), r = read (), v = read ();
			modify (1, 1, n, l, r, v);
		}
		else {
			int l = read (), r = read ();
			printf ("%.1lf\n", query (1, 1, n, l, r));
		}
	}
	return 0;
}