Dec. 13th 2025

Dec. 13th 2025

A. 动态开点

为了解决空间不够的情况

我们就可以到了该用这个节点的时候再申请空间

代码：

void update(int &id, int d, int posi, int left, int right) {
    if (!id) id = ++cnt; // cnt if the number of nodes;
    if (left == right) {
        tree[id] = d; // +=
        return ;
    }
    int mid = (left + right) >> 1;
    if (posi <= mid) update(ls[id], d, posi, left, mid); // ls[] : left son, rs[] : right son
    else update(rs[id], d, mid + 1, right);
    maitain(id);
}

比较正常的 update ：

void update(ll posi, ll d, ll id = 1, ll left = 1, ll right = n) {
	if (left == right && left == posi) {
		tree[id] += d;
		return ;
	}
	ll mid = (left + right) >> 1;
	if (posi <= mid) update(posi, d, ls(id), left, mid);
	else 			 update(posi, d, rs(id), mid + 1, right);
	maintain(id);
}

就多了一行：

if (!id) id = ++cnt

然后将 ls()、rs() 改成数组

!\(\textcolor{red}{\text{注意}}\)!：因为在 update 中 id 传的是地址，所以不能访问一个常量

例题

【模板】普通平衡树

P3369 【模板】普通平衡树
题目描述

您需要动态地维护一个可重集合 \(M\)，并且提供以下操作：

  1. 向 \(M\) 中插入一个数 \(x\)。
  2. 从 \(M\) 中删除一个数 \(x\)（若有多个相同的数，应只删除一个）。
  3. 查询 \(M\) 中有多少个数比 \(x\) 小，并且将得到的答案加一。
  4. 查询如果将 \(M\) 从小到大排列后，排名位于第 \(x\) 位的数。
  5. 查询 \(M\) 中 \(x\) 的前驱（前驱定义为小于 \(x\)，且最大的数）。
  6. 查询 \(M\) 中 \(x\) 的后继（后继定义为大于 \(x\)，且最小的数）。

对于操作 \(3,5,6\)，不保证当前可重集中存在数 \(x\)。

对于操作 \(5,6\)，保证答案一定存在。

输入格式

第一行为 \(n\)，表示操作的个数，下面 \(n\) 行每行有两个数 \(\text{opt}\) 和 \(x\)，\(\text{opt}\) 表示操作的序号（$ 1 \leq \text{opt} \leq 6 $）。

输出格式

对于操作 \(3,4,5,6\) 每行输出一个数，表示对应答案。

输入输出样例 #1

输入 #1

10
1 106465
4 1
1 317721
1 460929
1 644985
1 84185
1 89851
6 81968
1 492737
5 493598

输出 #1

106465
84185
492737
说明/提示

【数据范围】

对于 \(100\%\) 的数据，\(1\le n \le 10^5\)，\(|x| \le 10^7\)。

来源：Tyvj1728，原名：普通平衡树。

在此鸣谢！

View Code 
#include<bits/stdc++.h>
using namespace std;

using lf = double;
using ll = long long;
using ull = unsigned long long;

const int maxn = 1e5 + 5;
const ll MIN_X = -1e7;
const ll MAX_X = 1e7;

int n;
ll cnt = 1, root = 1;
ll a[maxn];
ll tree[maxn * 40];
ll ls[maxn * 40], rs[maxn * 40];


void maintain(ll id) {
	tree[id] = tree[ls[id]] + tree[rs[id]];
}

void update(ll posi, ll d, ll &id, ll left, ll right) {
	if (!id) id = ++cnt;
	if (left == right && left == posi) {
		tree[id] += d;
		return ;
	}
	ll mid = (left + right) >> 1;
	if (posi <= mid) update(posi, d, ls[id], left, mid);
	else 			 update(posi, d, rs[id], mid + 1, right);
	maintain(id);
}

ll query(ll L, ll R, ll id, ll left, ll right) {
	if (!id) return 0;
	if (L <= left && right <= R) return tree[id];
	ll mid = (left + right) >> 1;
	ll res = 0;
	if (L <= mid) res += query(L, R, ls[id], left, mid);
	if (R > mid)  res += query(L, R, rs[id], mid + 1, right);
	return res;
}

ll kth_element(ll k, ll id, ll left, ll right) {
	if (left == right) return left;
	ll mid = (left + right) >> 1;
	if (k <= tree[ls[id]]) return kth_element(k, ls[id], left, mid);
	else            	   return kth_element(k - tree[ls[id]], rs[id], mid + 1, right);
}

ll frank(ll x) {
	return query(MIN_X, x - 1, 1, MIN_X, MAX_X) + 1;
}

ll prev(ll x) {
	return kth_element(frank(x) - 1, 1, MIN_X, MAX_X);
}

ll suf(ll x) {
	return kth_element(frank(x + 1), 1, MIN_X, MAX_X);
}

void solve() {
	
	cin >> n;
	
	for (int i = 1; i <= n; i++) {
		int opt, x;
		cin >> opt >> x;
		if (opt == 1) update(x, 1, root, MIN_X, MAX_X);
		else if (opt == 2) update(x, -1, root, MIN_X, MAX_X);
		else if (opt == 3) cout << frank(x) << "\n";
		else if (opt == 4) cout << kth_element(x, 1, MIN_X, MAX_X) << "\n";
		else if (opt == 5) cout << prev(x) << "\n";
		else cout << suf(x) << "\n";
	}
	
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	
	//	freopen("T1.in", "r", stdin);
	//	freopen("T1.out", "w", stdout);
	
	int T = 1;
	//	cin >> T;
	while (T--) solve();
	
	return 0;
}

CF915E Physical Education Lessons

View Code 
#include<bits/stdc++.h>
using namespace std;

using lf = double;
using ll = long long;
using ull = unsigned long long;

const int maxn = 3e5 + 5;
const int MIN_X = 0;
const int MAX_X = 1e9;

int n, q;
int cnt = 1, root = 1;
int a[maxn];
int tree[maxn  * 70];
int lazy_tag[maxn * 70];
int ls[maxn * 70], rs[maxn * 70];

void maintain(int id) {
	tree[id] = tree[ls[id]] + tree[rs[id]];
}

void addtag(int &id, int d, int left, int right) {
	if (!id) id = ++cnt;
	tree[id] = d * (right - left + 1);
	lazy_tag[id] = d;
}

void pushdown(int id, int left, int right) {
	if (lazy_tag[id]) {
		int mid = left + ((right - left) >> 1);
		addtag(ls[id], lazy_tag[id], left, mid);
		addtag(rs[id], lazy_tag[id], mid + 1, right);
		lazy_tag[id] = 0;
	}
}

void update(int L, int R, int d, int &id, int left, int right) {
	if (!id) id = ++cnt;
	if (L <= left && right <= R) {
//		tree[id] += d;
		addtag(id, d, left, right);
		return ;
	}
	pushdown(id, left, right);
	int mid = left + ((right - left) >> 1);
	if (L <= mid) update(L, R, d, ls[id], left, mid);
	if (R > mid)  update(L, R, d, rs[id], mid + 1, right);
	maintain(id);
}

int query(int L, int R, int id, int left, int right) {
	if (!id) return 0;
	if (L <= left && right <= R) return tree[id];
	pushdown(id, left, right);
	int mid = left + ((right - left) >> 1);
	int res = 0;
	if (L <= mid) res += query(L, R, ls[id], left, mid);
	if (R > mid)  res += query(L, R, rs[id], mid + 1, right);
	return res;
}

int kth_element(int k, int id, int left, int right) {
	if (left == right) return left;
	int mid = left + (right - left) >> 1;
	if (k <= tree[ls[id]]) return kth_element(k, ls[id], left, mid);
	else            	   return kth_element(k - tree[ls[id]], rs[id], mid + 1, right);
}

int frank(int x) {
	return query(MIN_X, x - 1, 1, MIN_X, MAX_X) + 1;
}

int prev(int x) {
	return kth_element(frank(x) - 1, 1, MIN_X, MAX_X);
}

int suf(int x) {
	return kth_element(frank(x + 1), 1, MIN_X, MAX_X);
}

void solve() {
	
	cin >> n >> q;
	
	update(1, n, 1, root, MIN_X, MAX_X);
	
	for (int i = 1; i <= q; i++) {
		int l, r, opt;
		cin >> l >> r >> opt;
		
		update(l, r, opt - 1, root, MIN_X, MAX_X);
		
		cout << query(1, n, 1, MIN_X, MAX_X) << "\n";
	}
	
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	
	//	freopen("T1.in", "r", stdin);
	//	freopen("T1.out", "w", stdout);
	
	int T = 1;
	//	cin >> T;
	while (T--) solve();
	
	return 0;
}

【模板】线段树 1.5

View Code 
#include<bits/stdc++.h>
using namespace std;

using lf = double;
using ll = unsigned long long;
//using ull = long long;

const int maxn = 1e5 + 5;
const ll MIN_X = 1;
const ll MAX_X = 1e9;

int n, m;
ll cnt = 1, root = 1;
ll a[maxn];
ll tree[maxn * 70];
ll lazy_tag[maxn * 70];
ll ls[maxn * 70], rs[maxn * 70];

void maintain(ll id) {
	tree[id] = tree[ls[id]] + tree[rs[id]];
}

void addtag(ll &id, ll d, ll left, ll right) {
	if (!id) 
		id = ++cnt;
//		tree[id] = (left + right) * (right - left + 1) >> 1;
	
	tree[id] += (right - left + 1) * d;
	lazy_tag[id] += d;
}

void pushdown(ll id, ll left, ll right) {
	if (lazy_tag[id]) {
		ll mid = (left + right) >> 1;
		addtag(ls[id], lazy_tag[id], left, mid);
		addtag(rs[id], lazy_tag[id], mid + 1, right);
		lazy_tag[id] = 0;
	}
}

void update(ll L, ll R, ll d, ll &id, ll left, ll right) {
	if (!id) 
		id = ++cnt;
//		tree[id] = (left + right) * (right - left + 1) >> 1;
	
	if (L <= left && right <= R) {
		addtag(id, d, left, right);
		return ;
	}
	pushdown(id, left, right);
	ll mid = (left + right) >> 1;
	if (L <= mid) update(L, R, d, ls[id], left, mid);
	if (R > mid)  update(L, R, d, rs[id], mid + 1, right);
	maintain(id);
}

ll query(ll L, ll R, ll &id, ll left, ll right) {
	if (!id) 
		id = ++cnt;
//		tree[id] = (left + right) * (right - left + 1) >> 1;
	
	if (L <= left && right <= R) return tree[id];
	pushdown(id, left, right);
	ll mid = (left + right) >> 1;
	ll res = 0;
	if (L <= mid) res += query(L, R, ls[id], left, mid);
	if (R > mid)  res += query(L, R, rs[id], mid + 1, right);
	return res;
}

void solve() {

	cin >> n >> m;

	for (int i = 1; i <= m; i++) {
		int opt;
		cin >> opt;
		if (opt == 1) {
			ll l, r, k;
			cin >> l >> r >> k;
			update(l, r, k, root, 1, n);
		} else {
			ll l, r;
			cin >> l >> r;
			cout << ((l + r) * (r - l + 1) >> 1) + query(l, r, root, 1, n) << "\n";
		}
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);

	//	freopen("T1.in", "r", stdin);
	//	freopen("T1.out", "w", stdout);

	int T = 1;
	//	cin >> T;
	while (T--) solve();

	return 0;
}

B.周测

T1 hotel

组内链接

题目描述
奶牛们正北上加拿大雷湾进行文化之旅，准备在苏必利尔湖阳光明媚的湖畔享受假期。作为称职的旅行代理，贝茜将著名的坎伯兰街公牛麋鹿酒店定为它们的下榻之处。这座巨型酒店拥有N间客房（1 ≤ N ≤ 50,000），所有房间都位于一条极长走廊的同一侧（当然是为了更好地欣赏湖景）。

奶牛和其他旅客以Di（1 ≤ Di ≤ N）为团体规模抵达前台办理入住。每个团体i会向当值的麋鹿管理员坎穆申请Di间连续客房。若有可用连续房间，坎穆会分配房号r至r+Di-1的序列；若无可用连续房间，则会礼貌推荐其他住宿。坎穆总是尽可能选择最小的r值。

旅客也会以连续房间为单位退房。每次退房请求包含参数Xi和Di，表示腾空Xi至Xi+Di-1号房间（1 ≤ Xi ≤ N-Di+1）。这些房间在退房前可能部分或全部本就空置。

你的任务是协助坎穆处理M次（1 ≤ M < 50,000）入住/退房请求。酒店初始处于全空状态。

输入格式

  • 第1行：两个空格分隔的整数：N和M
  • 第2..M+1行：第i+1行包含以下两种格式之一的请求：

(a) 两个空格分隔的整数表示入住请求：1和Di

(b) 三个空格分隔的整数表示退房请求：2、Xi和Di

输出格式

  • 第1...行：对每个入住请求，输出一行整数r表示分配到的连续房间起始房号。若无法满足请求，则输出0。

原题

View Code 
#include<bits/stdc++.h>
using namespace std;

using lf = double;
using ll = long long;
using ull = unsigned long long;

const int maxn = 5e4 + 5;

struct node{
	ll l, r;
	ll sum;
};

int n, m;
node tree[maxn << 2];
ll lazy_tag[maxn << 2];

ll ls(ll id) {return id << 1;}
ll rs(ll id) {return id << 1 | 1;}

void maitain(ll id, ll left, ll right) {
	ll mid = (left + right) >> 1;
	tree[id].l = tree[ls(id)].l;
	if (tree[ls(id)].l == mid - left + 1)
		tree[id].l += tree[rs(id)].l;
	tree[id].r = tree[rs(id)].r;
	if (tree[rs(id)].r == right - mid)
		tree[id].r += tree[ls(id)].r;
	tree[id].sum = max({tree[ls(id)].sum, tree[rs(id)].sum, tree[ls(id)].r + tree[rs(id)].l});
}

void pushdown(ll id, ll left, ll right) {
	if (lazy_tag[id] != -1) {
		lazy_tag[ls(id)] = lazy_tag[rs(id)] = lazy_tag[id];
		if (lazy_tag[id] == 1) {
			tree[ls(id)].l = tree[ls(id)].r = tree[ls(id)].sum = 0;
			tree[rs(id)].l = tree[rs(id)].r = tree[rs(id)].sum = 0;
		} else {
			ll mid = (left + right) >> 1;
			tree[ls(id)].l = tree[ls(id)].r = tree[ls(id)].sum = mid - left + 1;
			tree[rs(id)].l = tree[rs(id)].r = tree[rs(id)].sum = right - mid;
		}
		lazy_tag[id] = -1;
	}
}

void build(ll id, ll left, ll right) {
	lazy_tag[id] = -1;
	if (left == right) return tree[id].l = tree[id].r = tree[id].sum = 1, void();
	
	ll mid = (left + right) >> 1;
	build(ls(id), left, mid);
	build(rs(id), mid + 1, right);
	maitain(id, left, right);
}

void upadate(ll L, ll R, ll d, ll id = 1, ll left = 1, ll right = n) {
	if (L <= left && right <= R) {
		ll what = 0;
		if (d == 0) what = right - left + 1;
		lazy_tag[id] = d;
		tree[id].l = tree[id].r = tree[id].sum = what;
		return ;
	}
	pushdown(id, left, right);
	ll mid = (left + right) >> 1;
	if (L <= mid) upadate(L, R, d, ls(id), left, mid);
	if (R > mid) upadate(L, R, d, rs(id), mid + 1, right);
	maitain(id, left, right);
}

ll query(ll d, ll id = 1, ll left = 1, ll right = n) {
	if (left == right) return left;
	ll mid = (left + right) >> 1;
	if (tree[ls(id)].sum >= d) return query(d, ls(id), left, mid);
	else if (tree[ls(id)].r + tree[rs(id)].l >= d) return mid - tree[ls(id)].r + 1;
	else return query(d, rs(id), mid + 1, right);
	return 0;
}

void solve() {
	cin >> n >> m;
	build(1, 1, n);
	for (int i = 1; i <= m; i++) {
		int opt;
		cin >> opt;
		
		if (opt == 1) {
			int d;
			cin >> d;
			if (d > tree[1].sum) {
				cout << "0\n";
				continue;
			}
			int r = query(d);
			cout << r << "\n";
			upadate(r, r + d - 1, 1);
		} else {
			int x, d;
			cin >> x >> d;
			upadate(x, x + d - 1, 0);
		}
		
	}
}

int main() {

	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	
//	freopen("hotel.in", "r", stdin);
////	freopen("hotel1.in", "r", stdin);
//	freopen("hotel.out", "w", stdout);
	
	int T = 1;
//	cin >> T;
	while (T--) solve();

	return 0;
}

T2 Palindrome Query

哈哈哈，打的暴力比某些Hash得的分还要高

View BF Code 
#include<bits/stdc++.h>
using namespace std;

using lf = double;
using ll = long long;
using ull = unsigned long long;

void solve() {

	int n, m;
	cin >> n >> m;
	string str;
	cin >> str;

	for (int i = 1; i <= m; i++){
		int opt;
		cin >> opt;
		if (opt == 1) {
			int x;
			char c;
			cin >> x >> c;
			str[x - 1] = c;
		} else {
			int l, r;
			cin >> l >> r;
			int i = l, j = r;
			bool flag = 1;
			if ((r - l + 1) & 1) {
				while (j - i > 1) {
					if (str[i - 1] != str[j - 1]) {
						cout << "No\n";
						flag = 0;
						break;
					}
					i++, j--;
				}
				if (flag) cout << "Yes\n";
			} else {
				while (i < j) {
					if (str[i - 1] != str[j - 1]) {
						cout << "No\n";
						flag = 0;
						break;
					}
					i++, j--;
				}
				if (flag) cout << "Yes\n";
			}
		}
	}
}

int main() {

	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);

//	freopen("palindrome.in", "r", stdin);
//	freopen("palindrome.out", "w", stdout);

	int T = 1;
	//	cin >> T;
	while (T--) solve();

	return 0;
}

思路：
一眼 Hash
听到有人双Hash 哈希冲突了
还有个单Hash 冲突的(模数用的1333331)

Segment Tree 的结构体设计：

struct node {
	long long L, R;
	long long len;
}

\(\rarr\) 结构体中 + 的实现：

node operator+(node a,node b){
	long long L=a.L*Pow[b.len]%mod+b.L;
	L%=mod;
	long long R=b.R*Pow[a.len]%mod+a.R;
	R%=mod;
	int len=a.len+b.len;
	return {L,R,len};
}

模数用 1e9 + 7 就好

代码：

View Code 
#include<bits/stdc++.h>
using namespace std;

using lf = double;
using ll = long long;
using ull = unsigned long long;

const int maxn = 1e6 + 5;
ll ls(ll i) {
	return i << 1;
}
ll rs(ll i) {
	return i << 1 | 1;
}

const ll M = 1e9 + 7;

#define mod %

struct node{
	ll len;
	ll left_hash, right_hash;
	node(){}
	node(char c) {
		len = 1;
		left_hash = c - 'a' + 1;
		right_hash = c - 'a' + 1;
	}
	node(ll l, ll lh, ll rh) : len(l), left_hash(lh), right_hash(rh) {}
};

string str;
node tree[maxn << 2];
ll base = 131;
ll Pow[maxn];

node operator+(node a, node b) {
	node res;
	res.len = a.len + b.len;
	res.left_hash = (a.left_hash * Pow[b.len] mod M + b.left_hash) mod M;
	res.left_hash = (res.left_hash + M) mod M;
	res.right_hash = (b.right_hash * Pow[a.len] mod M + a.right_hash) mod M;
	res.right_hash = (res.right_hash + M) mod M;
	return res;
}

void maintain(ll id) { // push up
	tree[id] = tree[ls(id)] + tree[rs(id)];
}

void build(ll id, ll left_, ll right_) {
	if (left_ == right_) {
		tree[id] = node(str[left_]);
		return;
	}
	ll mid = (left_ + right_) >> 1;
	build(ls(id), left_, mid);
	build(rs(id), mid + 1, right_);
	maintain(id);
}

void update_range(ll posi, ll id, ll left_, ll right_, char d) {
	if (left_ == right_) {
		tree[id] = node(d);
		return;
	}
	ll mid = (left_ + right_) >> 1;
	if (posi <= mid) update_range(posi, ls(id), left_, mid, d);
	else if (posi > mid) update_range(posi, rs(id), mid + 1, right_, d);
	maintain(id);
}

node query_range(ll L, ll R, ll id, ll left_, ll right_) {
	if (L <= left_ && right_ <= R) return tree[id];
	ll mid = (left_ + right_) >> 1;
	node res(0, 0, 0);
	if (L <= mid) res = res + query_range(L, R, ls(id), left_, mid);
	if (R > mid) res = res + query_range(L, R, rs(id), mid + 1, right_);
	return res;
}

node query(ll L, ll R, int n) {
	return query_range(L, R, 1, 1, n);
}

void update(ll posi, char d, int n) {
	update_range(posi, 1, 1, n, d);
}


void solve() {
	int n, m;
	cin >> n >> m;
	
	cin >> str;
	
	str = " " + str;
	
	Pow[0] = 1;
	for (int i = 1; i <= maxn - 5; i++)
		Pow[i] = (Pow[i - 1] * base mod M + M) mod M;
	
	build(1, 1, n);
	
	while (m--) {
		int oper;
		cin >> oper;
		
		if (oper == 1) {
			int x;
			char c;
			cin >> x >> c;
			update(x, c, n);
			str[x] = c;
		} else {
			int l, r;
			cin >> l >> r;
			node res = query(l, r, n);
			res.left_hash = (res.left_hash + M) mod M;
			res.right_hash = (res.right_hash + M) mod M;
			if (res.left_hash == res.right_hash) cout << "Yes\n";
			else								 cout << "No\n";
			
//			for (int i = l; i <= r; i++) cout << str[i];
//			cout << "\n";
//			
//			cout << res.left_hash << " " << res.right_hash << "\n";
		}
	}
	
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	
	int T = 1;
	while (T--) solve();
	
	return 0;
}

T3 回转寿司

权值线段树+动态开点

对于每一个 sum[i] ，查询 [sum[i] - r, sum[i] - l] 内的值得个数

View Code 
#include <bits/stdc++.h>
using namespace std;

using lf = double;
using ll = long long;
using ull = unsigned long long;

const int maxn = 1e5 + 5;
const ll MIN_X = -1e10;
const ll MAX_X = 1e10;

int n;
ll cnt = 1, root = 1;
ll a[maxn];
ll tree[maxn * 70];
ll ls[maxn * 70], rs[maxn * 70];

void maintain(ll id) { tree[id] = tree[ls[id]] + tree[rs[id]]; }

void update(ll posi, ll d, ll &id, ll left = MIN_X, ll right = MAX_X) {
	if (!id)
		id = ++cnt;
	if (left == right && left == posi) {
		tree[id] += d;
		return;
	}
	ll mid = (left + right) >> 1;
	if (posi <= mid)
		update(posi, d, ls[id], left, mid);
	else
		update(posi, d, rs[id], mid + 1, right);
	maintain(id);
}

ll query(ll L, ll R, ll &id, ll left = MIN_X, ll right = MAX_X) {
	if (!id)
		id = ++cnt;
	if (L <= left && right <= R)
		return tree[id];
	ll mid = (left + right) >> 1;
	ll res = 0;
	if (L <= mid)
		res += query(L, R, ls[id], left, mid);
	if (R > mid)
		res += query(L, R, rs[id], mid + 1, right);
	return res;
}

void solve() {
	
	int n, l, r;
	cin >> n >> l >> r;
	
	for (int i = 1; i <= n; i++) {
		int x;
		cin >> x;
		a[i] = a[i - 1] + x;
	}
	
	update(0, 1, root);
	
	ll ans = 0;
	for (int i = 1; i <= n; i++) {
		ans += query(a[i] - r, a[i] - l, root);
		update(a[i], 1, root);
	}
	
	cout << ans << "\n";
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	
	//	freopen("T1.in", "r", stdin);
	//	freopen("T1.out", "w", stdout);
	
	int T = 1;
	//	cin >> T;
	while (T--) solve();
	
	return 0;
}
posted @ 2025-12-13 10:35  Yangyihao  阅读(3)  评论(1)    收藏  举报