题意

Problem Link
\(n \leq 5 \times 10^5\)

题解

第一条注意的就是这个是条件概率！
所以每个位置分开算概率是不对的， 所以这不是sb题。
所以得维护\((x, y)\)表示\(a_x = a_y\)的概率。
然后分类讨论一下就好了。
所以还是sb题

/*
	QiuQiu /qq
  ____    _           _                 __                
  / __ \  (_)         | |               / /                
 | |  | |  _   _   _  | |  _   _       / /    __ _    __ _ 
 | |  | | | | | | | | | | | | | |     / /    / _` |  / _` |
 | |__| | | | | |_| | | | | |_| |    / /    | (_| | | (_| |
  \___\_\ |_|  \__,_| |_|  \__, |   /_/      \__, |  \__, |
                            __/ |               | |     | |
                           |___/                |_|     |_|
*/

#include <bits/stdc++.h>

using namespace std;

class Input {
	#define MX 1000000
	private :
		char buf[MX], *p1 = buf, *p2 = buf;
		inline char gc() {
			if(p1 == p2) p2 = (p1 = buf) + fread(buf, 1, MX, stdin);
			return p1 == p2 ? EOF : *(p1 ++);
		}
	public :
		Input() {
			#ifdef Open_File
				freopen("a.in", "r", stdin);
				freopen("a.out", "w", stdout);
			#endif
		}
		template <typename T>
		inline Input& operator >>(T &x) {
			x = 0; int f = 1; char a = gc();
			for(; ! isdigit(a); a = gc()) if(a == '-') f = -1;
			for(; isdigit(a); a = gc()) 
				x = x * 10 + a - '0';
			x *= f;
			return *this;
		}
		inline Input& operator >>(char &ch) {
			while(1) {
				ch = gc();
				if(ch != '\n' && ch != ' ') return *this;
			}
		}
		inline Input& operator >>(char *s) {
			int p = 0;
			while(1) {
				s[p] = gc();
				if(s[p] == '\n' || s[p] == ' ' || s[p] == EOF) break;
				p ++; 
			}
			s[p] = '\0';
			return *this;
		}
	#undef MX
} Fin;

class Output {
	#define MX 1000000
	private :
		char ouf[MX], *p1 = ouf, *p2 = ouf;
		char Of[105], *o1 = Of, *o2 = Of;
		void flush() { fwrite(ouf, 1, p2 - p1, stdout); p2 = p1; }
		inline void pc(char ch) {
			* (p2 ++) = ch;
			if(p2 == p1 + MX) flush();
		}
	public :
		template <typename T> 
		inline Output& operator << (T n) {
			if(n < 0) pc('-'), n = -n;
			if(n == 0) pc('0');
			while(n) *(o1 ++) = (n % 10) ^ 48, n /= 10;
			while(o1 != o2) pc(* (--o1));
			return *this; 
		}
		inline Output & operator << (char ch) {
			pc(ch); return *this; 
		}
		inline Output & operator <<(const char *ch) {
			const char *p = ch;
			while( *p != '\0' ) pc(* p ++);
			return * this;
		}
		~Output() { flush(); } 
	#undef MX
} Fout;

#define cin Fin
#define cout Fout
#define endl '\n'

using LL = long long;
using pii = pair<int, int>;

const int P = 998244353;

int power(int x, int k) {
	int res = 1;
	while(k) {
		if(k & 1) res = 1ll * res * x % P; 
		x = 1ll * x * x % P; k >>= 1;
	}
	return res;
}

inline int Mod(int x) { return x + ((x >> 31) & P); }
inline void pls(int &x, int v) { x = Mod(x + v - P); }
inline void dec(int &x, int v) { x = Mod(x - v); }

const int N = 1e5 + 10;

int n, m;

struct Node {
	Node *ls, *rs;
	int l, r, p;
	Node() {}
	Node(int _l, int _r) : l(_l), r(_r), p(0), ls(NULL), rs(NULL) {}
	inline void down(int q) {
		p = (1ll * p * (P + 1 - q) + 1ll * q * (P + 1 - p)) % P;
	}
} ;

struct SegmentTree1 {
	Node *root;
	SegmentTree1() { root = NULL; }
	void modify(Node *&x, int l, int r, int L, int R, int p) {
		if(x == NULL) x = new Node(l, r);
		if(L <= l && r <= R) { x -> down(p); return ; }
		int mid = (l + r) >> 1;
		if(L <= mid) modify(x -> ls, l, mid, L, R, p);
		if(R > mid) modify(x -> rs, mid + 1, r, L, R, p);
		return ;
	}
	void query(Node *x, int l, int r, int pos, int &p) {
		if(x == NULL) return ;
	//	cerr << x -> p << endl;
		p = (1ll * p * (P + 1 - x -> p) + 1ll * x -> p * (P + 1 - p) ) % P;
		int mid = (l + r) >> 1;
		if(pos <= mid) query(x -> ls, l, mid, pos, p);
		else query(x -> rs, mid + 1, r, pos, p);
	}
	void Modify(int l, int r, int L, int R, int p) {
		modify(root, l, r, L, R, p);
	}
	void Query(int l, int r, int pos, int &p) {
		query(root, l, r, pos, p);
	}
} ;

struct SegmentTree2 {
	#define ls(x) (x << 1)
	#define rs(x) (x << 1 | 1)
	SegmentTree1 tr[N << 2];
	void modify(int x, int l, int r, int L1, int R1, int L2, int R2, int p) {
		if(L1 <= l && r <= R1) {
			//cerr << l << ' ' << r << endl;
			//cerr << x << endl;
			tr[x].Modify(1, n, L2, R2, p); return ;
		}
		int mid = (l + r) >> 1;
		if(L1 <= mid) modify(ls(x), l, mid, L1, R1, L2, R2, p);
		if(R1 > mid) modify(rs(x), mid + 1, r, L1, R1, L2, R2, p);
	}
	void query(int x, int l, int r, int px, int py, int &p) {
		tr[x].Query(1, n, py, p);
		if(l == r) return ;
		int mid = (l + r) >> 1;
		if(px <= mid) query(ls(x), l, mid, px, py, p);
		else query(rs(x), mid + 1, r, px, py, p);
	}
} ;

SegmentTree1 Tree1;
SegmentTree2 Tree2;

int main() {
	cin >> n >> m;
	while(m --) {
		int opt, l, r; cin >> opt >> l >> r;
		if(opt == 1) {
			int p = power(r - l + 1, P - 2);
			if(l > 1) Tree2.modify(1, 1, n, 1, l - 1, l, r, p);
			if(r < n) Tree2.modify(1, 1, n, l, r, r + 1, n, p);
			Tree1.Modify(1, n, l, r, P + 1 - p);
			pls(p, p);
			Tree2.modify(1, 1, n, l, r, l, r, p);
			if(l > 1) Tree1.Modify(1, n, 1, l - 1, 1);
			if(r < n) Tree1.Modify(1, n, r + 1, n, 1);
		}
		else {
			int ans = 0;
			if(l == 1) 
				Tree1.Query(1, n, r, ans);
			else 
				Tree2.query(1, 1, n, l - 1, r, ans);
			cout << (P + 1 - ans) % P << endl;
		}
	}
	return 0;
}