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給定 \(1\to 2 \to \cdots \to n \to n+1\) 的邊和 \(m\) 條往回走的有向邊，長度都是 \(1\)。站在一個點時等概率選擇一條出邊，求 \(1\to n+1\) 期望長度。

\(1 \le n ,m \le 10^6\)。

Editorial

考慮一個套路：設 \(f_{i}\) 表示從 \(i\to i+1\) 行走距離的期望，答案就是 \(\sum\limits_{i=1}^n f_i\)。

這個東西也很好求……

Code

#include <bits/stdc++.h>

#define LL long long

int read(){
	char k = getchar(); int x = 0;
	while(k < '0' || k > '9') k = getchar();
	while(k >= '0' && k <= '9') x = x * 10 + k - '0' ,k = getchar();
	return x;
}

const LL MOD = 998244353;
LL qpow(LL a ,LL b ,LL p = MOD){
	LL Ans = 1;
	while(b){if(b & 1) Ans = Ans * a % p;
		a = a * a % p ,b >>= 1;
	}return Ans;
}

const int MX = 1e6 + 233;
LL dp[MX] ,zh[MX] ,qzh[MX];
int n ,m;
std::vector<int> fz[MX];

int main(){
	read();
	n = read() ,m = read();
	for(int i = 1 ,u ,v ; i <= m ; ++i){
		u = read() ,v = read();
		if(u != v)	fz[u].push_back(v);
		else zh[u]++;
	}
	LL Ans = 0;
	for(int i = 1 ; i <= n ; ++i){
		int ch = fz[i].size() + 1 + zh[i];
		LL inv = qpow(ch ,MOD - 2);
		dp[i] = 1;
		for(auto j : fz[i]){
			dp[i] = (dp[i] + inv * (qzh[i - 1] - qzh[j - 1] + MOD)) % MOD;
		}
		dp[i] = dp[i] * ch % MOD;
		// printf("f[%d] = %lld\n" ,i ,dp[i]);
		Ans = (Ans + dp[i]) % MOD;
		qzh[i] = (qzh[i - 1] + dp[i]) % MOD;
	}
	using namespace std;
	cout << Ans << endl;
	return 0;
}