逆元打表

#include<bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n ;
long long inv[10005], f[10005];
void ni(int n)
{
	f[1] = inv[1] = 1;
	for(int i = 2; i <= n; i++)
	{
		f[i] = (i * f[i-1]) % mod;
		inv[i] = (mod - (mod/i))*inv[mod%i]%mod; 
	}
}
int main()
{
	cin >> n;
	ni(n);
	return 0;
}

inv[m ≡-(p/m)*inv[p%m](mod p)