CodeForces - 1312D Count the Arrays （组合数，方案数）

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然后套上逆元计算组合数和快速幂就行了

const int MOD = 998244353;
ll fac[MAXN];
void init()
{
	fac[0] = 1;
	for (ll i = 1; i < MAXN; i++) fac[i] = fac[i - 1] * i % MOD;
}
ll exgcd(ll a, ll b, ll &x, ll &y)
{
	if (!b)
	{
		x = 1,y = 0;
		return a;
	}
	ll d = exgcd(b, a % b, y, x);
	y -= a / b * x;
	return d;
}
ll inv(ll a, ll n)
{
	ll x, y;
	exgcd(a, n, x, y);
	return (x + n) % n;
}
ll C(ll n, ll m){return fac[n] * inv(fac[m] * fac[n - m] % MOD, MOD) % MOD;}
ll qpow(ll a,ll b)//a^b//快速幂
{
    if(b==0) return 1;
    a%=MOD;
    ll ans=1,temp=a;
	while(b)
	{
		if(b&1)	 ans=(ans*temp)%MOD;
		temp=(temp*temp)%MOD;
		b>>=1;
	}
	return ans%MOD;
}
void solve()
{
    init();
    ll n,m;cin>>n>>m;
    if(n==2) cout<<0<<endl;
    else cout<<((C(m,n-1)%MOD)*(n-2)%MOD)*(qpow(2,n-3)%MOD)%MOD<<endl;
}