arc106F - Figures

题目大意

有n个点集，第i个点集的大小为ai，所有点互不相同，一条边连接两个点集中的某一对点，求以点集为单位的生成树个数同时满足每个点只被连最多一次

2<=n<=2e5

题解

直接爬

考虑prufer序上计数，一个度数为x的数的出现次数为x-1，数i的度数为j的方案为A(ai,j)

\(\prod_i\sum_{j>=0} \frac{x^j}{j!} A_{ai}^{j+1}=\prod_i ai(x+1)^{ai-1}\)

乘起来算即可

code

#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define mod 998244353
#define Mod 998244351
#define ll long long
//#define file
using namespace std;

int a[200001],n,i,j,k,l;
ll ans,sum;

ll qpower(ll a,int b) {ll ans=1; while (b) {if (b&1) ans=ans*a%mod;a=a*a%mod;b>>=1;} return ans;}
ll C(ll n,int m)
{
	ll s1=1,s2=1,i;
	fd(i,n,n-m+1) s1=s1*(i%mod)%mod;
	fo(i,1,m) s2=s2*i%mod;
	return s1*qpower(s2,Mod)%mod;
}

int main()
{
	#ifdef file
	freopen("f.in","r",stdin);
	#endif
	
	scanf("%d",&n),ans=1;
	fo(i,1,n) scanf("%d",&a[i]),sum+=a[i]-1,ans=ans*a[i]%mod;
	
	ans=ans*C(sum,n-2)%mod;
	fo(i,1,n-2) ans=ans*i%mod;
	printf("%lld\n",ans);
	
	fclose(stdin);
	fclose(stdout);
	return 0;
}