abc231g

\(\frac{1}{n^k} \sum\frac{k!}{\prod b_i!} \prod (a_i+b_i)\)，其中 \(\sum b_i=k\)

构造生成函数 \(f_i=\sum \frac{a_i+j}{j!}x^j=e^x(a_i+x)\)，欲求式为 \(k![x^k]\prod f_i=k![x^k] e^{nx}\prod (a_i+x)\)

预处理 \(g_i\) 为任选 \(i\) 个乘积的和，原式为 \(\frac{1}{n^k} k!\sum\limits_{i=0}^n g_{n-i} n^{k-i} \frac{1}{(k-i)!}=\sum\limits_{i=0}^n g_{n-i} \frac{k^{\underline i}}{n^i}\)

#include<bits/stdc++.h>
using namespace std;
#define inf 1e9
const int maxn=2e5+10;
const int mod=998244353;
inline int read(){
	int x=0,f=1;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
	return x*f;
}
const int N=2e3+5;
int n,k,a[N],dp[N][N],ans;
inline int ksm(int x,int y){
	int res=1;
	while(y){
		if(y&1)res=1ll*res*x%mod;
		x=1ll*x*x%mod;y>>=1;
	}return res;
}
int main(){
	n=read(),k=read();
	for(int i=1;i<=n;i++)a[i]=read();
	dp[0][0]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=i;j++)
			dp[i][j]=(dp[i-1][j]+1ll*a[i]*dp[i-1][j-1])%mod;
	int K=ksm(n,mod-2);
	for(int i=0,k1=1,k2=1;i<=n;i++){
		ans=(ans+1ll*dp[n][n-i]*k1%mod*k2)%mod;
		k1=1ll*k1*K%mod;k2=1ll*k2*(k-i)%mod;
	}printf("%d\n",ans);
    return 0;
}