【Codeforces 961 G】Partitions（組合數學）

傳送門

每個集合權值有一個$|S|$
考慮一個集合權值的組合意義
可以看做對於每個數，集合裏每個數都對他計算一次
於是對於每個數考慮他被計算多少次
如果自己對自己，顯然就是$S(n,k)$次，$S$爲第二類斯特林數
否則別的$n-1$個和自己一個集合的方案就是$S(n-1,k)*(n-1)$次
$O(n)$算一下第二類斯特林數就完了

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline ll readll(){
    char ch=gc();
    ll res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline int readstring(char *s){
	int top=0;char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	return top;
}
template<typename tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<typename tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=1e9+7;
inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=(r>=mod)?(r%mod):r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline int fix(int x){return (x<0)?x+mod:x;}
cs int N=200005;
int fac[N],ifac[N];
inline void init_inv(){
	fac[0]=ifac[0]=1;
	for(int i=1;i<N;i++)fac[i]=mul(fac[i-1],i);
	ifac[N-1]=Inv(fac[N-1]);
	for(int i=N-2;i;i--)ifac[i]=mul(ifac[i+1],i+1);
}
inline int C(int n,int m){return n<m?0:mul(fac[n],mul(ifac[m],ifac[n-m]));}
inline int S(int n,int k){
	int res=0;
	for(int i=0;i<=k;i++){
		if((k-i)&1)Dec(res,mul(C(k,i),ksm(i,n)));
		else Add(res,mul(C(k,i),ksm(i,n)));
	}
	Mul(res,ifac[k]);return res;
}
int n,k;
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	init_inv();
	n=read(),k=read();
	int res=0;
	for(int i=1;i<=n;i++)Add(res,read());
	Mul(res,add(S(n,k),mul(n-1,S(n-1,k))));
	cout<<res<<'\n';
	return 0;
}