【Codeforces 891 E】 Lust（生成函数）

传送门

考虑每次贡献可以差分成$a_j{\prod }_{i\ne j}{a}_i-(a_j-1){\prod }_{i\ne j}{a}_i$
那么可以发现最后答案就是${\prod }_i{a}_i-E({\prod }_i{d}_i)，d_i表示操作完后的a_i$
考虑求后面一块
显然可以构造生成函数
$$f_t(x)=\sum _{i=0}^{\infty }\frac{(a_t-i)x^i}{i!}$$
求得就是就是$\frac{k!}{n^k}[x^k]{\prod }_i{f}_i(x)$
考虑$f_t(x)=a_t{\sum }_i\frac{x^i}{i!}-{\sum }_i\frac{x^{i+1}}{i!}=(a_t-x)e^x$
那么考虑最后乘出来的多项式为$e^{nx}({\prod }_t(a_t-x))=e^{nx}{\sum }_i{g}_i{x}^i$
那么就是$\frac{k!}{n^k}{\sum }_{i=0}^n{f}_i\ast \frac{n^{k-i}}{(k-i)!}=k^{\underline{i}}{f}_i/n^i$

暴力做多项式乘法是$O(n^2)$
如果写$mtt$可以做到$O(nlo{g}^{2}n)$

#include
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair
#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=5005;
int f[N],n,k;
int a[N];
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read(),k=read();
	f[0]=1;int res=1;
	for(int i=1;i<=n;i++){
		int x=read();Mul(res,x);
		for(int j=i;~j;j--){
			Mul(f[j],x);
			if(j)Dec(f[j],f[j-1]);
		}
	}
	for(int pw1=1,pw2=1,mt=Inv(n),i=0;i<=n;i++){
		Dec(res,mul(pw1,mul(pw2,f[i])));
		Mul(pw1,k-i),Mul(pw2,mt);
	}
	cout<<res<<'\n';
	return 0;
}