\[
\sum _{\sum v_i = n-2} \prod (a_i ^{v_i+1} * (v_i+1) ^ m /v_i!) *(\sum (v_i+1)^m)
\]
将 \(\sum (v_i+1)^m\) 中的贡献分开算。
我们有两个生成函数。
第一个:
\[
\sum _i a^{i+1} * (v_i+1)^m x^i / i!
\]
第二个:
\[
\sum _i a^{i+1} * (i+1)^{2m} x^i / i!
\]
先将 \(\prod a\) 算到前面。
令:
\[
A(x) =\sum x^i * (i+1)^{m} /i!
\B(x) =\sum x^i * (i+1)^{2m} /i!
\]
我们要算的是:
\[
\sum _i B(a_ix)/A(a_ix) * \prod A(a_jx)
\]
后面可以ln,exp。
然后发现泥算个等幂和就行了。
补充一下等幂和怎么算:
生成函数是:
\[
\sum \frac 1 {1-a_ix}
\]
因为
\[
\ln' (f(x)) = \frac {f'(x)}{f(x)}
\\ln' (1-a_ix) = \frac {-a_i} {1-a_ix} = -a_i - a_i^2 x
\]
所以可以一个分治FFT算。
最终复杂度$n\log ^2n $
#include<bits/stdc++.h>
using namespace std;
const int N=5e5+5;
typedef long long ll;
const int mod=998244353;
int add(int a,int b){a+=b;return a>=mod?a-mod:a;}
int sub(int a,int b){a-=b;return a<0?a+mod:a;}
int mul(int a,int b){return (ll)a*b%mod;}
int qpow(int a,int b){int ret=1;for(;b;b>>=1,a=mul(a,a))if(b&1)ret=mul(ret,a);return ret;}
/*math*/
namespace Template_Poly{
typedef vector<int> Poly;
int rev[N];
Poly Poly_add(Poly A,Poly B){
A.resize(max(A.size(),B.size()));
for(size_t i=0;i<B.size();i++)A[i]=add(A[i],B[i]);
return A;
}
Poly Poly_sub(Poly A,Poly B){
A.resize(max(A.size(),B.size()));
for(size_t i=0;i<B.size();i++)A[i]=sub(A[i],B[i]);
return A;
}
void DFT(int *t,int n,int type){
int l=0;while(1<<l<n)++l;
for(int i=0;i<n;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
for(int i=0;i<n;i++)if(rev[i]>i)swap(t[rev[i]],t[i]);
for(int step=1;step<n;step<<=1){
int wn=qpow(3,(mod-1)/(step<<1));
for(int i=0;i<n;i+=step<<1){
int w=1;
for(int k=0;k<step;k++,w=mul(w,wn)){
int x=t[i+k],y=mul(t[i+k+step],w);
t[i+k]=add(x,y),t[i+k+step]=sub(x,y);
}
}
}
if(type==1)return;
for(int i=1;i<n-i;i++)swap(t[i],t[n-i]);
int inv=qpow(n,mod-2);
for(int i=0;i<n;i++)t[i]=mul(t[i],inv);
}
Poly NTT(Poly A,int n,Poly B,int m){
static Poly res,PolA,PolB;
PolA=A,PolB=B;
int len=1;while(len < n+m)len<<=1;
res.resize(len);
PolA.resize(len),PolB.resize(len);
DFT(&PolA[0],len,1);DFT(&PolB[0],len,1);
for(int i=0;i<len;i++) res[i]= mul(PolA[i],PolB[i]);
DFT(&res[0],len,-1);
res.resize(n+m-1);
return res;
}
Poly NTT(Poly A,Poly B){
return NTT(A,A.size(),B,B.size());
}
Poly Poly_inv(Poly A,int n){
if(n==1)return Poly(1,qpow(A[0],mod-2));
int len=1<<((int)ceil(log2(n))+1);
Poly x=Poly_inv(A,(n+1)>>1),y;
x.resize(len),y.resize(len);
for(int i=0;i<n;i++)y[i]=A[i];
DFT(&x[0],len,1),DFT(&y[0],len,1);
for(int i=0;i<len;i++)x[i]=mul(x[i],sub(2,mul(x[i],y[i])));
DFT(&x[0],len,-1);
x.resize(n);
return x;
}
Poly Poly_inv(Poly A){
return Poly_inv(A,A.size());
}
Poly Deri(Poly A){
int n=A.size();
for(int i=1;i<n;i++)A[i-1]=mul(A[i],i);
A.resize(n-1);
return A;
}
Poly Inte(Poly A){
int n=A.size();
A.resize(n+1);
for(int i=n;i;i--)A[i]=mul(A[i-1],qpow(i,mod-2));
A[0]=0;
return A;
}
Poly ln(Poly A){
int len=A.size();
A=Inte(NTT(Deri(A),Poly_inv(A)));
A.resize(len);
return A;
}
Poly exp(Poly A,int n){
if(n==1)return Poly(1,1);
Poly x=exp(A,(n+1)>>1),y;
x.resize(n);
y=ln(x);
for(int i=0;i<n;i++)y[i]=sub(A[i],y[i]);
y[0]++;
x=NTT(x,y);
x.resize(n);
return x;
}
Poly exp(Poly A){
return exp(A,A.size());
}
Poly sqrt(Poly A,int n){
if(n==1)return Poly(1,1);
Poly x=sqrt(A,(n+1)>>1),y;
x.resize(n),y.resize(n);
for(int i=0;i<n;i++)y[i]=A[i];
x=Poly_add(NTT(Poly_inv(x),y),x);
int inv2=qpow(2,mod-2);
for(int i=0;i<n;i++)
x[i]=mul(x[i],inv2);
x.resize(n);
return x;
}
Poly sqrt(Poly A){
return sqrt(A,A.size());
}
Poly rever(Poly A){
reverse(A.begin(),A.end());
return A;
}
void div(Poly A,Poly B,Poly &C,Poly &D){
int n=A.size(),m=B.size();
Poly ra=rever(A),rb=rever(B);
ra.resize(n-m+1),rb.resize(n-m+1);
C=NTT(ra,Poly_inv(rb));
C.resize(n-m+1);
C=rever(C);
D=Poly_sub(A,NTT(B,C));
D.resize(m);
}
}
using namespace Template_Poly;
typedef Poly poly;
int n,m;
int a[N],fac[N],ifac[N];
inline void init(int n = 100010){
fac[0]=ifac[0]=1;for(int i=1;i<=n;i++)fac[i] = mul(fac[i-1],i);
ifac[n]=qpow(fac[n],mod-2);for(int i=n-1;i;i--)ifac[i] = mul(ifac[i+1],i+1);
}
poly solve(int l,int r){
if(l==r){
poly ret(2,1);
ret[1]=sub(0,a[l]);
return ret;
}
int mid=(l+r)>>1;
return NTT(solve(l,mid),solve(mid+1,r));
}
poly func;
inline void calc(){
func = solve(1,n);
func = ln(func);
// for(int i=0;i<func.size();i++)cout << func[i] << " ";puts("");
func = Deri(func);
func.push_back(0);
for(int i=(int)func.size()-1;i;i--)func[i] = sub(0,func[i-1]);
func[0]=n;
// for(int i=0;i<func.size();i++)cout << func[i] << " ";puts("");
}
poly A,B,C;
poly ret;
void Doit(){
A.resize(n-1),B.resize(n-1);
for(int i=0;i<=n-2;i++){
A[i] = mul(qpow(i+1, m),ifac[i]);
B[i] = mul(qpow(i+1, 2*m),ifac[i]);
}
//A[0] = 1
// B = Poly_inv(B);
C = NTT(B,Poly_inv(A));
C.resize(n-1);
A = ln(A);
for(int i=0;i<=n-2;i++){
A[i] = mul(A[i], func[i]);
C[i] = mul(C[i], func[i]);
}
A = exp(A);
ret = NTT(A,C);
int ans=ret[n-2];
for(int i=1;i<=n;i++)ans=mul(ans,a[i]);
ans = mul(ans, fac[n-2]);
printf("%d\n",ans);
}
int main()
{
init();
cin >> n >> m;
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
calc();
Doit();
}
原文地址:https://www.cnblogs.com/weiyanpeng/p/12028247.html
时间: 2024-10-12 02:12:26