题面:https://www.luogu.org/problem/P4091
题解:[egin{array}{l}
f(n) = sumlimits_{i = 0}^n {sumlimits_{j = 0}^i {{
m{S}}(i,j) cdot {2^{
m{j}}} cdot j!} } \
= sumlimits_{i = 0}^n {sumlimits_{j = 0}^n {{
m{S}}(i,j) cdot {2^{
m{j}}} cdot j!} }
end{array}]
把公式[S(n,m) = frac{{sumlimits_{i = 0}^m {{{( - 1)}^i}{ m{cdot}}C_m^i} { m{cdot}}{{(m - i)}^n}}}{{m!}}]带进去
则[{ m{ackslash begin{ array} { l} nackslash begin{ array} { *{ 20} { l} } }}f(n) = sumlimits_{i = 0}^n {sumlimits_{j = 0}^n {{ m{S}}(i,j) cdot {2^{ m{j}}} cdot j!} = sumlimits_{i = 0}^n {sumlimits_{j = 0}^n {sumlimits_{k = 0}^j {frac{{{{( - 1)}^k}}}{{k!}} cdot frac{{{{(j - k)}^i}}}{{(j - k)!}}} cdot {{ m{2}}^{ m{j}}}{ m{cdotj}}!} } } { m{n{ }} = sum { m{\_}}j = { m{^}}n{ m{ { }}{2^j}{ m{cdotj}}!sum { m{\_}}k = 0{ m{^}}j{ m{ { }}sum { m{\_}}i = 0{ m{^}}n{ m{ { }}frac{{{{( - 1)}^k}}}{{k!}} cdot frac{{{{(j - k)}^i}}}{{(j - k)!}} = sum { m{\_}}j = { m{^}}n{ m{ { }}{2^j}{ m{cdot}}j!sum { m{\_}}k = 0{ m{^}}j{ m{ { ackslash frac{ { { }}( - 1){ m{^{ ^}}k{ m{} } } } }}k! cdot { m{} }}frac{{sumlimits_{i = 0}^n {{{(j - k)}^i}} }}{{(j - k)!}}{ m{} } } } } nackslash end{ array} ackslash ackslash = }}sum { m{\_}}j = { m{^}}n{ m{ { }}{2^j}{ m{cdot}}j!sum { m{\_}}k = 0{ m{^}}j{ m{ { ackslash frac{ { { }}( - 1){ m{^{ ^}}k{ m{} } } } }}k! cdot { m{} }}frac{{{{(j - k)}^{n + 1}} - 1}}{{(j - k)! cdot (j - k - 1)}}{ m{} nackslash end{ array} }}]
令[egin{array}{l}
h(x) = frac{{{{( - 1)}^x}}}{{x!}}\
g(x) = frac{{{x^{n + 1}} - 1}}{{x!{
m{cdot(x - 1)}}}}
end{array}]
后面的就是这两个的卷积，直接上ntt就行了
#include<bits/stdc++.h> #define ms(x) memset(x,0,sizeof(x)) #define sws ios::sync_with_stdio(false) using namespace std; typedef long long ll; const int maxn=5e5+5; const double pi=acos(-1.0); const ll mod=998244353;///通常情况下的模数, const ll g=3;///模数的原根998244353,1004535809,469762049 ll qpow(ll a,ll n,ll p){ ll ans=1; while(n){ if(n&1) ans=ans*a%p; n>>=1; a=a*a%p; } return ans; } int rev[maxn]; void ntt(ll a[],int n,int len,int pd){ rev[0]=0; for(int i=1;i<n;i++){ rev[i]=(rev[i>>1]>>1 | ((i&1)<<(len-1))); if(i<rev[i]) swap(a[i],a[rev[i]]); } for(int mid=1;mid<n;mid<<=1){ ll wn=qpow(g,(mod-1)/(mid*2),mod);///原根代替单位根 if(pd==-1) wn=qpow(wn,mod-2,mod);///逆变换则改成逆元 for(int j=0;j<n;j+=2*mid){ ll w=1; for(int k=0;k<mid;k++){ ll x=a[j+k],y=w*a[j+k+mid]%mod; a[j+k]=(x+y)%mod; a[j+k+mid]=(x-y+mod)%mod; w=w*wn%mod; } } } if(pd==-1){ ll inv=qpow(n,mod-2,mod); for(int i=0;i<n;i++){ a[i]=a[i]*inv%mod; } } } ll a[maxn],b[maxn],c[maxn]; void solve(int n,int m){ int len=0,up=1; while(up<=n+m) up<<=1,len++; ntt(a,up,len,1); ntt(b,up,len,1); for(int i=0;i<up;i++) c[i]=1ll*a[i]*b[i]%mod; ntt(c,up,len,-1); } ll fa[maxn]; int main(){ int n,m; sws; cin>>n; fa[0]=1; a[0]=1; b[0]=1; for(int i=1;i<=n;i++){ fa[i]=1ll*fa[i-1]*i%mod; int t=(i&1)==1?-1:1; a[i]=(t*qpow(fa[i],mod-2,mod)+mod)%mod; if(i==1) b[1]=n+1; else { b[i]=(qpow(i,n+1,mod)-1+mod)%mod*qpow(1ll*(i-1)*fa[i]%mod,mod-2,mod)%mod; } } solve(n,n); ll ans=0; for(ll i=0;i<=n;i++){ ans=(ans+qpow(2,i,mod)*fa[i]%mod*c[i]%mod)%mod; } cout<<ans<<endl; }