[LOJ#3120][Luogu5401][CTS2019]珍珠(容斥+生成函数)

https://www.luogu.org/blog/user50971/solution-p5401

 #include<cstdio>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int N=,mod=,i2=;
 int D,n,m,ans,fac[N],inv[N],ip2[N],f[N],g[N],rev[N],a[N],b[N];

 int ksm(int a,int b){
     int res=;
     for (; b; a=1ll*a*a%mod,b>>=)
         if (b & ) res=1ll*res*a%mod;
     return res;
 }

 int C(int n,int m){ return n<m ?  : 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod; }

 void NTT(int a[],int n,bool f){
     for (int i=; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
     for (int i=; i<n; i<<=){
         int wn=ksm(,f?(mod-)/(i<<):(mod-)-(mod-)/(i<<));
         for (int p=i<<,j=; j<n; j+=p){
             int w=;
             for (int k=; k<i; k++,w=1ll*w*wn%mod){
                 int x=a[j+k],y=1ll*w*a[i+j+k]%mod;
                 a[j+k]=(x+y)%mod; a[i+j+k]=(x-y+mod)%mod;
             }
         }
     }
     if (f) return;
     int inv=ksm(n,mod-);
     for (int i=; i<n; i++) a[i]=1ll*a[i]*inv%mod;
 }

 int main(){
     freopen("pearl.in","r",stdin);
     freopen("pearl.out","w",stdout);
     scanf("%d%d%d",&D,&n,&m);
     if (n-*m<){ puts(""); return ; }
     if (n-*m>=D){ printf("%d\n",ksm(D,n)); return ; }
     fac[]=ip2[]=;
     rep(i,,D) fac[i]=1ll*fac[i-]*i%mod,ip2[i]=1ll*ip2[i-]*i2%mod;
     inv[D]=ksm(fac[D],mod-);
     for (int i=D; i; i--) inv[i-]=1ll*inv[i]*i%mod;

     int l=,L=,t=;
     while (l<=(D<<)) l<<=,L++;
     for (int i=; i<l; i++) rev[i]=(rev[i>>]>>)|((i&)<<(L-));
     rep(i,,D) a[i]=1ll*t*ksm(D-*i+mod+mod,n)%mod*inv[i]%mod,b[i]=inv[i],t=mod-t;
     NTT(a,l,); NTT(b,l,);
     for (int i=; i<l; i++) a[i]=1ll*a[i]*b[i]%mod;
     NTT(a,l,);
     rep(i,,D) f[i]=1ll*ip2[i]*fac[i]%mod*C(D,i)%mod*a[i]%mod;
     for (int i=; i<l; i++) a[i]=b[i]=;

     t=(D&)?mod-:;
     rep(i,,D) a[i]=1ll*f[i]*fac[i]%mod,b[i]=1ll*t*inv[D-i]%mod,t=mod-t;
     NTT(a,l,); NTT(b,l,);
     for (int i=; i<l; i++) a[i]=1ll*a[i]*b[i]%mod;
     NTT(a,l,);
     rep(i,,D) g[i]=1ll*a[D+i]*inv[i]%mod;
     rep(i,,n-*m) ans=(ans+g[i])%mod;
     printf("%d\n",ans);
     return ;
 }