Description
Solution
写了个傻逼的 (O(n+m*log^2)) 的做法,卡了一下午才过 (bzoj)
首先设 (f[i]) 表示至少有 (i) 种颜色数量为 (s)
显然 (f[i]=C_{m}^{i}*C_{n}^{i*s}*frac{(i*s)!}{s!^i}*m^{n-i*s})
设 (g[i]) 表示至少有 (i) 种颜色数量为 (s)
然后写出了个傻逼式子:
(g[n]=f[n])
(g[i]=f[i]-sum_{j=i+1}^{n}g[j]*C_j^i)
然后分治(NTT),强行多一个 (log),实际上对 (f) 数组直接容斥就行了
#include<bits/stdc++.h>
#define RG register
using namespace std;
template<class T>void gi(T &x){
int f;char c;
for(f=1,c=getchar();c<'0'||c>'9';c=getchar())if(c=='-')f=-1;
for(x=0;c<='9'&&c>='0';c=getchar())x=x*10+(c&15);x*=f;
}
const int N=1e7+10,M=1e5+10,mod=1004535809;
int n,m,s,w[M],f[M],Fac[N],inv[N],g[M],lim,D,INV[N];
inline int qm(int x,int k){
int sum=1;
while(k){
if(k&1)sum=1ll*sum*x%mod;
x=1ll*x*x%mod;k>>=1;
}
return sum;
}
inline int C(int a,int b){
return 1ll*Fac[a]*inv[b]%mod*inv[a-b]%mod;
}
int E,L=0,R[M*4],G[M*4];
inline void NTT(int *A){
for(RG int i=0;i<E;i++)if(i<R[i])swap(A[i],A[R[i]]);
for(RG int i=1;i<E;i<<=1){
int t0=G[i],x,y;
for(RG int j=0;j<E;j+=i<<1){
int t=1;
for(RG int k=0;k<i;k++,t=1ll*t*t0%mod){
x=A[j+k];y=1ll*A[j+k+i]*t%mod;
A[j+k]=(x+y)%mod;A[j+k+i]=(x-y+mod)%mod;
}
}
}
}
inline void mul(int *A,int *B){
NTT(A);NTT(B);
for(RG int i=0;i<=E;i++)A[i]=1ll*A[i]*B[i]%mod;
NTT(A);
reverse(A+1,A+E);
for(RG int i=0,t=INV[E];i<=E;i++)A[i]=1ll*A[i]*t%mod;
}
int A[M*4],B[M*4];
inline void solve(int l,int r){
if(l==r){
f[lim-l]=(f[lim-l]-1ll*g[l]*inv[lim-l]%mod+mod)%mod;
return ;
}
int mid=(l+r)>>1,len=(r-l+1);
solve(l,mid);
L=0;
for(E=1;E<=len;E<<=1)L++;
for(RG int i=0;i<E;i++)R[i]=(R[i>>1]>>1)|((i&1)<<(L-1)),A[i]=0,B[i]=inv[i];
for(RG int i=l;i<=mid;i++)A[i-l]=1ll*f[lim-i]*Fac[lim-i]%mod;
mul(A,B);
for(RG int i=mid+1;i<=r;i++)g[i]=(g[i]+A[i-l])%mod;
solve(mid+1,r);
}
int main(){
freopen("pp.in","r",stdin);
freopen("pp.out","w",stdout);
cin>>n>>m>>s;INV[0]=INV[1]=1;G[1]=qm(3,(mod-1)/2);
for(RG int i=2;i<400000;i++){
G[i]=qm(3,(mod-1)/(i<<1));
INV[i]=(mod-1ll*(mod/i)*INV[mod%i]%mod)%mod;
}
for(RG int i=0;i<=m;i++)gi(w[i]);
lim=min(n/s,m),D=max(n,m);
Fac[0]=inv[0]=inv[1]=INV[0]=INV[1]=1;
for(RG int i=1;i<=D;i++)Fac[i]=1ll*Fac[i-1]*i%mod;
inv[D]=qm(Fac[D],mod-2);
for(RG int i=D-1;i>=2;i--)inv[i]=1ll*inv[i+1]*(i+1)%mod;
for(RG int i=0;i<=lim;i++)
f[i]=1ll*C(m,i)*C(n,i*s)%mod*Fac[i*s]%mod*qm(qm(Fac[s],i),mod-2)%mod*qm(m-i,n-i*s)%mod;
solve(0,lim);
int ans=0;
for(RG int i=0;i<=lim;i++)ans=(ans+1ll*f[i]*w[i])%mod;
cout<<ans<<endl;
return 0;
}