[HAOI 2018]染色

传送门

Description

一个长度为\(N\)的序列, 每个位置都可以被染成 \(M\)种颜色中的某一种.

出现次数恰好为 \(S\)的颜色种数有\(i\)种, 会产生\(w_i\)的愉悦度.

对于所有染色方案, 能获得的愉悦度的和对\(1004535809\)取模的结果.

Solution?

\[
ans=\sum_{i=0}^{lim} w_i\cdot num_i
\]

how to get \(num_i\)?

\(f_i\) : the number of occurrences of at least i colors is exactly the number of S

so \(f_i=\binom{m}{i}\cdot \frac{n!}{(s!)^i(n-iS)!}\cdot(m-i)^{n-iS}\)

According to the binomial inversion, we can know that:
\[
num_i=\sum_{j=i}^{lim}(-1)^{j-i}\binom{j}{i}f[j]
\]
so
\[
num_i=\frac{1}{i!}\sum_{j=i}^{lim} \frac{(-1)^{j-i}}{(j-i)!}\cdot(f[j]\cdot j!)
\]
we can use NTT.

Code?

#include<bits/stdc++.h>
#define reg register
#define ll long long
#define db double
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}
const int P=1004535809,G[2]={3,334845270},NN=5e5+5;
int Mul(int x,int y){return 1ll*x*y%P;}
int Add(int x,int y){return (x+y)%P;}
const int MN=1e7+5,MM=1e5+5,MS=155;
int N,M,S,W[MM],f[MM],fac[MN],inv[MN];
int fpow(int x,int y){int r=1;for(;y;y>>=1,x=Mul(x,x))if(y&1)r=Mul(r,x);return r;}
int C(int x,int y){if(x<0||y<0||x<y)return 0;return Mul(fac[x],Mul(inv[y],inv[x-y]));}
int a[NN],b[NN],pos[NN];
void NTT(int *a,bool ty,int L)
{
    reg int i,j,k,w,wn,x,y;
    for(i=0;i<L;++i) if(pos[i]<i) swap(a[i],a[pos[i]]);
    for(i=1;i<L;i<<=1)
    {
        wn=fpow(G[ty],(P-1)/(i<<1));
        for(j=0;j<L;j+=(i<<1))
            for(w=1,k=0;k<i;++k,w=Mul(w,wn))
            {
                x=a[j+k],y=Mul(a[j+i+k],w);
                a[j+k]=Add(x,y);a[j+i+k]=Add(x,P-y);
            }
    }
    if(ty)for(j=fpow(L,P-2),i=0;i<L;++i)a[i]=Mul(a[i],j);
}
int ans;
int main()
{
    N=read(),M=read(),S=read();
    reg int i,lim;
    for(i=0;i<=M;++i) W[i]=read();
    lim=max(N,M);
    for(fac[0]=i=1;i<=lim;++i) fac[i]=Mul(fac[i-1],i);
    for(inv[0]=inv[1]=1,i=2;i<=lim;++i) inv[i]=Mul(inv[P%i],(P-P/i));
    for(i=1;i<=lim;++i) inv[i]=Mul(inv[i],inv[i-1]);
    lim=min(M,N/S);
    for(i=0;i<=lim;++i)
        f[i]=Mul(Mul(C(M,i),Mul(fac[N],fpow(inv[S],i))),Mul(inv[N-i*S],fpow(M-i,N-i*S)));
    for(i=0;i<=lim;++i) b[lim-i+1]=Mul(f[i],fac[i]);
    for(i=0;i<=lim;++i) a[i]=(i&1)?(P-inv[i]):inv[i];
    int MA;
    for(MA=1;MA<=(lim<<1);MA<<=1);
    for(i=0;i<MA;++i) pos[i]=(pos[i>>1]>>1)|((i&1)*(MA>>1));
    NTT(a,0,MA);NTT(b,0,MA);
    for(i=0;i<MA;++i) a[i]=Mul(a[i],b[i]);
    NTT(a,1,MA);
    for(i=0;i<=lim;++i) ans=Add(ans,Mul(W[i],Mul(a[lim-i+1],inv[i])));
    return 0*printf("%d\n",ans);
}

原文地址：https://www.cnblogs.com/PaperCloud/p/10925578.html