傳送門(mén)

Description

一個(gè)長(zhǎng)度為\(N\)的序列, 每個(gè)位置都可以被染成 \(M\)種顏色中的某一種.

出現(xiàn)次數(shù)恰好為 \(S\)的顏色種數(shù)有\(i\)種, 會(huì)產(chǎn)生\(w_i\)的愉悅度.

對(duì)于所有染色方案, 能獲得的愉悅度的和對(duì)\(1004535809\)取模的結(jié)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); }

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轉(zhuǎn)載于:https://www.cnblogs.com/PaperCloud/p/10925578.html

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