题目大意
思路
设(f_i)表示至少出现了i种颜色的方案数
[egin{aligned}
f_i&={m choose i} imes frac{(s imes i)!}{(s!)^{i}} imes {nchoose s imes i} imes (m-i)^{n-s imes i}\
f_i&={m choose i} imes frac{n!}{(s!)^{i} imes (n-s imes i)!} imes (m-i)^{n-s imes i}\
end{aligned}
]
设(g_i)表示恰好出现了i种颜色的方案数，不难得到(g_i)的容斥式子。
[egin{aligned}
g_i=&f_i-sum_{j=i+1}^{m}g_{j} imes {jchoose i}\
g_i=&f_i-frac{sum_{j=i+1}^{m} imes frac{g_j imes j!}{(j-i)!}}{i!}
end{aligned}
]
上面是一个经典的分治FFT形式的式子，使用分治FFT可以做到(nlog^2n)。
其实可以更加优化，上面的式子同样也是一个经典的二项式反演。
[egin{aligned}
f_i&=sum_{j=i}^{m}{jchoose i} imes g_{j}\
g_i&=sum_{j=i}^{m}(-1)^{j-i} imes {jchoose i} imes f_{j}
end{aligned}
]
直接用NTT一遍求出g的值即可。
/*=======================================
* Author : ylsoi
* Time : 2019.3.22
* Problem : loj2527_ntt
* E-mail : ylsoi@foxmail.com
* ====================================*/
#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a,i##_end_=b;i<=i##_end_;++i)
#define DREP(i,a,b) for(int i=a,i##_end_=b;i>=i##_end_;--i)
#define debug(x) cout<<#x<<"="<<x<<" "
#define fi first
#define se second
#define mk make_pair
#define pb push_back
typedef long long ll;
using namespace std;
void File(){
freopen("loj2527_ntt.in","r",stdin);
freopen("loj2527_ntt.out","w",stdout);
}
template<typename T>void read(T &_){
_=0; T f=1; char c=getchar();
for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
for(;isdigit(c);c=getchar())_=(_<<1)+(_<<3)+(c^'0');
_*=f;
}
string proc(){
ifstream f("/proc/self/status");
return string(istreambuf_iterator<char>(f),istreambuf_iterator<char>());
}
const int maxn=1e7+10;
const int maxm=1e5+10;
const int mod=1004535809;
int n,m,S,w[maxm],ans;
int fac[maxn],ifac[maxn];
int qpow(int x,int y=mod-2){
int ret=1; x%=mod;
while(y){
if(y&1)ret=1ll*ret*x%mod;
x=1ll*x*x%mod;
y>>=1;
}
return ret;
}
void inc(int &x,int y){
x+=y;
if(x>=mod)x-=mod;
else if(x<0)x+=mod;
}
int C(int x,int y){
if(x<0 || y<0 || x<y)return 0;
return 1ll*fac[x]*ifac[y]%mod*ifac[x-y]%mod;
}
namespace Poly{
int om[maxm<<2],dn[maxm<<2],lim,cnt;
void dft(int *A,int ty){
if(ty==-1)reverse(A+1,A+lim);
REP(i,0,lim-1)if(i<dn[i])swap(A[i],A[dn[i]]);
for(int len=1;len<lim;len<<=1){
int w=om[len<<1];
for(int L=0;L<lim;L+=len<<1){
int wk=1;
REP(i,L,L+len-1){
int u=A[i],v=1ll*A[i+len]*wk%mod;
A[i]=(u+v)%mod;
A[i+len]=(u-v)%mod;
wk=1ll*wk*w%mod;
}
}
}
if(ty==-1){
int inv=qpow(lim);
REP(i,0,lim-1)A[i]=(1ll*A[i]*inv%mod+mod)%mod;
}
}
}
void init(){
read(n),read(m),read(S);
REP(i,0,m)read(w[i]);
fac[0]=1;
int lim=max(n,m);
REP(i,1,lim)fac[i]=1ll*fac[i-1]*i%mod;
ifac[lim]=qpow(fac[lim]);
DREP(i,lim-1,0)ifac[i]=1ll*ifac[i+1]*(i+1)%mod;
}
int f[maxm<<2],g[maxm<<2];
using namespace Poly;
int main(){
// File();
init();
REP(i,0,m)if(S*i<=n)f[i]=1ll*C(m,i)*fac[S*i]%mod*
qpow(ifac[S],i)%mod*C(n,S*i)%mod*qpow(m-i,n-S*i)%mod*fac[i]%mod;
REP(i,0,m)g[i]=1ll*(i&1 ? -1 : 1)*ifac[i]%mod;
reverse(g,g+m+1);
lim=1,cnt=0;
while(lim<=m+m)lim<<=1,++cnt;
if(!cnt)cnt=1;
om[lim]=qpow(3,(mod-1)/lim);
for(int i=lim>>1;i;i>>=1)
om[i]=1ll*om[i<<1]*om[i<<1]%mod;
REP(i,0,lim-1)dn[i]=dn[i>>1]>>1|((i&1)<<(cnt-1));
dft(f,1),dft(g,1);
REP(i,0,lim-1)g[i]=1ll*g[i]*f[i]%mod;
dft(g,-1);
REP(i,0,m)inc(ans,1ll*g[i+m]*w[i]%mod*ifac[i]%mod);
printf("%d
",ans);
return 0;
}
/*=======================================
* Author : ylsoi
* Time : 2019.3.21
* Problem : loj2527
* E-mail : ylsoi@foxmail.com
* ====================================*/
#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a,i##_end_=b;i<=i##_end_;++i)
#define DREP(i,a,b) for(int i=a,i##_end_=b;i>=i##_end_;--i)
#define debug(x) cout<<#x<<"="<<x<<" "
#define fi first
#define se second
#define mk make_pair
#define pb push_back
typedef long long ll;
using namespace std;
void File(){
freopen("loj2527.in","r",stdin);
freopen("loj2527.out","w",stdout);
}
template<typename T>void read(T &_){
_=0; T f=1; char c=getchar();
for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
for(;isdigit(c);c=getchar())_=(_<<1)+(_<<3)+(c^'0');
_*=f;
}
string proc(){
ifstream f("/proc/self/status");
return string(istreambuf_iterator<char>(f),istreambuf_iterator<char>());
}
const int maxn=1e7+10;
const int maxm=1e5+10;
const int mod=1004535809;
int n,m,S,w[maxm],ans;
int fac[maxn],ifac[maxn];
int qpow(int x,int y=mod-2){
int ret=1; x%=mod;
while(y){
if(y&1)ret=1ll*ret*x%mod;
x=1ll*x*x%mod;
y>>=1;
}
return ret;
}
int C(int x,int y){
if(x<0 || y<0 || x<y)return 0;
return 1ll*fac[x]*ifac[y]%mod*ifac[x-y]%mod;
}
void inc(int &x,int y){
x+=y;
if(x>=mod)x-=mod;
else if(x<0)x+=mod;
}
namespace Poly{
int om[maxm<<2],dn[maxm<<2],lim,cnt;
void dft(int *A,int ty){
if(ty==-1)reverse(A+1,A+lim);
REP(i,0,lim-1)if(i<dn[i])swap(A[i],A[dn[i]]);
for(int len=1;len<lim;len<<=1){
int w=om[len<<1];
for(int L=0;L<lim;L+=len<<1){
int wk=1;
REP(i,L,L+len-1){
int u=A[i],v=1ll*A[i+len]*wk%mod;
A[i]=(u+v)%mod;
A[i+len]=(u-v)%mod;
wk=1ll*wk*w%mod;
}
}
}
if(ty==-1){
int inv=qpow(lim);
REP(i,0,lim-1)A[i]=1ll*A[i]*inv%mod;
}
}
void multi(int *A,int *B,int *C,int la,int lb){
lim=1,cnt=0;
while(lim<=la+lb)lim<<=1,++cnt;
REP(i,0,lim-1){
dn[i]=dn[i>>1]>>1|((i&1)<<(cnt-1));
if(i>la)A[i]=0;
if(i>lb)B[i]=0;
}
dft(A,1),dft(B,1);
REP(i,0,lim-1)C[i]=1ll*A[i]*B[i]%mod;
dft(C,-1);
}
}
void math_init(){
fac[0]=1;
int lim=max(n,m);
REP(i,1,lim)fac[i]=1ll*fac[i-1]*i%mod;
ifac[lim]=qpow(fac[lim],mod-2);
DREP(i,lim-1,0)ifac[i]=1ll*ifac[i+1]*(i+1)%mod;
using namespace Poly;
lim=1;
while(lim<=m+m)lim<<=1;
om[lim]=qpow(3,(mod-1)/lim);
for(int i=lim>>1;i;i>>=1)
om[i]=1ll*om[i<<1]*om[i<<1]%mod;
}
int f[maxm<<2],g[maxm<<2],ap[maxm<<2],bp[maxm<<2];
void divide(int l,int r){
if(l==r){
g[l]=(f[l]-1ll*g[l]*ifac[l]%mod)%mod;
inc(ans,1ll*g[l]*w[l]%mod);
return;
}
using namespace Poly;
int mid=(l+r)>>1;
divide(mid+1,r);
REP(i,0,r-mid-1)ap[i]=1ll*g[i+mid+1]*fac[i+mid+1]%mod;
REP(i,0,r-l-1)bp[i]=ifac[i+1];
reverse(bp,bp+r-l);
multi(ap,bp,ap,r-mid-1,r-l-1);
REP(i,l,mid)inc(g[i],ap[i-mid-1+r-l]);
divide(l,mid);
}
int main(){
//File();
read(n),read(m),read(S);
REP(i,0,m)read(w[i]);
math_init();
REP(i,0,m)if(S*i<=n)
f[i]=1ll*C(m,i)*fac[n]%mod*qpow(ifac[S],i)%mod*ifac[n-S*i]%mod*qpow(m-i,n-S*i)%mod;
divide(0,m);
printf("%d
",ans);
return 0;
}