BZOJ 3622: there is nothing to fear binomial inversion of the dynamic programming +

code:

#include <cstdio> 
#include <algorithm>           
#define ll long long    
#define N 2007 
#define mod 1000000009 
#define setIO(s) freopen(s".in","r",stdin)  
using namespace std;    
int qpow(int x,int y) 
{
    int tmp=1; 
    for(;y;y>>=1,x=(ll)x*x%mod) 
        if(y&1) tmp=(ll)tmp*x%mod;  
    return tmp; 
}
int f[N][N],a[N],b[N],l[N],fac[N],inv[N];   
int C(int x,int y) { return (ll)fac[x]*inv[y]%mod*inv[x-y]%mod; }
int main () 
{ 
    // setIO("input"); 
    int i,j,n,k,tmp;   
    scanf("%d%d",&n,&k),tmp=(n+k)/2;     
    if((n+k)&1) { printf("0\n");  return 0; }   
    for(i=1;i<=n;++i) scanf("%d",&a[i]); 
    for(i=1;i<=n;++i) scanf("%d",&b[i]);  
    sort(a+1,a+1+n),sort(b+1,b+1+n);      
    for(i=1;i<=n;++i) l[i]=lower_bound(b+1,b+1+n,a[i])-b-1;                 
    f[0][0]=1,fac[0]=inv[0]=1;   
    for(i=1;i<=n;++i) fac[i]=(ll)i*fac[i-1]%mod,inv[i]=qpow(fac[i],mod-2); 
    for(i=1;i<=n;++i) 
    {
        for(j=0;j<=l[i];++j)   
        { 
            f[i][j]=f[i-1][j];    
            if(j) (f[i][j]+=(ll)(l[i]-j+1)*f[i-1][j-1]%mod)%=mod;     
        }
    }                 
    int years = 0;           
    for(i=tmp;i<=n;++i) 
    {
        int d=((i-tmp)&1)?mod-1:1;                                         
        (+ = Ans (II) d * C (i, tmp)% v * f [n] [i] * v% fac [Ni]% v) = v%; 
    }       
    printf("%d\n",ans);  
    return 0;
}