P4859 has nothing to be afraid of the

The meaning of problems

Set \ (a_i \) represents the \ (I \) candy energy, \ (B_i \) represents the \ (I \) energy of pills

First may satisfy Conditions, \ (A> B \) to the number of \ (\ + K FRAC {n-2}} {\) .

Because just as (i \) \ the number of poor program requirements, we first find the least \ (i \) number of the program, followed by the binomial inversion find the answer.

First \ (a, b \) in ascending order.

Set \ (h_ {i, j} \) representing the forward \ (I \) candy, at least chose \ (J \) of \ (a> b \) of the number of programs, \ (cnt_i \) represents the ratio of \ (i \) a small number of pills a candy.

有：
\(h_{i,j}=h_{i-1,j}+(cnt_i-(j-1))*h_{i-1,j-1}\)

Set \ (F_i \) represents exactly (I \) \ to \ (a> b \) of the number of programs, \ (G_i \) represents at least \ (I \) of \ (a> b \) of the number of programs.

Apparently \ (n-G_i = F_ {,} * I (Ni)! \) , Then the binomial inversion can.

code:

#include<bits/stdc++.h>
using namespace std;
const int maxn=2010;
const int mod=1e9+9;
int n,m,ans;
int a[maxn],b[maxn],cnt[maxn],g[maxn],fac[maxn],inv[maxn];
int f[maxn][maxn];
inline int power(int x,int k)
{
    int res=1;
    while(k)
    {
        if(k&1)res=1ll*res*x%mod;
        x=1ll*x*x%mod;k>>=1;
    }
    return res;
}
inline int C(int n,int m)
{
    if(n<m)return 0;
    return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int main()
{
    scanf("%d%d",&n,&m);
    if((n+m)&1){puts("0");return 0;}
    m=(n+m)>>1;
    fac[0]=1;
    for(int i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
    inv[n]=power(fac[n],mod-2);
    for(int i=n;i;i--)inv[i-1]=1ll*inv[i]*i%mod;
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)scanf("%d",&b[i]);
    sort(a+1,a+n+1);sort(b+1,b+n+1);
    for(int i=1,j=0;i<=n;i++)
    {
        while(b[j+1]<a[i]&&j<n)j++;
        cnt[i]=j;
    }
    for(int i=0;i<=n;i++)f[i][0]=1;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=i;j++)
            f[i][j]=(f[i-1][j]+1ll*max(0,cnt[i]-(j-1))*f[i-1][j-1]%mod)%mod;
    for(int i=0;i<=n;i++)g[i]=1ll*f[n][i]*fac[n-i]%mod;
    for(int i=m;i<=n;i++)
        if((i-m)&1)ans=(ans-1ll*C(i,m)*g[i]%mod+mod)%mod;
        else ans=(ans+1ll*C(i,m)*g[i]%mod)%mod;
    printf("%d",ans);
    return 0;
}