hdu 4372 Count the Buildings

【分析】
斯特林数

【代码】

//hdu 4372 Count the Buildings
#include<cstdio>
#include<cstring>
#include<iostream>
#define ll long long
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std; 
const ll mod=1e9+7;
const int mxn=3005; 
int n,f,b,T;
ll c[mxn][mxn],s[mxn][mxn],ans;   //组合数，斯特林数 
inline void init()
{
    ll i,j;
    s[0][0]=1;
    fo(i,0,2000) c[i][0]=1;
    fo(i,1,2000)
      fo(j,1,i)
      {
            c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
            s[i][j]=(s[i-1][j-1]+(i-1)*s[i-1][j])%mod;
      }
}
int main()
{
    init();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&f,&b);
        if(f+b-2<=2000) ans=(s[n-1][f+b-2]*c[f+b-2][f-1])%mod;
        else ans=0%mod;
        printf("%lld\n",ans);
    }
    return 0;
}