BZOJ4710 JSOI2011分特产（容斥原理+组合数学）

　　显然可以容斥去掉每人都不为空的限制。每种物品分配方式独立，各自算一个可重组合乘起来即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define P 1000000007
#define N 1010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,m,a[N],C[N<<][N<<],f[N],ans;
void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4710.in","r",stdin);
    freopen("bzoj4710.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    m=read(),n=read();
    for (int i=;i<=n;i++) a[i]=read();
    C[][]=;
    for (int i=;i<=;i++)
    {
        C[i][]=C[i][i]=;
        for (int j=;j<i;j++) C[i][j]=(C[i-][j-]+C[i-][j])%P;
    }
    for (int i=;i<=m;i++)
    {
        f[i]=;
        for (int j=;j<=n;j++)
        f[i]=1ll*f[i]*C[a[j]+i-][i-]%P;
    }
    for (int i=m;i>=;i--)
    if (m-i&) inc(ans,P-1ll*C[m][i]*f[i]%P);
    else inc(ans,1ll*C[m][i]*f[i]%P);
    cout<<ans;
    return ;
}