【JSOI2011】bzoj4710 分特产

枚举分给几个人进行容斥，可以得到答案是

∑i=1n(ni)∏j=1m(aj+i−1i−1)[n−i&1?−1:1]

#include<cstdio>#include<algorithm>using namespace std;#define LL long longconst int p=1000000007,maxn=2000;int pow(int base,int k){    int ret=1;    for (;k;k>>=1,base=(LL)base*base%p)        if (k&1) ret=(LL)ret*base%p;    return ret;}int a[maxn+10],fac[maxn+10],inv[maxn+10];int main(){    int n,m,mx=1,ans=0,tem;    scanf("%d%d",&n,&m);    for (int i=1;i<=m;i++)    {        scanf("%d",&a[i]);        mx=max(mx,a[i]);    }    mx+=n-1;    fac[0]=inv[0]=1;    for (int i=1;i<=mx;i++) fac[i]=(LL)fac[i-1]*i%p;    inv[mx]=pow(fac[mx],p-2);    for (int i=mx-1;i>=1;i--) inv[i]=(LL)inv[i+1]*(i+1)%p;    for (int i=1;i<=n;i++)    {        tem=(LL)fac[n]*inv[i]%p*inv[n-i]%p;        for (int j=1;j<=m;j++)            tem=(LL)tem*fac[a[j]+i-1]%p*inv[a[j]]%p*inv[i-1]%p;        if (n-i&1) ans=(ans-tem+p)%p;        else ans=(ans+tem)%p;    }    printf("%d\n",ans);}