bzoj 2142 礼物

题目大意：

$n$件礼物，送给$m$个人，其中送给第$i$个人礼物数量为$w_i$，求送礼物的方案数

思路：

显然答案为$inom{n}{w_1} inom{n-w_1}{w_2} cdots inom{n-w_1 ... w_{m-1}}{w_m}$

化简得到$frac{n!}{w_1 ! w_2 ! cdots w_m! (n-sum)!}$

然后扩展卢卡斯即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define ll long long
#define db double
#define inf 2139062143
#define MAXN 400100
#define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
#define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
#define ren for(register int i=fst[x];i;i=nxt[i])
#define pb(i,x) vec[i].push_back(x)
#define pls(a,b) ((a+b)%MOD+MOD)%MOD
#define mns(a,b) ((a%MOD-(b)%MOD)%MOD+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,a[10],ans,MOD;
int exgcd(int a,int b,int &x,int &y)
{
    if(!b) {x=1,y=0;return a;}
    int d=exgcd(b,a%b,y,x);y-=(a/b)*x;return d;
}
int q_pow(int bas,int t,int p,int res=1) 
{
    for(;t;t>>=1,bas=(1LL*bas*bas)%p) if(t&1) res=(1LL*res*bas)%p;return res;
}
int inv(int n,int p){int x,y;exgcd(n,p,x,y);return (x+p)%p;}
int fac(int n,int p,int pk)
{
    if(!n) return 1;int res=1;
    rep(i,2,pk) if(i%p) res=(1LL*res*i)%pk;res=q_pow(res,n/pk,pk);
    rep(i,2,n%pk) if(i%p) res=(1LL*res*i)%pk;return (1LL*res*fac(n/p,p,pk))%pk;
}
int work(int p,int pk)
{
    int res=fac(n,p,pk),s=0;for(int i=n;i;i/=p) s+=i/p;
    rep(i,1,m) {for(int j=a[i];j;j/=p) s-=j/p;res=mul(res,inv(fac(a[i],p,pk),pk));}
    return mul(q_pow(p,s,pk),res);
}
void exlucas()
{
    int rst=MOD,pk;rep(i,2,sqrt(MOD)) if(rst%i==0)
    {
        pk=1;while(rst%i==0) rst/=i,pk*=i;
        ans=pls(mul(mul(work(i,pk),inv(MOD/pk,pk)),MOD/pk),ans);
    }
    if(rst!=1) ans=pls(mul(mul(work(rst,rst),inv(MOD/rst,rst)),MOD/rst),ans);
    printf("%d
",ans);
}
int main()
{
    MOD=read();n=read(),m=read();int s=0;rep(i,1,m) a[i]=read(),s+=a[i];
    if(s>n)puts("Impossible");else {if(n-s) a[++m]=n-s;exlucas();}
}