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$\binom{n}{w_1} \binom{n-w_1}{w_2} \cdots \binom{n-w_1 ... w_{m-1}}{w_m}$

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

然后扩展卢卡斯即可

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<vector>
 9 #include<map>
10 #include<set>
11 #define ll long long
12 #define db double
13 #define inf 2139062143
14 #define MAXN 400100
15 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
16 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
17 #define ren for(register int i=fst[x];i;i=nxt[i])
18 #define pb(i,x) vec[i].push_back(x)
19 #define pls(a,b) ((a+b)%MOD+MOD)%MOD
20 #define mns(a,b) ((a%MOD-(b)%MOD)%MOD+MOD)%MOD
21 #define mul(a,b) (1LL*(a)*(b))%MOD
22 using namespace std;
23 inline int read()
24 {
25     int x=0,f=1;char ch=getchar();
26     while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
27     while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
28     return x*f;
29 }
30 int n,m,a[10],ans,MOD;
31 int exgcd(int a,int b,int &x,int &y)
32 {
33     if(!b) {x=1,y=0;return a;}
34     int d=exgcd(b,a%b,y,x);y-=(a/b)*x;return d;
35 }
36 int q_pow(int bas,int t,int p,int res=1) 
37 {
38     for(;t;t>>=1,bas=(1LL*bas*bas)%p) if(t&1) res=(1LL*res*bas)%p;return res;
39 }
40 int inv(int n,int p){int x,y;exgcd(n,p,x,y);return (x+p)%p;}
41 int fac(int n,int p,int pk)
42 {
43     if(!n) return 1;int res=1;
44     rep(i,2,pk) if(i%p) res=(1LL*res*i)%pk;res=q_pow(res,n/pk,pk);
45     rep(i,2,n%pk) if(i%p) res=(1LL*res*i)%pk;return (1LL*res*fac(n/p,p,pk))%pk;
46 }
47 int work(int p,int pk)
48 {
49     int res=fac(n,p,pk),s=0;for(int i=n;i;i/=p) s+=i/p;
50     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));}
51     return mul(q_pow(p,s,pk),res);
52 }
53 void exlucas()
54 {
55     int rst=MOD,pk;rep(i,2,sqrt(MOD)) if(rst%i==0)
56     {
57         pk=1;while(rst%i==0) rst/=i,pk*=i;
58         ans=pls(mul(mul(work(i,pk),inv(MOD/pk,pk)),MOD/pk),ans);
59     }
60     if(rst!=1) ans=pls(mul(mul(work(rst,rst),inv(MOD/rst,rst)),MOD/rst),ans);
61     printf("%d\n",ans);
62 }
63 int main()
64 {
65     MOD=read();n=read(),m=read();int s=0;rep(i,1,m) a[i]=read(),s+=a[i];
66     if(s>n)puts("Impossible");else {if(n-s) a[++m]=n-s;exlucas();}
67 }
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posted @ 2019-04-04 18:10  jack_yyc  阅读(105)  评论(0编辑  收藏  举报