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 }