「BZOJ 2142」礼物

题目链接

戳这

Title Solution

这一道题显然可以看出公式为：

\[ans=C_{n}^{w_1}*C_{n-w}^{w_2}*...*C_{w_m}^{w_m}
\]

然后就可以用扩展Lucas求解了。

至于扩展Lucas:戳这

code

#include<bits/stdc++.h>

#define rg register

#define int long long

#define file(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout);

using namespace std;

int read(){

    int x=0,f=1;

    char c=getchar();

    while(c<'0'||c>'9') f=(c=='-')?-1:1,c=getchar();

    while(c>='0'&&c<='9') x=x*10+c-48,c=getchar();

    return f*x;

}

void exgcd(int a,int b ,int &x,int &y){

    if(!b){x=1,y=0;return;}

    exgcd(b,a%b,x,y);

    int t=x;

    x=y,y=t-(a/b)*y;

}

int inv(int a,int b){

    int x,y;

    return exgcd(a,b,x,y),(x%b+b)%b;

}

int ksm(int a,int b,int p){

    int ans=1;

    while(b){

        if(b&1)

            ans=a*ans%p;

        a=a*a%p;

        b>>=1;

    }

    return ans%p;

}

int crt(int x,int p,int mod){

    return inv(p/mod,mod)*(p/mod)*x;

}

int fac(int x,int a,int b){

    if(!x)

        return 1;

    int ans=1;

    for(int i=1;i<=b;i++)

        if(i%a)

            ans*=i,ans%=b;

    ans=ksm(ans,x/b,b);

    for(int i=1;i<=x%b;i++)

        if(i%a)

            ans*=i,ans%=b;

    return ans*fac(x/a,a,b)%b;

}

int C(int n,int m,int a,int b){

    int N=fac(n,a,b),M=fac(m,a,b),Z=fac(n-m,a,b),sum=0;

    for(int i=n;i;i=i/a)

        sum+=i/a;

    for(int i=m;i;i=i/a)

        sum-=i/a;

    for(int i=n-m;i;i=i/a)

        sum-=i/a;

    return N*ksm(a,sum,b)%b*inv(M,b)%b*inv(Z,b)%b;

}

int exlucas(int n,int m,int p){

    int t=p,ans=0;

    for(int i=2;i*i<=p;i++){

        int k=1;

        while(t%i==0)

            k*=i,t/=i;

        ans+=crt(C(n,m,i,k),p,k),ans%=p;

    }

    if(t>1)

        ans+=crt(C(n,m,t,t),p,t),ans%=p;

    return ans%p;

}

int a[11];

void slove(){

    int p=read(),n=read(),m=read(),sum=0;

    for(int i=1;i<=m;i++)

        a[i]=read(),sum+=a[i];

    if(n<sum)

        printf("Impossible\n"),exit(0);

    int ans=1;

    for(int i=1;i<=m;i++)

        ans*=exlucas(n,a[i],p),ans%=p,n-=a[i];

    printf("%lld",ans%p);

}

main(){

    slove();

    return 0;

}