「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
"),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;
}