codeforces396A

sol：很显然就是找出所有质因数，然后分别塞进去就行了，怎么塞就是组合数。感觉就是道小学奥数题

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=; bool f=; char ch=' ';
    while(!isdigit(ch))    {f|=(ch=='-'); ch=getchar();}
    while(isdigit(ch)) {s=(s<<)+(s<<)+(ch^); ch=getchar();}
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<) {putchar('-'); x=-x;}
    if(x<) {putchar(x+''); return;}
    write(x/); putchar((x%)+'');
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int Mod=;
int n;
map<int,int>Map;
int num[],cnt[];
int fac[],invf[];
inline int ksm(int x,int y)
{
    int ans=;
    while(y)
    {
        if(y&) ans=1LL*ans*x%Mod;
        x=1LL*x*x%Mod;
        y>>=;
    }return ans;
}
inline int C(int n,int m)
{
    int oo;
    oo=1LL*fac[n]*invf[m]%Mod*invf[n-m]%Mod;
    return oo;
}
int main()
{
    freopen("codeforces396A_data.in","r",stdin);
    int i,j,k,x;
    R(n);
    fac[]=invf[]=;
    for(i=;i<=;i++) fac[i]=1LL*fac[i-]*i%Mod;
    invf[]=ksm(fac[],Mod-);
    for(i=;i>=;i--) invf[i]=1LL*invf[i+]*(i+)%Mod;
    for(i=;i<=n;i++)
    {
        R(x);
        for(j=;j<=sqrt(x);j++) if(x%j==)
        {
            int oo=;
            while(x%j==) {x/=j; oo++;}
            if(!Map[j])
            {
                Map[j]=++(*num); num[*num]=j; cnt[*num]=oo;
            }else cnt[Map[j]]+=oo;
        }
        if(x>)
        {
            if(!Map[x])
            {
                Map[x]=++(*num); num[*num]=x; cnt[*num]=;
            }else cnt[Map[x]]++;
        }
    }
    int ans=;
    for(i=;i<=*num;i++)
    {
        ans=1LL*ans*C(cnt[i]+n-,n-)%Mod;
    }Wl(ans);
    return ;
}