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=0; bool f=0; char ch=' ';
    while(!isdigit(ch))    {f|=(ch=='-'); ch=getchar();}
    while(isdigit(ch)) {s=(s<<3)+(s<<1)+(ch^48); ch=getchar();}
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0) {putchar('-'); x=-x;}
    if(x<10) {putchar(x+'0'); return;}
    write(x/10); putchar((x%10)+'0');
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int Mod=1000000007;
int n;
map<int,int>Map;
int num[1000005],cnt[1000005];
int fac[15505],invf[15505];
inline int ksm(int x,int y)
{
    int ans=1;
    while(y)
    {
        if(y&1) ans=1LL*ans*x%Mod;
        x=1LL*x*x%Mod;
        y>>=1;
    }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[0]=invf[0]=1;
    for(i=1;i<=15500;i++) fac[i]=1LL*fac[i-1]*i%Mod;
    invf[15500]=ksm(fac[15500],Mod-2);
    for(i=15499;i>=1;i--) invf[i]=1LL*invf[i+1]*(i+1)%Mod;
    for(i=1;i<=n;i++)
    {
        R(x);
        for(j=2;j<=sqrt(x);j++) if(x%j==0)
        {
            int oo=0;
            while(x%j==0) {x/=j; oo++;}
            if(!Map[j])
            {
                Map[j]=++(*num); num[*num]=j; cnt[*num]=oo;
            }else cnt[Map[j]]+=oo;
        }
        if(x>1)
        {
            if(!Map[x])
            {
                Map[x]=++(*num); num[*num]=x; cnt[*num]=1;
            }else cnt[Map[x]]++;
        }
    }
    int ans=1;
    for(i=1;i<=*num;i++)
    {
        ans=1LL*ans*C(cnt[i]+n-1,n-1)%Mod;
    }Wl(ans);
    return 0;
}