CF 615D Multipliers

题目：http://codeforces.com/contest/615/problem/D

求n的约数乘积。

设d(x)为x的约数个数，x=p1^a1+p2^a2+……+pn^an，f(x)为x的约数乘积。

若a,b互质，有f(ab)=f(a)^d(b)*f(b)^d(a)，d(ab)=d(a)*d(b)

然后f(p^k)=p^(k*(k+1)/2)

（其实就是拆出来算贡献。。

#include<cstring>
#include<iostream>
#include<cstdio>
#include<queue>
#include<cmath>
#include<algorithm>
#define rep(i,l,r) for (ll i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define low(x) (x&(-x)) 
#define maxn 200500
#define inf int(1e9)
#define mm 1000000007
#define ll long long
using namespace std;
ll x;
ll a[maxn],n,d[maxn],now;
ll read(){
    ll x=0,f=1; char ch=getchar();
    while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
ll Pow(ll x,ll y){
    ll ans=1;
    while (y){
        if (y&1) ans=ans*x%mm;
        x=x*x%mm;
        y>>=1;
    }
    return ans;
}
int main(){
    n=read();
    rep(i,1,n){
        x=read(); d[x]++;
    }
    ll ans=1,D=1;
    rep(i,2,200000) if (d[i]){
        ll fb=Pow(i,d[i]*(d[i]+1)/2)%mm;
        ans=Pow(ans,d[i]+1)*Pow(fb,D)%mm;
        D=D*(d[i]+1)%(mm-1);
    }
    printf("%lld
",ans);
    return 0;
}