hdu5525Product+约数的乘积

Problem Description
Given a number sequence A1,A2….An,indicating N=∏ni=1iAi.What is the product of all the divisors of N?

Input
There are multiple test cases.
First line of each case contains a single integer n.(1≤n≤105)
Next line contains n integers A1,A2….An,it’s guaranteed not all Ai=0.(0≤Ai≤105).
It’s guaranteed that ∑n≤500000.

Output
For each test case, please print the answer module 109+7 in a line.

Sample Input

4
0 1 1 0
5
1 2 3 4 5

Sample Output

36
473272463

Source
BestCoder Round #61 (div.2)

Recommend
hujie | We have carefully selected several similar problems for you: 5906 5905 5903 5900 5899

方法同http://blog.youkuaiyun.com/xtulollipop/article/details/52706262
http://codeforces.com/problemset/problem/615/D

$设这个数为X=p^{a1}_1⋅p^{a2}_2⋯p^{an}_n$
考虑$p^{a_1}_1$对于约数积的贡献,首先会在后面所有$∏^n_{i=2} (a_i+1)$个约数中出现
$且p^{a_1}_1取值有p^0_1、p^1_1⋯p^{a1}_1,那么p1的贡献:$
$Ep1=(p^0_1*p^1_1⋯p^{a1}_1)^{∏^n_{i=2} (a_i+1)}$
　　 $=(p^\frac{{a_1⋅(a_1+1)}}{2})^{∏^n_{i=2}(ai+1)}$
　　 $=p^{\frac{a_1⋅∏^n_{i=1}(a_i+1)}{2}}$
$同理Ep_i=p_i^\frac{{ai}⋅∏^n_{i=1}(a_i+1)}{2}$
so $ans=∏Ep_i$
同样的有连个地方要注意，一个是$∏^n_{i=2} (a_i+1)$可能会很大，，得用费马小定理$a^p\%m=a^{φ(m)+p\%φ(m)}\%m$
第二个就是有除法的膜。$\frac{a}{b}\%c=\frac{a\%bc}{c}$

打个表。。然后每输入一个数，直接质因数分解就好了。。
orz..坑质因数的个数可能很大。。求约数个数的时候，注意细节处理，，（不知所错的。。wa了6,7发。。）

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define maxn 200000
const LL mod1=1000000007;
const LL mod2=2*(mod1-1);
LL quick_mod(LL a,LL n,LL mod){
    LL ans=1;
    while(n){
        if(n&1) ans=ans*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return ans;
}
int prime[100005];
bool visit[100005];
int cnt=0;
void getprime(){
    memset(visit,false,sizeof(visit));
    for(int i=2;i<=100005;i++){
        if(visit[i]==false){
            prime[cnt++]=i;
            for(int j=i+i;j<=100005;j+=i) visit[j]=true;
        }
    }
}

map<int,LL>Q;
map<int,LL>::iterator it;

LL a[100006];

void add(int x,LL p){
    for(int i=0;prime[i]*prime[i]<=x;i++){
        while(x%prime[i]==0){
            a[prime[i]]+=p;
            x/=prime[i];
        }
    }
    if(x>1) a[x]+=p;
}
int main(){
    int n;
    LL x;
    getprime();
    while(cin>>n){
        Q.clear();
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++){
            cin>>x;
            add(i,x);
        }
        for(int i=2;i<=n;i++) {
            if(a[i]!=0) Q[i]=a[i];
        }
       /* for(it=Q.begin();it!=Q.end();it++){
            cout<<it->first<<" "<<it->second<<endl;
        }*/
        LL all=1;
        for(it=Q.begin();it!=Q.end();it++){
            all*=((it->second)+1)%mod2;//wa了5,6发，取膜啊！
            all%=mod2;
        }
        LL ans=1;
        for(it=Q.begin();it!=Q.end();it++){
            LL p=1LL*(it->second)%mod2;//取膜啊！
            //cout<<p<<endl;
            LL temp=p*all%mod2;
            temp/=2;
            temp+=mod1-1;
            //cout<<temp<<endl;
            ans*=quick_mod((it->first),temp,mod1);
            ans%=mod1;
            //cout<<ans<<endl;
        }
        cout<<ans<<endl;
    }
    return 0;
}