hdu 5525 Product 数论算贡献

Product

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

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)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5525

 

题意：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e5+10,M=4e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1e9+7;

///   数组大小
int prime(int n)
{
    if(n<=1)
    return 0;
    if(n==2)
    return 1;
    if(n%2==0)
    return 0;
    int k, upperBound=n/2;
    for(k=3; k<=upperBound; k+=2)
    {
        upperBound=n/k;
        if(n%k==0)
            return 0;
    }
    return 1;
}

vector<int>p;
int si[N];
void init()
{
    for(int i=2;i<=100000;i++)
        if(prime(i))
        p.push_back(i);
}
ll quick(ll a,ll b,ll c)
{
    ll ans=1;
    while(b)
    {
        if(b&1)ans=(ans*a)%c;
        b>>=1;
        a=(a*a)%c;
    }
    return ans;
}
int main()
{
    init();
    int n;
    while(~scanf("%d",&n))
    {
        memset(si,0,sizeof(si));
        for(int i=1;i<=n;i++)
        {
            int z;
            int x=i;
            scanf("%d",&z);
            for(int j=0;j<p.size();j++)
            {
                if(1LL*p[j]*p[j]>x)break;
                if(x==1)break;
                while(x%p[j]==0)
                {
                    si[j]+=z;
                    si[j]=(si[j])%(2*(mod-1));
                    x/=p[j];
                }
            }
            if(x!=1)
            {
                int pos=lower_bound(p.begin(),p.end(),x)-p.begin();
                si[pos]+=z;
            }
        }
        ll sum=1;
        for(int i=0;i<p.size();i++)
        {
            sum*=(si[i]+1);
            sum%=(2*(mod-1));
        }
        //cout<<sum<<endl;
        ll ans=1;
        for(int i=0;i<p.size();i++)
        {
            ans=(ans*quick(p[i],((sum*si[i])%(2*(mod-1)))/2,mod))%(mod);
        }
        printf("%lld
",ans);
    }
    return 0;
}