莫欺少年穷，修仙之旅在这开始—>算法基础课题解

思路：

$ N = \prod\limits_{i=1}^k p_i^{a_i}=p_1^{a_1}\cdot p_2^{a_2}… p_k^{a_k} $

$ 约数个数：\prod\limits_{i=1}^k(a_i+1)=(a_1+1)(a_2+1)…(a_k+1) $

$\begin{aligned}约数之和：\prod\limits_{i=1}^k \sum\limits_{j=0}^{a_i} p_i^j & = \prod\limits_{i=1}^k (p_i^0+p_i^1+…+p_i^{a_i}) \\\\ & = (p_1^0+p_1^1+…+p_1^{a_1})(p_2^0+p_2^1+…+p_2^{a_2})…(p_k^0+p_k^1+…+p_k^{a_k})\end{aligned}$

可参考： 约数之和

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int mod = 1e9 + 7;

int main()
{
    int n;
    cin>>n;

    unordered_map<int,int> primes;
    while(n--)
    {
        int x;
        cin>>x;

        //找出所有小于等于 sqrt(x) 的质因子
        for(int i=2;i<=x/i;i++)
            while(x%i==0)
            {
                x/=i;
                primes[i]++;
            }

        //唯一大于 sqrt(x) 的质因子
        if(x>1) primes[x]++;
    }

    LL res=1;
    for(auto prime : primes) res=res*(prime.second+1)%mod;

    cout<<res<<endl;

    return 0;
}