一個數 N 的約數個數:
設 N = p1^a1 * p2 ^ a2 * p3 ^ a3 … pn ^ an;
約數個數 sum = (a1 + 1) * (a2 + 1) * ***** ( an + 1).
約數之和: (p1^0 + p1 ^ 1 + … p1^a1) * (p2^0 + p2 ^ 1 + …p2^a2) …
求幾個數的乘積的約數之和
#include <iostream>
#include <cstring>
#include <unordered_map>
#include <algorithm>
using namespace std;
const int mod = 1e9 + 7;
int n;
long long ans = 1;
unordered_map<int, int> m;
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
cin >> n;
while (n--)
{
int x;
cin >> x;
for (int i = 2; i <= x / i; i++)
{
while (x % i == 0)
{
x /= i;
m[i]++;
}
}
if (x > 1)
m[x]++;
}
for (unordered_map<int, int>::iterator it = m.begin(); it != m.end(); it++)
{
int p = it->first;
int k = it->second;
long long t = 1;
while (k--)
t = (t * p + 1) % mod;
ans = (ans * t) % mod;
}
cout << ans << endl;
}