如果 N = p1^c1 * p2^c2 * … *pk^ck(p为质因子)
N的约数个数:(c1 + 1) * (c2 + 1) * … * (ck + 1)
N的约数之和:(p1^0 + p1^1 + … + p1^c1) * … * (pk^0 + pk^1 + … + pk^ck)

最大公约数
辗转相除法
O(logn)

int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}

更相减损术
O(n)

int gcd_sub(int a, int b)
{
    if (a < b)
    {
        swap(a, b);
    }
    if (b == 1)
    {
        return a;
    }
    return gcd_sub(b, a / b);
}

最小公倍数
O(logn)

int lcm(int a, int b)
{
    return (ll)a * b / gcd(a, b);
}

n个数乘积的约数个数

unordered_map<int, int> primes;

int main()
{
    std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    while (n--)
    {
        int x;
        cin >> x;
        for (int i = 2; i <= x / i; i++)
        {
            while (x % i == 0)
            {
                x /= i;
                primes[i]++;
            }
        }
        if (x > 1)
        {
            {
                primes[x]++;
            }
        }
    }
    ll res = 1;
    for (auto prime : primes)
    {
        res = res * (prime.second + 1) % 1000000007;
    }
    cout << res << endl;
    return 0;
}

n个数的乘积的约数之和

unordered_map<int, int> primes;

void divide(int x)
{
    for (int i = 2; i <= x / i; i++)
    {
        if (x % i == 0)
        {
            while (x % i == 0)
            {
                x /= i;
                primes[i]++;
            }
        }
    }
    if (x > 1)
    {
        primes[x]++;
    }
}

int main()
{
    std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        int x;
        cin>>x;
        divide(x);
    }
    ll res=1;
    for(auto prime:primes)
    {
        int p=prime.first,a=prime.second;
        ll t=1;
        while(a--)
        {
            t=(t*p+1)%mod;
        }
        res=res*t%mod;
    }
    cout<<res<<endl;
    return 0;
}