B - Function

\(f(n)=\left\{\begin{array}{ll} 1, & n \in\{1\} \bigcup \text { Prime } \\ p f\left(p^{k-2}\right), & n=p^{k}(p \in \text { Prime; } k>1) \\ f\left(p_{1}^{e 1}\right) \prod_{i=2}^{r} p_{i}^{e_{i}}, & n=\prod_{i=1}^{r} p_{i}{ }^{e_{i}}\left(p_{1}<p_{2}<\cdots<p_{r} ; p_{i} \in \text { Prime; } r \geq 2\right) \end{array}\right.\)

求\(\sum_{i=1}^{n} f(i)\)​

假設一個合數\(n\)，代入第三個式子，可得\(f(n)=p_1f(p_1^{e_1-2})\prod_{i=1}^{r} p_{i}{ }^{e_{i}}\)，再將\(p_1^{e_1-2}\)代入第二個式子，可得\(f(n)=p_1^2f(p_1^{e_1-4})\prod_{i=1}^{r} p_{i}{ }^{e_{i}}\)，以此類推，最後可推出\(f(n)=p_1^{\frac{e_1}{2}}\prod_{i=1}^{r} p_{i}{ }^{e_{i}}\).

Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define gcd __gcd
#define endl '\n'
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int maxn = 1e7 + 5;
int prime[maxn], vis[maxn] = {0}, cnt = 0;
ll power(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = ans * a % mod;
        }
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}
ll Euler(int n)
{
    ll res = 1;
    for (int i = 2; i <= n; i++)
    {
        if (!vis[i])
        {
            prime[cnt++] = i;
            res++;
        }
        for (int j = 0; prime[j] <= n / i; j++)
        {
            int tmp = prime[j] * i;
            vis[tmp] = 1;
            int count = 0;
            while (tmp % prime[j] == 0)
            {
                count++;
                tmp /= prime[j];
            }
            res += power(prime[j], count / 2) * tmp;
            if (i % prime[j] == 0)
            {
                break;
            }
        }
    }
    return res;
}

signed main()
{
    ll n;
    cin >> n;
    printf("%lld\n", Euler(n));
    return 0;
}