rin和快速迭代

#include <bits/stdc++.h>

using namespace std;

long long n;
long long sum;
int a[1000005];

int main(){
    cin >> n;
    while (n != 2){
        int i = 2;
        long long x = n;

        for(i = 2; i <= x/i; ++i)
            while(x % i == 0){
                x /= i;
                a[i] ++;
            }

        long long ans = 1;
        if (x != 1) ans *= 2;
       for(int j = 2; j <= i; ++j) if (a[j] != 0) ans *= (a[j] + 1);
       sum ++;
       n = ans;
       memset(a, 0, sizeof(a));
    }
    printf("%lld", sum);
    return 0;
}