题解 | rin和快速迭代

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;


ll fx(ll a){
	ll num = 0;
	
	for(ll i=1 ; i*i <= a ; i++){
		if(i*i == a){
			num++;
		}
		else if(a % i == 0){
			num += 2;
		}
	}
	return num;
	
}

int main(){
	ll n;  cin >> n;
	ll ans = 0;
	
	while(fx(n) != 2){
		ans++;
		n = fx(n);
		//cout << n << endl;
	}
	
	cout << ans + 1;
	
	return 0;
}