题解 | rin和快速迭代

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        int count = 0;
        
        while (n > 2) {
            n = countFactors(n);
            count++;
        }
        
        System.out.println(count);
    }

    // 计算 n 的因子个数
    private static long countFactors(long n) {
        long count = 0;
        for (long i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                if (i * i == n) {
                    count += 1; // 如果 i*i == n, 只加一次
                } else {
                    count += 2; // 否则加两次
                }
            }
        }
        return count;
    }
}