题解 | rin和快速迭代

import sys
def countFactor(num):
    count=2
    for i in range(2,int(num**0.5)+1):
        #若i为num的因子，则num/i同样为其因子，所以因子数加2
        if num%i==0:
            count+=2
        #若i**2=num,此时因子相同，则需-1
        if i*i==num:
            count-=1
    return count
n = int(input())
res=0
while n>2:
    n=countFactor(n)
    res+=1
print(res)