#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
int i = 0;
while(n!=1)
{
i++;
if(n%2==1)
{
n = (3 * n + 1)/2;
}
else
{
n = n / 2;
}
//cout << n << endl;
}
cout << i << endl;
return 0;
}
//callatz假设的1000以内的数验证
原文：http://www.cnblogs.com/1915884031A-qqcom/p/7230642.html