1001. 害死人不偿命的(3n+1)猜想 (15)

1001. 害死人不偿命的(3n+1)猜想 (15)

较简单，直接代码实现：

#include <cstdio>

int main()
{
    int n;
    scanf("%d",&n);
    int k =0; 
    while(n!=1)
    {   
        if(n%2 == 0)
        {   
            k++;
            n = n/2;
        }   
        else
        {   
            n = 3*n+1;
        }   
    }   
    printf("%d
",k);
    return 0;
}