有一个数字魔法,给你一个正整数n,如果n为偶数,就将他变为n/2, 如果n为奇数,就将他变为乘3加1
不断重复这样的运算,经过有限步之后,一定可以得到1
牛牛为了验证这个魔法,决定用一个整数来计算几步能变成1
number = int(input()) times = 0 while number != 1: if number % 2 == 0: number = number / 2 elif number % 2 != 0: number = number * 3 + 1 times = times + 1 print(times)
n=int(input()) s=0 while True: if n ==1: break elif n % 2 ==0: n=n/2 s+=1 else: n=n*3+1 s+=1 print(s)AI说用while True 比较优雅。
n = int(input()) steps = 0 while n != 1: try: if n%2 ==0 : n = n/2 steps += 1 else: n = 3*n+1 steps += 1 except: break print (steps)