需要一些帮助来编写Python中的持久数字程序,以确定具有持久性的最小非负整数 (715 ---> 35 ---> 15 ---> 5)3,然后是4,然后是5,然后是6,然后是7.输出格式如下:
The smallest integer with a persistence of 3 is: 39
The smallest integer with a persistence of 4 is: xx
The smallest integer with a persistence of 5 is: xxx
The smallest integer with a persistence of 6 is: xxxx
The smallest integer with a persistence of 7 is: xxxxx
任何代码链接或线索都将不胜感激,谢谢。
答案 0 :(得分:-1)
def main():
low_limit = 3
high_limit = 7
for limit in range(low_limit, high_limit+1):
number = 1
while persistence(number) < limit:
number += 1
print('The smallest integer with a persistence of', limit, 'is:', number)
def persistence(x):
pcount = 0 # counting the p from 0
while x>=10:
y=x # temp copy of x
z=1 # z holds the value of y as being broken down and multplied
while (y!=0):
z=z*(y%10) # stripping last number and mult by last z value
y=y//10 # integer division, dropping off the last digit
x=z
pcount +=1
return pcount
main()