字符串处理求助

请问该如何做，将下列字符串

vlan 3041 profile single

  bindinterface s192.168.111 cisco

  ip host192.168.111.230

  ip host192.168.111.231

  ip host192.168.111.232/30

转成

vlan 3041 profile single

  bindinterface s192.168.111 cisco

  ip host192.168.111.230

  ip host192.168.111.231

  ip host192.168.111.232

  ip host 192.168.111.233

  ip host 192.168.111.234

  ip host 192.168.111.235

烦请协助，谢谢

可变长子网掩码，掩码是30，也就是4个地址，第一个是地址段地址，最后一个是广播地址，就剩两个host地址，是不是这样？

代表四个 ip ，不管 netmask

問題解決了

import re

with open('example.txt', 'r', encoding='utf-8') as rf:
    lines = rf.readlines()
with open('example-aft.txt', 'w', encoding='utf-8') as wf:
    for index, line in enumerate(lines):
        if line.startswith('  ip host'):
            data = re.search(r'(.*)\W(\d+)/30', line)
            if data:
                prefix_ip = data.group(1)
                multi_1 = int(data.group(2))
                multi_2 = int(data.group(2)) + 1
                multi_3 = int(data.group(2)) + 2
                multi_4 = int(data.group(2)) + 3

                ip_1 = prefix_ip + "." + str(multi_1) + "\n"
                ip_2 = prefix_ip + "." + str(multi_2) + "\n"
                ip_3 = prefix_ip + "." + str(multi_3) + "\n"
                ip_4 = prefix_ip + "." + str(multi_4) + "\n"
                ip_sum = ip_1+ip_2+ip_3+ip_4
                lines[index] = '%s' %ip_sum
    wf.writelines(lines)

    print('END')