题解 | #识别有效的IP地址和掩码并进行分类统计#

def get_lst(s):
    lst = []
    for i in s.split('.'):
        if i.isdigit():
            lst.append(int(i))
    return lst
    
def check_ip(lst):
    if len(lst) != 4:
        return False
    # 检查每段数字是否在0～255之间
    for i in lst:
        if i not in range(0, 256):
            return False
    return True

def check_mask(lst):
    if len(lst) != 4:
        return False
    res = ''
    for i in lst:
        temp = bin(i).replace('0b', '')
        res += (8 - len(temp)) * '0' + temp # 不足8位用‘0’补全
    if '01' in res or '1' not in res or '0' not in res:
        return False
    return True
    
dic = {
    'a' : 0, 'b' : 0, 'c' : 0, 'd' : 0, 'e' : 0, 'illegal' : 0, 'private' : 0
}

while True:
    try:
        ip, mask = input().split('~')
        lst_ip, lst_mask = get_lst(ip), get_lst(mask)
        # 忽略0和127开头的IP地址
        if lst_ip[0] in [0, 127]:
            continue
        if not check_mask(lst_mask):
            dic['illegal'] += 1
            continue
        if not check_ip(lst_ip):
            dic['illegal'] += 1
            continue
        # 检查是否为私网
        if (lst_ip[0] == 10) or (lst_ip[0] == 172 and lst_ip[1] in range(16, 32)) or (lst_ip[0] == 192 and lst_ip[1] == 168):
            dic['private'] += 1 # 私网
        # 继续检查第一段
        if lst_ip[0] in range(1, 127):
            dic['a'] += 1
        elif lst_ip[0] in range(128, 192):
            dic['b'] += 1
        elif lst_ip[0] in range(192, 224):
            dic['c'] += 1
        elif lst_ip[0] in range(224, 240):
            dic['d'] += 1
        elif lst_ip[0] in range(240, 256):
            dic['e'] += 1
    except:
        break

for i in dic.values():
    print(i, end=' ')
print()