题解 | #识别有效的IP地址和掩码并进行分类统计#

import sys

Dict = {"A": 0, "B": 0, "C": 0, "D": 0, "E": 0, "ERROR": 0, "PRIVATE": 0}


def ERROR_ip(ip: str):
    try:
        ips = list(map(int, ip.split(".")))
        for item in ips:
            if int(item) < 0 or int(item) > 255:
                return True
        return False
    except:
        return True


def ERROR_mask(mask: str):
    if ERROR_ip(mask):
        return True
    masks = list(map(int, mask.split(".")))
    T = [128, 192, 224, 240, 248, 252, 254]
    if masks[0] not in T + [255] or masks[3] == 255:
        return True
    for i in range(1, 4):
        if masks[i - 1] in [0] + T:
            if masks[i] != 0:
                return True
        if masks[i - 1] == 255:
            if masks[i] not in [0] + T + [255]:
                return True
    return False


def PRIVATE_ip(ip: str):
    ip0 = int(ip.split(".")[0])
    ip1 = int(ip.split(".")[1])
    if ip0 == 10:
        return True
    if ip0 == 172 and ip1 >= 16 and ip1 <= 16:
        return True
    if ip0 == 192 and ip1 == 168:
        return True
    return False


for line in sys.stdin:
    a = line.split("~")
    ip, mask = a[0], a[1].split()[0]
    ip0 = int(ip.split(".")[0])
    if ip0 == 0 or ip0 == 127:
        continue
    if not ERROR_ip(ip):
        if not ERROR_mask(mask):
            if ip0 >= 1 and ip0 <= 126:
                Dict["A"] += 1
            elif ip0 >= 128 and ip0 <= 191:
                Dict["B"] += 1
            elif ip0 >= 192 and ip0 <= 223:
                Dict["C"] += 1
            elif ip0 >= 224 and ip0 <= 239:
                Dict["D"] += 1
            elif ip0 >= 240 and ip0 <= 255:
                Dict["E"] += 1
            if PRIVATE_ip(ip):
                Dict["PRIVATE"] += 1
        else:
            Dict["ERROR"] += 1
    else:
        Dict["ERROR"] += 1

for v in Dict.values():
    print(v, end=" ")