删掉默认之后

（每道题开头都有同一段：net_02 的 to_int / mask、判「前缀是否包含地址」的 contains，和本机的路由表 ROUTES——每行 (前缀, 下一跳或 None 表示直连, 接口, 度量)。）

本机的路由表：

0.0.0.0/0        via 192.168.10.1   eth0  100   ← 默认
192.168.10.0/24  直连               eth0  100
10.0.0.0/8       via 192.168.10.254 eth0  100   ← 机房
10.20.0.0/16     via 192.168.10.253 eth0  100   ← 机房里更具体的一段
127.0.0.0/8      直连               lo    0

把默认路由那一行去掉，再对同样六个地址查表，看几个查不到（None）：

def to_int(ip):
    a, b, c, d = [int(x) for x in ip.split(".")]
    return (a << 24) | (b << 16) | (c << 8) | d


def mask(prefix):
    return (0xFFFFFFFF << (32 - prefix)) & 0xFFFFFFFF


def contains(cidr, ip):
    net, p = cidr.split("/")
    return to_int(net) & mask(int(p)) == to_int(ip) & mask(int(p))


ROUTES = [
    ("0.0.0.0/0", "192.168.10.1", "eth0", 100),
    ("192.168.10.0/24", None, "eth0", 100),
    ("10.0.0.0/8", "192.168.10.254", "eth0", 100),
    ("10.20.0.0/16", "192.168.10.253", "eth0", 100),
    ("127.0.0.0/8", None, "lo", 0),
]


def lookup(routes, ip):
    best = None
    for cidr, via, dev, metric in routes:
        if not contains(cidr, ip):
            continue
        plen = int(cidr.split("/")[1])
        if best is None or plen > best[0] or (plen == best[0] and metric < best[1]):
            best = (plen, metric, cidr, via, dev)
    return best

no_default = [r for r in ROUTES if r[0] != "0.0.0.0/0"]
ips = ("8.8.8.8", "1.1.1.1", "10.7.7.7", "192.168.11.5", "172.16.0.9", "192.168.10.1")
lost = [ip for ip in ips if lookup(no_default, ip) is None]
print(str(len(lost)) + "/" + ",".join(lost))