题解 | #识别有效的IP地址和掩码并进行分类统计#

import sys

A =[]
B =[]
C=[]
D=[]
E=[]
wrong=[]
pri= []

for line in sys.stdin:
    
    cnt=line
    is_wrong = 0
    a = line.strip().split("~")
    ip,mask = a[0], a[1]
    iplist = ip.split('.')
    masklist= mask.split('.')
    if iplist[0] =='0' or iplist[0]=='127':
        continue
    if len(iplist)!=4 or len(masklist)!=4:
        wrong.append(cnt)
        is_wrong =1
        continue

    for i in iplist:
        if not i.isdigit():
            wrong.append(cnt)
            is_wrong =1
            break
        if int(i)>255 or int(i)<0:
            wrong.append(cnt)
            is_wrong =1
            break
    if is_wrong:
        continue
    twostr = ''
    for i in masklist:
        if not i.isdigit():
            wrong.append(cnt)
            is_wrong =1
            break
        if int(i)>255 or int(i)<0:
            wrong.append(cnt)
            is_wrong =1
            break
        intwo = bin(int(i))[2:]
        twostr += intwo.zfill(8)
    is_one=1
    if twostr[0]==0:
        wrong.append(cnt)
        is_wrong =1
        continue
    for i in twostr:
        if i=='0':
            is_one=0
        if is_one==0 and i=='1':
            wrong.append(cnt)
            is_wrong =1
            break
    if is_one:
        wrong.append(cnt)
        is_wrong =1
        continue

    if is_wrong:
        continue
    if int(iplist[0])>=1 and int(iplist[0])<=126:
        A.append(cnt)
    if int(iplist[0])>=128 and int(iplist[0])<=191:    
        B.append(cnt)
    if int(iplist[0])>=192 and int(iplist[0])<=223:
        C.append(cnt)
    if int(iplist[0])>=224 and int(iplist[0])<=239:    
        D.append(cnt)
    if int(iplist[0])>=240 and int(iplist[0])<=255:
        E.append(cnt)

    if (int(iplist[0])==10) or  (int(iplist[0])==172 and int(iplist[1])>=16 and int(iplist[1])<=31) or (int(iplist[0])==192 and int(iplist[1])==168):
        pri.append(cnt)

print(len(A),len(B),len(C),len(D),len(E),len(wrong),len(pri))