题解 | #识别有效的IP地址和掩码并进行分类统计#

#这题主要是对题目的理解，出题描述写的不是很清楚
#首先对于0,127开头的ip都不参加统计
#然后对于掩码错误 或IP错误的地址行进行计数
#最后对剩下的地址行进行分类统计和私有ip地址统计
import sys
A,B,C,D,E,F,G=0,0,0,0,0,0,0
def ill(s):
    if "0" not in s or "1" not in s:
        return True
    else:
        for i in range(len(s)):
            if s[i]=='0':
                s=s[i:]
                for i in range(len(s)):
                    if s[i]=="1":
                        return True
                return False

for line in sys.stdin:
    b=[]
    a = line.split()[0].split("~")
    a[0]=a[0].split(".")
    a[1]=a[1].split(".")
    b.append(['' if i == '' else int(i) for i in a[0]])
    b.append(['' if i == '' else int(i) for i in a[1]])
    s=""
    if b[0][0] in [0, 127]:
        continue
    for i in range(len(b[1])):
        s=s+bin(b[1][i])[2:].rjust(8,'0')
    if ill(s):
        F=F+1
        continue
    if '' in a[0]:
        F=F+1
        continue
    if b[0]<=[126,255,255,255] and b[0]>=[1,0,0,0]:
        A=A+1
    if b[0] >=[128,0,0,0] and b[0]<=[191,255,255,255]:
        B = B+ 1
    if b[0]>=[192,0,0,0] and b[0]<=[223,255,255,255]:
        C = C+ 1
    if b[0]>=[224,0,0,0] and b[0]<=[239,255,255,255]:
        D = D + 1
    if b[0]>=[240,0,0,0] and b[0]<=[255,255,255,255]:
        E = E + 1
    if ((b[0]>=[10,0,0,0] and b[0]<=[10,255,255,255])
            or (b[0]>=[172,16,0,0] and b[0]<=[172,31,255,255])
            or (b[0]>=[192,168,0,0] and b[0]<=[192,168,255,255])):
        G=G+1


print(A,B,C,D,E,F,G)