题解 | #识别有效的IP地址和掩码并进行分类统计#

A,B,C,D,E,I,P=0,0,0,0,0,0,0

def ill(s): #判断掩码是否非法 非法返回true 跳出循环
    if ('0' not in s) or ('1' not in s):
        return True
    for i in range(len(s)):
        if s[i]=='0':
            for each in s1[i:]:
                if each != '0':
                    return True
                
while True:
    try:
        s=list(map(str,input().split('~')))
        b,a=[],[]
        s1=''
        
        for each in s:
            b.append(list(map(str,each.split('.'))))
        if b[0][0]=='0' or b[0][0]=='127':
            continue
        if '' in b[0] or '' in b[1]:
            I+=1
            continue
        for each in s: 
            a.append(list(map(int,each.split('.'))))
        for i in range(len(a[1])):
            s1+=bin(a[1][i])[2:].rjust(8,'0') #右对齐补零避免把类似255.255.255.32的掩码判断为合法
        if ill(s1):
            I+=1
            continue
            
        if [1,0,0,0]<=a[0]<=[126,255,255,255]:
                A+=1
        elif [128,0,0,0]<=a[0]<=[191,255,255,255]:
                B+=1
        elif [192,0,0,0]<=a[0]<=[223,255,255,255]:
                C+=1
        elif [224,0,0,0]<=a[0]<=[239,255,255,255]:
                D+=1
        elif [240,0,0,0]<=a[0]<=[255,255,255,255]:
                E+=1
        if [10,0,0,0]<=a[0]<=[10,255,255,255] or [172,16,0,0]<=a[0]<=[172,3,255,255] or [192,168,0,0]<=a[0]<=[192,168,255,255]:
                P+=1
    except:
        break
print(A,B,C,D,E,I,P)