题解 | #识别有效的IP地址和掩码并进行分类统计#

L = [0] * 7  # result: A-E:0-4 Error:5 Private:6


def isip(lst):  # 基础: 判断是否为ip 并成为ip列表 [d,d,d,d]
    if len(lst) != 4 or '' in lst:
        return 0
    for i in lst:
        if int(i) < 0 or int(i) > 255:
            return 0
    return 1


def ismask(lst):
    if isip(lst):
        lst01 = []
        for i in lst:
            lst01.append(bin(int(i))[2:].zfill(8))
        str01 = ''.join(lst01)
        t1 = abs(str01.rfind('1'))  # 避免全0无法排查
        t2 = str01.find('0')
        if t1 + 1 == t2:
            return 1
        else:
            return 0
    else:
        return 0


def Ip_classification(lst):  # 主要
    f = int(lst[0])
    if 0 < f <= 126:  # A
        L[0] += 1
    elif 128 <= f <= 191:  # B
        L[1] += 1
    elif 192 <= f <= 223:  # C
        L[2] += 1
    elif 224 <= f <= 239:  # D
        L[3] += 1
    elif 240 <= f <= 255:  # E
        L[4] += 1

def isprivate(lst):  # 基础
    f = int(lst[0])
    if f == 10: return 1
    if f == 172 and 16 <= int(lst[1]) <= 31: return 1
    if f == 192 and int(lst[1]) == 168:
        return 1
    else:
        return 0


def isaddprivate(lst):  # 主要
    if isprivate(lst):
        L[6] += 1

if __name__ == '__main__':
    while True:
        try:
            ip, mask = input().split('~')
        except:
            break
        ip_lst = ip.split('.') # [d,d,d,d]
        mask_lst = mask.split('.')
        if ip_lst[0] == '0' or ip_lst[0] =='127':
            continue    # ip 不分类就不再考虑掩码问题
        r1 = isip(ip_lst)
        r2 = ismask(mask_lst)
        if r1 ==1 and r2 ==1:  # 都合法
            isaddprivate(ip_lst)
            Ip_classification(ip_lst)
        else:
            L[5] += 1
    for i in L:
        print(i,end=' ')