题解 | #识别有效的IP地址和掩码并进行分类统计#
识别有效的IP地址和掩码并进行分类统计
https://www.nowcoder.com/practice/de538edd6f7e4bc3a5689723a7435682
A,B,C,D,E,NO,__ = 0,0,0,0,0,0,0
while True:
try:
n = input().split('~')
zi = n[0]
yan = n[1]
res_list = zi.split('.')
res_list1 = yan.split('.')
if res_list[0] != '0' and res_list[0] != '127':
if '' in res_list or '' in res_list1 or len(res_list) != 4 or len(res_list1) != 4:
NO += 1
else:
res_list = list(map(int,res_list))
res_list1 = list(map(int,res_list1))
res_str = ''
for i in res_list1:
res_str += str(bin(i))[2:].rjust(8,'0')
res_str1 = sorted(res_str,reverse=True)
if list(res_str) != res_str1 or len(set(res_str1)) != 2 or res_str1.count('1') == 1:
NO += 1
else:
for i in range(4):
if res_list[i] not in range(0,256) or res_list1[i] not in range(0,256):
NO +=1
break
else:
if res_list[0] in range(1,127):
A += 1
if res_list[0] == 10:
__ += 1
elif res_list[0] in range(128,192):
B += 1
if res_list[0] == 172 and res_list[1] in range(16,32):
__ += 1
elif res_list[0] in range(192,224):
C += 1
if res_list[0] == 192 and res_list[1] == 168:
__ += 1
elif res_list[0] in range(224,240):
D += 1
elif res_list[0] in range(240,256):
E += 1
except:
break
print(A,end=' ')
print(B,end=' ')
print(C,end=' ')
print(D,end=' ')
print(E,end=' ')
print(NO,end=' ')
print(__)
