题解 | #整数与IP地址间的转换#
整数与IP地址间的转换
https://www.nowcoder.com/practice/66ca0e28f90c42a196afd78cc9c496ea?tpId=37&tqId=21256&rp=1&ru=/exam/oj/ta&qru=/exam/oj/ta&sourceUrl=%2Fexam%2Foj%2Fta%3Fpage%3D1%26tpId%3D37%26type%3D37&difficulty=undefined&judgeStatus=undefined&tags=&title=
ls=input().split('.')
num=int(input())
#num-->bin_num
def turn_to_bin(num):
nums=[]
while num!=0:
nums.append(num%2)
num=num//2
nums=nums[::-1]
s=''
for i in nums:
s+=str(i)
other_zero_len=8-len(s)
while other_zero_len!=0:
s='0'+s
other_zero_len-=1
return s
#str-->num 2进制字符串--》10进制数
def str_turn_to_num(s):
sum=0
n=0
for i in s[::-1]:
sum+=int(i)*2**n
n+=1
return (sum)
#ip--> num
sum_str=''
for item in ls:
sum_str=sum_str+turn_to_bin(int(item))
sum=0
n=0
for i in sum_str[::-1]:
sum+=int(i)*2**n
n+=1
#num-->ip
num=bin(num)
num=str(num[2:])
if len(num)%8!=0:
zero_len=8-len(num)%8
num=zero_len*'0'+num
index=0
ip=''
while index!=len(num):
a=''
a+=num[index]
index+=1
while index%8!=0:
a+=num[index]
index+=1
ip+=str(str_turn_to_num(a))+'.'
print(sum)
print(ip[:-1])
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