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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