iscc2016-basic-明察秋毫

 

 

 

 

查看源代码,找到
maybe not flag : Jr1p0zr2VfPp

移位密码,注意判断字母大小写,并且数字无变化

 

s = "Jr1p0zr2VfPp"
p = list(s)
for i in range(26):
    for j in range(len(s)):
        if s[j].isupper():
            p[j] = chr((ord(s[j])-65+i)%26+65)
        elif s[j].islower():
            p[j] = chr((ord(s[j])-97+i)%26+97)
    print ''.join(p)

  

posted @ 2016-05-30 21:32  ailx10  阅读(174)  评论(0编辑  收藏  举报