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)
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