buu Reverse学习记录(27) [FlareOn4]login

题目链接：https://buuoj.cn/challenges#[FlareOn4]login

题目是个html网页，打开查看源码

“rotFlag = flag.replace(/[a-zA-Z]/g, function©{return String.fromCharCode((c <= “Z” ? 90 : 122) >= (c = c.charCodeAt(0) + 13) ? c : c - 26);});”
这是关键的加密地方，就是一个rot13加密，写个脚本解密下就行。

s = 'PyvragFvqrYbtvafNerRnfl@syner-ba.pbz'
flag = ''
for i in s:
    if (i >= 'A' and i <= 'Z'):
        x = ord(i) - 13
        if x < 65:
            x += 26
        flag += chr(x)
    elif (i >= 'a' and i <= 'z'):
        x = ord(i) - 13
        if x < 97:
            x += 26
        flag += chr(x)
    else:
        flag += i
print(flag)

flag:ClientSideLoginsAreEasy@flare-on.com