[GUET-CTF2019]re
例行检查 ，64bit，upx
去壳，用64位IDA打开,进入函数
可以看到让我输入的字符串符合sub_4009AE就输出correct，直接看sub_4009AE
里面是一长串的比较，对我们输入的每一位判断
因此只要对每一位逆向算一下就可以了
a1=chr(166163712//1629056)
a2=chr(731332800//6771600)
a3=chr(357245568//3682944)
a4=chr(1074393000//10431000)
a5=chr(489211344//3977328)
a6=chr(518971936//5138336)
a8=chr(406741500//7532250)
a9=chr(294236496//5551632)
a10=chr(177305856//3409728)
a11=chr(650683500//13013670)
a12=chr(298351053//6088797)
a13=chr(386348487//7884663)
a14=chr(438258597//8944053)
a15=chr(249527520//5198490)
a16=chr(445362764//4544518)
a17=chr(981182160//10115280)
a18=chr(174988800//3645600)
a19=chr(493042704//9667504)
a20=chr(257493600//5364450)
a21=chr(767478780//13464540)
a22=chr(312840624//5488432)
a23=chr(1404511500//14479500)
a24=chr(316139670//6451830)
a25=chr(619005024//6252576)
a26=chr(372641472//7763364)
a27=chr(373693320//7327320)
a28=chr(498266640//8741520)
a29=chr(452465676//8871876)
a30=chr(208422720//4086720)
a31=chr(515592000//9374400)
a32=chr(719890500//5759124)
print(a1,a2,a3,a4,a5,a6,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18,a19,a20,a21,a22,a23,a24,a25,a26,a27,a28,a29,a30,a31,a32)
a7并没有给我们，那就从0开始一个一个试，到1的时候成功了
flag{e165421110ba03099a1c039337}