CRYPTO
1，NO SOS
题目给了一段由.和-构成的密码
由于题目提示不是摩斯码，将.和-化为0和1，长度为65位无法与8或7整除，无法转换为ascii，但可以被5整除，猜测为培根密码，将0化为a，1化为b
解密得flagisguetkkp
以格式flag{guetkkp}
MISC
1，KO
题目给了一段由.?!构成的txt
猜测为OOK用
(http://www.splitbrain.org/sercives/ook)
运行得到flag：welcome to CTF
RE
1,RE
题目给了一个ELF，拖进IDA，发现大段未能反汇编的，猜测为压缩壳 用upx -d
解压缩，将得到的文件再次拖入IDA
由此写脚本
a1=166163712//1629056 a2=731332800//6771600 a3=357245568//3682944 a4=1074393000//10431000 a5=489211344//3977328 a6=518971936//5138336 a7=406741500//7532250 print(chr(a1)+chr(a2)+chr(a3)+chr(a4)+chr(a5)+chr(a6)+'A'+chr(a7),end='') a8=294236496//5551632 a9=177305856//3409728 a10=650683500//13013670 a11=298351053//6088797 a12=386348487//7884663 a13=438258597//8944053 a14=249527520//5198490 a15=445362764//4544518 a16=981182160//10115280 a17=174988800//3645600 print(chr(a8)+chr(a9)+chr(a10)+chr(a11)+chr(a12)+chr(a13)+chr(a14)+chr(a15)+chr(a16)+chr(a17),end='') a18=493042704//9667504 a19=257493600//5364450 a20=767478780//13464540 a21=312840624//5488432 a22=1404511500//14479500 a23=316139670//6451830 a24=619005024//6252576 a25=372641472//7763364 a26=373693320//7327320 a27=498266640//8741520 a28=452465676//8871876 a29=208422720//4086720 a30=515592000//9374400 a31=719890500//5759124 print(chr(a18)+chr(a19)+chr(a20)+chr(a21)+chr(a22)+chr(a23)+chr(a24)+chr(a25)+chr(a26)+chr(a27)+chr(a28)+chr(a29)+chr(a30)+chr(a31))
得到flag
ps：特别注意没有a1[6],即没有第七位，因此flag不唯一，在脚本中用A代替