Sure eno ugh, you still r emember what we talked about at that time! This is part of the s ecret: `hgame{U_ Kn0w_LSB&W`.I pu t the rest of th e content here, https://drive.go ogle.com/file/d/ 13kBos3Ixlfwkf3e 0z0kJTEqBxm7RUk- G/view?usp=shari ng, if you direc tly access the g oogle drive clou d disk download in China, it wil l be very slow, you can try to u se Scientific In ternet access so lves the problem of slow or inac cessible access to external netw ork resources. T his is my favori te music, there is another part of the secret in the music, I us e Steghide to en crypt, the passw ord is also the 6-digit password we agreed at th e time, even if someone else fin ds out here, it should not be so easy to crack ( ( hope so
for line in f.readlines(): cmd=cmdFormat %(stegoFile,extractFile,line.strip()) p=Popen(cmd,shell=True,stdout=PIPE,stderr=STDOUT) content=str(p.stdout.read(),'gbk') for err in errors: if err in content: break else: print (content), print ('the passphrase is %s' %(line.strip())) f.close() return
if __name__ == '__main__': foo() print ('ok') pass
p = 11239134987804993586763559028187245057652550219515201768644770733869088185320740938450178816138394844329723311433549899499795775655921261664087997097294813 q = 12022912661420941592569751731802639375088427463430162252113082619617837010913002515450223656942836378041122163833359097910935638423464006252814266959128953
知道p、q、e,解密
import gmpy2
p = 11239134987804993586763559028187245057652550219515201768644770733869088185320740938450178816138394844329723311433549899499795775655921261664087997097294813 q = 12022912661420941592569751731802639375088427463430162252113082619617837010913002515450223656942836378041122163833359097910935638423464006252814266959128953 e = 65537 c = 110674792674017748243232351185896019660434718342001686906527789876264976328686134101972125493938434992787002915562500475480693297360867681000092725583284616353543422388489208114545007138606543678040798651836027433383282177081034151589935024292017207209056829250152219183518400364871109559825679273502274955582
n=p*q phi=(p-1) * (q-1) d = gmpy2.invert(e,phi) # 通过e和phi求d m = gmpy2.powmod(c,d,n) # 通过d和n解密 print(m) print(hex(m)[2:]) #16进制明文 print(bytes.fromhex(hex(m)[2:])) #16进制转文本
for i inrange(100): url1="http://week-1.hgame.lwsec.cn:30128/api/verifyAnswer" data = {"id": res} resp1 = session.post(url1, cookies = cookies, data = data) print(resp1.text) cookies = resp1.cookies.get_dict() url2="http://week-1.hgame.lwsec.cn:30128/api/getScore" resp2 = session.get(url2, cookies = cookies) print(resp2.text) cookies = resp2.cookies.get_dict() url='http://week-1.hgame.lwsec.cn:30128/api/getQuestion' resp = session.get(url,cookies = cookies) cookies = resp.cookies.get_dict() res=dic[resp.content.decode()[12:-2].replace("\\u0026","&").replace(' ','')]