做公司网站怎么推广/公众号代运营

这回跟着个队伍跑，不过还是2X以后的成绩，前边太卷了。

自己会的部分，有些是别人已经提交了的。记录一下。

Crypto

crypto 1

给了一些数据，像这样就没有别的了

ct = [0, 1, 1, 2, 5, 10, 20, 40, 79, 159, 317, 635, 1269, 2538, 5077, 10154, 20307, 40615, 81229, 162458, 324916, 649832, 1299665, 2599330, 5198659, 10397319, 20794638, 41589276, 83178552, 166357103, 332714207, 665428414, 1330856827, 2661713655, 5323427309, 10646854619, 21293709237, 42587418474, 85174836949, 170349673898, 340699347795, 681398695591, 1362797391181, 2725594782363, 5451189564725, 10902379129451, 21804758258901, 43609516517803, 87219033035605, 174438066071211, 348876132142421, 697752264284843, 1395504528569685, 2791009057139370, 5582018114278740, 11164036228557480, 22328072457114960, 44656144914229920, 89312289828459841, 178624579656919682, 357249159313839363, 714498318627678726, 1428996637255357453, 2857993274510714906, 5715986549021429811, 11431973098042859623, 22863946196085719246, 45727892392171438492, 91455784784342876983, 182911569568685753966, 365823139137371507933, 731646278274743015865, 1463292556549486031730, ...]

看了会发现一个规律：

ct[i] = bit + sum(ct[:i])

把这些0，1弄出来就OK了。我来得早，这时候还没人上线，两个crypto都是1血，跟签到抢到1血一样。

#Alice and Bob decided to create their own super cryptosystem. However.. it has a super flaw
k = 0
f = ''
for i in range(1,len(ct)):f += str(ct[i]-k)k += ct[i]print(f)#前面的和+bit 
f = '0'+f
for i in range(0, len(f),8):print(chr(int(f[i:i+8],2)), end='')#OSC{SUP3r!NCr3451NG_53QU3NC3}

crypto 2

一个e=3的RSA题

from Crypto.Util.number import bytes_to_long, getPrime
from secret import messagesdef RSA_encrypt(message):m = bytes_to_long(message)p = getPrime(1024)q = getPrime(1024)N = p * qe = 3c = pow(m, e, N)return N, e, cfor m in messages:N, e, c = RSA_encrypt(m)print(f"n = {N}")print(f"e = {e}")print(f"c = {c}")

给了好多的n,c显然，感觉如果flag不是很大可以爆破，不过只有第1行爆破出来了，正好是flag

from gmpy2 import iroot
from Crypto.Util.number import long_to_bytesmsg = open('output.txt').readlines()for i in range(11):n = int(msg[i*3][4:])e = int(msg[i*3+1][4:])c = int(msg[i*3+2][4:])print(n)print(e)print(c)while True:a,b = iroot(c,3)if b:print(long_to_bytes(a))break c+=n #OSC{C0N6r47U14710N5!_Y0U_UND3r574ND_H0W_70_U53_H4574D5_8r04DC457_4774CK_______0xL4ugh}

Rev

snake

python 字节码

  2           0 LOAD_CONST               1 (0)2 LOAD_CONST               0 (None)4 IMPORT_NAME              0 (base64)6 STORE_FAST               0 (base64)3           8 LOAD_CONST               1 (0)10 LOAD_CONST               2 (('Fernet',))12 IMPORT_NAME              1 (cryptography.fernet)14 IMPORT_FROM              2 (Fernet)16 STORE_FAST               1 (Fernet)18 POP_TOP4          20 LOAD_CONST               3 (b'gAAAAABj7Xd90ySo11DSFyX8t-9QIQvAPmU40mWQfpq856jFl1rpwvm1kyE1w23fyyAAd9riXt-JJA9v6BEcsq6LNroZTnjExjFur_tEp0OLJv0c_8BD3bg=')22 STORE_FAST               2 (encMessage)5          24 LOAD_FAST                0 (base64)26 LOAD_METHOD              3 (b64decode)28 LOAD_CONST               4 (b'7PXy9PSZmf/r5pXB79LW1cj/7JT6ltPEmfjk8sHljfr6x/LyyfjymNXR5Z0=')30 CALL_METHOD              132 STORE_FAST               3 (key_bytes)6          34 BUILD_LIST               036 STORE_FAST               4 (key)7          38 LOAD_FAST                3 (key_bytes)40 GET_ITER>>   42 FOR_ITER                 9 (to 62)44 STORE_FAST               5 (k_b)8          46 LOAD_FAST                4 (key)48 LOAD_METHOD              4 (append)50 LOAD_FAST                5 (k_b)52 LOAD_CONST               5 (160)54 BINARY_XOR56 CALL_METHOD              158 POP_TOP60 JUMP_ABSOLUTE           21 (to 42)10     >>   62 LOAD_GLOBAL              5 (bytes)64 LOAD_FAST                4 (key)66 CALL_FUNCTION            168 STORE_FAST               4 (key)11          70 LOAD_FAST                1 (Fernet)72 LOAD_FAST                4 (key)74 CALL_FUNCTION            176 STORE_FAST               6 (fernet)12          78 LOAD_FAST                6 (fernet)80 LOAD_METHOD              6 (decrypt)82 LOAD_FAST                2 (encMessage)84 CALL_METHOD              186 LOAD_METHOD              7 (decode)88 CALL_METHOD              090 STORE_FAST               7 (decMessage)13          92 LOAD_GLOBAL              8 (print)94 LOAD_FAST                7 (decMessage)96 CALL_FUNCTION            198 POP_TOP100 LOAD_CONST               0 (None)102 RETURN_VALUE
None

手扣

import base64
from cryptography.fernet import FernetencMessage = b'gAAAAABj7Xd90ySo11DSFyX8t-9QIQvAPmU40mWQfpq856jFl1rpwvm1kyE1w23fyyAAd9riXt-JJA9v6BEcsq6LNroZTnjExjFur_tEp0OLJv0c_8BD3bg='key_bytes = base64.b64decode(b'7PXy9PSZmf/r5pXB79LW1cj/7JT6ltPEmfjk8sHljfr6x/LyyfjymNXR5Z0=')key = []
for k_b in key_bytes:key.append(k_b^160)
key = bytes(key)decMessage = Fernet(key).decrypt(encMessage)
print(decMessage)#FLAG{FLY_L1k3_0xR4V3N}
#0xL4ugh{FLY_L1k3_0xR4V3N}

easy-Peasy

高低4位互换

  v11[0] = 1947518052;v11[1] = 84227255;v11[2] = -181070859;v11[3] = -972881100;v11[4] = 1396909045;v11[5] = 1396929315;v12 = -10397;v13 = 0;v3 = 0i64;v16 = 0i64;v17 = 15i64;LOBYTE(Block[0]) = 0;sub_140001350(Block);sub_1400015D0(std::cout, (__int64)"Enter The Flag: ");sub_140001A50(std::cin, Block);               // 读入Blockif ( v16 == 26 ){while ( 1 ){v4 = Block;if ( v17 >= 0x10 )v4 = (void **)Block[0];v5 = Block;if ( v17 >= 0x10 )v5 = (void **)Block[0];if ( *((unsigned __int8 *)v11 + v3) != ((*((char *)v4 + v3) >> 4) | (16 * (*((_BYTE *)v5 + v3) & 0xF))) )// 高低互换break;if ( ++v3 >= 26 ){v6 = sub_1400015D0(std::cout, (__int64)"The Flag is: ");v7 = Block;if ( v17 >= 0x10 )v7 = (void **)Block[0];sub_140001C50(v6, v7, v16);goto LABEL_12;}}}

解

va = [1947518052,84227255,-181070859,-972881100,1396909045,1396929315,-10397]from pwn import p32v11 = b''.join([p32(v&0xffffffff) for v in va])
f = ''
for v in v11:f += chr(((v&0xf)<<4)+(v>>4))print(f)
#FLAG{CPP_1S_C00l_24527456}
#0xL4ugh{CPP_1S_C00l_24527456}

lets go

代码看着很复杂，但明显能看出是字节变换

void __cdecl main_main()
{__int64 v0; // r14__int128 v1; // xmm15__int64 *v2; // rax__int64 v3; // r8__int64 v4; // r9__int64 v5; // rax__int64 v6; // rcxunsigned __int64 v7; // rbx__int64 v8; // rsiint v9; // r10d__int64 v10; // r11int v11; // r10d__int64 v12; // raxunsigned __int64 v13; // rcxint v14; // r10d__int64 v15; // raxunsigned __int64 v16; // rcx__int64 v17; // raxunsigned __int64 v18; // rcx__int64 v19; // rax__int64 v20; // [rsp-46h] [rbp-D8h]char v21; // [rsp+0h] [rbp-92h]char v22; // [rsp+1h] [rbp-91h]char v23; // [rsp+1h] [rbp-91h]__int64 v24; // [rsp+2h] [rbp-90h]__int64 v25; // [rsp+Ah] [rbp-88h]__int64 v26; // [rsp+12h] [rbp-80h]__int64 v27; // [rsp+3Ah] [rbp-58h] BYREF__int64 *v28; // [rsp+42h] [rbp-50h]void *v29; // [rsp+4Ah] [rbp-48h]char **v30; // [rsp+52h] [rbp-40h]__int128 v31; // [rsp+5Ah] [rbp-38h]const char *v32; // [rsp+6Ah] [rbp-28h]__int64 *v33; // [rsp+72h] [rbp-20h]void *v34; // [rsp+7Ah] [rbp-18h]char **v35; // [rsp+82h] [rbp-10h]if ( (unsigned __int64)&v27 <= *(_QWORD *)(v0 + 16) )runtime_morestack_noctxt_abi0();v34 = &unk_494360;v35 = &off_4C3F50;fmt_Fprint();runtime_newobject();v28 = v2;*v2 = 0LL;v32 = "\b";v33 = v2;fmt_Fscanln();v3 = *v28;v27 = *v28;v4 = v28[1];v25 = v4;v5 = 0LL;v6 = 0LL;v7 = 0LL;v8 = 0LL;while ( v5 < v4 ){v24 = v6;v26 = v5;v9 = *(unsigned __int8 *)(v3 + v5);if ( (unsigned __int8)(v9 - 65) > 0x19u ){if ( (unsigned __int8)(v9 - 97) > 0x19u ) // 符号{v10 = v6 + 1;if ( v7 < v6 + 1 ){v21 = *(_BYTE *)(v3 + v5);runtime_growslice(v20);v10 = v8 + 1;v3 = v27;v4 = v25;LOBYTE(v9) = v21;v8 = v17;v7 = v18;v5 = v26;v6 = v24;}*(_BYTE *)(v8 + v6) = v9;}else{v10 = v6 + 1;v14 = v9 - 26 * ((unsigned __int8)(v9 - 81) / 0x1Au);if ( v7 < v6 + 1 ){v23 = v14;runtime_growslice(v20);v10 = v8 + 1;v3 = v27;v4 = v25;LOBYTE(v14) = v23;v8 = v15;v7 = v16;v5 = v26;v6 = v24;}*(_BYTE *)(v8 + v6) = v14 + 16;}}else{v10 = v6 + 1;v11 = v9 - 26 * ((unsigned __int8)(v9 - 49) / 0x1Au);if ( v7 < v6 + 1 ){v22 = v11;runtime_growslice(v20);v10 = v8 + 1;v3 = v27;v4 = v25;LOBYTE(v11) = v22;v8 = v12;v7 = v13;v5 = v26;v6 = v24;}*(_BYTE *)(v8 + v6) = v11 + 16;}++v5;v6 = v10;}runtime_slicebytetostring();if ( v8 == 32 && (unsigned __int8)runtime_memequal() ){v31 = v1;v19 = runtime_convTstring();*(_QWORD *)&v31 = &unk_494360;*((_QWORD *)&v31 + 1) = v19;fmt_Fprintf();}else{v29 = &unk_494360;v30 = &off_4C3F60;fmt_Fprint();}
}

这个只需要个码表就行，输入字母得到的就是码表，然后手工对应填一下

输入得到对应关系
abcdefghijklmnopqrstuvwxyz012345
qrstuvwxyzabcdefghijklmnop012345ABCDEFGHIJKLMNOPQRSTUVWXYZ012345
QRSTUVWXYZABCDEFGHIJKLMNOP01234567890{_} 数字符号不变c = b"u507rv78qr5t6q99941422uursv94464"u507rv78qr5t6q99941422uursv94464
e507bf78ab5d6a99941422eebcf94464手工输入得到flag
ef➤  
Correct:)
FLAG{e507bf78ab5d6a99941422eebcf94464}
52      in example/user/hello/hello.go换头
0xL4ugh{e507bf78ab5d6a99941422eebcf94464}

XPacker 未完成

看上去是运行时释放一个程序再找，但怎么运行都不释放，也许想错了

Misc

ATT-IP

流量题，wireShark打开一点点到结果

57包：
tcp://91.243.59.76:23927/
0xL4ugh{91.243.59.76_23927}

PVE

这个是别的提交的，只因为第1步没作出来后边的就没想弄。

给的一vmware的临时文件.vmem 一直想用什么软件找到。其实只有第1题用软件，其它都是在时差搜文本。

后来下了volatility 3才弄好这一步

┌──(kali㉿kali)-[~/volatility3]
└─$ py vol.py -f ~/ctf/PVE.vmem Banners   
Volatility 3 Framework 2.4.1
Progress:  100.00               PDB scanning finished                  
Offset  Banner0x1a00180       Linux version 4.4.0-186-generic (buildd@lcy01-amd64-002) (gcc version 5.4.0 20160609 (Ubuntu 5.4.0-6ubuntu1~16.04.12) ) #216-Ubuntu SMP Wed Jul 1 05:34:05 UTC 2020 (Ubuntu 4.4.0-186.216-generic 4.4.228)
0x211e6a4       Linux version 4.4.0-186-generic (buildd@lcy01-amd64-002) (gcc version 5.4.0 20160609 (Ubuntu 5.4.0-6ubuntu1~16.04.12) ) #216-Ubuntu SMP Wed Jul 1 05:34:05 UTC 2020 (Ubuntu 4.4.0-186.216-generic 4.4.228)
0x1aaf7338      Linux version 4.4.0-186-generic (buildd@lcy01-amd64-002) (gcc version 5.4.0 20160609 (Ubuntu 5.4.0-6ubuntu1~16.04.12) ) #216-Ubuntu SMP Wed Jul 1 05:34:05 UTC 2020 (Ubuntu 4.4.0-186.216-generic 4.4.228)
0x1fde00a8      Linux version 4.4.0-186-generic (buildd@lcy01-amd64-002) (gcc version 5.4.0 20160609 (Ubuntu 5.4.0-6ubuntu1~16.04.12) ) #216-Ubuntu SMP Wed Jul 1 05:34:05 UTC 2020 (Ubuntu 4.4.0-186.216-generic 4.4.228)flag1:0xL4ugh{Ubuntu_4.4.0-186-generic}

第二步是Apache的版本，搜apache有很多，主要是两个版本，第2个=的是

0xL4ugh{2.2.14}

第三步找 flag 搜0xL4ugh

sudo echo "0xL4ugh{S4D_Y0U_G07_M3}" > flag.txt
0xL4ugh{S4D_Y0U_G07_M3}

第四个找程序里隐写的flag

int main(void){char flag[] = "0xL4ugh{H1DD3N_1N_PR0CE$$}";sleep(6969696969);return 0;
}0xL4ugh{H1DD3N_1N_PR0CE$$}

第五个找攻击求密码，没弄成，linux的密码存在shadow文件但这里没密码，再找passwd也没有，按密码的特征 $1$salt$enc_pass 搜$1$得到一个密码

Name: passwd/user-password-crypted
Template: passwd/user-password-crypted
Value: $1$JULXDu5H$c/QbtxOmwiX0rR7f0NXxj.
Owners: ubiquity
Flags: seen

从网站上解密得到 mrx 提交不正确。

队友还作了一些，刚开始已经狂飙到第2了，后来慢慢又多了1位成2x了