BeginCTF 解题记录

2024-02-08 · 10 分钟阅读 · WP MISC Web Reverse Crypto Forensics ·

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题量大，题目难度适中，我挑选关键题目写写 WP

Misc

Tupper

先把文件内容提取并拼接起来：

1txts = []
2for i in range(0, 673, 4):
3    path = f"{i}.txt"
4    with open(path, 'r') as file:
5        txts.append(file.read())
6txts = ''.join(txts) 
7print(txts)

得到一段 Base64： 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

解码得到一串数：

14278193432728026049298574575557534321062349352543562656766469704092874688354679371212444382298821342093450398907096976002458807598535735172126657504131171684907173086659505143920300085808809647256790384378553780282894239751898620041143383317064727136903634770936398518547900512548419486364915399253941245911205262493591158497708219126453587456637302888701303382210748629800081821684283187368543601559778431735006794761542413006621219207322808449232050578852431361678745355776921132352419931907838205001184

根据题目名称可以知道与塔伯自指公式有关。用 Tupper's self-referential fomula 解一下：

where is crazyman 系列

三道社工题，前两道用谷歌识图可以直接找到地点；第三道的图片：

矿泉水瓶子上有 Boudl Apart' Hotel 字样，根据提示在谷歌地图里找到在 Boudl Al Munsiyah 旁的 Starbucks。flag 在谷歌地图里这一家 Starbucks 的评论区里，按时间顺序查看能找到。

devil's word

一查是温州话，听音频把“魔鬼的语言”转成数字 0-9，最后十六进制转字符得到 flag。

发音	数字
leng	0
lia	2
sa	3
sii	4
ng	5
leu	6
cai	7
bo	8
jau	9

使用某些文本编辑器的 Ctrl+H 一键替换的时候注意，不要把 leng 里的 ng 替换成 5。

real check in

MJSWO2LOPNLUKTCDJ5GWKX3UN5PUEM2HNFXEGVCGL4ZDAMRUL5EDAUDFL5MU6VK7O5UUYMK7GEYWWZK7NE3X2=== 一眼 Base32

Web

zupload

Web 做不了一点，查资料只做了个签到题。本题的后端没有保护，直接改 ?action=/flag 访问 flag 所在目录。

Reverse

红白机

读 6502 汇编。这玩应有现成的在线工具：Easy 6502

不过还是自己写了个脚本：

  1def op_LDA(arg):
  2    global reg_acc, line_ptr
  3    reg_acc = int(arg[-3:], 16)
  4    line_ptr += 1
  5
  6
  7def op_LDX(arg):
  8    global x_index, line_ptr
  9    x_index = int(arg[-3:], 16)
 10    line_ptr += 1
 11
 12
 13def op_LDY(arg):
 14    global y_index, line_ptr
 15    y_index = int(arg[-3:], 16)
 16    line_ptr += 1
 17
 18
 19def op_STA(arg):
 20    global reg_acc, mem, line_ptr, x_index
 21    addr = int(arg.split(',')[0][1:], 16) + x_index - 0x200
 22    mem[addr] = reg_acc
 23    line_ptr += 1
 24
 25
 26def op_INX(arg):
 27    global x_index, line_ptr
 28    x_index += 1
 29    line_ptr += 1
 30
 31
 32def op_CPX(arg):
 33    global x_index, line_ptr, reg_cmp
 34    param = int(arg[-3:], 16)
 35    if param == x_index:
 36        reg_cmp = 1
 37    else:
 38        reg_cmp = 0
 39    line_ptr += 1    
 40
 41
 42def op_BNE(arg):
 43    global reg_cmp, line_ptr, seg_addr, seg_name
 44    if reg_cmp:
 45        line_ptr += 1
 46    else:
 47        ind = seg_name.index(arg[0])
 48        line_ptr = seg_addr[ind]
 49
 50
 51op_dict = {  # 指令
 52    "LDA": op_LDA,
 53    "LDX": op_LDX,
 54    "LDY": op_LDY,
 55    "STA": op_STA,
 56    "INX": op_INX,
 57    "CPX": op_CPX,
 58    "BNE": op_BNE,
 59}
 60
 61
 62line_ptr = 0   # 指令地址指针
 63seg_addr = []  # 段地址
 64seg_name = []  # 段名称
 65mem = [0] * 0x400  # 初始化内存
 66x_index = 0  # X 索引寄存器
 67y_index = 0  # Y 索引寄存器
 68reg_acc = 0  # 累加器
 69reg_cmp = 0  # 比较标志位
 70
 71with open("6502.txt", 'r') as asm_6502:
 72    
 73    for line in asm_6502:  # 读取段标识
 74        if line.strip()[0].islower():
 75            seg_addr.append(line_ptr)
 76            seg_name.append(line[0])
 77        line_ptr += 1
 78    
 79    line_ptr = 0  # 从头读取
 80    asm_6502.seek(0)
 81    lines = asm_6502.readlines()
 82
 83    while line_ptr < 407:  # 运行
 84        if lines[line_ptr][0].islower():  # 跳过段标识
 85            line_ptr += 1
 86        else:
 87            opcode = lines[line_ptr][0:3]
 88            arg = lines[line_ptr][4:]
 89            op_dict[opcode](arg)
 90
 91    for i in range(len(mem)):  # 显示
 92        if i % 32 == 0:  # 显示器宽度
 93            print('')
 94        if mem[i] == 0:
 95            print('.', end = ' ')
 96        else:
 97            print('■', end = ' ')
 98
 99'''output:
100. ■ ■ . ■ . . . . . . . . . . . ■ ■ . . . . . . . . . . . . . .
101. ■ . . ■ . . . . . . . . . . . ■ . . . ■ . . ■ ■ ■ . . ■ . . .
102■ ■ ■ . ■ . ■ ■ ■ . . ■ ■ ■ . . ■ . . ■ . ■ . ■ . . . ■ . ■ . .
103. ■ . . ■ . ■ . ■ . . ■ . ■ . ■ ■ . . ■ . . . ■ ■ . . ■ . ■ . .
104. ■ . . ■ . ■ ■ ■ ■ . ■ ■ ■ . . ■ . . ■ ■ ■ . . . ■ . ■ . ■ . .
105. . . . . . . . . . . . . ■ . . ■ . . ■ . ■ . . . ■ . ■ . ■ . .
106. . . . . . . . . . . ■ ■ ■ . . ■ ■ . . ■ . . ■ ■ . . . ■ . . .
107. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
108■ ■ ■ . . . . . ■ ■ ■ . . . . . ■ . . . ■ ■ ■ . ■ . ■ . ■ ■ ■ .
109. . ■ . . . . . . ■ . . . . . . ■ . . . ■ . ■ . ■ . ■ . ■ . . .
110. ■ ■ . . . . . . ■ . . . . . . ■ . . . ■ . ■ . ■ . ■ . ■ ■ ■ .
111■ ■ . . . . . . . ■ . . . . . . ■ . . . ■ . ■ . ■ . ■ . ■ . . .
112■ . . . . . . . . ■ . . . . . . ■ . . . ■ . ■ . ■ . ■ . ■ . . .
113■ ■ ■ . ■ ■ ■ . ■ ■ ■ . ■ ■ ■ . ■ ■ ■ . ■ ■ ■ . . ■ . . ■ ■ ■ .
114. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
115. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
116. . . . . . . . ■ ■ . . . . . . . . . . . . . . . . . . . . . .
117. . . . . . . . . ■ . . . . . . . . . . . . . . . . . . . . . .
118. . . . . . . . . ■ . . . . . . . . . . . . . . . . . . . . . .
119. . . . ■ . ■ . . ■ ■ . . . . . . . . . . . . . . . . . . . . .
120. . . . ■ . ■ . . ■ . . . . . . . . . . . . . . . . . . . . . .
121. . . . ■ . ■ . . ■ . . . . . . . . . . . . . . . . . . . . . .
122■ ■ ■ . ■ ■ ■ . ■ ■ . . . . . . . . . . . . . . . . . . . . . .
123. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
124. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
125. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
126. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
127. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
128. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
129. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
130. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
131. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
132'''

算是手搓个小小解释器吧。

Xor

查出来有 UPX 壳，先脱壳放进 IDA。

在最后比较字符串的地方找到了一个加密过的 flag：

加密过程如下：

……

加密过程过于繁琐，豁免还有几个类似加密逻辑。先尝试动调，试着输入密文，竟发现：

难绷，非预期了（）

俄语学习

最开始有 30 道俄语题目，不会用 pwntools，遂手自笔录到最后一步。

最后这里对输入的内容有 sub_43AFAA() 和 sub_43A555() 两次操作。

sub_43AFAA() 中调用 sub_4419E0() 生成一个船新的字符串 byte_4CB1E8，随后在 sub_441B00() 中做了一次 Xor Swap，RC4 中的第三个步骤加密即是 Xor Swap。

sub_43A555() 中有一个字符串比较的操作，随后在 sub_43CBC0() 中同样调用 sub_441B00()。

关键就在 byte_4CB1E8[i] = Str[i] + byte_4CAEE8[i] - 112; 这一句，动调之后能看见 byte_4CAEE8，于是上脚本：

1>>> cip1 = "+i&[@Y:g8[&l$f8S8v$Y&e>{"
2>>> cip2 = [0x35, 0x6D, 0x35, 0x64, 0x35, 0x77, 0x35, 0x64, 0x35, 0x62, 0x35, 0x6E, 0x35, 0x6D, 0x35, 0x64, 0x35, 0x77, 0x35, 0x64, 0x35, 0x62, 0x35, 0x6E, 0x35, 0x6D, 0x35, 0x64, 0x35, 0x77, 0x35, 0x64, 0x35, 0x62, 0x35, 0x6E, 0x8E]
3>>> for i in range(len(cip1)):
4...     print(chr(ord(cip1[i]) + 112 - cip2[i]), end = '')
5...
6flag{Russian_is_so_easy}

stick game

最绷不住的一题，本来是一血的，结果题目下了（）

附件里的 Javascript 脚本里边有这么一坨：

1function _0x3339(_0xc5a7d3,_0x197349){const _0x4be1b2=_0x271a();return _0x3339=function(_0x5f266e,_0x306e60){_0x5f266e=_0x5f266e-(0x1*-0x159c+-0x5*-0x69b+0x9*-0x10e);let _0x97d2a2=_0x4be1b2[_0x5f266e];if(_0x3339['vsEbbX']===undefined){var _0x4e47ab=function(_0x504f68){const _0x55694b='abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789+/=';let _0x4868cb='',_0x5f5158='';for(let _0x288bde=-0x1275+-0x96f*-0x4+0xeb*-0x15,_0x5400b3,_0x1abc9a,_0x5b8a10=-0x1715*-0x1+-0x2413+0xcfe;_0x1abc9a=_0x504f68['charAt'](_0x5b8a10++);~_0x1abc9a&&(_0x5400b3=_0x288bde%(0xc*-0x27e+0x2*0xc11+-0x2*-0x2e5)?_0x5400b3*(0x1795*0x1+0x3*0x3b+-0x802*0x3)+_0x1abc9a:_0x1abc9a,_0x288bde++%(0x611+-0x70e+0x101))?_0x4868cb+=String['fromCharCode'](-0x1bd9+-0x114a*0x2+-0x386*-0x12&_0x5400b3>>(-(-0xdcf+-0x5*0x51b+0x2758)*_0x288bde&0x22b7+-0x2548+0x3*0xdd)):-0x4*0x765+-0xe07+0x2b9b){_0x1abc9a=_0x55694b['indexOf'](_0x1abc9a);}for(let _0x4deb0f=-0x11cf+-0x1b73*0x1+0x2*0x16a1,_0x529665=_0x4868cb['length'];_0x4deb0f<_0x529665;_0x4deb0f++){_0x5f5158+='%'+('00'+_0x4868cb['charCodeAt'](_0x4deb0f)['toString'](0x14b*-0xa+0x1bd8+-0x76d*0x2))['slice'](-(-0x10d5+0x1*0x265+0xe72));}return decodeURIComponent(_0x5f5158);};const _0x11fa5a=function(_0x4173dc,_0x5da0bd){let _0x48d668=[],_0x20d9fc=0xa31*0x1+0x252d*0x1+-0x2f5e,_0x1ab256,_0x59d767='';_0x4173dc=_0x4e47ab(_0x4173dc);let _0x15c31e;for(_0x15c31e=0x18e3*-0x1+0x12ff+0x5e4;_0x15c31e<-0x26f9+-0xb*-0x1fd+0x121a;_0x15c31e++){_0x48d668[_0x15c31e]=_0x15c31e;}for(_0x15c31e=-0x17bc+0x415+0x13a7;_0x15c31e<-0xa46+-0x1207+-0x1*-0x1d4d;_0x15c31e++){_0x20d9fc=(_0x20d9fc+_0x48d668[_0x15c31e]+_0x5da0bd['charCodeAt'](_0x15c31e%_0x5da0bd['length']))%(0x1c5f+0x267b+-0x41da),_0x1ab256=_0x48d668[_0x15c31e],_0x48d668[_0x15c31e]=_0x48d668[_0x20d9fc],_0x48d668[_0x20d9fc]=_0x1ab256;}_0x15c31e=0x1450+0x139*0x5+-0x1a6d,_0x20d9fc=0x250e+-0x1*0x12fd+-0x1211;for(let _0x786500=-0x4*0x2e8+-0x452*-0x4+-0x5a8;_0x786500<_0x4173dc['length'];_0x786500++){_0x15c31e=(_0x15c31e+(0x127a+-0xd*0x2f1+0x16*0xe6))%(0x1b69+0x10d*-0x25+0x54*0x26),_0x20d9fc=(_0x20d9fc+_0x48d668[_0x15c31e])%(0x144b+0xbf8+0x1f43*-0x1),_0x1ab256=_0x48d668[_0x15c31e],_0x48d668[_0x15c31e]=_0x48d668[_0x20d9fc],_0x48d668[_0x20d9fc]=_0x1ab256,_0x59d767+=String['fromCharCode'](_0x4173dc['charCodeAt'](_0x786500)^_0x48d668[(_0x48d668[_0x15c31e]+_0x48d668[_0x20d9fc])%(0x21b5+0x780+-0x2835)]);}return _0x59d767;};_0x3339['fKEsSz']=_0x11fa5a,_0xc5a7d3=arguments,_0x3339['vsEbbX']=!![];}const _0xdaf7a0=_0x4be1b2[0x1a7c+0x13a3*-0x1+-0x6d9],_0xe679c5=_0x5f266e+_0xdaf7a0,_0x223335=_0xc5a7d3[_0xe679c5];return!_0x223335?(_0x3339['wQPYZX']===undefined&&(_0x3339['wQPYZX']=!![]),_0x97d2a2=_0x3339['fKEsSz'](_0x97d2a2,_0x306e60),_0xc5a7d3[_0xe679c5]=_0x97d2a2):_0x97d2a2=_0x223335,_0x97d2a2;},_0x3339(_0xc5a7d3,_0x197349);}(function(_0x56cb23,_0xed8547){const _0x18ee5d=_0x3339,_0x1e59ec=_0x56cb23();while(!![]){try{const _0x1325e2=-parseInt(_0x18ee5d(0x20d,'*!up'))/(-0x6a3*0x1+0x7b6+0x2*-0x89)*(parseInt(_0x18ee5d(0x2b8,'RTq]'))/(-0x1a9a+-0x19ac+-0x4*-0xd12))+parseInt(_0x18ee5d(0x276,'Ur3M'))/(0x90e+-0x47*0x30+-0x1*-0x445)*(-parseInt(_0x18ee5d(0x299,'Lj5i'))/(-0x22bd+-0x1*-0xdf+-0x10f1*-0x2))+parseInt(_0x18ee5d(0x286,'bK)('))/(-0x1190+0x1*-0x220f+0x33a4)*(-parseInt(_0x18ee5d(0x287,'P0I8'))/(0xae+-0x7e5*0x2+0xf22))+-parseInt(_0x18ee5d(0x292,'*!up'))/(-0x1*-0x545+0x146a+0xcd4*-0x2)+-parseInt(_0x18ee5d(0x291,'B*#j'))/(-0xc83*0x1+0x17*-0x11a+0x25e1)*(-parseInt(_0x18ee5d(0x247,'a8v%'))/(-0x1ed3+0xb69*-0x2+0x1ad7*0x2))+-parseInt(_0x18ee5d(0x2a0,'D93x'))/(0x22bb+-0xa34*0x1+-0x187d)+parseInt(_0x18ee5d(0x23a,'euu1'))/(-0x10db+0xadd+-0x1*-0x609);if(_0x1325e2===_0xed8547)break;else _0x1e59ec['push'](_0x1e59ec['shift']());}catch(_0x386714){_0x1e59ec['push'](_0x1e59ec['shift']());}}}(_0x271a,-0xa4*-0x2aeb+-0x107909+0x3ec97),(function(){const 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_0x51c506;};return _0x271a();}

显然是被混淆了。试了几个反混淆工具，最终选择了这个 JS Deobfuscator

然后 flag 就被明文显示了

这是初代 flag，后来第二遍上的题目用的解决方法和这个一模一样。最开始这道题难度是 hard，重新上来就成了 easy。

real check in xor

略（真没啥好写的）

Crypto

fake_n

已知的 fake_n 由 17 个质数相乘得到，未知的 really_n 由其中的 15 个质数相乘得到。really_n 共有 $C_{17}^{15}$ 种可能。不妨爆破：

 1import gmpy2
 2from Crypto.Util.number import *
 3
 4c = 6451324417011540096371899193595274967584961629958072589442231753539333785715373417620914700292158431998640787575661170945478654203892533418902
 5primelst = [2215221821, 2290486867, 2333428577, 2361589081, 2446301969, 2507934301, 2590663067, 3107210929, 3278987191, 3389689241, 3417707929, 3429664037, 3716624207, 3859354699, 3965529989, 4098704749, 4267348123]
 6
 7for i in range(1, 17):
 8	for j in range(i):
 9		tmplst = [2215221821, 2290486867, 2333428577, 2361589081, 2446301969, 2507934301, 2590663067, 3107210929, 3278987191, 3389689241, 3417707929, 3429664037, 3716624207, 3859354699, 3965529989, 4098704749, 4267348123]
10		n = 1
11		phi = 1
12		del tmplst[i]
13		del tmplst[j]
14		n = 1
15		phi = 1
16		for k in tmplst:
17			n *= k
18			phi *= k - 1
19		e = 65537
20		d = gmpy2.invert(e, phi)
21		m = pow(c, d, n)
22		print(long_to_bytes(m), end = '\n\n')

我玩青水的

后来知道，下面这个方法叫低指数加密攻击：

 1from Crypto.Util.number import *
 2import gmpy2
 3
 4p = 7709388356791362098686964537734555579863438117190798798028727762878684782880904322549856912344789781854618283939002621383390230228555920884200579836394161
 5c = 5573755468949553624452023926839820294500672937008992680281196534187840615851844091682946567434189657243627735469507175898662317628420037437385814152733456
 6e = 2
 7
 8jud = 1
 9k = 1
10while jud:
11	y = c + k * p
12	m, exact = gmpy2.iroot(y, 2)
13	if exact:
14		print(long_to_bytes(m))
15		print(k)
16		jud = 0
17	k += 1

OEIS2

改编自强网杯“OEIS”，那一个可以查表。

题目要计算 $(2^{28} + 5)!$ 各位和的 SHA256，NR289 师傅建议我使用 Sagemath 硬算：

1import hashlib
2upper = str(gamma(2**28 + 6))
3res = 0
4for i in upper:
5    res += int(i)
6print(hashlib.sha256(str(res).encode()).hexdigest())

吃顿饭的工夫就出了。

hard_ecc

浅看了一点 ECC，这道题已知的量有：圆锥曲线 ec、公钥 Q、基点 T，要求的是私钥。

 1A = [0, 3, 0, 973467756888603754244984534697613606855346504624, 864199516181393560796053875706729531134503137794]
 2p = 992366950031561379255380016673152446250935173367
 3t = [295622334572794306408950267006569138184895225554, 739097242015870070426694048559637981600496920065, 1]
 4q = [282367703408904350779510132139045982196580800466, 411950462764902930006129702137150443195710071159, 1]
 5flag_bytes = b''
 6
 7ec = EllipticCurve(GF(p), [A[0], A[1], A[2], A[3], A[4]])
 8
 9T = ec((t[0], t[1], t[2]))
10Q = ec((q[0], q[1], q[2]))
11
12secret = discrete_log(Q, T, operation= '+')
13
14flag_bytes = int(secret).to_bytes((secret.bit_length() + 7) // 8, 'little')
15flag = flag_bytes.decode('utf-8')
16
17print(flag)

Forensics

学取证咯系列

做出来的前面五道分别使用 cmdscan iehistory mimikatz filescan 可以直接出

逆向工程(reverse)入门指南

Linux 里使用 pdftotxt，然后可以找到 flag。

beginner_Forensics!!!!

用 010 打开，看到是一个 Batch Encryption 混淆，我使用 https://blog.csdn.net/Hunter98234/article/details/108672926 中提供的脚本还原。