HWS2023 冬令营

CRYPTO

math

题目信息

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from Crypto.Util.number import *
from secret import flag,a,b
from random import shuffle
 
D = 0x1337
assert a**2 - D*b**2 == 1
m = bytes_to_long(flag)
p = getPrime(512)
x = (a * m + b) % p
y=bin(x)[2:]
c=[y[i:i+85] for i in range(0,len(y),85)]
shuffle(c)
print(p)
print(c)
 
# p=11199186558148426014734492874345932099384932524559726349180064588241518696390926354448476750322781683505497782040786332571272422812867731614415519029276349
# c=['0010101111100011101101011111111001011000100110001001000000010001111011110101110011111', '0011010010010010110010011011001100110001100010101110001010001101110001100000111011010', '0110101101011101110000100001000000010001110110001010000000010110010101100100101110000', '0100111001011010000101100111100110101100011100100111011000110001111101000110110101101', '1100100110011101010011011111000101011011010000101100011001110100101000101101111110100', '1110111110001110000101100000000100111010110000001111010001101001100001010110101010001']

D = 0x1337 a2 - D*b2 == 1为Pell方程 网站在线（[http://www.numbertheory.org/php/pell.html](http://www.numbertheory.org/php/pell.html)）解一下，得到方程a，b的最小整数解 a = 1809338099956399320，b = 25797719546469589

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from Crypto.Util.number import *
from itertools import permutations

D = 0x1337
a = [1809338099956399320]
b = [25797719546469589]

for i in range(20):
    a, b = a + [a[-1] * a[0] + D * b[-1] * b[0]], b + [a[-1] * b[0] + b[-1] * a[0]]

p=11199186558148426014734492874345932099384932524559726349180064588241518696390926354448476750322781683505497782040786332571272422812867731614415519029276349
c=['0010101111100011101101011111111001011000100110001001000000010001111011110101110011111', '0011010010010010110010011011001100110001100010101110001010001101110001100000111011010', '0110101101011101110000100001000000010001110110001010000000010110010101100100101110000', '0100111001011010000101100111100110101100011100100111011000110001111101000110110101101', '1100100110011101010011011111000101011011010000101100011001110100101000101101111110100', '1110111110001110000101100000000100111010110000001111010001101001100001010110101010001']
res = [int(''.join(res), 2) for res in permutations(c)]
for i in range(len(a)):
    ai, bi = a[i], b[i]
    for c in res:
        m = (c - bi) * inverse(ai, p) % p
        flag = long_to_bytes(m)
        if b'flag' in flag:
            print(flag)

#b'flag{5b80aaa2-2bb2-0ef1-4aa0-a5a9387239d5}'

Numbers Game

题目不难，比赛的时候没搞出来很难受呜呜呜
MT19937还原，根据624组32bit的数据还原state之后逆推上一组state进而得到前12个32bit的随机数值。参考: https://blog.csdn.net/zippo1234/article/details/109279944

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from randcrack import *
from random import Random
from hashlib import md5
# right shift inverse
def inverse_right(res,shift,bits=32):
    tmp = res
    for i in range(bits//shift):
        tmp = res ^ tmp >> shift
    return tmp

# right shift with mask inverse
def inverse_right_values(res,shift,mask,bits=32):
    tmp = res
    for i in range(bits//shift):
        tmp = res ^ tmp>>shift & mask
    return tmp

# left shift inverse
def inverse_left(res,shift,bits=32):
    tmp = res
    for i in range(bits//shift):
        tmp = res ^ tmp << shift
    return tmp

# left shift with mask inverse
def inverse_left_values(res,shift,mask,bits=32):
    tmp = res
    for i in range(bits//shift):
        tmp = res ^ tmp << shift & mask
    return tmp

def backtrace(cur):
    high = 0x80000000
    low = 0x7fffffff
    mask = 0x9908b0df
    state = cur
    for i in range(11,-1,-1): #逆求的32bit 组数 此处需要逆求12组
        tmp = state[i+624]^state[i+397]
        # recover Y,tmp = Y
        if tmp & high == high:
            tmp ^= mask
            tmp <<= 1
            tmp |= 1
        else:
            tmp <<=1
        # recover highest bit
        res = tmp&high
        # recover other 31 bits,when i =0,it just use the method again it so beautiful!!!!
        tmp = state[i-1+624]^state[i+396]
        # recover Y,tmp = Y
        if tmp & high == high:
            tmp ^= mask
            tmp <<= 1
            tmp |= 1
        else:
            tmp <<=1
        res |= (tmp)&low
        state[i] = res
    return state

def recover_state(out):
    state = []
    for i in out:
        i = inverse_right(i,18)
        i = inverse_left_values(i,15,0xefc60000)
        i = inverse_left_values(i,7,0x9d2c5680)
        i = inverse_right(i,11)
        state.append(i)
    return state

if __name__ == '__main__':
    id=['d5d97afad7ef619b4badd8d2da10ee24', '67f4660a8335fb9f4152a9fbc44c9c77', '8cd43c85ebe9cc7036a37f47ccd1d1e4',
     'ee3e8c62e8b0100027589d6de82677ef', '463bb2f3731ad0e786302bf78da08330', 'e245b1852a3b92734e46eb3421bd76c9',
     '0b74736786b4ab94651e3b706a548e55', '79cb596b28c2b4e02738f93b5bfbe0d3', '5a9c46837952952045564b5b395acad1',
     'd3c2d90a05d1a059fbeba4a05a608798', '4da0306c8ab58097d2fef9114e6fcb6d', '9707fabbd3c96de66917f15998ac9201',
     '9dd3e46fc930abfb523fe31e8ee8a658', '3716a8fd05f7388e7151d09431e61ee1', '9acf027679a6d7a674a43dee4f5bea35',
     '78702a2b125a940519337b1bf50aff8a', '262cf3b8c8072a602048a24756c83fcf', '092f8d227ec583c4734a6f449de521a3',
     '712aa300302b57fed458553426348fce', '834d4a0ea451cc04b469636b18c56435', '754b5284b14402c61e3b1e56cb2d41e9',
     '51d742ca6a341032afcb5dc645f54bfa', 'c1a33d104f47e33d6932905b483a2018', '3def7c2a14cff6b2864a2100956df07f',
     '7e3606e4ec1c99fe5a8593ae44f25a70', '404c6139570883dbab8e5a299d7a5017', 'f07597079e1b68ebb4e2d16b83b7b484',
     '0723daf5c65f8ba6cd6e43fcdf9d18dc', '4c54db12829f165837384b66978e8438', 'cd056e64e1f31461cab2e66ece9d3278',
     '2f6ae16fab122cbce240e32464a1ab57', 'f86e0e9ee23498340f62d8617f6f5218', 'af4ebe2535885c783c89f8d8c4815076',
     'c8eae5b5c7aca2c5fdb4f284e2cf65c5', 'bec0d8ecabedd9811a8ecb6052b21d8c', '731bd3421b6517aa101357fe1c49caf3',
     '344f4a26cdaf1a782d9b32208e1a3e92', '892b1741d304878461dd0774a335ea3d', '56e2a484ca40f43e059bf5f0bd822bdc',
     'd7c0762df71b31c14654147fb9a0595c', '57016a179ba5509f6b04a161ac628b34', 'e49a2d573522c1ee3e8348ceca0295a4',
     '4b0f49c7e6469e82832e9cc90b9e17c0', 'eadc9d0c8b75127fe0a7f71881de1ea7', 'db9fe5537768207bd8cdf770bcd42dfa',
     'ba2f57578752628d1ecac419b3a8bc36', '3752e70b5d2b578a8d412d84aab43705', '54e97795df8781c776cbb1ce4f5f31fc',
     '32794880abc9f68102c24e92ad9c7cd5',
     'b5eb7e651ca298f6873694c47d1cd3da', 'a188934777d2c67e3d59692135005497', '34c308fa1644b387169ea88c1b575490']
    code=['2eebed894580fb900c3615d4866150e68322ff4d48e7509f85ff4543969b0cf6',
     '017aefd63b4ce14eb376161902d92894a15f680e7b055ce25c3b02c6b49db0a6',
     'e12d945904032887c967ae03a48c8b096abc79dc64134d872693599d4f6c91cc',
     '0f8957f365f53a7209baa852905c5da5dba54ddb403ab17a4c9b3a051540d49c',
     '091e3e41815cd32f482f2cf54ac3338fc918c2a657896af1aa1b23ea528664f4',
     '5916cd18e8c48c545232112f2179aa7a722350a8a0ced4f1363cc61bc9d83630',
     '37702710d47a1c17278743253123f6eac85b16d9393432ec65100143bc8657e8',
     'c3a7006293c48957cba5c010f945483cfdc47650a79d0e8a8a9a52c174244a10',
     'e1a90a475dd21da64d64ea1dd50a82a622061d08e9d4b3016982c4a0b42f1251',
     '2f40ff85024765e58ed175927c53e0a279cda96ce755b602f89bfc171108ba3d',
     'f4f8337dd7267d02638a9cc531c8a02fe0316dd5ff6f8c8aec898c060c6fb217',
     '0df2f3fe1eb976944a2de5729fca4a12b83c8329b4f514869856ebb94b7d7bf9',
     'abb2f1277b1cb5ad07254b7f7ed346bcdb73282b306123ce0b5befe42a9e796d',
     '6ef31ed6a2a465bcd146c2438bf391bfd9f3187cf54afae512220a7a94714bf3',
     '6981f99ba288923a5cd68908ecc8795bf301f1e7d081ac6580a63902fd52da01',
     'b6b07975697de1c0342e3711b59165b849125794ef2541c6c60dcf40e689df7b',
     '20e3af3a2bcedfd15199975cc9edcde14cb13fdff3ea0607a4747601e500aede',
     '606aa1daf188b9dd5fd6141312f1828846f92baa519e70e5c6923a352421fe2f',
     '1dc6a60112e99c5b884c0bb5430a7a54eba8aef34fada9cb96bce79a22456cf2',
     'fd81e36f576119a19185cd12e87544b42f9fcc3dcb5e7ba282b1a128d73d63c5',
     '7a01e947ca012a5c9a8dd20e693a7788cd6157a5c3fce5fa7c7e09014ea3266e',
     'ccd162b182c773062514ffc3551ed47f32293700083782d902efc55b1e9795c6',
     'dbb6115aefd2638eebf44d3be6e9525e09ff036269d954469a0f925b496327a7',
     'ac08696fe64bb5e58cd3e558463213ca08ab7805e33f45459ae14b35fc5858c6',
     '6e1594596ad1c656635af29d14b22a5ed10e545442eea5f4e056e20d11aa33b3',
     '4afd17e64562603c66ce0ca42d544ca48511c3d560f9180c231a9ccb28a0e55e',
     '06b2d2b24ca626e18fdcb3196e989e3b05150500815511f48200ddea9aacda48',
     '947a11e8258fd161d02b0eb1b2e8fcd9ff1684a7c75f7c506ddee91f08316f56',
     'a0e58b15736cab055be1d03860dedb6b8136f6739308c2e0ef6a7490c6b4a1f1',
     '31476f64f96d72a398e7eb789068b8cfd61ed9cac95d76824a2bbf5b682ea72d',
     '974259198846a4849d318afd325d860a6e40dfd39e1ed8c7b6d87c990e35efcf',
     'e93d02ba7079c488c30e377794b4fcc62eeafb6c3f02197f1ebc059e3e5f7f07',
     'ded96eec0d25c05a54671a001bcd99f5c6d3991d2fdf80afb8f0861a44f3fd64',
     '86f34c1da65f07634de7c302a6df306dd545806022411a318900bd33b0aff9bd',
     '59baf4cf3d3b85fcdefd1a9b60c3d78926ff73650ac375c616b30fe9063b377d',
     '02f2ee251ecc19fa52dd42c0e0b609bc2e7a8ae11ce6ca35396a4bc74d12ac63',
     '597d09b4c43012b1b4b040d8c62d5fb02d1c4249de4eea06e1447000ba53f50c',
     'bb87bd8e1903db2df41c349914e9f3591bb032400be6e86ed10e7d292243a374',
     'cec38960069fdc8090cfbdd166e15ec8a77ce5b4d6b350805a63d2b54cbf0187',
     '1439d35a9c0caa9cacabe8e6179a02d51ebb9fb51125d5eeaba47ff6b8abcc97',
     '820d72d49e0fd86ef47b022a3091b326be8b0d2a42f87cbc918737b9972ea62f',
     'e625e82100031fa4a4410700034859cd4b35c086f5ac2870c6909c16a6831bfc',
     'f0e22f5167d29331351e1718c17b5420f6a29d84357273e1bfb24c2ff34fa675',
     'a5a7991f0c5c6a44f68c5c18661611057dbabeff7847623315ce784095645f75',
     '07cd3f5c5c690214af5feda27b3aec257543a8f96bc509805028a0e95c5d98c6',
     '6e5dd4405c1b446b9530e4d9356c05b71d1bfcf8a79778588143c3b6fec3fded',
     'a33f821d277e6f73c09a5ecd14cef80bac29ffe2b917225c27725d2447ac0489',
     'c18a4a02dd1f7ee65bf5596c53549c286a754afb0e3b90b2369cdd1e43c0b986',
     '6b8df66aa27b40ca275bc133958b5d543167edf919e7e623a496c9afbf7594d6',
     'c48124eeef995c7bb705bd240a26dbf6bdbe42ba29addf7ac78fa28dbed3debe',
     '040eef11dc6e0f3bb3e58e12d2a57ee274071a9c6224f27db70e19a8aa7d4df2',
     '4ecd0a955e52657fab8dfdac2df4d805f1d08b031df8bce22ac01fe20ca72c5b']

    data = []
    for i in range(len(id)):
        t1 = int(id[i], 16)
        for j in range(4):
            data.append(t1 & 0xffffffff)
            t1 >>= 32
        t2 = int(code[i], 16)
        for j in range(8):
            data.append(t2 & 0xffffffff)
            t2 >>= 32
    print(len(data))
state1=recover_state(data)
state0=backtrace([0]*12+state1)[:624]
ran=Random()
ran.setstate((3,tuple(state0+[0]),None))
id0=ran.getrandbits(128)
code0=ran.getrandbits(256)
assert ran.getrandbits(32)==data[0]
flag=md5((hex(id0)[2:].zfill(32)+hex(code0)[2:].zfill(64)).encode()).hexdigest()
print("flag{%s}"%flag)

#flag{22b307a4ac14c89888d5a6c79f7f963c}

REVERSE

babyre

jadx打开在MainActivity中发现判断逻辑

查看check函数

因为看到了check函数，根据逻辑去解发现解不出来

看到了c这个类中对enc这个文件进行了异或。将apk解压发现assets文件夹下有enc这个文件。按逻辑复原数据。

发现这是一个dex文件，接着用jadx打开

发现aes加密，网站直接解

MISC

sound from somewhere

一眼顶针为SSTV 使用MSSTV转换

CTF > Write Up
#Write Up
HWS2023 冬令营
https://g1at.github.io/2023/02/06/2023HWS/
作者
g0at
发布于
2023年2月6日
许可协议