0xGame2024部分Reverse与Crypto

Re

week2

Xor_Random

动调获取随机数直接异或

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#include<stdio.h>
int main(){
    srand(0x77);
    rand();
    char v21=0x7b;
    char v8;
    char a[]={0x0C, 0x4F, 0x10, 0x1F, 0x4E, 0x16, 0x21, 0x12, 0x4B, 0x24, 
              0x10, 0x4B, 0x0A, 0x24, 0x1F, 0x17, 0x09, 0x4F, 0x07, 0x08, 
              0x21, 0x5C, 0x2C, 0x1A, 0x10, 0x1F, 0x11, 0x16, 0x59, 0x5A};
    for(int i=0;i<30;i++){
        if (i&1){
            v8=v21;
        }
        else{
            v8=v21+3;
        }
        printf("%c",a[i]^v8);
    } 
    return 0;
}

Zzz

断点上面扩展了位数，且调试的时候发现输入的v5,v6,v7会被颠倒

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from Crypto.Util.number import long_to_bytes
from libnum import n2s
from z3 import *
v10 = BitVec('v10', 32)
v11 = BitVec('v11', 32)
v12 = BitVec('v12', 32)

res=[]
solver = Solver()
solver.add(11 * v11 + 14 * v10 - v12 == 0x48FB41DDD)
solver.add(9 * v10 - 3 * v11 + 4 * v12 == 0x2BA692AD7)
solver.add((((v12 - v11)) >> 1) + (v10 ^ 0x87654321) == 3451779756)
if solver.check()==sat:
    model=solver.model()
    res.append(n2s(model[v10].as_long()))
    res.append(n2s(model[v11].as_long()))
    res.append(n2s(model[v12].as_long()))
print(res)

#0xGame{e544267d-7812-44b3-a35d-d085a85201a4}

week3

BabyAsm

直接正向输出flag

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a=[20,
   92,
   43,
   69,
   81,
   73,
   95,
   23,
   72,
   22,
   24,
   69,
   25,
   27,
   22,
   17,
   23,
   29,
   24,
   73,
   17,
   24,
   85,
   27,
   112,
   76,
   15,
   92,
   24,
   1,
   73,
   84,
   13,
   81,
   12,
   0,
   84,
   73,
   82,
   8,
   82,
   81,
   76,
   125]
print(len(a))
for i in range(22):
    a[i]+=28
for i in range(22,43):
    a[i]=a[i]^a[i-22]
s=''
for i in range(len(a)):
    s+=chr(a[i])
print(s)

Puzzle

9*9的数组，大概率是数独题

手算比较麻烦，建议直接套模板

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from z3 import *

# 定义数独的9x9网格
board = [[5, 7, 0, 9, 4, 0, 8, 0, 0], 
         [0, 0, 8, 0, 3, 0, 0, 0, 5], 
         [0, 1, 0, 2, 0, 0, 0, 3, 7], 
         [0, 0, 9, 7, 2, 0, 0, 0, 0], 
         [7, 3, 4, 0, 0, 8, 0, 0, 0], 
         [0, 0, 0, 0, 0, 0, 7, 5, 1], 
         [3, 0, 0, 0, 1, 4, 2, 0, 0], 
         [0, 6, 0, 0, 0, 2, 0, 4, 0], 
         [0, 2, 7, 0, 0, 9, 5, 0, 0]]

# 创建一个9x9的矩阵，存放Z3变量（每个格子一个变量）
X = [[Int(f'x_{i}_{j}') for j in range(9)] for i in range(9)]

# 约束1：所有变量取值范围在1到9之间
cells_c = [And(X[i][j] >= 1, X[i][j] <= 9) for i in range(9) for j in range(9)]

# 约束2：每一行的数字不能重复
rows_c = [Distinct(X[i]) for i in range(9)]

# 约束3：每一列的数字不能重复
cols_c = [Distinct([X[i][j] for i in range(9)]) for j in range(9)]

# 约束4：每个3x3的子方格中的数字不能重复
squares_c = [Distinct([X[3 * i + di][3 * j + dj] for di in range(3) for dj in range(3)])
             for i in range(3) for j in range(3)]

# 约束5：已知的棋盘数字需要保持不变
instance_c = [If(board[i][j] == 0, True, X[i][j] == board[i][j]) for i in range(9) for j in range(9)]

# 创建求解器
solver = Solver()
solver.add(cells_c + rows_c + cols_c + squares_c + instance_c)

# 检查是否有解
if solver.check() == sat:
    model = solver.model()
    result_board = [[model.evaluate(X[i][j]) for j in range(9)] for i in range(9)]
    # 打印填充好的数独
    for row in result_board:
        print(row)
    
    # 将解转换为字符串形式，用于生成输入
    nextLine = ""
    for i in range(9):
        for j in range(9):
            if board[i][j] == 0:
                nextLine += str(model.evaluate(X[i][j]))
    
    print(nextLine)
else:
    print("无解")

求出nextLine之后6个数字一组，10转16之后作为字符加到sb中输出

tea

一开始把密文前后交换，直接解密发现中间错误

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#include <stdio.h>
#include <stdint.h>

int main() {
    uint32_t v[2] = {1, 2};
    uint32_t v0 = v[0], v1 = v[1], sum, i; /* set up */
    uint32_t delta = 0x61C88647;
    uint32_t v5[10] = {0x85336dd3,0x2a7a7c3b,0x64306238,0x36396434,0x62336364,0x38376533,0x37323664,0x33363463,0xf8ee8ea2,0xc9b65cce};
    unsigned int k[4] = {2,0,2,4}, l = 0, r = 0;
    uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];
    for (int m = 0; m < 10; m += 2) {
        sum = delta*(-32);
        for (i = 0; i < 32; i++) { 
            v5[m + 1] -= ((v5[m] << 4) + k2) ^ (v5[m] + sum) ^ ((v5[m] >> 5) + k3);
            v5[m] -= ((v5[m + 1] << 4) + k0) ^ (v5[m + 1] + sum) ^ ((v5[m + 1] >> 5) + k1);
            sum += delta;
        } 
    }

    for (int i = 0; i < 10; i++) {
        for (int m = 0; m <= 3; m++) {
            printf("%c", (v5[i] >> (8 * m)) & 0xff);
        }
    }
    return 0;
}

调试发现之后前8位和最后8位进行了加密，也就是说中间24位不变

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0xGame{e8b0d4d96dc3b3e78d627c463c9953a1}

ezAndroid

直接搜索secret找到密文

week4

PyPro

python3.12无法完全反编译，要用pydas

知道先aes_ecb再base64

cy好像解不了用脚本解密

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import base64
from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad

key = 0x554B134A029DE539438BD18604BF114

key_bytes = key.to_bytes(16, byteorder='big')
print(key_bytes)

encrypted_data = "2e8Ugcv8lKVhL3gkv3grJGNE3UqkjlvKqCgJSGRNHHEk98Kd0wv6s60GpAUsU+8Q"

encrypted_bytes = base64.b64decode(encrypted_data)


cipher = AES.new(key_bytes, AES.MODE_ECB)

try:
    decrypted_data = unpad(cipher.decrypt(encrypted_bytes), AES.block_size)

    print("解密结果:", decrypted_data.decode('utf-8'))
except (ValueError, KeyError):
    print("解密失败或填充错误！")

发现密钥是奇数位，15.5个字节，前面补一个0

tea2.0

main一个tea

交叉引用密文发现有tls

发现也是一次tea，但是不用管直接动调获取最后的密文

有个奇怪的点是明明密文有48位，为什么比较的时候只有44位

经过调试发现最后一轮的v1初始值是0

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#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
void decrypt(uint32_t v[2], uint32_t const key[4]) {
    unsigned int i;
    uint32_t v0 = v[0], v1 = v[1], delta = 0x61C88647, sum = delta * (-64);
    for (i=0; i < 64; i++) {
        v1 -= (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum>>11) & 3]);
        sum += delta;
        v0 -= (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);

    }
    v[0] = v0; v[1] = v1;
}

int main() {
    uint32_t const k[4] = {0x4512,0x9832,0x5647,0x6314};
    uint32_t enc[] = {0xd04bb2c4,0xa625cf11,0x57686f2b,0x3b181680,0x6fc33f75,0x1779b334,0xd95ca4f6,0xc1ca9fa2,0x651a6aff,0xc9ddba37,0x3087d163,0x187d7e10};
    for(int i=0; i<12; i+=2){
        decrypt(&enc[i], k);
    }
    for (int i = 0; i < 12; i++) {
        for (int m = 0; m <= 3; m++) {
            printf("%c", (enc[i] >> (8 * m)) & 0xff);
        }
    }
    return 0;
}

MineSweeper

先是交换后面异或，另外从资源段加载密文

imgimg

haha就是KEY

enc从resources.assets中加载

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key='0xoX0XOxOXoxGAME'
arr=[0x45,0x21,0x3E,0x08,0x57,0x31,0x09,0x4D,0x42,0x45,0x42,0x44,0x5D,0x5A,0x4B,0x4B,
     0x52,0x56,0x16,0x44,0x66,0x45,0x6C,0x40,0x57,0x44,0x33,0x35,0x51,0x75,0x0D,0x58,
     0x15,0x71,0x11,0x1B,0x0B,0x08,0x76,0x04,0x4F,0x5C,0x68,0x3C]
num=0
s='This is: True_KEY!for #0xgAmE_Unity~Cryption'
print(len(s))
KEY=[]
for i in range(len(s)):
    KEY.append(ord(s[i]))
for i in range(44):
    num=(num+ord(key[i%len(key)]))%44
    KEY[i],KEY[num]=KEY[num],KEY[i]
for i in range(44):
    arr[i]^=KEY[i]
    print(chr(arr[i]),end="")

#0xGame{36ecd059-b3e7-73c8-fa80-0a2abef3c757}

crack

经过调试发现异或的数一直为6

问gpt发现是sha256

调试发现buf1就是输入的地址，也就是输入特定hash值即可

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0xGame{1b4e549d12ccb4bb4936f95fedecebf55494dec8}

JustSoso

getKey在native层

直接在资源段搜索密文

得到key后rc4，有一处魔改

之后再base64

getKey的部分静态有点复杂

直接尝试用objection注入得到rc4的key，android hooking watch class_method com.ctf.justsoso.MainActivity.getKey –dump-return

用frida也行，frida -U -l 1.js -f com.ctf.justsoso

最后解密即可

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import base64
def rc4(data, key):
    S = [0]*256
    for i in range(256):
        S[i]=256-i
    j = 0
    out = []

    for i in range(256):
        j = (j + S[i] + key[i % len(key)]) % 256
        S[i], S[j] = S[j], S[i]

    i = j = 0
    for char in data:
        i = (i + 1) % 256
        j = (j + S[i]) % 256
        S[i], S[j] = S[j], S[i]
        out.append(char ^ S[(S[i] + S[j]) % 256])

    return bytes(out)

s='nB9RCjwReif5P1H1MYO6m/hucCGjI6EE9wWEx/E4N+bO5k5ior6MnqAGQfc='
out=base64.b64decode(s)


data = bytes([0x9c, 0x1f, 0x51, 0x0a, 0x3c, 0x11, 0x7a, 0x27, 0xf9, 0x3f, 0x51, 0xf5, 0x31, 0x83, 0xba, 0x9b, 0xf8, 0x6e, 0x70, 0x21, 0xa3, 0x23, 0xa1, 0x04, 0xf7, 0x05, 0x84, 0xc7, 0xf1, 0x38, 0x37, 0xe6, 0xce, 0xe6, 0x4e, 0x62, 0xa2, 0xbe, 0x8c, 0x9e, 0xa0, 0x06, 0x41, 0xf7])
key = bytes([171,149,153,151,193,31,143,241,189,165,181,193,153,161])

decrypted = rc4(out, key)
print(decrypted)
#0xGame{fd51ce4b-4556-4cf9-9430-67480614e43b}

Crypto

week1

crt-1

e和phi不互素，先求出e//2对应的m，再开根号，答案就是其中一个

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n,e = (1022053332886345327, 294200073186305890)
c = 107033510346108389
 
phi = 970868178 * 1052721012
#print(gcd(e,phi))
d = invese_mod(e//2, phi)
m2 = pow(c,d,n)
 
P.<x> = PolynomialRing(Zmod(n))
f = x^2 - m2
res = f.roots(multiplicities=False)
 
for i in res:
    MD5(int(i))

'127016d0be858ef48a99723710ad4d49'
'f4107420d94cc7037114376d8566d4ef'
'3932f6728585abbf751a212f69276d3e'
'15820afdb9a129e89e40e57f40ff8de9'

week2

rc4

前半部分求得四个字符，后半部分注意先fromhex

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from random import choice
from hashlib import sha256
from string import ascii_letters, digits
while(1):
    s4 = ''.join([choice(ascii_letters+digits) for _ in range(4)])
    #print(s4)
    s20=s4+'5mcbb8TCD9muMJy0'
    s=sha256(s20.encode()).hexdigest()
    if s=='08d287b38cb45c90dd1a8ab20484d8122119e2e50e50b60300ee30b74a86f456':
        print(s4)
        break

a='edd186458e9679384d4623dcc37845c803f5e70cf8c60fbe5a339de88bbf93540b8ba4e407d97cadaf2fa4499d88dea4a87471'
a=bytes.fromhex(a)
b='01234567890abcdef'*6
b=bytes.fromhex(b)
print(b)
m=[]
for i in range(len(b)):
   m.append(a[i]^b[i])
print(m)

m=[236, 242, 195, 34, 7, 156, 197, 230, 189, 84, 23, 138, 187, 232, 238, 5, 236, 244, 196, 73, 159, 79, 5, 2, 132, 195, 143, 220, 221, 199, 3, 255, 198, 100, 165, 199, 66, 190, 245, 167, 19, 241, 84, 91, 169, 222, 166, 52, 3, 185, 158]

enc='dc8a84436af9bede8c622fb888dbd928dac3f778b2763461b3eeebecebf72ecca302c6a324dbc4c572c43226'
enc=bytes.fromhex(enc)
flag=''
for i in range(len(enc)):
    flag+=chr(enc[i]^m[i%len(m)])
print(flag)
#0xGame{81682337-6731-91c7-d060-3efcdfe1ba5f}

RSAIV

因为交互的总时间足够长，challenge2适合yafu手动解后输入

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from math import isqrt
from random import choice
from hashlib import sha256
from string import ascii_letters, digits
from Crypto.Util.number import *
from gmpy2 import gmpy2, gcdext
from pwn import *

io= remote('118.195.138.159',10003)

def s(a) : io.send(a)
def sa(a, b) : io.sendafter(a, b)
def sl(a) : io.sendline(a)
def sla(a, b) : io.sendlineafter(a, b)
def r() : return io.recv()
def pr() : print(io.recv())
def ru(a) : return io.recvuntil(a)
def inter() : io.interactive()



def ch0(s):
    n,e,c=eval(s)
    # n=1748540272127031261953491227523826413186688228587730569034148979069843516521168331238821457
    # e=3
    # c=73664223980107312196785660420184405948303349111082519137365225806576685634069671168357
    k = 0
    t = gmpy2.iroot(c,e)
    if t[1]:
        #print(t[0])
        sl(str(t[0]).encode())
    else:
        while 1:
            m = gmpy2.iroot(k*n+c,e)
            if m[1]:
                #print(bytes_to_long(m[0]))
                sl(str(bytes_to_long(m[0])).encode())
                break
            else:
                k += 1

def ch1(s):
    n,e,c,dp=eval(s)
    # n=174434352212179643958855935833036063001
    # e=0x10001
    # c=39466599612748960141905412873983556787
    # dp=6171497056913161467

    for i in range(1,e):
        t = (dp * e - 1) % i        #这是p-1前面的系数
        if t == 0:
            p = (dp * e - 1) // i + 1
            if n % p == 0:
                q = n // p
                d = gmpy2.invert(e,(p-1)*(q-1))
                #print(str(pow(c,d,n)).encode())
                sl(str(pow(c,d,n)).encode())

def ch2(s):
    n,e,c=eval(s)
    # n=319642561199483600677262826186024249019
    # e=170254189003332402684383900760365999599
    # c=315096655202591609017165088376826841837
    # p,q=fermat(n)
    # phi=(p-1)*(q-1)
    # d=inverse(e,phi)
    # m=pow(c,d,n)
    # print(m)
    # sl(str(m).encode())

def ch3(s):
    n,e1,c1,e2,c2=eval(s)
    # n = 123859953327496855816310504481883257257
    # c1 = 115331678509320635403714895034425874310
    # c2 = 19516419686488748417408440624144239780
    # e1 = 162726227029902673640708475886431248549
    # e2 = 92095361035743610350805143550253960191
    _ ,s1 ,s2 = gcdext(e1,e2)
    m = pow(c1,s1,n) * pow(c2,s2,n) % n
    print(str(m).encode())
    sl(str(m).encode())

s=ru('sha256(XXXX+')
print(s)
s1=io.recvuntil(b') == ',drop=True)
print(s1)
s2=io.recvuntil(b'\n',drop=True)
print(s2)
while(1):
    s4 = ''.join([choice(ascii_letters+digits) for _ in range(4)])
    #print(s4)
    s20=s4+s1.decode()
    s=sha256(s20.encode()).hexdigest()
    #print(type(s))
    if s==s2.decode():
        print(s4)
        sl(s4.encode())
        break

sla('>',b'0')
s=ru(')')
ch0(s)
sla('input choice:',b'1')
io.recvuntil('>',drop=True)
s=ru(')')
ch1(s)
# sla('input choice:',b'2')
# io.recvuntil('>',drop=True)
# s=ru(')')
# ch2(s)
sla('input choice:',b'3')
io.recvuntil('>',drop=True)
s=ru(')')
ch3(s)
inter()
比赛wp
#伪随机数 #z3 #tea魔改 #pydas #unity逆向 #rc4魔改 #中国剩余定理 #RSA
0xGame2024部分Reverse与Crypto
https://j1ya-22.github.io/2024/10/23/0xGame2024部分Reverse与Crypto/
作者
j1ya
发布于
2024年10月23日
许可协议