0xGame 2024 W1&2 wp|Annms

Rev[week1 AK]

Binary Master

IDA喵了

解法二：

理论上输入0x114514也行

signsign

Babybase

Str为输入字符串，v4为加密后字符串，在checkflag内与

比较，回到encode函数

不难发现这是个标准base64

xor_beginning

密文在v5内，异或78-i

exp:

#include<bits/stdc++.h>

int main()
{
      char v5[40];
     strcpy(v5, "~5\v*',3");
    v5[7] = 31;
  v5[8] = 118;
  v5[9] = 55;
  v5[10] = 27;
  v5[11] = 114;
  v5[12] = 49;
  v5[13] = 30;
  v5[14] = 54;
  v5[15] = 12;
  v5[16] = 76;
  v5[17] = 68;
  v5[18] = 99;
  v5[19] = 114;
  v5[20] = 87;
  v5[21] = 73;
  v5[22] = 8;
  v5[23] = 69;
  v5[24] = 66;
  v5[25] = 1;
  v5[26] = 90;
  v5[27] = 4;
  v5[28] = 19;
  v5[29] = 76;
  for(int i=0;i<=29;i++)
  {
    printf("%c",v5[i]^78-i);
    
  }
    return 0;
 }

xor_Endian

v6为密文，encrypt函数：

以Key0xGame2024为密钥依次异或，结合题目所述Endian，小端序排列密文解密即可

[W2]BabyUPX

加密方式为交换高低四位

exp

#include<bits/stdc++.h>

int main()
{
    int enc[44]={0x03, 0x87, 0x74, 0x16, 0xD6, 0x56, 0xB7, 0x63, 0x83, 0x46, 0x66, 0x66, 0x43, 0x53, 0x83, 0xD2, 
    0x23, 0x93, 0x56, 0x53, 0xD2, 0x43, 0x36, 0x36, 0x03, 0xD2, 0x16, 0x93, 0x36, 0x26, 0xD2, 0x93, 
    0x73, 0x13, 0x66, 0x56, 0x36, 0x33, 0x33, 0x83, 0x56, 0x23, 0x66};
    for(int i=0;i<=43;i++)
    {
       
        printf("%c",(((enc[i]&0x0f)<<4)|((enc[i]&0xf0)>>4)));
    }
    return 0;
 }

0xGame{68dff458-29e5-4cc0-a9cb-971fec338e2f}

EzPyc

import hashlib
user_input = input('请输入神秘代号：')
if user_input != 'Ciallo~':
    print('代号不是这个哦')
    exit()
input_hash = hashlib.md5(user_input.encode()).hexdigest()
input_hash = list(input_hash)
for i in range(len(input_hash)):
    if ord(input_hash[i]) in range(48, 58):
        original_num = int(input_hash[i])
        new_num = (original_num + 5) % 10
        input_hash[i] = str(new_num)
    input_hash = ''.join(input_hash)
    print('0xGame{{{}}}'.format(input_hash))

md5:7f5ef5762bf8a2c043d836b522127e54

exp:

fakeflag='7f5ef5762bf8a2c043d836b522127e54'
fakeflag=list(fakeflag)
for i in range(len(fakeflag)):
    if ord(fakeflag[i]) in range(48, 58):
        original_num = int(fakeflag[i])
        new_num = (original_num + 5) % 10
        fakeflag[i] = str(new_num)
print(fakeflag)
# //2f0ef0217bf3a7c598d381b077672e09

EzJar

def find_multiplicative_inverse(a, m):
    for x in range(1, m):
        if (a * x) % m == 1:
            return x
    return None

def decrypt(ciphertext):
    Alphabat = "0123456789abcdef"
    decrypted_string = ""
    
    # Find the multiplicative inverse of 5 modulo 16
    inverse = find_multiplicative_inverse(5, 16)
    
    for char in ciphertext:
        if char in Alphabat:
            # Find the index of the character in Alphabat
            i = Alphabat.index(char)
            # Reverse the transformation using the multiplicative inverse
            original_index = (i - 3 + 16) % 16  # Adding 16 to ensure non-negative result
            original_index = (original_index * inverse) % 16
            decrypted_string += Alphabat[original_index]
        else:
            decrypted_string += char
    
    return decrypted_string

# Given ciphertext
ciphertext = "ab50e920-4a97-70d1-b646-cdac5c873376"
# Decrypt the ciphertext
decrypted_text = decrypt(ciphertext)
print(decrypted_text)