game

64位elf，ollvm混淆，直接用deflat就能去混淆

展示的是去除之后

arr=[ 1,
    0,
    5,
    3,
    2,
    7,
    0,
    0,
    8,
    8,
    0,
    9,
    0,
    5,
    0,
    0,
    2,
    0,
    0,
    7,
    0,
    0,
    1,
    0,
    5,
    0,
    3,
    4,
    9,
    0,
    1,
    0,
    0,
    3,
    0,
    0,
    0,
    1,
    0,
    0,
    7,
    0,
    9,
    0,
    6,
    7,
    0,
    3,
    2,
    9,
    0,
    4,
    8,
    0,
    0,
    6,
    0,
    5,
    4,
    0,
    8,
    0,
    9,
    0,
    0,
    4,
    0,
    0,
    1,
    0,
    3,
    0,
    0,
    2,
    1,
    0,
    3,
    0,
    7,
    0,
    4]
for i in range(9):
    for j in range(9):
        print(arr[i*9+j],end="")
    print()
dic= {}
for i in range(32,128):
    a=((i & 0xF3 | ~i & 0xC) - 20)
    dic.update({a:i})
print(dic)
print(- 232084296 + 232084248)
arr=[1,
  0,
  5,
  3,
  2,
  7,
  0,
  0,
  8,
  8,
  0,
  9,
  0,
  5,
  0,
  0,
  2,
  0,
  0,
  7,
  0,
  0,
  1,
  0,
  5,
  0,
  3,
  4,
  9,
  0,
  1,
  0,
  0,
  3,
  0,
  0,
  0,
  1,
  0,
  0,
  7,
  0,
  9,
  0,
  6,
  7,
  0,
  3,
  2,
  9,
  0,
  4,
  8,
  0,
  0,
  6,
  0,
  5,
  4,
  0,
  8,
  0,
  9,
  0,
  0,
  4,
  0,
  0,
  1,
  0,
  3,
  0,
  0,
  2,
  1,
  0,
  3,
  0,
  7,
  0,
  4]
enc=[1, 4, 5, 3, 2, 7, 6, 9, 8, 8, 3, 9, 6, 5, 4, 1, 2, 7, 6, 7, 2, 8, 1, 9, 5, 4, 3, 4, 9, 6, 1, 8, 5, 3, 7, 2, 2, 1, 8, 4, 7, 3, 9, 5, 6, 7, 5, 3, 2, 9, 6, 4, 8, 1, 3, 6, 7, 5, 4, 2, 8, 1, 9, 9, 8, 4,7, 6, 1, 2, 3, 5, 5, 2, 1, 9, 3, 8, 7, 6, 4]
encode=[]
for i in range(81):
    if arr[i]==0:
        encode.append(enc[i])
arr=encode
try:
    for i in range(100):
        arr.remove(0)
except:
    pass
print(len(arr))
print(arr)
for i in range(len(arr)):
    arr[i]+=48
    arr[i]=dic[arr[i]]
print(arr)
for i in range(len(arr)//2):
    i*=2
    tmp=arr[i]
    arr[i]=arr[i+1]
    arr[i+1]=tmp
flag=""
for i in arr:
    flag+=chr(i)
print(flag[len(flag)//2:],end="")
print(flag[:len(flag)//2],end="")

KDEEIFGKIJ@AFGEJAEF@FDKADFGIJFA@FDE@JG@J

先贴出解密代码，最开始还以为是一个数独游戏，结果和数独关系不大，这道题用deflat脚本我没恢复trace函数，但是不影响，直接看加密代码

check1函数中一共有三个操作，分别是前一半和后一半换位，奇偶换位，和一个对称的加密，

check2中让D0g3和给定的sudoku比较，其中，D0g3中的0元素是自己输入的经过check1加密后再减48得到的

直接写解密脚本就行