easy_python

python字节码

逻辑整理后就给flag

flag = [204, 141, 44, 236, 111, 140, 140, 76, 44, 172, 7, 7, 39, 165, 70, 7, 39, 166, 165, 134, 134, 140, 204, 165, 7, 39, 230, 140, 165, 70, 44, 172, 102, 6, 140, 204, 230, 230, 76, 198, 38, 175]
for i in range(42):
    flag[i] = ((flag[i] >> 5) | (flag[i] << 3) )& 0xff
print(bytes(flag))
#flag{ddbae889-2895-44df-897d-2ae30df77b61}

godeep

题目提示是找字符串，我们交叉引用的Rigth只有一个地方。

接着梳理程序流程。

在 在main_convert(v4, v7);中，把我们输入的字符串的每个值转二进制

 而下面的godeep_tree函数中相当于一个完全二叉树，左0右1。拿着转换二进制来进行走，一直要走到Right这个叶子。

 此时*v4的值为1，所以最后8个字是xxxx xxx1 8个一组恢复一个字节

所以思路就是从Right叶子逆推到main函数，一直对函数调用进行交叉引用

s=[0b01100110,0b01100011,0b00110000,0b00110011,0b01100010,0b01100100,0b00111001,0b00110111,0b00101101,0b01100110,0b01100110,0b00110111,0b01100010,0b00101101,0b00110100,0b00110001,0b00111001,0b01100110,0b00101101,0b00111000,0b00111001,0b00111000,0b00110111,0b00101101,0b00110111,0b00111000,0b01100010,0b01100011,0b00110111,0b00110100,0b00110101,0b01100100,0b00110011,0b01100010,0b00110000,0b01100001]
print(bytes(s))
print(len(s))
#b'fc03bd97-ff7b-419f-8987-78bc745d3b0a'
#36

reindeer game

玩游戏通关就行，懒得py打包逆向了

flag{82a2acb6-9803-4936-92db-f1431d90c6d1}

baby_transform（未完成）

这个程序逻辑也不复杂

void __fastcall FFT(char *a1, double *a2, int len)
{
  double v3; // [rsp+0h] [rbp-40h]
  int i; // [rsp+20h] [rbp-20h]
  int v6; // [rsp+24h] [rbp-1Ch]
  double v8; // [rsp+30h] [rbp-10h]
  double v9; // [rsp+38h] [rbp-8h]

  for ( i = 0; i < len; ++i )
  {
    v6 = 0;
    v8 = 0.0;
    v9 = 0.0;
    while ( v6 < len )
    {
      v8 = (double)a1[v6] * cos((double)v6 * (-6.283185307179586 * (double)i) / (double)len) + v8;
      v3 = (double)a1[v6];
      v9 = sin((double)v6++ * (-6.283185307179586 * (double)i) / (double)len) * v3 + v9;
    }
    *a2 = v9;
    a2[1] = v8;
    a2 += 2;
  }
}

好像是FFT算法，但我找了几个恢复的脚本。都恢复不出来。

不知道咋回事