Lihua's for

使用ida打开，f5反编译

分析代码可知：
加密逻辑就是b[i]=i ^ flag[i]
然后验证a[i]==b[i]-->a[i]=i ^ flag[i]
所以解密flag:flag[i]=i^a[i]
求a[i]的数据就是&unk_403040处的数据

双击进入&unk_403040

写一个解密脚本

a = [
    0x66, 0x6D, 0x63, 0x64, 0x7F, 0x64, 0x32, 0x36, 0x6A, 0x6C,
    0x3E, 0x3D, 0x39, 0x20, 0x6F, 0x3A, 0x20, 0x77, 0x3F, 0x27,
    0x25, 0x27, 0x22, 0x3A, 0x7A, 0x2E, 0x78, 0x7A, 0x31, 0x2F,
    0x29, 0x29, 0x16, 0x40, 0x44, 0x45, 0x12, 0x47, 0x47, 0x41,
    0x1A, 0x54
]

flag = ''
for i in range(42):
    flag += chr(i ^ a[i])
print(flag)