【逆向分析】BUUCTF 逆向题目 [GXYCTF2019]luck_guy

https://buuoj.cn/challenges#[GXYCTF2019]luck_guy

https://files.buuoj.cn/files/c4ae5d67227e2354c3b512d6bd8b180a/attachment.zip

信息:     文件名: H:/第七届“强网杯”全国网络安全挑战赛/BUUCTF/[GXYCTF2019]luck_guy/luck_guy    大小: 9176(8.96 KiB)    操作系统: Ubuntu Linux(16.04.12,ABI: 2.6.32)    架构: AMD64    模式: 64 位    类型: EXEC    字节序: LE

unsigned __int64 get_flag(){  unsigned int v0; // eax  int i; // [rsp+4h] [rbp-3Ch]  int j; // [rsp+8h] [rbp-38h]  __int64 s[5]; // [rsp+10h] [rbp-30h] BYREF  unsigned __int64 v5; // [rsp+38h] [rbp-8h]
  v5 = __readfsqword(0x28u);  v0 = time(0LL);  srand(v0);  for ( i = 0; i <= 4; ++i )  {    switch ( rand() % 200 )    {      case 1:        puts("OK, it's flag:");        memset(s, 0, sizeof(s));        strcat((char *)s, f1);        strcat((char *)s, &f2);        printf("%s", (const char *)s);        break;      case 2:        printf("Solar not like you");        break;      case 3:        printf("Solar want a girlfriend");        break;      case 4:        strcpy((char *)s, "icug`ofx7F");        strcat(&f2, (const char *)s);        break;      case 5:        for ( j = 0; j <= 7; ++j )        {          if ( j % 2 == 1 )            *(&f2 + j) -= 2;          else            --*(&f2 + j);        }        break;      default:        puts("emmm,you can't find flag 23333");        break;    }  }  return __readfsqword(0x28u) ^ v5;}

flag = 'GXY{do_not_'f2 = [0x7F, 0x66, 0x6F, 0x60, 0x67, 0x75, 0x63, 0x69][::-1]
for i, value in enumerate(f2):    flag += chr(value - (2 if i % 2 == 1 else 1))
print(flag)

flag{do_not_hate_me}