【逆向分析】BUUCTF 逆向题目 SimpleRev

https://buuoj.cn/challenges#SimpleRev

https://files.buuoj.cn/files/7458c5c0ce999ac491df13cf7a7ed9f1/SimpleRev

信息:     文件名: H://BUUCTF/SimpleRev/SimpleRev    大小: 13128(12.82 KiB)    操作系统: Ubuntu Linux(ABI: 3.2.0)    架构: AMD64    模式: 64 位    类型: DYN    字节序: LE

unsigned __int64 Decry(){  char v1; // [rsp+Fh] [rbp-51h]  int v2; // [rsp+10h] [rbp-50h]  int v3; // [rsp+14h] [rbp-4Ch]  int i; // [rsp+18h] [rbp-48h]  int v5; // [rsp+1Ch] [rbp-44h]  char src[8]; // [rsp+20h] [rbp-40h] BYREF  __int64 v7; // [rsp+28h] [rbp-38h]  int v8; // [rsp+30h] [rbp-30h]  __int64 v9[2]; // [rsp+40h] [rbp-20h] BYREF  int v10; // [rsp+50h] [rbp-10h]  unsigned __int64 v11; // [rsp+58h] [rbp-8h]
  v11 = __readfsqword('(');  *(_QWORD *)src = 'SLCDN';  v7 = 0LL;  v8 = 0;  v9[0] = 'wodah';  v9[1] = 0LL;  v10 = 0;  text = join(key3, (const char *)v9);  strcpy(key, key1);  strcat(key, src);  v2 = 0;  v3 = 0;  getchar();  v5 = strlen(key);  for ( i = 0; i < v5; ++i )  {    if ( key[v3 % v5] > 64 && key[v3 % v5] <= 90 )      key[i] = key[v3 % v5] + 32;    ++v3;  }  printf("Please input your flag:");  while ( 1 )  {    v1 = getchar();    if ( v1 == 'n' )      break;    if ( v1 == ' ' )    {      ++v2;    }    else    {      if ( v1 <= '`' || v1 > 'z' )      {        if ( v1 > '@' && v1 <= 'Z' )        {          str2[v2] = (v1 - 39 - key[v3 % v5] + 97) % 26 + 97;          ++v3;        }      }      else      {        str2[v2] = (v1 - 39 - key[v3 % v5] + 97) % 26 + 97;        ++v3;      }      if ( !(v3 % v5) )        putchar(32);      ++v2;    }  }  if ( !strcmp(text, str2) )    puts("Congratulation!n");  else    puts("Try again!n");  return __readfsqword(0x28u) ^ v11;}

#include <stdio.h>#include <string.h>
int main() {    int v3 = 0;    char key[] = "adsfkndcls";    char str2[] = "killshadow";    int v5 = strlen(key);
    for (int v2 = 0; v2 < v5; v2++) {        for (int v1 = 'A'; v1 <= 'z'; v1++) {            if (str2[v2] == (v1 - 39 - key[v3 % v5] + 97) % 26 + 97) {                printf("%c", v1);                ++v3;                break;            }        }    }        return 0;}

key = 'adsfkndcls'text = 'killshadow'v5 = len(key)flag = ''
for v3, text_char in enumerate(text):    for v1 in range(65, 91):         if (v1 - 39 - ord(key[v3 % v5]) + 97) % 26 + 97 == ord(text_char):            flag += chr(v1)
print(flag)

flag{KLDQCUDFZO}