Reverse Cipher

We have recovered a binary and a text file. Can you reverse the flag.

We’re given an ELF binary rev and the encrypted flag rev_this. I used Dogbolt to decompile. Here’s the relevant code:

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int __fastcall main(int argc, const char **argv, const char **envp)
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{
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  char ptr[23]; // [rsp+0h] [rbp-50h] BYREF
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  char v5; // [rsp+17h] [rbp-39h]
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  int v6; // [rsp+2Ch] [rbp-24h]
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  FILE *v7; // [rsp+30h] [rbp-20h]
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  FILE *stream; // [rsp+38h] [rbp-18h]
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  int j; // [rsp+44h] [rbp-Ch]
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  int i; // [rsp+48h] [rbp-8h]
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  char v11; // [rsp+4Fh] [rbp-1h]
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  stream = fopen("flag.txt", "r");
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  v7 = fopen("rev_this", "a");
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  if ( !stream )
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    puts("No flag found, please make sure this is run on the server");
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  if ( !v7 )
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    puts("please run this on the server");
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  v6 = fread(ptr, 0x18uLL, 1uLL, stream);
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  if ( v6 <= 0 )
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    exit(0);
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  for ( i = 0; i <= 7; ++i )
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  {
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    v11 = ptr[i];
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    fputc(v11, v7);
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  }
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  for ( j = 8; j <= 22; ++j )
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  {
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    v11 = ptr[j];
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    if ( (j & 1) != 0 )
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      v11 -= 2;
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    else
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      v11 += 5;
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    fputc(v11, v7);
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  }
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  v11 = v5;
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  fputc(v5, v7);
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  fclose(v7);
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  return fclose(stream);
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}

Seems like it’s just a simple encryption function that subtracts 2 from the character if it’s at an odd index, and adds 5 to the character if it’s at an even index. This should be pretty easily reversible!

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s = "w1{1wq84fb<1>49" # encrypted flag within the {}
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for i in range(8, 23):
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        if (i & 1) != 0:
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                print(chr(ord(s[i - 8]) + 2),end='')
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        else:
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                print(chr(ord(s[i - 8]) - 5),end='')

RUn the script to get the flag!

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picoCTF{r3v3rs36ad73964}