$ file easy_rev flag
easy_rev: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=d6ab8e0c86636e8331cc465ae54a5013598dd79e, not stripped
flag:     openssl enc'd data with salted password

void __cdecl main(int argc, const char **argv, const char **envp)
{
  signed int i; // [sp+8h] [bp-48h]@1
  int v6[10]; // [sp+10h] [bp-40h]@1

  puts("==========================================");   // banner
  puts("       NEWBIE REV1 right here !!");
  puts("solve the magic I putted, and get the flag");
  puts("==========================================");
  printf(">> ");
  for ( i = 0; i < 10; ++i )
    scanf("%d", &v6[i]);                                // 10개의 정수 입력
  puts("++++++++++++++++++++++++++++++++++++++++++");
  puts("Let's See the result!!!!");
  puts("++++++++++++++++++++++++++++++++++++++++++");
  if ( swap(v6) == 10 )                                 // 성공
  {
    puts(">> Yes, You got right ( IF YOU CERTAINLY INSERTED EXACTLY 10 NUMBERS )");
    puts(">> You just need to 'add' all the no for every index. That sum is key for flag file !!");
    puts(">> (flag file is encryted aes-256-cbc of openssl)");
  }
  else                                                  // 실패
  {
    puts(">> Nope :< BETTER TRY HARDER THAN THAT");
  }
}

int __fastcall swap(int *a1)
{
  int result = 0; // [sp+18h] [bp-48h]@1
  int v4 = 3; // [sp+1Ch] [bp-44h]@1
  int i; // [sp+24h] [bp-3Ch]@1
  int j; // [sp+28h] [bp-38h]@9
  int k; // [sp+2Ch] [bp-34h]@12
  int v9[10] = {0x4f, 4, 0x24, 0x274, 0x75, 0x3e, 0x99a, 0xffffff9b, 0x29, 0xef}; // [sp+30h] [bp-30h]@1

  for ( i = 0; i < 10; ++i )    // a1의 인덱스i에 따라 서로 다른 값과 xor
  {
    if ( i % 3 == 1)
      v4 -= i;
    else if ( i % 3 == 2 )
      v4 += i;
    else
      v4 *= i;
    a1[i] ^= v4;
  }

  for ( j = 0; j < 10; ++j )    // 0xF 와 xor
    a1[j] ^= 0xF;

  for ( k = 0; k < 10; ++k )    // v9와 모두 같아야 함
  {
    if ( a1[k] == v9[k] )
      ++result;
  }
  return result;
}

$ ./easy_rev
==========================================
       NEWBIE REV1 right here !!
solve the magic I putted, and get the flag
==========================================
>> 64 -12 42 632 -123 53 2445 -123 63 1
++++++++++++++++++++++++++++++++++++++++++
Let's See the result!!!!
++++++++++++++++++++++++++++++++++++++++++
>> Yes, You got right ( IF YOU CERTAINLY INSERTED EXACTLY 10 NUMBERS )
>> You just need to 'add' all the no for every index. That sum is key for flag file !!
>> (flag file is encryted aes-256-cbc of openssl)

>>> sum([64, -12, 42, 632, -123, 53, 2445, -123, 63, 1])
3042

$ openssl enc -d -aes-256-cbc -in flag -k 3042
flag{R2versing_1s_b4sed_0n_H4cking_:)}