r2C +17 d17
rF0 r2C -17 u17
r33 +0B d0B
r43 +0C d0C
rF0 r33 -0B u0B
rF0 r43 -0C u0C
r1B +16 d16
rF0 r1B -16 u16 r29 +2C d2C
r43 +0C d0C rF0 r29 -2C u2C
rF0 r43 -0C u0C r1B +16 d16
rF0 r1B -16 u16 r29 +2C d2C
rF0 r29 -2C u2C
r35 +1C d1C
rF0 r35 -1C u1C r44 +12 d12
r3C +18 d18
rF0 r44 -12 u12
rF0 r3C -18 u18 r2D +15 d15
r29 +2C d2C rF0 r2D -15 u15
rF0 r29 -2C u2C r2B +09 d09
rF0 r2B -09 u09 r4B +0F d0F
rF0 r4B -0F u0F r1C +04 d04
r34 +0A d0A rF0 r1C -04 u04
rF0 r34 -0A u0A
r29 +2C d2C
rF0 r29 -2C u2C
r59 +E5 dE5
r24 +08 d08
rF0 r24 -08 u08
r42 +0E d0E
r44 +12 d12 rF0 r42 -0E u0E
rF0 r44 -12 u12
rF0 r59 -E5 uE5
r12 +E1 dE1
r54 +2F d2F
rF0 r54 -2F u2F
rF0 r12 -E1 uE1
r33 +0B d0B
r44 +12 d12 rF0 r33 -0B u0B
rF0 r44 -12 u12
r4B +0F d0F
rF0 r4B -0F u0F r1C +04 d04
rF0 r1C -04 u04
r4D +13 d13
r43 +0C d0C
rF0 r4D -13 u13 rF0 r43 -0C u0C r1C +04 d04
rF0 r1C -04 u04 r31 +11 d11 r44 +12 d12
rF0 r31 -11 u11 rF0 r44 -12 u12
r4B +0F d0F
r1C +04 d04 rF0 r4B -0F u0F
rF0 r1C -04 u04
r12 +E1 dE1
r5B +30 d30
rF0 r5B -30 u30
rF0 r12 -E1 uE1
r11 +E0 dE0
r21 +06 d06
The key element in solving this task is obtainting a HID keyboard mapping like this one
The flag format EKO{, encoded, looks like this: 08 0e 12 2f after a quick investigation we find a part of the ciphertext that corresponds to it:
rF0 r24 -08 u08 //E
r42 +0E d0E //K
r44 +12 d12 rF0 r42 -0E u0E //K
rF0 r44 -12 u12 //O
rF0 r59 -E5 uE5 //shift
r12 +E1 dE1 //
r54 +2F d2F //{
rF0 r54 -2F u2F
Using some trivial deduction, we can figure out that dHEX signals a key-down and a uHEX key-up
We were able to get the flag using a simple script in python
EKO{holapianola}