Rating:
# Tux's Kitchen
## Description
I need to bake it!
nc crypto.hsctf.com 8112
## Solution
The server print a cute penguin and a list of numbers.
We can see that the numbers are generated by function final_bake()
```
print(final_baking(flag,key))
```
The function's body is:
```
def final_baking(food,key):
baked = rand0m_mess(food,key)
treasure = []
for i in range(len(baked)):
treasure.append(ord(food[i])*baked[i])
print(ord(food[i])*baked[i]))
treasure = prepare(treasure)
return treasure
```
I noticed that the numbers in treasure are multiple of the order of each character in the flag.
Knowing the number, I can make a list of printable candidate that is factor of the number.
And prepare() is:
```
def prepare(food):
good_food = []
for i in range(len(food)):
good_food.append(food[i]^MY_LUCKY_NUMBER)
print(food[i]^MY_LUCKY_NUMBER)
for k in range(len(good_food)):
good_food[i] += MY_LUCKY_NUMBER
return good_food
```
So I first get 3 lists of numbers and reverse the prepare process.
```
for i in range(len(ans)):
ans[-1] = ans[-1] - MY_LUCKY_NUMBER
ans[i] = ans[i] ^ MY_LUCKY_NUMBER
````
For all printable characters, if it is the factor of the number, put it in the candidate list.
```
for j in range(32,127):
if i // j * j == i:
pos += chr(j)
```
If the character is in all three of the candidate, it's big chance that it is the answer.
Running the [script](solve.py), I get the result:
```
4hs!c:t"3f){:t4h1s_1s_07ne_;v3&9ry_$6l07ng_"3f$6l 0@`g_1b3!ca8s3_:t5(
