```textPrison Break119 Points
Author: FeDEX
Your friend has been captured by some country's secret services and is being held in a prison. Having reached the prison, you realise there is a code there that you need to break.The enemies have been kind enough to leave a large file containing 3 numbers on each line and the following message for you:"Start with a list of 10^7 zeros and for every line containing a,b,c separated by a space in the given file, add c modulo 10 to every number in your list between indices a and b (a included only).Indices start at 1 in the list. At the end, compute the product modulo 999999937 of the nonzero digits in your list and you will obtain the password needed to free your friend".The problem is that your friend needs medication within the next 3 days so can you break the password soon enough?
Flag Format: HackTM{CODE}
Challenge Files: https://drive.google.com/file/d/1CNwGf_lKq8wHA8qYQJFkqP5HCybmYBel/view```
We got a huge file (138Mb) with a lot of lines:
```bash$ wc -l Given_File.txt9999999 Given_File.txt```
Each line is composed by three numbers as evoqued in the description (a, b and c):
```text183 183 0548 3000548 591 8000091 541 2000041 895 1000095 1296 296 4625 625 2```
With the same description, we can clean the file to got less lines and speed up the computation time to get code the code:
```bash$ cat Given_File.txt | sort -n -k 1 | uniq -c | sed 's/^[ \t]*//;s/[ \t]*$//' | grep -v ' 0' > clean_file```
* sort the line by the first number (and not the entire line)* count each line to add before it the occurence number* replace the tabulation added before* remove all the line ended by 0 since no loop will be done with it
```pythonfile = open("clean_file", 'r')l = []for i in range(9000999):    l.append(0)
for line in file:    o,a,b,c = line.split(' ')    value = (int(c) % 10) * int(o)    if value == 0:        continue    for i in range(int(a), int(b)):        l[i] += value
f = []f.append(0)for v in l:    current = int(v) % 10    if current != 0:        f.append(int(v) % 10)
final = 1for a in f:    if a == 0:        continue    final = (a * final) % 999999937
print("Flag is : HackTM{" + str(final) + "}")```
Since Python is not really fast for doing it, I choose to use [pypy](https://pypy.org/). You can use it with your python script but pypy handle it with a different approach than regular Python interpretor.
```bash$ pypy3 prison_break.pyFlag is : HackTM{585778044}```
My script took `948,21s user 0,69s system 99% cpu 15:50,45 total`. I guess I can make it better and improve the input file too.
PS : I tried to compile my python script to C code with [cython](https://cython.org/) but I can't get a working binary :(.