Your code fails for the following case:
Plaintext: A B A B A B C ........
Ciphertext: 2 2 2 2 2 6 ..
The solution is to backtrace from where the sequence ends.
This is what your code produces
1
5 6
2 2 2 2 2 3
Case #1: BABABAC

1
5 6
2 2 2 2 2 3
Case #1: BABABAC

Same Result in Both the cases. 


On Tuesday, April 9, 2019 at 8:38:14 PM UTC+5:30, [email protected] wrote:
> import math
> 
> def solve():
>   _, l = map(int, input().split())
>   p = [int(i) for i in input().split()]
> 
>   fs, sc = p[0:2]  
>   pr = math.gcd(fs, sc) if fs != sc else int(math.sqrt(fs))
> 
>   vals = []
>   vals.append(fs / pr)
>   vals.append(pr)
>   vals.append(sc / pr)
> 
>   for i in range(2, l):
>     vals.append(p[i] / vals[i])
> 
>   prs = list(sorted(set(vals)))
>   
>   return ''.join([chr(65 + prs.index(i)) for i in vals])
> 
> for t in range(1, 1 + int(input())):
>   print('Case #{}: {}'.format(t, solve()))

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