Editorial for this problem seems to be resulting an unnecessarly
complicated code.
Solution is O(R*S) and 15 LOC (python 3). [I missed it during the contest
"of course", as I spent much time on the interactive problem.]
The idea is same but it is easy to prove that resulting sizes for A and B
can be calculated by the fact, that one [actually the first]
2-difference-cancelling move is: A ends before the 3rd rank appears, and
the last card in B is always increased by 2, starting with R+1. End of A
can be calculated by just maintaining the longest same-rank so far for all
ranks.
Special case is the same [ending B for the last is R*S-2, not R*S-1], but
no need to count, it's always R-1, (S-1)*R.
Axioma / Kata
spoiler alert
-----
T = int(input())
for tst in range(T):
R, S = map(int, input().split())
nums=[1]*R
i=0
todo=R*(S-1)
print("Case #" + str(tst + 1) + ": " + str((todo+1)//2))
for lst in range(R+1,R*S,2):
fst=nums[i]+nums[(i+1)%R]
print(fst,lst-fst)
nums[i]+=1
nums[(i+1)%R]+=1
i=(i+2)%R
if lst==R*S-2:
print(S-1,(R-1)*S)
'Samantha' via Google Code Jam <[email protected]> ezt írta
(időpont: 2020. ápr. 18., Szo 1:28):
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