Again, someone hacked my B solution coz of the Python Dictionary (this can happen also when using CPP unordered_map)↵
↵
But I noticed that it can be done using an array because the constraint bi <= k which will allow to avoid Py dict↵
↵
So, I wonder did the problem setter put this constraint for Python users to be able to solve it also in Python as they can avoid dictionary because of it, or the constraint exists for any other reason?↵
↵
↵
↵
UPD1: Now, I have used d[str(key)] = value and it works and got ACC not TLE like the previous↵
↵
Submission: [submission:289783097]↵
↵
↵
~~~~~↵
brands = defaultdict(int)↵
for _ in range(k):↵
bi, ci = read_numbers()↵
brands[str(bi)] += ci↵
↵
brands = list(brands.values())↵
brands.sort(reverse=True)↵
ans = 0↵
for i in range(min(n, len(brands))):↵
ans += brands[i]↵
↵
print(ans)↵
~~~~~↵
↵
↵
↵
↵
↵
the only change is the str(bi) instead of bi, **can anyone try to hack it or show me something that will let this not work**↵
↵
coz I really wants to use Python all time (recursion problem has no idea how to solve it yet, but I am speaking now about this problem)↵
↵
But I noticed that it can be done using an array because the constraint bi <= k which will allow to avoid Py dict↵
↵
So, I wonder did the problem setter put this constraint for Python users to be able to solve it also in Python as they can avoid dictionary because of it, or the constraint exists for any other reason?↵
↵
↵
↵
UPD1: Now, I have used d[str(key)] = value and it works and got ACC not TLE like the previous↵
↵
Submission: [submission:289783097]↵
↵
↵
~~~~~↵
brands = defaultdict(int)↵
for _ in range(k):↵
bi, ci = read_numbers()↵
brands[str(bi)] += ci↵
↵
brands = list(brands.values())↵
brands.sort(reverse=True)↵
ans = 0↵
for i in range(min(n, len(brands))):↵
ans += brands[i]↵
↵
print(ans)↵
~~~~~↵
↵
↵
↵
↵
↵
the only change is the str(bi) instead of bi, **can anyone try to hack it or show me something that will let this not work**↵
↵
coz I really wants to use Python all time (recursion problem has no idea how to solve it yet, but I am speaking now about this problem)↵