Calculates sum of divisors.
O(lim * log(lim)).
for(int i = 2; i <= lim; ++i){
    ++s[i];
    for(int j = i + i; j <= lim; j += i)
        s[j] += i;
}

Maybe I don't understand it clear, but I think it will get TL.
Difficult of this algorithm = O(n^2)

No, it's O(n log n). Learn some math.