Comments
+13

Thanks :}, congrats to you too on becoming master
+8

yes ;)
-11

yeah it should be, I think by the time I reached C, the problem statements were already changed, so I didn't notice that.
+10

I just hope, i get a +1 after that :}
0

Aapna kya lena dena
+19

Kuch bhi xD?? Show me a single global/div1 contest where an Indian was the winner(Rank1)
0

got it, but the constant (c) can be too large, i think
0

log(n) factor can be avoided while precomputation,and powers of two can be calculated in O(1), using the left shift (1<<i), constraints were fine, but the TL was too tight, my code barely passed. Btw i don't see that 1600ms code in your solution.
0

How is no. of divisors N bounded by N^(1/3)?, Like 24 has 8 divisors, which is much more than 24^(1/3) https://math.stackexchange.com/questions/1699330/upper-bound-on-number-of-factors#:~:text=Let%20d(n)%20denote%20the,nloglogn)).
0

Wow
0

https://www.spoj.com/problems/MSTS/ In this question, number of spanning tree is 16 while number of minimum spanning tree is 8, how are they same??...maybe its applicable for only unweighted graph or...maybe I didn't get what you said :'(