Solution for p2 k=3
DP[i] =dp[i-1]+2*dp2[i-2]+dp[i-3]
Dp2[i]=dp3[i-1]+dp2[i-3]
dp3[i]=dp3[i-3]+dp[i-3]
answer is dp[n]
dp2[i] is the number of ways of to tile a 3i grid with one corner removed and dp3[i] is the number of ways to tile a 3i grid with the rightmost cell removed from first and second row
Getting 135 in pretests.
Cutoff for 12 will be 135 or 200. Gof 132 
Cutoff for class 8th in INOI 2020
100 or 132
Cutoff for 10th?
I am in 10th and got 132 in pretests
Cutoff for 9th?
Enter score here
See all scores entered here
INOI 2020 was far easier than last year!I think the cutoff is going to be 200 for class 12! But sadly enough my computer hanged today and I spotted the constraint 1<=k<=3, 20 mins after wasting time on a 2D dp solution involving dp[][]! Hence I endend up getting only 132!!
I should have attempted the tiling problem first!
Please sort it and add the totalling function
It’s not mine
How many people get selected from 12th?
I guess 200 or 135 will be the cutoff for all classes
What is the password?
YouShallNotPass
Apparently for the first question, if you try to make 2 DFS calls for each component of the graph - 1 to determine which node will serve as the root and 1 to calculate strength of that component with respect to this node, the program gives you a TLE in 2 cases of subtask 2.
What’s the right logic then ?
Root of the tree can be determined by the minimum/maximum value node according to the parity of the number of vertices of the tree.
My approach was to run DFS on the graph to seperate its connected components out. Then, on each connected component run a DFS to determine its strength.
strength_even = strength of even sized connected components
strength_odd = strength of odd sized connected components
Print strength_even and strength_odd
This gave AC(100 pts) on pretests. Hope it passes system tests.