### chokudai's blog

By chokudai, history, 4 weeks ago, ,

We will hold AtCoder Beginner Contest 145.

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

• +28

 » 3 weeks ago, # |   0 How To Solve 'E'?
•  » » 3 weeks ago, # ^ |   +1 It's a standard knapsack problem, the only difference being that the order you want to eat the food in is increasing by A (you want the longest food you'll eat to be at the end)
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 I still didn't get , why sorting with respect to time gives AC. Can you please,explain?
•  » » » » 3 weeks ago, # ^ |   0 Because you want to eat a couple, and then the last one you eat can go over. Since it can go over and is not bounded by T, then the BEST one to go over is the largest one from the subset you selected. Formally if you have a subset of times A1, A2, A3 ... Ak, you can eat them iff, you have a subset of k-1 that is <= (T-1). Now you see that you want to remove the maximum one from the subset. Sorting it and doing a knapsack will guarantee that the last one is the added is the maximum one in the subset
•  » » 3 weeks ago, # ^ |   0 Let $dp[i][j]$ be maximal happiness you can achieve taking some of first i dishes during j minutes. Let $backdp[i][j]$ be maximal happiness you can achieve taking some of last dishes (with indices from i to n) during j minutes. Now let's go through the dish that we will eat last ($i$) and time we spent on eating first $i - 1$ dishes ($t'$). Try to update answer as $min(answer, dp[i - 1][t'] + backdp[i + 1][(t - 1) - t'] + b[i]$. It's easy to calc $dp$ : $dp[i][j] = max(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - a[i]] + b[i]$ (if such exist)$)$. Same with $backdp$.
 » 3 weeks ago, # |   0 How to Solve D ?
•  » » 3 weeks ago, # ^ |   +1 If you print the matrix you notice that every third diagonal is part of pascals triangle (so just implement combinations and get the right row and column in the triangle)
•  » » » 3 weeks ago, # ^ |   0 Can you please detail more the idea and solution ?? Thanks
•  » » » » 3 weeks ago, # ^ |   0 To get to x, y from 0, 0 we have to do some (i + 1, j + 2) and some (i + 2, j + 1) moves. So, let's iterate how many times we will do (i + 1, j + 2) moves and check whether if moves left can be done with some number of (i + 2, j + 1). If the total number of moves is N, and we did K (i + 1, j + 2) type moves, number of ways to do it will be $C_{N}^{K}$
•  » » » 3 weeks ago, # ^ |   0 But how to implement combinations.I think it can't work when using mod.This worries me so much.
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   +9 You can calculate factorials modulo n, store them in fact[x]; And then inverse of factorials modulo N. (since MOD is prime, invFact[n] = POW(fact[n], MOD-2))And since comb(a,b) = fact[a] / (fact[b] * fact[a-b]) modulo this meansfact[a] * invfact[a] * invfact[a-b].Edit: Tutorial that explains https://www.geeksforgeeks.org/compute-ncr-p-set-3-using-fermat-little-theorem/
•  » » » » » 3 weeks ago, # ^ |   0 Thanx
•  » » » » » 3 weeks ago, # ^ |   0 You also can use Extended Euclidean to find modular inverse. I prefer it more than fermat.link: You only need to read equations, so don't worry about language.
•  » » » 3 weeks ago, # ^ |   +1 Matrix
 » 3 weeks ago, # | ← Rev. 2 →   +13 DPcoder rip rating
 » 3 weeks ago, # |   0 How to Solve D ? plz explain
 » 3 weeks ago, # |   0 what's the dp at F? plase
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +6 best[i][j][k] -> best cost if you only consider the first i columns, the i-th column has value j and you've done k changes already. (j is really big, but you notice that there's no use in changing in something that's not already in the input, or 0). (Note, since you only look 1 in the past with i, you can only store 0/1 for it) The big observation to make this fit in time is that if the previous one has height X, then if you change the current one, you don't want to make it shorter, since that won't decrease the cost, but might increase it in the future.
 » 3 weeks ago, # |   0 Hello, I need guidance please for E. For problem E I have an approach that gave me wrong answer and I don't understand why. I just do a normal knapsack problem and I stop at time T-1 means knapsack(weight=T-1,n) then by backtracing the elements of my solution knapsack[t-1][n], I delete them from the val set and choose the maximum left and add it to my result. it gave me 5 wrong answers. Can somebody help me ?? My solution