### Mahdi_Hajibeygi's blog

By Mahdi_Hajibeygi, history, 5 years ago,

Hi friends it's not a professional Editorial I just wrote it myself and i hope it can help you.

A: you see the only way we can decrease number of prime factors 2 is to divide n by 2, and the same for 3 and 5;

so for decreasing prime a factor 5 we have to use operation 3 and after that two times operation 1.

for decreasing prime a factor 3 we have to use operation 2 and after that operation 1.

so if we define f(x) to be number of prime factors x in n answer will be f(2) + 2 * f(3) + 3 * f(5);

my_solution

B:

for each query let cnt[x] (0 <= x && x < 3) be number of elements like y that y mod 3 = x. then easily we can prove that we chould not do anything with numbers divisible by 3. and for cnt[1] and cnt[2] we can merge min(cnt[1], cnt[2]) this operation "merge a number x(x mod 3 = 1) with a number y(y mod 3 = 2)" that these make numbers divisible by 3 again and for remaining numbers that all are equal mod 3 if they are x elements we can make [x / 3] numbers divisible by 3 with them

my_solution

C: imagine numbers arr [0, 1, 2, 3, 4, 5] now each round find first 0 1 2 3 4 5 as a subsequence greedy this way "find first 0 after that first 1 and so on" then put them in a group and don't use them in next rounds.

whenever you couldn't find any new subsequence you must delete all remaining elements

my_solution

D: first of all let's call prime numbers in a type_1, others in a type_2, and primes in b type_3, and others in b type_4

now let's solve problem recursive see biggest number(x).

if it is prime it's type_3 cause for each i (i >= 2) Pi > i so if it's type_1 we should have a bigger prime in our numbers that it is not true. so now delete x and y(that P(y) = x) and solve remaining elements recursive.

else it's type_2 cause if it's type_4 imagine it's type_2 so we have inserted x / y that (y > 1) in b that it's false cause x is biggest number. so now delete x and y(maximum divisor of x that is smaller than x) and solve remaining elements recursive.

my_solution

E: let's make a dfs and solve problem with dfs tree.

these tree is bipartite and as we have n >= 2 for each vertex v, d(v) > 1 so if we choose any of these parts we will have condition that we need.

now let's choose smaller one.

my_solution

F:

I suggest you read my_solution for problem F(Destroy it!). I have dp[N][10] that dp[i][j] is maximum sum if we have get x cards from first i round and x mod 10 = j. and it's easy to see that in each round we may use only 3 strongest card that cost 1, strongest card cost 2 and strongest card cost 3 so by a mask on these(at most 5)cards you can update dp easily.

I HOPE IT WAS USEFUL FOR YOU :)

• +82

 » 5 years ago, # |   +4 If there is any problem (except that it's too short) tell me to fix it that others can use better blog.or if you have better solutions for any problem tell me.
 » 5 years ago, # |   +23 you forgot to mention contest code "565(Div3)" in the title. btw thanks for editorial
•  » » 5 years ago, # ^ |   +9 yes you're right i'll fix it.thanks.
 » 5 years ago, # |   +9 Auto comment: topic has been updated by Mahdi_Hajibeygi (previous revision, new revision, compare).
 » 5 years ago, # | ← Rev. 3 →   0 hab noh ebbiunigga you gay
 » 5 years ago, # | ← Rev. 3 →   +17 my_solution for problem F dp[N][10] :| O(N)
•  » » 5 years ago, # ^ | ← Rev. 2 →   +3 Wow!if you want write your solution if it's faster than mine and i'll edit blog so others can use it
 » 5 years ago, # |   +14 What I Need Is WEED
•  » » 5 years ago, # ^ |   +1 funny :)
 » 5 years ago, # |   0 Can u explain C?
•  » » 5 years ago, # ^ |   0 did you under stand part I find a subsequence and make a group?
•  » » » 5 years ago, # ^ |   0 how to find first 4 than first 8 ans so on?
•  » » » » 5 years ago, # ^ |   0 Have indexes of 0s 1s ... in sets and use upper bound
•  » » » » » 5 years ago, # ^ |   +8 you mean i store indexes of all those 6 numbers in a set?? and then check the sequence?
•  » » » » » » 5 years ago, # ^ |   +3 A set for each type of numbersread my code you will get what i mean code is neat you will understand it
•  » » » 5 years ago, # ^ |   +3 No matter. If you got that part do it and you will get ACC.If no i ment get first 0 after that first 1 after that 0 after that first 2 after that 1 and so on.
 » 5 years ago, # |   +4 Thanks!
•  » » 5 years ago, # ^ |   +6 Your welcome :)