Note that the number of hacks is palindromic!

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**Code:** 35036267

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**Code:** 35036334

**Code** (bless pragmas): 35036420

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**Code:** 35036537

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**Code:** 35053688

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**Code** (backward dp): 35036591

**Code** (forward dp): 35036674

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Note that the solutions might not fully match the editorial, though the ideas are still the same.

**Code** (GreenGrape): 35055945, 35055961

**Code** (KAN): 35055995

**Code** (xen): 35055977

Thanks for the pretty early editorial!!!

In solution for problem E you wrote that time complexity is . Can you explain why ? Because for every n we are doing operations so I think it's more like

It was possible to solve this problem faster — we can use standard trick and write

c_{i}usingO(logc_{i}) nominals. Than complexity is .Instead we can use two pointers

Here To achieve previous complexity.

What do you mean by nominals?

Say

c= 2^{n}+k, wherek< 2^{n}. Then, these 2 are equivalent: choosing an element [0...c] times and choosing subsets of {2^{0}, 2^{1}...2^{n - 1},k+ 1} where an element of subset is the number of times we choose the element.So then we can compute the maximum value of

min(dp[i—1][k] +X,W+k*B) +cost[i] *k, wherektakesc_{i}different values by splitting the interval intoO(logc_{i}) intervals and taking max from each?Isn't that slower than the solution using a queue?

Saw your submission — 35045018, got the idea. Thanks :)

Can you suggest a problem where this ideia is necessary?

It's indeed as you said, but we can achieve the previously said complexity:

We have dp[i][j] = max(dp[i][j], min(dp[i — 1][j — k] + X, W + (j — k) * B) — cost[i] * k). Let's get a new k = j — old k. Now, dp[i][j] = max(dp[i][j], min(dp[i — 1][k] + X, W + (k) * B) + cost[i] * (k — j)). We can take -cost[i] * j from inside to get

dp[i][j] = -cost[i] * j + max(dp[i][j], min(dp[i — 1][k] + X, W + (k) * B) + cost[i] * k)

Now we can use a Max_queue (push element, pop element as in a queue but answers what's the max inside the queue) to solve it in O(n * sum(birds)). Edit: since nothing inside the max/min depends on j

Code: http://codeforces.com/contest/922/submission/35036545

how to write the Mathematical formula like above?

read about LaTex formulas. On codeforces they must be in $$ signs.

example:

`x_i - a^{3} \le \sqrt[5]{ \frac{1}{f(x)} }`

I learned it from this blog: http://codeforces.com/blog/entry/20729.

For 922B - Magic Forest it is possible to write a bruteforce solution and just calculate answer for all possible input dynamically (O(n^3) with no optimisation). And then use pre-calculated answers in final submission.

No, it's not possible, it will be tle because n is too large for O(n^3) to fit in a 1s time limit.

It's even possible to pass with O(N^3) without compiler optimisation. http://codeforces.com/contest/922/submission/35010602

I think your code is optimised because c < a+b

`for (c = b; c <= n && c < a + b; c++)`

Yeah I mean, it's not a "compiler" optimisation (pragma) unlike the editorial, and is still an O(N^3) algorithm.

Very good contest ! The problems were all short and I enjoyed solving it immensely !

editorial of problem D is not available. How to solve Problem D.

Briefly: it is not hard to proof that

cnt_{s}andcnt_{h}in (s[i] +s[j]) and in (s[j] +s[i]) equals. So answer will depend only on their relative location.We can sort all

s_{i}with anyn·lognsort algorithm, using`ans(s[i] + s[j]) > ans(s[j] + s[i])`

as compare function.Then build string

tfrom sorted arrays_{1},s_{2}, ...,s_{n}and outputans(t).http://codeforces.com/contest/922/submission/35039116

In C,my brute force solution passes. i can't understand why,can anyone explain??

I think it means that condition

`m[n % i] == 1`

becomes true at smallk(why? theory of numbers/modules and etc.), else you stop iterating and output "Yes".When I did not read that in

Dand for an hour was trying to find the way to calculate dp for 10^{7}birds :DThere were so many hacks in problem A that I've thought this will be a regular unrated contest)))

Why? 99% author planned (and waited) many hacks in

A.What is an official solution for problem F?

What about the editorial of problem F? I want to know how to solve it, thank you!

I think this tournament was a good tournament! The problem was neat and fun. I have a question in Problem C I know that K is always above 43, but I do not know why. When K is 6t, 6t + 1, 6t + 2, 6t + 3, 6t + 4, we prove it all, but when K is 6t + 5, we can not prove it. Please help me! Please!

What's wrong with the solution of the editorial? Have you read it?

If there are no repeated remainders in

`n % 1`

,`n % 2`

, ...,`n % 43`

, then it has to be the case that`n % 1 = 0`

,`n % 2 = 1`

,`n % 3 = 2`

, ...,`n % 43 = 42`

. This is equivalent to`1 | n + 1`

,`2 | n + 1`

, ...,`43 | n + 1`

. Therefore also`LCM(1, 2, ..., 43) | n + 1`

, so`9419588158802421600 | n + 1`

. Therefore this is only possible if`n >= 9419588158802421599`

.Thank you! I think my thinking was short on this part. Thank you so kindly.

Can someone explain me why in Magic Forest we have to include conditionals like x < j and i+j <= x (x = i ^ j) I understand why there's x > n but I can't figure out meaning of other conditionals.

`x < j`

: in that case the third number is smaller than the second, which is not allowed. The statement say that we have to count tripples (a,b,c) witha≤b≤c.`i + j < x`

: if this is the case than we don't have a triangle. E.g. try to draw a triangle with the side lengths 1, 1, and 3. Impossible. For more information: Existance of a triangeCould anyone tell, in question "C", how does complexity

O(min(k,log(n))is calculated? Is that an approximation(to achieve >43) or can we calculate maximum number through LCM in O(log(number)).Look up the sequence of lcm of first n integers on the Internet. (> 43) then lcm > 10^18.

I solved 922C in O(1) time. Although, the solution in the editorial seems straight forward, it didn't strike me during the contest! Solution link here.

Your gcd function doesn't work in constant time so it is O(log n)

Can anyone please briefly explain the claim

e^{m}-k< =C.m^{1 / 3}that is made in the editorial for problem F.Number of divisors of n is approximately

Possibly a dumb question but how do we show that? It's a cool fact but I'd liek to understand it better.

See Growth Rate

Thanks

m is the least number such that e(m) >= k. so e(m-1) < k. between m and m-1 you removed edges of divisors of m. so e(m)-number of divisors of m < k, so e(m)-k < number of divisors of m. and number of divisors of m has (approximately) an upper bound of m^(1/3)

I got it now. Thank You everyone.

My solution drops out all

i(starting from 2) ifcnt_{pairs}(i) ≤current_{pairs}-k.I'm not sure if it is correct solution but it got AC: http://codeforces.com/contest/922/submission/35051349

my idea is like yours, but i'm not sure too, do u kown the reason now? can u tell me?

I've found that a lot of people pass the problem but they don't know why it works.

Can any one provide the topdown dp solution for problem E?

It's just a knapsack problem

Can someone elaborate C? I don't get how k is approximated to 43?

Explanation

My AC solution to F: http://codeforces.com/contest/922/submission/35125488

First I find the least m where e(m) >= k (like in the editorial). then I see how many edges should be removed to reach k, let their name be "excess". I followed a greedy approach where: while excess != 0, I remove the node from the interval [floor(m/2)+1, m] which has the greatest number of divisors and this number of divisors is <= excess, then subtract this number from excess. I know that the initial value of excess is O(m^(1/3)), but why does this greedy approach always succeed ?

For the cases where the number of primes in the interval [floor(m/2)+1, m] is >= excess it succeeds because the primes will always make it possible to reach k. but for the 16 possible counters mentioned in the editorial, why does this approach succeed ?

For Problem F, " You could've even written recursive bruteforce " is mentioned in the editorial. What would be the time complexity for the same ?

can someone please explain why (n % i = i-1 : for all valid i) implies (n+1) % i = 0!

Explanation

thanks for this :)

My solution 35116450 for Problem F uses the idea in GreenGrape's 35055945 solution with only difference in that I did upper_bound over all numbers (not only over primes) and surprisingly it got AC. Fortunately it happened after the contest, to be fair with others. Seems like the acceptance criteria is much less intimidating than the problem statement looks.

What is wrong in my solution for D

http://codeforces.com/contest/922/submission/35161046

Please help

I think this is the opposite of what you wanted to do? If c1 equals to 0, that means there are no 's', so it should go to the end of the final string.

For example, the answer of 2 sh h

should be 2 with the string "shh", but your code make it "hsh" so it prints 1 instead.

Also, the answer could exceed the range of int, like in case of you have 50000 's' first then 50000 'h', which makes the answer 2.5 billion.

Can E be done recursively?

If the knapsack problem can be recursion, then it can also, obviously this is a knapsack problem

If you're talking about a memory search, of course

how are u relating it with knapsack?

can u please tell in detail?

Let us solve this problem step by step.

First, take the bird as an item, the number of birds is the amount of this item，the bag volume in this problem is not necessary to defined accurately, for your convenienc,you can take the sum(c) as the bag volume :D

In this problem, the choice of birds is in a certain order.“Moving only forward",but in the knapsack,we actually consider each item in the order we specify.

right? So it doesn't affect us to solve this problem.

Now, this problem turns into a multiple knapsack problem（I don't know the exact English name of this problem，but i think u can understand it , right?）

Let's study how to solve this simple problem before adding other conditions，one of the ways is to dismantling the same items into different items (but their volume is the same, in this problem each bird's volume is 1,value is -cost[i]).

The problem becomes a common knapsack problem.

For instance，you have You have 6 A items and 4 B items,You can think of them as 10 things, but some things are the same volume and same value.

By the way,The best way to split it is this：

for example , you have 16 A items you can dismantle them into 1,2,4,8,1(2 ^ n ). You just have to think about the 5 pieces that are split up.

Next, it's time to solve the problem completely.

mana capacity indicates that the current backpack value cannot exceed this upper limit,and after putting the bird in the backpack, the value of the knapsack cannot be negative (mana exhausted).You need to solve the maximum volume of knapsack that can be filled up under this restriction( Of course, it can't exceed the total number of birds, that is sum (c).）

my english not very well , if u cant understand,you can see my code http://codeforces.com/contest/922/submission/35191705

Of course, in fact, just dp[maxc] is enough,A bit of redundancy in this code

It's not recursively, and if you need it, I can write a code to show u

Maybe some details I forgot to answer

Thnx alot,it really helped..

I accept the problem F just now, but i still don't kown why i can accept. first i find the min m satisfy e(m) >= k, then i choose the prime from 2 to m, Obviously, if I delete a prime number, it will reduce the M / P pair. If I delete the prime number, the total pair is still larger than k, and I'll delete it。

here is my mainly code （I didn't do any other extra judgment as solution.）

i want to kown how to prove this idea is correct？？？

In addition, I want to know the details of the practice of recursion. Can anyone explain to me more about the principles and proof of F in more detail? I admit that I might be stupid: D

I have a randomized solution for F which differs from the editorial. The main idea is that if we have a set $$$S$$$, then adding an element $$$x$$$ to the set and recomputing $$$I(S)$$$ can be done quickly. So we can randomly insert an element if $$$I(S) < K$$$ and remove a random element if $$$I(S) \ge K$$$. A set data structure does not support querying a random element, so we can use priority based data structure instead to allow for random removal/increments.

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