#### Div2 A Colorful Stones (Simplified Edition) (Author: rng_58)

In this problem you just need to implement what is written in the statement. Make a variable that holds the position of Liss, and simulate the instructions one by one.

#### Div2 B Roadside Trees (Simplified Edition) (Author: snuke)

The optimal path of Liss is as follows: First she starts from the root of tree 1. Walk up the tree to the top and eat a nut. Walk down to the height *min*(*h*_{1}, *h*_{2}). Jump to the tree 2. Walk up the tree to the top and eat a nut. Walk down to the height *min*(*h*_{2}, *h*_{3}), ... and so on.

#### Div1 A / Div2 C Escape from Stones (Author: DEGwer)

In this problem, there are many simple algorithms which works in *O*(*n*). One of them (which I intended) is following:

You should prepare 2 vectors. If *s*[*i*] = '*l*', you should push i to the first vector, and if *s*[*i*] = '*r*', you should push i to the second vector. Finally, you should print the integers in the second vector by default order, after that, you should print the integers in the first vector in the reverse order.

This algorithm works because if Liss divides an interval into two intervals *A* and *B* and she enters *A*, she will never enter *B*.

#### Div1 B / Div2 D Good Sequences (Author: DEGwer)

The main idea is DP. Let's define *dp*[*x*] as the maximal value of the length of the good sequence whose last element is *x*, and define *d*[*i*] as the (maximal value of *dp*[*x*] where *x* is divisible by *i*).

You should calculate *dp*[*x*] in the increasing order of x. The value of *dp*[*x*] is (maximal value of *d*[*i*] where *i* is a divisor of *x*) + 1. After you calculate *dp*[*x*], for each divisor *i* of *x*, you should update *d*[*i*] too.

This algorithm works in *O*(*nlogn*) because the sum of the number of the divisor from 1 to *n* is *O*(*nlogn*).

Note that there is a corner case. When the set is {1}, you should output 1.

#### Div1 C / Div2 E Choosing Balls (Author: hogloid)

There are many *O*(*Q* * *N* * *logN*) solutions using segment trees or other data structures, but probably they will get time limit exceeded.

We can solve each query independently. First, let's consider the following DP algorithm.

*dp*[*c*] := the maximal value of a sequence whose last ball's color is *c*

For each ball *i*, we want to update the array. Let the *i*-th ball's color be *col*[*i*], the *i*-th ball's value be *val*[*i*], and the maximal value of dp array other than *dp*[*col*[*i*]] be *otherMAX*. We can update the value of *dp*[*col*[*i*]] to *dp*[*col*[*i*]] + *val*[*i*] × *a* or *otherMAX* + *val*[*i*] × *b*. Here, we only need to know *dp*[*col*[*i*]] and *otherMAX*. If we remember the biggest two values of *dp* array in that time and their indexes in the array, *otherMAX* can be calculated using the biggest two values, which always include maximal values of dp array other than any particular color.

Since the values of dp array don't decrease, we can update the biggest two values in *O*(1). Finally, the answer for the query is the maximal value of *dp* array.

The complexity of the described algorithm is *O*(*QN*).

#### Div1 D Colorful Stones (Author: rng_58)

First, let's consider a simpler version of the problem: You are given a start state and a goal state. Check whether the goal state is reachable from the start state.

Define *A*, *B*, *C*, and *D* as in the picture below, and let *I* be the string of your instructions. *A* and *B* are substrings of *s*, and *C* and *D* are substrings of *t*.

It is possible to reach the goal state from the start state if there exists an instruction *I* such that:

- 1
*A*is a subsequence of*I*. - 2
*B*is not a subsequence of*I*. - 3
*C*is a subsequence of*I*. - 4
*D*is not a subsequence of*I*.

So we want to check if such string *I* exists. (string *s*1 is called a subsequence of *s*2 if it is possible to get *s*2 by removing some characters of *s*1)

There are some obvious "NO" cases. When *D* is a subsequence of *A*, it is impossible to satisfy both conditions 1 and 4. Similarly, *B* must not be a subsequence of *C*. Are these sufficient conditions? Let's try to prove this hypothesis.

To simplify the description we will introduce some new variables. Let *A*', *B*', *C*', and *D*' be strings that can be obtained by removing the first characters of *A*, *B*, *C*, and *D*. Let *c*1 and *c*2 be the first characters of *A* and *C*.

Suppose that currently the conditions are satisfied (i.e. *D* is not a subsequence of *A* and *B* is not a subsequence of *C*).

- If
*c*1 =*c*2, you should perform the instruction*c*1 =*c*2. The new quatruplet will be (*A*',*B*',*C*',*D*') and this also satisies the conditions. - If
*c*1 ≠*c*2 and*B*' is not a subsequnce of*C*, you should perform the instruction*c*1. The new quatruplet will be (*A*',*B*',*C*,*D*) and this also satisies the conditions. - If
*c*1 ≠*c*2 and*D*' is not a subsequnce of*A*, you should perform the instruction*c*2. The new quatruplet will be (*A*,*B*,*C*',*D*') and this also satisies the conditions.

What happens if all of the above three conditions don't hold? In this case *A* and *C* have the same length and *A* = *c*1*c*2*c*1*c*2..., *B* = *c*2*c*1*c*2*c*1. In particular the last two characters of *A* and *B* are swapped: there are different characters *x* and *y* and *A* = ...*xy*, *B* = ...*yx*. Now you found a new necessary condition! Generally, if *A* and *B* are of the form *A* = ...*xy* and *B* = ...*yx*, the goal state is unreachable. If the last instruction is *x*, Vasya must be in the goal before the last instruction, but then Vasya will go further after the last instruction. If the last instruction is *y*, we will also get a contradiction.

Finally we have a solution. The goal state is reachable from the start state if and only if *D* is not a subsequence of *A*, *B* is not a subsequnce of *C*, and *A* and *C* are not of the form *A* = ...*xy*, *C* = ...*yx*. The remaining part is relatively easy, so I'll leave it as an exercise for readers.

#### Div1 E Roadside Trees (Author: snuke)

For this problem there are slides: Please check here.

UPD: Thank you for pointing out mistakes. They are fixed.

It's nice to know that you idea for hard problem (Div 1 C) (for me certainly) is absolutely right=) But not enough time for debugging and AC=(

I have solved the div1 A problem using the mentioned algorithm, but I still receive that TLE error on the 31st test. I use Ruby. Is there a Ruby solution, which will pass the time test?

Here is my algorithm:

Thanks in advance!

Ruby is slow. Use C++, FPC, Java, C# and it will be accepted.

My C++ Solution got also TLE. I used vectors and Microsoft Visual C++ Compiler! But using Gnu C++, It got Accepted!

A citation from here:

Some folks managed to pass with Python, but it may happen that in Ruby I/O functions are much too slow to get AC.

Thank you. Tommorow there will be a match, and now I'm deciding which language to use. I have looked through all the submissions to that tasks, way most of them are written with C++, some Java, and even two Python solutions. Nothing else. My "second favourite" language is C#, and I have written a solution to the same problem in C#, but...

`Превышено ограничение времени на тесте 35`

(TLE on 35th). So, C# is too slow too? Here is my C# codeIn one of Java solutions I've seen class "FastScanner" with only method "FastRead", which reads from STDIO. If you think that the time issue is in IO, maybe there is a possibility to write some wrappers aroung Ruby's gets and puts?

I don't now C# well but I think

`WriteLine`

method flushes the output. I inserted a hack into your solution and it passed: 2981286But it's better to write your own input/output class. I know user it4.kp uses C# — look over his submissions.

Wow, your advice is EXTREMELY useful! Thank you.

I have tried same in Ruby, and now it fails on 45th, not 31st test. I just replaced that

`puts left`

with puts`left.join("\n")`

, and the perfomance went up dramaticly. So, it seems that the weakest part in Ruby is the IO. Will think about it more.Hello. Just use StringBuilder. But don't forget that it uses a lot of memory. Example:http://codeforces.ru/contest/265/submission/2974651 Also you can use a small buffer: http://codeforces.ru/contest/265/submission/2974575 That is eduardische's advice. And.. Console.Write() works faster than Console.WriteLine(). http://codeforces.ru/contest/265/submission/2973339

Can't view images from the problem Div1 D: 403 forbitten. :-(

Fixed now.

It seems in problem D should be "A = c1c2c1c2..., C = c2c1c2c1." instead of "A = c1c2c1c2..., B = c2c1c2c1." And further should be C instead of B.

I assume you meant “when

iis adivisorofx” instead of “wheniis amultipleofx.”, in Good Sequences.For Div2 D Good Sequences, I don't understand why the sum of the number of the diviser from 1 to n is O(nlogn). Can someone prove it?

Let p be a number in the range 1 — n Now, p will divide n/p no.s Sum of no. of divisors — {Summation(p = 1 to n)} n/p <= n * {Summation(p = 1 to n)} 1/p = n * H(n) ( H is sum of harmonic series ) and since ln(n) < H(n) < ln(n) + 1 thus, n * ln(n) < Answer < n * (ln(n) + 1)

Thank you :)

Div1 A / Div2 C Escape from Stones (Author: DEGwer) :

My solution is in O(2n), But code is so simple like it,

nice! But actually O(2n) = O(n) by the definition.

Div1 A/Div2 C

`Finally, you should print the integers in the first vector by default order, after that, you should print the integers in the second vector in the reverse order.`

should be: Finally, you should print the integers in the

secondvector by default order, after that, you should print the integers in thefirstvector in the reverse order.2 problems of 7 are made by round author)

So what?

I solved DIV 2 C by looking at the input and output, I never understood the problem statement !!

Could someone explain DIV1-C in more detail? Thanks.

Div2 C

Escape from Stonescan also be solved usingDoubly Linked Lists.Maintain a pointer

`operatingPointer`

which is initialized to the`head`

of list (head of the list contains 1).if we encounter 'l' at

`ith`

position of input add i+1 to the`left`

of`operatingPointer`

and update`operatingPointer`

to newly created node.Similarly if we encounter 'r' at

`ith`

position of the string add i+1 to the`right`

of`operatingPointer`

and update`operatingPointer`

to newly created node.Then print the whole list.

Implementation using Doubly Linked Lists

I thick c++ has builtin implemented doubly linked list.

For Div 2 D/ Div 1 B, why do we have to update d[i] for all divisors of x? Can someone explain? Thanks in advance.

because "You should calculate dp[x] in the increasing order of x" and we are using divisors not all integers that have a common divisor with x ... you have also to imagine that for each new x , we are searching for the highest d [ i ] and inserting this x in all possible sequences ... that 's why "update" is necessary

Could a top-down approach be used for div1 B/div2 D? I tried a top down approach but I am not able to memoize it.

In Div A/Div2 C Escape from Stones the algorithm is easy to understand if someone explains it,

but I don't understand how this solution came to be, which trail of thought led to this solution.

can someone please explain it.

Read my comment below.

I don't think editorial nor comments explanation on solution 264A - Escape from Stones are intuitive so I'll try to explain how I came up with a solution.

You can visualize the problem as a binary tree. Each node is responsable for a interval

[a, b]. Starting from the root, responsable for[0, 1], if you go left then root will have a left child responsable for interval[0, 1 / 2]. Since you're a left child, last time the stone fell on position 1 / 2, the end point of the interval.Note that from that child, it doesn't matter if you go left or right: the stone will never fall on a position bigger than 1 / 2 (the end point). The ideia is: since a stone always fall on the midpoint

(a + b) / 2of the lower and end points, it can never fall after the endpoint. In fact, the stone must fall infinitely many times to the right to reach that endpoint. So, we were on a node [0, 1], the stone fell and we jumped to the left, creating a left child [0, 1 / 2] and we now know that there is no way the stone will every fall to a position > 1 / 2. This means that in the final solution, the current node [0, 1 / 2] will come AFTER all nodes (descendents) that come after this one.If we jumped to the right instead, creating a right child responsable for interval [1 / 2, 1] we can show the same way that the stone will never fall before the starting point 1 / 2. This means that the current node will com BEFORE all nodes (descendents) that come after this one, since we know that the stone can't fall on a postion < 1 / 2 from now on.

This is an algorithm: if in the i'th stone we jumped to the left, we now that stone i will become AFTER all next stones. If we jumped right, stone i will become BEFORE all next stones.

Code: 23071895

elegant... :)

Excellent explanation. Thanks!

Awesome explanation thanks! :')

This is the Two Pointer solution that I was looking for. Thanks!

Strange! My C++ code got accepted with printf but got TLE with cout. This is too harsh

printf is faster than cout. if u want to use cout its better type ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); in your code this make cout fast

I'm unable to solve Escape from Stones in java. In spite of using fast io, it always gives me a TLE now matter how much i optimize it sol did anyone get it AC in java?

Hi akhi29

Two points-

1. Try to use

`StringBuilder`

(popular practice among Java user's) instead of directly printing to the output console.2. When you call,

`close()`

of`PrintStream`

, it isn't obliged to flush your output to the console. You need to call`flush()`

explicitly to ensure that the data, if present in the buffer is flushed to the output console. You can refer the modified submission 30322735In DIV 2 C I took 2 variables hi = 1 and lo = 0, if character is 'l' then I just change hi = (hi + lo) / 2 otherwise lo = (hi + lo) / 2 and mapped the value, sort them and print them, but got WA on Testcase 11, could you please tell me what is I'm missing?

I hope it may be because of the precision problem. I am also facing the same and below is my solution

30185936

in DIV2 D what will be answer in case 5 : 2 4 8 6 9. My solution that got AC. outputs 5, but how? 8 is bigger than 6 they can't form a sequence

In

Div 1 AStones falls in 'k' position and print the stone position from leftI calculated 'k' and 'd' as k=(X+Y)/2 and d=(Y-X)/2 And if 'l' then x-=d otherwise y+=d. then sorted the array which contains 'k' s value and its position no. but my implementation isn't matching with the test case..

Did I misunderstood the ques or my idea is wrong?

CODE

The thing is that, suppose you only go left then it means that the stones would be at 1/2,1/4,1/8,...till 1/(2^(10^6)) (considering maximum length of the string) which can't be stored in the float value.

For those facing TLE in the problem Div1 A (Div2 C Escape from Stones) in C++ even with

`O(N)`

time complexity the reason can be`endl vs '\n'`

.`'\n`

' inserts a new line to the output stream while`endl`

inserts a new line as well as flushes the output stream.So using

`'\n'`

can solve the TLE issue.Also you can use fast IO with cin/cout too instead of using scanf/printf by adding these lines in your code.

`ios::sync_with_stdio(false);`

`cin.tie(0);`

`cout.tie(0);`

Believe it or not this actually solved the problem for me. I was using a linked list and basically pushing to the front and back and was confused how I was getting TLE with 1,000,000 O(1) operations. I ended up writing my own code for a linked list and still failed case 31, changing the endl to \n made it accepted. This made it go from 2000 ms TLE to 233 ms, thank you for your post.

Div 2C can also be done using a doubly linked list submission