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# Priority queue

In computer science, a priority queue is an abstract data type which is like a regular queue or stack data structure, but where additionally each element has a "priority" associated with it. In a priority queue, an element with high priority is served before an element with low priority. If two elements have the same priority, they are served according to their order in the queue.

It is a common misconception that a priority queue is a heap. A priority queue is an abstract concept like "a list" or "a map"; just as a list can be implemented with a linked list or an array, a priority queue can be implemented with a heap or a variety of other methods.

A priority queue must at least support the following operations:

• insert_with_priority: add an element to the queue with an associated priority
• pull_highest_priority_element: remove the element from the queue that has the highest priority, and return it
This is also known as "pop_element(Off)", "get_maximum_element", or "get_front(most)_element".
Some conventions reverse the order of priorities, considering lower values to be higher priority, so this may also be known as "get_minimum_element", and is often referred to as "get-min" in the literature.
This may instead be specified as separate "peek_at_highest_priority_element" and "delete_element" functions, which can be combined to produce "pull_highest_priority_element".

In addition, peek (in this context often called find-max or find-min), which returns the highest priority element but does not modify the queue, is very frequently implemented, and nearly always executes in O(1) time. This operation and its O(1) performance is crucial to many applications of priority queues.

More advanced implementations may support more complicated operations, such as pull_lowest_priority_element, inspecting the first few highest- or lowest-priority elements, clearing the queue, clearing subsets of the queue, performing a batch insert, merging two or more queues into one, incrementing priority of any element, etc.

## Similarity to queues

One can imagine a priority queue as a modified queue, but when one would get the next element off the queue, the highest-priority element is retrieved first.

Stacks and queues may be modeled as particular kinds of priority queues. In a stack, the priority of each inserted element is monotonically increasing; thus, the last element inserted is always the first retrieved. In a queue, the priority of each inserted element is monotonically decreasing; thus, the first element inserted is always the first retrieved.

## Implementation

### Naive implementations

There are a variety of simple, usually inefficient, ways to implement a priority queue. They provide an analogy to help one understand what a priority queue is. For instance, one can keep all the elements in an unsorted list. Whenever the highest-priority element is requested, search through all elements for the one with the highest priority. (In big O notation: O(1) insertion time, O(n) pull time due to search.)

### Usual implementation

To improve performance, priority queues typically use a heap as their backbone, giving O(log n) performance for inserts and removals, and O(n) to build initially. Alternatively, when a self-balancing binary search tree is used, insertion and removal also take O(log n) time, although building trees from existing sequences of elements takes O(n log n) time; this is typical where one might already have access to these data structures, such as with third-party or standard libraries.

Note that from a computational-complexity standpoint, priority queues are congruent to sorting algorithms. See the next section for how efficient sorting algorithms can create efficient priority queues.

There are several specialized heap data structures that either supply additional operations or outperform these approaches. The binary heap uses O(log n) time for both operations, but also allow queries of the element of highest priority without removing it in constant time. Binomial heaps add several more operations, but require O(log n) time for requests. Fibonacci heaps can insert elements, query the highest priority element, and increase an element's priority in amortized constant time[1] though deletions are still O(log n)). Brodal queues can do this in worst-case constant time.

While relying on a heap is a common way to implement priority queues, for integer data, faster implementations exist. This can even apply to data-types that have a finite range, such as floats:

For applications that do many "peek" operations for every "extract-min" operation, the time complexity for peek actions can be reduced to O(1) in all tree and heap implementations by caching the highest priority element after every insertion and removal. For insertion, this adds at most a constant cost, since the newly inserted element is compared only to the previously cached minimum element. For deletion, this at most adds an additional "peek" cost, which is typically cheaper than the deletion cost, so overall time complexity is not significantly impacted.

## Equivalence of priority queues and sorting algorithms

### Using a priority queue to sort

The semantics of priority queues naturally suggest a sorting method: insert all the elements to be sorted into a priority queue, and sequentially remove them; they will come out in sorted order. This is actually the procedure used by several sorting algorithms, once the layer of abstraction provided by the priority queue is removed. This sorting method is equivalent to the following sorting algorithms:

### Using a sorting algorithm to make a priority queue

A sorting algorithm can also be used to implement a priority queue. Specifically, Thorup says:[4]

We present a general deterministic linear space reduction from priority queues to sorting implying that if we can sort up to n keys in S(n) time per key, then there is a priority queue supporting delete and insert in O(S(n)) time and find-min in constant time.

That is, if there is a sorting algorithm which can sort in O(S) time per key, where S is some function of n and word size,[5] then one can use the given procedure to create a priority queue where pulling the highest-priority element is O(1) time, and inserting new elements (and deleting elements) is O(S) time. For example if one has an O(n log log n) sort algorithm, one can create a priority queue with O(1) pulling and O(log log n) insertion.

## Libraries

A priority queue is often considered to be a "container data structure".

The Standard Template Library (STL), and the C++ 1998 standard, specifies priority_queue as one of the STL container adaptor class templates. It implements a max-priority-queue. Unlike actual STL containers, it does not allow iteration of its elements (it strictly adheres to its abstract data type definition). STL also has utility functions for manipulating another random-access container as a binary max-heap. The Boost (C++ libraries) also have an implementation in the library heap.

Python's heapq module implements a binary min-heap on top of a list.

Java's library contains a PriorityQueue class, which implements a min-priority-queue.

Go's library contains a container/heap module, which implements a min-heap on top of any compatible data structure.

The Standard PHP Library extension contains the class SplPriorityQueue.

Apple's Core Foundation framework contains a CFBinaryHeap structure, which implements a min-heap.

## Applications

### Bandwidth management

Priority queuing can be used to manage limited resources such as bandwidth on a transmission line from a network router. In the event of outgoing traffic queuing due to insufficient bandwidth, all other queues can be halted to send the traffic from the highest priority queue upon arrival. This ensures that the prioritized traffic (such as real-time traffic, e.g. an RTP stream of a VoIP connection) is forwarded with the least delay and the least likelihood of being rejected due to a queue reaching its maximum capacity. All other traffic can be handled when the highest priority queue is empty. Another approach used is to send disproportionately more traffic from higher priority queues.

Many modern protocols for Local Area Networks also include the concept of Priority Queues at the Media Access Control (MAC) sub-layer to ensure that high-priority applications (such as VoIP or IPTV) experience lower latency than other applications which can be served with Best effort service. Examples include IEEE 802.11e (an amendment to IEEE 802.11 which provides Quality of Service) and ITU-T G.hn (a standard for high-speed Local area network using existing home wiring (power lines, phone lines and coaxial cables).

Usually a limitation (policer) is set to limit the bandwidth that traffic from the highest priority queue can take, in order to prevent high priority packets from choking off all other traffic. This limit is usually never reached due to high level control instances such as the Cisco Callmanager, which can be programmed to inhibit calls which would exceed the programmed bandwidth limit.

### Discrete event simulation

Another use of a priority queue is to manage the events in a discrete event simulation. The events are added to the queue with their simulation time used as the priority. The execution of the simulation proceeds by repeatedly pulling the top of the queue and executing the event thereon.

### Dijkstra's algorithm

When the graph is stored in the form of adjacency list or matrix, priority queue can be used to extract minimum efficiently when implementing Dijkstra's algorithm, although one also needs the ability to alter the priority of a particular vertex in the priority queue efficiently.

### Huffman coding

Huffman coding requires one to repeatedly obtain the two lowest-frequency trees. A priority queue makes this efficient.

### A* and SMA* search algorithms

The A* search algorithm finds the shortest path between two vertices or nodes of a weighted graph, trying out the most promising routes first. The priority queue (also known as the fringe) is used to keep track of unexplored routes; the one for which a lower bound on the total path length is smallest is given highest priority. If memory limitations make A* impractical, the SMA* algorithm can be used instead, with a double-ended priority queue to allow removal of low-priority items.

### ROAM triangulation algorithm

The Real-time Optimally Adapting Meshes (ROAM) algorithm computes a dynamically changing triangulation of a terrain. It works by splitting triangles where more detail is needed and merging them where less detail is needed. The algorithm assigns each triangle in the terrain a priority, usually related to the error decrease if that triangle would be split. The algorithm uses two priority queues, one for triangles that can be split and another for triangles that can be merged. In each step the triangle from the split queue with the highest priority is split, or the triangle from the merge queue with the lowest priority is merged with its neighbours.


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# Latex documentclass bachelor thesis

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This post is specifically for students of school of electrical engineering and computer sciences (seecs), national university of science and technology (nust) who are in their thesis phase and want to format their thesis document in latex. (master/phd thesis, latex report documentclass) thesis with x s thesis using latex is pre-set in the style beamer presentation template for bachelor. Latex class thesis – 133341 this \documentclass[a4paper]{mythesis} master, or bachelor (plus some more) at linköping psu latex thesis class. Latex bibtex bachelor thesis 439111 biblatex book caption chapter citation cite code color colortbl definecolor description documentclass draft enumerate.

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Most of them will already be using latex somehow but phd/master/bachelor thesis: which document class i used the book class from koma-script in my thesis. \documentclass[options] thesis lau optiontotypesetalaurea (bachelor’sdegree).

Latex documentclass bachelor thesis
Rated 5/5 based on 49 review

2018.

format: plain


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# Survey: What are people subscribed to on WeChat?

Note by Silas Hundt
5 years, 7 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Well I'm on skype but not on we chat

- 5 years, 7 months ago

Oh no, I don't know what is troll and what isn't. I guess we've entered real troll status.

Staff - 5 years, 7 months ago

Yeah, better watch out. Have you gotten any of my emails?

- 5 years, 7 months ago

For anybody who doesn't know, WeChat is like KiK+Skype. It's pretty cool.

- 5 years, 7 months ago

What is Kik?

- 5 years, 7 months ago

Well, Kik is WeChat - Skype. Does that help?

- 5 years, 7 months ago

HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA yeah KiK is just standard messages/texting.

- 5 years, 7 months ago

Then why don't you get a phone instead of using Kik?

- 5 years, 7 months ago

People use KiK AND text. It's kind of stupid, I know. :D

- 5 years, 7 months ago

What is WeChat???

- 5 years, 7 months ago

My favorite comment of all time too. But I can't even tell the difference between QQ and Wechat, except that Wechat is supposed to be targeted towards more Western audiences.

- 5 years, 7 months ago

Haha, was expecting this comment! :D

- 5 years, 7 months ago

Hahaha

Staff - 5 years, 7 months ago


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# Can the overuse of geothermal energy become an environmental concern?

At what power output would we be using so much geothermal energy that we cool the core enough to endanger the Earth's magnetic field and have to stop using it? Is this a conceivable concern for a future energy crisis?

-

No, this seems highly unlikely.

According to online sources:

Average human power consumption in 2008 was estimated at 15 TW.

Total annual heat loss from the Earth due to the surface heat flux is estimated at 44.2 TW.

Estimates of the electricity generating potential of geothermal energy around the world are consistently less than 2 TW.

So although the average power consumption is a surprisingly large fraction of the total geothermal heat flux, the part that might possibly be extracted for power is quite small.

There are much more concerning environmental impacts to geothermal power development than freezing the core and destroying the Earth's Magnetic field. These include increased rates of release of carbon and sulfur into the atmosphere and land use issues.

-
However unlikely, commercially invalid, or otherwise infeasible with today's technology there is a number of watts that if removed from the earth's core will solidify it. +1 for other 'more concerning environmental impacts'. –  Mazura Aug 1 '14 at 0:54
@Mazura, a key point though is you cannot draw an arbitrary wattage from the core with geothermal energy. You can only get some fraction of the heat flux that is already conducting up to the surface anyway. So you aren't really affecting the core cooling rate. –  ZSG Nov 19 '14 at 5:56
Mark, not sure what you mean by increased rates of carbon and sulphur released, can you say more? –  a different ben Nov 25 '14 at 22:45
Regarding land use issues, footprints per kW of current geothermal power plants are amongst the smallest of all electricity-producing industries. –  a different ben Nov 25 '14 at 22:47
@a different ben - Because C & S gas species are in volcanic/geothermal emissions, there are measurable emissions of these from a geothermal power plant. Should this be regulated in the same way as a carbon emission from an coal-fired plant. It is measured at a 'stack'. On the other hand it may be 'natural' in the sense that it was going to be emitted from a fumarole anyway. My understanding is that this is treated by regulatory agencies as the same. Perhaps this is a confusion, but I believe this has in fact had an impact on the licensing of geothermal power plants in places such as Hawaii. –  Mark Rovetta Nov 27 '14 at 2:00

This is a bit of a what-if kind of answer, with some rough estimates. Just considering the cooling aspect - let's pretend that Earth is solid for a momnet.

The time-dependent heat equation tells us how long it will take for a change in temperature to travel a certain distance by conduction (the 'characteristic timescale'). The distance to the outer core, where the magnetic field is generated, is about 3000 km.

Assume that the geothermal installation is at the surface, and that it loses its heat instantaneously. Then, the 'characteristic timescale' is given by $$\tau = \frac{l^2}{\kappa}$$ Taking $l=3 \times 10^{6}$m as the distance to travel, and $\kappa = 1 \times 10^{-6}$m2/s as the mean thermal diffusivity of the rocks, then it would take $2.85 \times 10^{11}$ years for the change in temperature due to extracting geothermal heat at Earth's surface to propagate to the outer core. The sun would have transitioned to a red giant by then.

The problem being of course that it's not all solid. Convection in the mantle would obliterate the temperature signal. I don't know how you'd go about calculating that.

More practically, a deep geothermal project of the EGS flavour would be managed as a mining project. The reservoir would be used up within 20 years or so, then either abandoned or mothballed (until perhaps it reheated?). The maximum temperature drawdown for economic feasibility might be of the order of 50°C or less, depending on a boatload of factors.

-


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# XOR cipher

In cryptography, the simple XOR cipher is a type of additive cipher,[1] an encryption algorithm that operates according to the principles:

A ${\displaystyle \oplus }$ 0 = A,
A ${\displaystyle \oplus }$ A = 0,
(A ${\displaystyle \oplus }$ B) ${\displaystyle \oplus }$ C = A ${\displaystyle \oplus }$ (B ${\displaystyle \oplus }$ C),
(B ${\displaystyle \oplus }$ A) ${\displaystyle \oplus }$ A = B ${\displaystyle \oplus }$ 0 = B,

where ${\displaystyle \oplus }$ denotes the exclusive disjunction (XOR) operation. This operation is sometimes called modulus 2 addition (or subtraction, which is identical).[2] With this logic, a string of text can be encrypted by applying the bitwise XOR operator to every character using a given key. To decrypt the output, merely reapplying the XOR function with the key will remove the cipher.

## Example

For example, the string "Wiki" (01010111 01101001 01101011 01101001 in 8-bit ASCII) can be encrypted with the repeating key 11110011 as follows:

 01010111 01101001 01101011 01101001 ${\displaystyle \oplus }$ 11110011 11110011 11110011 11110011 = 10100100 10011010 10011000 10011010

And conversely, for decryption:

 10100100 10011010 10011000 10011010 ${\displaystyle \oplus }$ 11110011 11110011 11110011 11110011 = 01010111 01101001 01101011 01101001

## Use and security

The XOR operator is extremely common as a component in more complex ciphers. By itself, using a constant repeating key, a simple XOR cipher can trivially be broken using frequency analysis. If the content of any message can be guessed or otherwise known then the key can be revealed. Its primary merit is that it is simple to implement, and that the XOR operation is computationally inexpensive. A simple repeating XOR (i.e. using the same key for xor operation on the whole data) cipher is therefore sometimes used for hiding information in cases where no particular security is required.

If the key is random and is at least as long as the message, the XOR cipher is much more secure than when there is key repetition within a message.[3] When the keystream is generated by a pseudo-random number generator, the result is a stream cipher. With a key that is truly random, the result is a one-time pad, which is unbreakable even in theory.

In any of these ciphers, the XOR operator is vulnerable to a known-plaintext attack, since plaintext ${\displaystyle \oplus }$ ciphertext = key. It is also trivial to flip arbitrary bits in the decrypted plaintext by manipulating the ciphertext. This is called malleability.

## Example implementation

Example using the Python programming language.[4]

```from __future__ import print_function, unicode_literals
from os import urandom

def genkey(length):
"""Generate key"""
return urandom(length)

def xor_strings(s, t):
"""xor two strings together"""
if isinstance(s, str):
# Text strings contain single characters
return "".join(chr(ord(a) ^ ord(b)) for a, b in zip(s, t))
else:
# Python 3 bytes objects contain integer values in the range 0-255
return bytes([a ^ b for a, b in zip(s, t)])

message = 'This is a secret message'
print('message:', message)

key = genkey(len(message))
print('key:', key)

cipherText = xor_strings(message.encode('utf8'), key)
print('cipherText:', cipherText)
print('decrypted:', xor_strings(cipherText, key).decode('utf8'))

# verify
if xor_strings(cipherText, key).decode('utf8') == message:
print('Unit test passed')
else:
print('Unit test failed')
```


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# barabasi.game

0th

Percentile

##### Generate scale-free graphs according to the Barabasi-Albert model

The BA-model is a very simple stochastic algorithm for building a graph.

Keywords
graphs
##### Usage
barabasi.game(n, m = NULL, out.dist = NULL, out.seq = NULL, out.pref = FALSE,
directed=TRUE)
##### Arguments
n
Number of vertices.
m
Numeric constant, the contant out-degree of the vertices.
out.dist
Numeric vector, the distiribution of the out-degree.
out.seq
Numeric vector, the out-degree sequence of the vertices
out.pref
Logical, if true the total degree is used for calculating the citation probability, otherwise the in-degree is used.
directed
Whether to create a directed graph.
##### Details

The out-degree of the vertices is determined by the m, out.dist and out.seq parameters, only at most one of these should be not NULL. If all of these are NULL, then constant out-degree one is used. If m is not NULL, it specifies the constant out-degree.

Otherwise if out.dist is not NULL, it is used to generate an out-degree sequence, the maximum out-degree is length(out.dist)-1 and the probability of having out-degree $k$ is out.dist[k+1]/sum(out.dist).

Otherwise if out.seq is not NULL it is used directly as the out-degrees of the vertices. Its length should be exactly n.

##### Value

• A graph object.

##### References

Barab�si, A.-L. and Albert R. 1999. Emergence of scaling in random networks Science, 286 509--512.

random.graph.game
g <- barabasi.game(1000)
degree.distribution(g)


text:
European Numerical Mathematics and
Advanced Applications Conference 2019
30th sep - 4th okt 2019, Egmond aan Zee, The Netherlands
15:45   Isogeometric Analysis and Fluid Structure Interaction (Part 3)
 15:45 25 mins A minimal set of unisolvent weights for high-order Whitney forms on simplices Ana Alonso Rodriguez, Ludovico Bruni Bruno, Francesca Rapetti Abstract: Whitney elements on simplices are widely used finite elements in computational electromagnetics (see [2] and [1]) as they offer the simplest construction of polynomial discrete differential forms on simplicial complexes. In the lowest order case (that is, coefficients are polynomials of degree one), the associated degrees of freedom are the integrals of such k-forms on the k-simplices of the mesh and consequently bear some physical information. However, the classical degrees of freedom used for high order Whitney forms (see, e.g., [4]), the so-called moments, lack this meaning. For this reason, in [5] an alternative set of degrees of freedom, which we refer to as weights, is introduced. Namely, such degrees of freedom read as the integrals of k-forms on the small k-simplices that are built upon the principal lattice of the elements. In [3] it is proved that this set of weights induces unisolvent degrees of freedom. However they are redundant: their number exceeds the dimension of the space of differential forms. In this work, we show that a particular subset of this weights with cardinality equal to the dimension of the space of differential forms is still unisolvent. 16:10 25 mins Goal-Oriented A Posteriori Error Estimates for Elliptic Problems discretized by the Discontinuous Galerkin Method Ondřej Bartoš, V. Dolejší Abstract: We deal with the numerical solution of linear elliptic problems using the discontinuous Galerkin method with focus on the goal-oriented a posteriori error estimates. The aim is to estimate the error of the quantity of interest represented by a linear functional. The abstract error estimate is based on the knowledge of the (exact) solution z of the dual problem corresponding to the primal one. In order to de_ne a computable estimate, the dual solution z has to be replaced by its approximation z+ h and the remaining term represented by di_erence z 􀀀 z+ h is usually neglected. In this presentation we propose an approach which estimates the neglected term. Consequently, we are able to derive a guaranteed error estimate. This type of estimates are known for the conforming _nite element approximation but it is new in the framework of discontinuous Galerkin method. Moreover, we introduce an e_cient implementation approach allowing a simultaneous solution of the primal and dual problems. Finally, several numerical examples will be presented. 16:35 25 mins Monotonicity considerations for stabilized DG cut cell schemes for the unsteady advection equation Florian Streitbürger, Christian Engwer, Sandra May, Andreas Nüßing Abstract: For solving unsteady hyperbolic conservation laws on cut cell meshes, the so called small cell problem is a big issue: one would like to use a time step that is chosen with respect to the background mesh and use the same time step on the potentially arbitrarily small cut cells as well. For explicit time stepping schemes this leads to instabilities. In a recent preprint [arXiv:1906.05642], we propose penalty terms for stabilizing a DG space discretization to overcome this issue for the unsteady linear advection equation. The usage of the proposed stabilization terms results in stable schemes of first and second order in one and two space dimensions. In one dimension, for piecewise constant data in space and explicit Euler in time, the stabilized scheme can even be shown to be monotone. In this contribution, we will examine the conditions for monotonicity in more detail. 17:00 25 mins Finite Element Approximation of the Spectrum of the Curl Operator Ana Alonso Rodriguez, Jessika Camano, Rodolfo Rodriguez, Alberto Valli, Pablo Venegas Abstract: The presentation concerns the formulation and analysis of the eigenvalue problem for the curl operator in a multiply connected domain and its numerical approximation by means of finite elements. Different boundary conditions that render the curl operator self-adjoint are taken into account. I will focus on those boundary conditions that in particular impose ${\rm curl} \, {\bf u} \cdot {\bf n}=0$. In the case of a multiply connected domain, different contributions from co-homology spaces distinguish different self-adjoint extensions. This distinction can be made by imposing the vanishing of the line integrals on suitable homological cycles lying on the boundary. A saddle-point variational formulation is devised and analyzed, and a finite element numerical scheme is proposed. It is proved that eigenvalues and eigenfunctions are efficiently approximated and some numerical results are presented in order to assess the performance of the method.


text:
• MOHIT TANNARANA

Articles written in Pramana – Journal of Physics

• Fabrication and photoresponse of $\rm{n-WS_{2}/p-V_{0.25}W_{0.75}Se_{2}}$ van der Waals heterojunction

Transition metal dichalcogenides (TMDCs) have shown tremendous potential for application in the field of optoelectronics owing to their extraordinary characteristics. The $\rm{WS_{2}/V_{0.25}W_{0.75}Se_{2}}$ van der Waals heterostructure was fabricated by layer transfer technique and its $I – V$ characteristic was measured at room temperature. The fabricated pn-junction heterostructure shows obvious current rectification with a rectification ratio of $\thicksim{39}$ at $\pm{1}$ V. The heterostructure was analysed in the dark and under polychromatic illumination. The noticeable rise in reverse current is observed at higher intensity of illumination. The photocurrent and photoresponsivity are found to be enhanced as intensity and bias voltage are increased. The higher value of the ideality factor of $\thicksim{2}$ is attributed to the inhomogeneity of the heterojunction.

• A study of optical band gap in GeS$_x$Se$_{1−x}$ ($x = 0, 0.5, 1$) single crystals grown using chemical vapour transport technique

Two-dimensional transition metal chalcogenides like GeSe, GeS and SnSe are widely used in a variety of electrical and optoelectrical applications. Alloying becomes the most important tool to alter the structural, optical and electrical properties of the material. Here, efforts have been applied to grow the crystals of GeS$_x$Se$_{1−x}$ ($x = 0, 0.5, 1$) using iodine (I$_2$) as a transporting agent by the chemical vapour transport technique. The elemental confirmation of grown crystals was done by the energy-dispersive analysis of X-rays. The lattice parameters were obtained from powder X-ray diffraction patterns of all the grown compounds. For the optical study of grown compounds, UV–Vis spectroscopy was performed in the wavelength range of 700–1450 nm. The optical absorption process was studied in detail using the direct and the indirect transitions from two- and three-dimensional models. Moreover, band-gap modification by incorporating sulphur in different concentrations is studied. It is found that the band gap increases with increasing sulphur content in germanium selenide crystals.

• # Pramana – Journal of Physics

Volume 95, 2021
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019


text:
### M - Matrix Parity

##### Languages: C, C++, Java, Python, ... (details)

We call a matrix of integers even if the sum of the numbers in each row is even and the sum of the numbers in each column is even.

In the following examples, matrices $A$, $B$ and $C$ are even. Matrix $D$ is not even because the second column and the last row have odd sums.  Matrix $E$ is not even because its two columns have odd sums.

Given a matrix $A$, convert it into an even matrix by performing as few as possible of the following type of operation:

• Select a cell of the matrix and increment its value by 1. You may select the same cell in different operations.

Print any final even matrix that results of executing the minimum number of operations.

#### Input

The first line contains two integers $r$ and $c$ $(1 \le r, c \le 50)$. Following $r$ lines define matrix $A$. Each line contains $c$ integers with values between $0$ and $100$. The $j$-th number of the $i$-th line is the cell $A_{i, j}$ of the matrix.

#### Output

Print any even matrix that results from performing the minimum number of operations on matrix $A$.  Then print $r$ rows, with $c$ numbers in each row, describing the matrix after applying the operations.

#### Sample test(s)

Input
2 3 1 9 3 2 8 9
Output
2 9 3 2 9 9
Input
2 5 1 2 3 4 5 3 1 4 1 5
Output
1 2 4 4 5 3 2 4 2 5
Input
1 2 2 6
Output
2 6


text:
11 months left
Glory!

# Problem description

Welcome to Taï National Park, data scientists! In this challenge, your goal is to classify the species that appear in camera trap images collected by our research partners at the Wild Chimpanzee Foundation and the Max Planck Institute for Evolutionary Anthropology. As mentioned in the about page, camera traps are one of the best tools available to study and monitor wildlife populations, and the enormous amounts of data they provide can be used to track different species for conservation efforts—once they are processed.

Thanks to our partners, we have a trove of images from camera traps located in different sites around Taï National Park. There are seven types of critters captured in this set of images: birds, civets, duikers, hogs, leopards, other monkeys, and rodents. There are also images that contain no animals. Your job is to build a model that can help researchers predict whether an image contains one of these seven types of species. Let's predict!

## The features in this dataset

For this challenge, you are provided with images, along with a few attributes of each image that might be helpful in setting up your training and testing sets. The images for the training and testing sets are in train_features and test_features respectively, and additional information about each image is in train_features.csv and test_features.csv. Let's take a look at a few of our featured critters!

### Dataset

Each record in the dataset corresponds to a single image captured by a camera trap. Each record has one .jpg file associated with it in either the train_features or test_features directory, depending on which dataset it is a part of. Each image is accompanied by additional information in train_features.csv and test_features.csv, which contain the following fields:

• id (string, unique identifier) - a unique identifier for each image
• filepath (string, feature) - image path including its split directory (train or test)
• site (string, feature) - the site in which the image was taken

There is no overlap between the sites in the training data and the sites in the testing data, which means that the model you build must perform well in new contexts. You may use this site feature to create your own train and test set with disjoint sites to make sure your model can make good predictions on sites it has not been trained on.

Note: External data not provided through the challenge is not allowed for use in this competition. However, participants can use pre-trained computer vision models as long as they are publicly available freely and openly without any use of challenge data.

### Feature data example

Here is an example of a single image in our dataset - ZJ000048, which captures a monkey in the act of looking cute:

Here is the associated metadata for this image in train_features.csv:

id ZJ000048 train_features/ZJ000048.jpg S0025

Here is an example of a blank image (ZJ000144), where there are no animals detected:

These are its accompanying characteristics in train_features.csv:

id ZJ000144 train_features/ZJ000144.jpg S0013

Finally, here is an image from test_features (ZJ016556)—unlike the previous two images, which were a part of the training set, there is no test_labels.csv file to look up for this one! Your model will have to do the work:

Here is the information we have from test_features.csv:

id ZJ016556 test_features/ZJ016556.jpg S0086

### Tips for working with images as features

Working with images as features can be an exciting challenge. Here are some tips to consider as you begin:

#### Generalizing to new sites

As noted above, a model may erroneously learn to predict a species class based on characteristics of the environment, rather than characteristics of the species itself.

In order to make sure that our models predict well in new contexts, it is important that train and test sets have entirely different environments. In this case, we ensure this by making sure that sites are entirely in the train set or the test set, but never in both. We recommend that you take a similar approach in setting up your own splits of the training set.

#### Image transformations

Further, there are many differences to account for for images taken even within the same site. For example, consider these different images of our leopard friends:

 Here is an image of a leopard on the prowl (ZJ000102): and here is an image of a leopard hiding out (ZJ000090). Here is an image of a leopard investigating the camera (ZJ000097), and here is an image of the leopard waving goodbye (ZJ000253).

As you can see, there are significant differences in our images of leopards. The animals may be close or far from the camera, in the sun or in the shadows, or facing toward or away from the lens, among other variations. There are also differences in the color of the image, the weather conditions it was taken, and the type of camera.

In order to teach the model that these images are all of leopards despite these differences, it can be helpful to perform a preprocessing step called image augmentation. Image augmentation involves transforming the training set in multiple ways—rotating, distorting in color or sharpness, zooming in or out, are a few examples. These manipulations of the image can help the model make correct predictions in contexts it has not been exposed to before.

## Labels

There are eight possible labels for every image. If the image does not contain any animals, it is labeled as blank. Otherwise it must be labeled as containing one of the seven species groups included in our dataset:

1. antelope_duiker
2. bird
3. civet_genet
4. hog
5. leopard
6. monkey_prosimian
7. rodent

Note that each image should be associated with one class, since each image contains at most one animal. (Taï National Park is of course a great place to make friends, but we have chosen only those images that capture a single species class). The train_labels.csv file therefore has a single row per image, containing the following values:

• id (string, identifier): the unique identifier corresponding to the image
• antelope_duiker, bird, blank, civet_genet, hog, leopard, monkey_prosimian, rodent (float, target variables): one column for each of the possible predicted classes that indicates whether the image is of that class (1) or not (0). Each row will add up to 1, since the classes are mutually exclusive (there are no images in this dataset that have more than one species class in them).

Here are the first few rows of train_labels.csv:

id antelope_
duiker
bird blank civet_
genet
hog leopard monkey_
prosimian
rodent
ZJ000000 0.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0
ZJ000001 0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.0
ZJ000002 0.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0
ZJ000003 0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.0
ZJ000004 0.0 0.0 0.0 0.0 0.0 1.0 0.0 0.0

## Submission and evaluation

For this challenge, you will submit the probability that an image belongs to each of the eight possible classes for each image in test_features. Your performance will be based on how correctly you are able to identify the animals and blanks in these images.

### Submission format

The submission format contains nine columns:

• id (string): unique identifier for each row from test_features.csv
• antelope_duiker, bird, blank, civet_genet, hog,leopard, monkey_prosimian, rodent (float, target variables): there is a column for each of the eight possible classes containing the probability that the image is of that class.

The predictions for the target variables should be float probabilities that range between 0.0 and 1.0. Each row should add up to 1, since the classes are mutually exclusive. In submission_format.csv, the target columns are currently filled in with a random placeholder probability. To create a submission, overwrite these eight columns with your predictions for the respective classes. Your final submission should resemble train_labels.csv in form, but will be for the ids in test_features.csv.

For example, if you made the following predictions:

id antelope_
duiker
bird blank civet_
genet
hog leopard monkey_
prosimian
rodent
ZJ016488 0.048233 0.189185 0.044914 0.199588 0.106118 0.132915 0.166410 0.112637
ZJ016489 0.097078 0.061400 0.026409 0.241530 0.144344 0.051780 0.287811 0.089648
ZJ016490 0.124658 0.089101 0.189225 0.174494 0.180540 0.079995 0.085672 0.076314
ZJ016491 0.109966 0.048397 0.055598 0.323600 0.322356 0.063252 0.008160 0.068671
ZJ016492 0.165742 0.184610 0.005431 0.136806 0.000389 0.122078 0.151521 0.233423
...
 ZJ020947 ZJ020948 ZJ020949 ZJ020950 ZJ020951 0.143675 0.185103 0.109074 0.158833 0.083497 0.010513 0.155293 0.154011 0.068913 0.150572 0.197232 0.124579 0.131486 0.152534 0.039096 0.135588 0.098239 0.149066 0.168635 0.086321 0.136863 0.077155 0.12148 0.162241 0.036345 0.115411 0.302392 0.072757 0.105491 0.183422 0.106307 0.077875 0.118298 0.093348 0.088412 0.110691 0.129448 0.009582 0.243225 0.206995

the .csv file that you submit would look like this:

id,antelope_duiker,bird,blank,civet_genet,hog,leopard,monkey_prosimian,rodent
ZJ016488,0.048233,0.189185,0.044914,0.199588,0.106118,0.132915,0.16641,0.112637
ZJ016489,0.097078,0.0614,0.026409,0.24153,0.144344,0.05178,0.287811,0.089648
ZJ016490,0.124658,0.089101,0.189225,0.174494,0.18054,0.079995,0.085672,0.076314
ZJ016491,0.109966,0.048397,0.055598,0.3236,0.322356,0.063252,0.00816,0.068671
ZJ016492,0.165742,0.18461,0.005431,0.136806,0.000389,0.122078,0.151521,0.233423

...

ZJ020947,0.143675,0.185103,0.109074,0.158833,0.083497,0.010513,0.155293,0.154011
ZJ020948,0.068913,0.150572,0.197232,0.124579,0.131486,0.152534,0.039096,0.135588
ZJ020949,0.098239,0.149066,0.168635,0.086321,0.136863,0.077155,0.12148,0.162241
ZJ020950,0.036345,0.115411,0.302392,0.072757,0.105491,0.183422,0.106307,0.077875
ZJ020951,0.118298,0.093348,0.088412,0.110691,0.129448,0.009582,0.243225,0.206995

### Performance metric

To measure your model's accuracy by looking at prediction error, well use a metric called log loss. This is an error metric, so a lower value is better (as opposed to an accuracy metric, where a higher value is better). Log loss can be calculated as follows:

$$loss = -\frac{1}{N}\cdot\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{M} y_{ij}\log p_{ij}$$

where $N$ is the number of observations, $M$ is the number of classes (in terms of our problem $M=5$), $y_{ij}$ is a binary variable indicating if classification for observation $i$ was correct, and $p_{ij}$ was the user-predicted probability that label $j$ applies to observation $i$.

In Python you can easily calculate log loss using the scikit-learn function sklearn.metrics.log_loss. R users may find the MultiLogLoss` function in the MLmetrics package.

## Good luck!

Good luck and enjoy this problem! If you have any questions you can always visit the user forum for this competition. Happy predicting!


text:
# connected space compact space

## Conneted space

If there are disjoint open subsets $$U \cup V = X$$. The space X is disconnected. Otherwise connected.

## Path connected space

Every elements $$x,y \in X$$ have image of $$f:[a,b] \rightarrow X$$ that $$f(a) = x, f(b)=y$$ is in $$X$$.

## Component

Relation class $$[x]$$ is a subset of of $$X$$ those elements is also elements of connected subspaces of containing $$x$$. Connected space has only one component. $$\mathbb{R}$$ except $${0}$$ is disjoint space and has two componets $${(-\infty,0)}$$ and $${(0, \infty)}$$.

## Compact space

Every open cover has at least one finite open subcovers. $$[0,a)$$ is open in subspace $$[0,1]$$ of $$\mathbb{R}$$. Subset $$A \subset X$$ topology $$\tau'$$ is $$\tau \cap A$$.


text:
# Math Help - Infinite Surd Help

1. ## Infinite Surd Help

Hi there, first time poster here.

I'm having trouble with infinite surds.
I understand that as $n$ approches infinity $an -an+1$ approaches zero.
I understand that the limit of the sequence is the exact value of the infinite surd.
I also know how to solve the surd by setting it equal to x and using quadratics.

But what I don't understand is how to connect any of it. I can't explain how to get from $an -an+1$ to the exact value.

In other words, I'm at point A and I know where point B is, I just don't know how to get there.

2. I'll simply redirect you toward the following thread: http://www.mathhelpforum.com/math-he...nite-surd.html

Hope this helps!

3. I appreciate the help, but I already knew about that method. What I don't understand is how I'm supposed to reach that method using only =0 as n approaches infinity.

4. Originally Posted by bookjunkie24
I appreciate the help, but I already knew about that method. What I don't understand is how I'm supposed to reach that method using only =0 as n approaches infinity.
The equation you're posting makes no sense: an - an = 0 ..... Please post the correct equation using appropriate notation.


text:
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latex author email affiliation

# The Blog

Latest news

### latex author email affiliation


text:
Lemma 8.5.2. Let $\mathcal{C}$ be a site. Let $p : \mathcal{S} \to \mathcal{C}$ be a category over $\mathcal{C}$. The following are equivalent

1. $\mathcal{S}$ is a stack in groupoids over $\mathcal{C}$,

2. $\mathcal{S}$ is a stack over $\mathcal{C}$ and all fibre categories are groupoids, and

3. $\mathcal{S}$ is fibred in groupoids over $\mathcal{C}$ and is a stack over $\mathcal{C}$.

Proof. Omitted, but see Categories, Lemma 4.34.2. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
## 3M Forms Strategic Relationship with Amperex Technologies for NMC Li-Ion Cathode Materials

##### 23 March 2009
 SEM images of particle morphology of 3M NMC cathode materials. Source: 3M. Click to enlarge.

3M has entered into a strategic relationship and an agreement with China-based Amperex Technologies Ltd. (ATL) that will expand the use of nickel-manganese-cobalt (NMC) cathode materials in lithium-ion batteries targeted at consumer electronics and automotive applications.

3M will become a strategic supplier of NMC cathode materials to ATL and will work jointly with ATL to develop new cathode, anode and electrolyte materials for the lithium-ion battery industry. Under the agreement, 3M granted ATL a license to 3M intellectual property for the use of 3M NMC cathode materials in lithium-ion batteries.

 Charge/discharge profiles of 3M NMC cathode materials. Source: 3M. Click to enlarge.

3M battery cathodes are mixed metal layered oxides composed of nickel, manganese, and cobalt. They have improved thermal stability and up to six times less cobalt for reduced price volatility when compared to the industry standard lithium cobalt oxide. 3M currently offers three cathode products: BC-618, BC-718 and BC-723. 3M has developed these proprietary cathode powder formulations with precise control of structure and compositions tailored for both energy and power applications to serve the consumer electronics and automobile industries.

In 2007, the US Department of Energy awarded 3M up to $1.14 million (total DOE/industry cost share:$ 2.28 million) for a two-year project to screen nickel manganese cobalt cathode materials through building and testing of small-sized cells. (Earlier post.) The award was part of a set of five targeted at plug-in hybrid electric vehicle (PHEV) research under DOE’s collaboration with the United States Advanced Battery Consortium (USABC).

Working with ATL is a great opportunity for 3M to apply its advanced materials in next-generation lithium-ion batteries for higher capacity and improved thermal stability, important attributes for electronics and transportation.

—Chris Milker, Business Manager, 3M Electronics Markets Materials Division

NMC materials in general offer a capacity of about 170 mAh g-1 when cycling to 4.3 V vs. lithium, but have the ability to increase their capacity to levels approaching 250 mAh g-1 at higher potential. Achieving these high capacities currently results in a decrease in the life of the battery, however. In a Letter of Inquiry seeking prospective researchers for FY 2008 work with the Batteries for Advanced Transportation Technologies Program, Lawrence Berkeley National Laboratory noted that:

While considerable progress has been made in identifying new cathodes that possess energy that exceeds those presently available, the enhanced capacity is only accessible at potentials that are outside the stability window of presently available electrolytes....A more stable cathode-electrolyte interface is crucial to achieve the capacities needed to bring PHEVs to the market.

Patents on the 3M composition have been issued in China, Taiwan and USA. Applications are pending in EP, Hong Kong, Korea, and Japan. 3M battery materials product and technology portfolio includes cathode powders, anode powders, specialty electrolyte salt/additives, and heat transfer fluids.

Amperex Technologies Limited supplies lithium-ion polymer battery to name-brand companies worldwide. ATL, headquartered in Hong Kong, has manufacturing facilities in China.

In 2002, ATL entered an OEM agreement with Valence to manufacture the company’s lithium-ion phosphate batteries. For fiscal year 2008, the ATL agreement was responsible for the bulk of Valence’s $586,000 in licensing and royalty revenue (down from$703,000 in FY 2007 and \$724,000 in FY 2006).


text:
Home > maths > Rules of Indices

## Rules of Indices

Indices are numbers that are “to the power of” another number often written in the form ab. This is usually taken to mean a x a b times eg
23 = 2 x 2 x 2 = 8
–a is multiplies by itself b times

Note//In more advanced maths ab is often taken to mean exp(b ln(a))

There are a number of rules regarding how to manipulate indices of which the most important are listed below:

1. ab x ac = ab+c
since we have axa b times time axa c times giving axa b+c times
2. ab ÷ ac = ab-c
by similar logic to point 1
3. (ab)c = abc
since (ab)c = ab x ab… c times…ab
but by (1) we get ab+b+…+b = abc
4. a1/b = b√a
Since by (3) (a1/b)b = ab/b = a
but re arranging we get the result
a1/b = b√a

If there any other rules that I haven’t included and aren’t immediately obvious from the above rules please leave them in the comments below

By David Woodford

Categories: maths Tags: , ,
1. October 2, 2009 at 2:18 pm

How would one solve the equation X to the power A multiplied by Y to the bower B

2. October 2, 2009 at 6:25 pm

you cant solve the equation unless it is equal to somthing. If you have
$X^A Y^B = Z$
then
$X = (ZY^{-B})^{\frac{1}{A}}$

3. October 12, 2009 at 11:30 pm

Btw, I forgot to say thanks for that. It really helped so I thought I’d come back and say thank you.


text:
# Suppose that a recursive routine were invoked to calculate F(4). How many times would a recursive call of F(1) occur?

Although you haven't said so explicitly, it seems as though you are talking about Fibonacci sequences, since the usual notation for that in Mathematics/Computer Science classes is F(n), and the routines are recursive.  The Fibonacci sequence is defined as:

`F(n)=F(n-1)+F(n-2)` , where `F(2)=1` and `F(1)=1` .

which means that

`F(4)=F(3)+F(2)`

And we can see that the recursive calls continue as

`F(3)=F(2)+F(1)`

This means that the call of `F(1)=1` is called only once.

Approved by eNotes Editorial Team


text:
### interlude's blog

By interlude, history, 12 months ago,

Hello Codeforces!

CQXYM and I are glad to invite you to Codeforces Round #745 (Div. 1) and Codeforces Round #745 (Div. 2), which will be held on Sep/30/2021 13:15 (Moscow time). Note the unusual time of the round.

Each division will have 6 problems and 2 hours to solve them. All problems were written and prepared by CQXYM and me. The round will be rated for both divisions.

We would like to thank:

This is our first round, and great efforts have been put into preparing this round. Were you to kindly participate in this round, we would be very grateful and hope you will enjoy it.

Good luck!

UPD: Here are the scoring distributions:

Div. 1: $500$ — $1000$ — $1750$ — $2000$ — $3500$ — $3500$

Div. 2: $500$ — $1000$ — $1500$ — $1750$ — $2500$ — $2750$

And we would also like to thank Um_nik for testing the round.

UPD2: Editorial is out!

UPD3:

## Winners

Congratulations to the winners:

Div1:

Div2:

• -1232

 » 12 months ago, # | ← Rev. 2 →   -155 As a tester I would encourage you to participate in this round and assure you that the problems are interesting and good .Update :- My review to testers
•  » » 12 months ago, # ^ |   0 Thanks.
•  » » 12 months ago, # ^ |   +195 As a participant I will never trust your word again :D
•  » » » 12 months ago, # ^ |   +122 As a Chinese participant I will never participate in Chinese rounds again :D
•  » » 12 months ago, # ^ |   +49 my face after reading problem A :
•  » » » 12 months ago, # ^ |   +13 You should use a spoiler
•  » » 12 months ago, # ^ |   -6 So u really Trolled us!.
•  » » 12 months ago, # ^ |   +46 it is surely the worst contest I have taken part in
•  » » 12 months ago, # ^ |   +6 You were wrong
•  » » 12 months ago, # ^ | ← Rev. 2 →   +19 Here is my review I gave to problem setters . Name is blurred due to security reasons .review
•  » » » 12 months ago, # ^ |   +27 Then you shouldn't have said all all these things (your comment)in the first place !
•  » » » 12 months ago, # ^ | ← Rev. 5 →   +5 your review is good.problem B has too many situations and what makes it worse is K (Diameter) might be negative.and presets weren't sufficient to show wrongs during contest time (intended).
•  » » 12 months ago, # ^ |   0 wtf???forces
 » 12 months ago, # |   +27 Here it is!
 » 12 months ago, # |   -57 Mathforces?
•  » » 12 months ago, # ^ |   -73 Mathforces!
•  » » » 12 months ago, # ^ |   +25 It was FSTforces!
•  » » 12 months ago, # ^ |   +54 Sorry but I don't understand.
•  » » » 12 months ago, # ^ |   -59 Just asking
•  » » » 12 months ago, # ^ | ← Rev. 2 →   +75 I think he might be implying that cf contests prepared by Chinese authors tend to have more math problems. :)
•  » » 12 months ago, # ^ |   -16 totally agree after reading 1st problem.
•  » » 12 months ago, # ^ |   +40 20k+ users registered and only around 3500 submissions, weird...
•  » » » 12 months ago, # ^ |   +25 They didnt have noticed the unusual time
•  » » 12 months ago, # ^ |   +6 You were so ******* right my guy. Damn
•  » » 12 months ago, # ^ |   0 Hardforces!
 » 12 months ago, # | ← Rev. 2 →   +40 Another Chinese round and friendly time for Chinese! Looking forward to great problems and having fun!
 » 12 months ago, # |   -61 Sad, why isn't Monogon a VIP tester :(
•  » » 12 months ago, # ^ |   +170 I'm only moderately important this time.
•  » » 12 months ago, # ^ |   -81 What? 1-gon's comment didn't hit 100 upvotes in a day?? Impossible!)
 » 12 months ago, # |   +3 This time is good for me.Orz CQXYM!
•  » » 12 months ago, # ^ |   -31 Orz,2000 scores for a round!
•  » » » 12 months ago, # ^ |   -25 Orz purple Hearth
•  » » 12 months ago, # ^ |   -29 Orz zqs!
 » 12 months ago, # | ← Rev. 2 →   -58 Second Shortest announcement after Tourist's Round announcement Since I started using CF.
 » 12 months ago, # |   -50 Bit manipulation problems?
 » 12 months ago, # |   +6 I hope.
 » 12 months ago, # |   +65 I don't have anything against Chinese rounds, but... really, morning on a weekday?
•  » » 12 months ago, # ^ |   +154 Probably people in different time zones exist.
•  » » » 12 months ago, # ^ |   +95 That's why Saturdays exist.
•  » » » » 12 months ago, # ^ |   +99 ICPC World Finals I guess.
•  » » 12 months ago, # ^ |   +30 In China it is 6:05p.m. on the day before the National Day holiday. You can imagine the experience of participating a CF round after school :)
•  » » » 12 months ago, # ^ |   +3 But (maybe?) most middle school students have to stay at school and study at night.
•  » » » » 12 months ago, # ^ |   +9 It often happens on weekdays. But if tomorrow is holiday, students can go back home in the afternoon.
•  » » 12 months ago, # ^ |   +23 I think that it's because the National Day holiday in China is coming.However authors don't know 6 p.m. is even worse than 10 p.m. for many Chinese because most of schools end at 9 :(
•  » » » 12 months ago, # ^ |   -34 However we must take the fact that ICPC World Final will be held today into consideration. Thus we choose 6 p.m. over 9 p.m.
 » 12 months ago, # | ← Rev. 2 →   +7 Rare time....but maybe just right for Chinese contestants like me. I'm a little EXCITED. Hope all of us can enjoy it!
 » 12 months ago, # |   +21 I finnish school 5 minutes before the round, can't join, but I wish those who read the greatest of luck.
•  » » 12 months ago, # ^ |   +71 I managed to convince the teacher to let me compete at school.
 » 12 months ago, # |   +3 Big thanks for the time of the round from Asia! :)
 » 12 months ago, # |   +12 Great, 3:05 AM.
•  » » 12 months ago, # ^ |   0 i guess that's an i for an i for us having to participate from 10:30pm to almost 1am before...
 » 12 months ago, # |   +27 interlude I think we need to update this blog to make this appear on top of the div.3 contest. Many people might not see it. Especially the unusual start time of the round. I didn't see this blog existed. Because the first post from the top was round 744. thanks for considering. :) good luck to the people who are participating.
 » 12 months ago, # |   +72 lighthearted meme
•  » » 12 months ago, # ^ | ← Rev. 3 →   -15 .
•  » » 12 months ago, # ^ |   -18 Hahahahhahahhahaa!!!Although no, it's not funny. You idiot.
•  » » » 12 months ago, # ^ |   +3 You High on Weed . :) __ | | | | | | | | ( | | | | )Hope you Understand my feeling.
•  » » » 12 months ago, # ^ |   -10 That's not a good idea to play jokes like this on the Internet. We can sure that there was something wrong with the education you get from your parents.
•  » » » » 12 months ago, # ^ |   -15 заткнись телочка.еще что-то скажешь про моих родителей, зубы выбью тебе, тварь.
•  » » 12 months ago, # ^ |   +66 Meanwhile me
•  » » » 12 months ago, # ^ | ← Rev. 2 →   -26 Hahahahhahahhahaa!!!Although no, it's not funny. You idiot.
•  » » » » 12 months ago, # ^ |   -27 Hahahahhahahhahaa!!!Although no, it's not funny. You idiot.
•  » » » » » 12 months ago, # ^ |   -26 Hahahahhahahhahaa!!!Although no, it's not funny. You idiot.
•  » » » » » » 12 months ago, # ^ |   +4 let me stop you guys :)
•  » » » » » » 12 months ago, # ^ |   0 break XD;
•  » » » 12 months ago, # ^ |   +10 Imagine having school online, and not offline
•  » » 12 months ago, # ^ |   +31 Me:
•  » » » 12 months ago, # ^ |   +1 Now that the contest is over, are you happy with your decision?
•  » » » » 12 months ago, # ^ |   +3 i guess yes as he got a very good rank in this one
•  » » 12 months ago, # ^ |   +11 I have a test in my class today that completely overlaps with the contest timings. Cant skip them even if I want to :sobb:
 » 12 months ago, # |   +6 Another great round!! Hope I can do better than before in it.
 » 12 months ago, # |   0 Hi there. Could the contest be at 18.35?
•  » » 12 months ago, # ^ |   +3 The round authors are Chinese; I assume starting at 22:35 is not comfortable to them.
•  » » » 12 months ago, # ^ |   +15 Oh, I am sorry I didn't think that. Thanks.
 » 12 months ago, # |   +6 Hoping to become Pupil after this round...Best of Luck everyone!!
•  » » 12 months ago, # ^ |   +8 you did it! Congrats :D
•  » » » 12 months ago, # ^ |   0 Thanks a lot!!!!
 » 12 months ago, # |   +5 As a Chinese,I usually leave school at 6:30 PM,so I cannot enjoy this round :(
•  » » 12 months ago, # ^ |   +11 Leave school, but don't leave a contest. (Don't take it seriously.)
•  » » 12 months ago, # ^ |   +10 And now I think I am lucky :D.
 » 12 months ago, # |   -26 i do remember that one of the past contest nearly at the same time as of this one and it went bad ...........
 » 12 months ago, # | ← Rev. 2 →   -19 Can you somehow postpone the competition for 2 — 3 hours? I think it would be more convenient to write everything in the evening than in the afternoon. usually the competitions are held at 17:30 Moscow time and this is very convenient.
 » 12 months ago, # |   0 Is 1900-2100 no longer in Div 2?
•  » » 12 months ago, # ^ |   +5 People rated <2100 can participate in div2, when it is div2Only round(so no parallel div1), but when there is a parallel div1 round, 1900+ have to participate there. It has been like this for a long time
•  » » » 12 months ago, # ^ |   0 Oh! I see
 » 12 months ago, # |   -51 A way to increase contribution from 0 to 100:0-20: leave stupid comments20-40: write a blog on how to increase contribution from 0 to 2040-60: write a blog on how to increase contribution from 20 to 4060-80: write a blog on how to increase contribution from 40 to 6080-100: write a blog on how to increase contribution from 60 to 80
•  » » 12 months ago, # ^ |   -9 A way to increase contribution from 0 to 126:0-126: leave stupid comments
•  » » » 12 months ago, # ^ |   +7 that leads in either ways UP or DOWN (+126 || -126)
 » 12 months ago, # |   -9 Big thanks for the friendly time of the round:)
•  » » 12 months ago, # ^ |   0 Actually, 9 p.m. is better than 6 p.m. for Chinese students.
 » 12 months ago, # |   0 Starting My Codeforces Journey From your round...Hope to have my first giving round and your first making round great!Thank you
 » 12 months ago, # |   +1 My first div2 round here. Hopefully it doesnt go downhill. fingers crossed. I like this time slot :)
•  » » 12 months ago, # ^ |   +3 This ages very well
 » 12 months ago, # |   0 All the best to all the participants. :))
 » 12 months ago, # | ← Rev. 2 →   +2 From contest authors and testers, I'm waiting really good contest!
 » 12 months ago, # |   -118 Greeting all chinese people!
•  » » 12 months ago, # ^ | ← Rev. 3 →   -27 Idk what's up with everyone, so retracting it for the better!
•  » » 12 months ago, # ^ |   +38 I think the content you posted is not appropriate, because we all know that the Great Firewall in Mainland China blocks almost all "user-generated content" websites, and codeforces as a "user-generated content website" is not blocked by it because There is very little political content on codeforces. If you post more content in this area, the result is likely to be that codeforces is blocked by the Great Firewall of China. As a result, it will cause great inconvenience to users in mainland China. They need to cross the Great Firewall to access codeforces.
•  » » » 12 months ago, # ^ |   -14 Good.
•  » » » 12 months ago, # ^ | ← Rev. 2 →   -87 However, according to some netizens in the QQ group Universal OJ Users, almost everyone in China mainland who use Codeforces can across the Great FirewallBut anyway, I don't think any political content posted on codeforces is appropirate because codeforces is a website for people who love programming. I think whether it will make codeforces blocked or not it's not good to post content which is not about algorithms on codeforces
•  » » » 12 months ago, # ^ | ← Rev. 2 →   -121 Although I heard some netizens say, and I also think whether Codeforces is blocked or not doesn't matter to many codeforces users, I think after all using websites not blocked is more convenient.
•  » » 12 months ago, # ^ |   +72 Lol, this is pretty good, don't know why it's being downvoted
 » 12 months ago, # |   0 can anyone tell me what is the process of penalty in DIV 2 round
•  » » 12 months ago, # ^ |   +5 Technically, there are no 'penalties' on time count. But yes, you get less points if you solve a problem after incorrect attempts.Each problem is assigned points (e.g., Problem A in this round is 500), the faster you solve a problem, the more points you get. The amount of points you can achieve gradually decreases over time.Problem A points deteriorates at a rate of 2 points per minute (you get 498 points if you solve after 1 minute, 496 points if you solve after 2 minutes, 494 points if you solve after 3 minutes and so on)Plus, every incorrect attempt costs you 50 points. Now let's see, if you solve that problem A after 10 minutes on your 2nd attempt, you get (500) - (10*2) for 10 minutes to solve - (50) for one incorrect attempt = 430.If you solve that problem A after 15 minutes on your 3rd attempt, you get (500) - (15*2) for 15 minutes to solve - (50*2) for two incorrect attempt = 370.Additionally, the minimum points you get on any problem is lower bounded by 30% of the max points on that particular problem. e.g., irrelevant of how many incorrect attempts or how much time, you will get a minimum of 500 * 30% = 150 for that problem A.
 » 12 months ago, # | ← Rev. 2 →   +26 Round delayed by 10 minutes
•  » » 12 months ago, # ^ | ← Rev. 2 →   -13 10 mins delay for my first contest...Bad vibes
 » 12 months ago, # | ← Rev. 5 →   0 deleted...
 » 12 months ago, # | ← Rev. 2 →   0 Time added 10 minutes!
 » 12 months ago, # |   0 contest extended by 10 min!
•  » » 12 months ago, # ^ |   0 does it change anything?
 » 12 months ago, # |   +4 Why this sudden addition of ten minutes in starting the round??
•  » » 12 months ago, # ^ |   0 may be due to some technical issues
•  » » 12 months ago, # ^ |   0 To reach 20K+ participations maybe.
 » 12 months ago, # |   +69 Hopefully the delay doesn't signify an unrated round.
 » 12 months ago, # |   +87 now what to do with 10 min of my life??
•  » » 12 months ago, # ^ |   +98 Think about what to do with 10 min of life
•  » » » 12 months ago, # ^ |   -18 Listening to your words is just like listening to your words lol
•  » » » 12 months ago, # ^ |   0 Listen to your words, so I heard your words
•  » » 12 months ago, # ^ | ← Rev. 2 →   0
 » 12 months ago, # |   -8 how many point i need to join to div 1 ???
•  » » 12 months ago, # ^ |   +1 1900 and above
•  » » 12 months ago, # ^ |   0 1900
 » 12 months ago, # | ← Rev. 3 →   +14 I hope it'll be the only problem of the contest.Wish all the best!!!P.S. The "problem" here doesn't mean the problems that we're going to solve in the contest.
 » 12 months ago, # |   +90 This doesn't really matter, but why would you use "obsidian" instead of "black"?
•  » » 12 months ago, # ^ |   +42 Minecraft
•  » » 12 months ago, # ^ |   +42 Because author wanted to complicate the problem statement without any reason.
 » 12 months ago, # | ← Rev. 4 →   +190 If you can't write a nice problemset for cf, just move the problemset to other websites.Terrible contest.edit: Chinese slang
•  » » 12 months ago, # ^ |   -15 ?______?
•  » » » 12 months ago, # ^ |   0 It means he hates the idea provider due to hard problems, but it may be Chinese slang that is very disruptive and not appropriate to show in public.
 » 12 months ago, # |   +31 First time not being able to solve (or not wanting to implement such an ugly solution for) even the first problem. Might just be me, but I think this contest needed a lot more tuning.
•  » » 12 months ago, # ^ | ← Rev. 2 →   -49 Thought it was one of best contests I competed in, besides B and C out of order. Finally some actual implementation and more ideas than just greedy/math with one for loop required...If you can't solve a 30 line solution due to implementation I'm not sure you should be in div1. I've long said there is severe lack of implementation skill for most low div1 users and below.
 » 12 months ago, # |   +14 Hardforces (: (at least for me )
 » 12 months ago, # |   +13 Coding is very hard
 » 12 months ago, # |   +21 div.1 too hard for me :(
 » 12 months ago, # |   +154 Why would you write statement like this: "and the diameter of the graph must be less than $k-1$?"Couldn't you write "diameter of the graph must be less than $k$"?Or even better, "diameter of the graph must be less than or equal to $k$?
•  » » 12 months ago, # ^ |   +13 For real, my code was so messed up I had to rewrite because I got confused in the middle.
 » 12 months ago, # | ← Rev. 5 →   +27 I'm sorry but i am happy because today it's not only me who can't solve a single problem.
 » 12 months ago, # |   +90 How to unregister?
 » 12 months ago, # |   +18 Not my cup of tea...
 » 12 months ago, # |   +13 Damn bro you didn't have to kill us with those problem statements!
 » 12 months ago, # |   +12 Man!! explain me the problems. Cant understand anything from the statements
 » 12 months ago, # |   +7 its my 2nd chinese round and i find it harder the usual div2 coz i can solve avg 2 problem but in this round i can't do even A. Chinese loves maths too much ≧ ﹏ ≦
 » 12 months ago, # |   +7 only 3.7k submissions to problem A, btw WAonpretest2forces :(
 » 12 months ago, # |   +72 At least I'm not alone, here we are :')
•  » » 12 months ago, # ^ |   +1 Getting demoralized when not able to solve any problem.then coming to comments-->turns out to be a natural therapy
 » 12 months ago, # |   0 Not even able to solve one quest, very tough round.but atleast doing no submission won't affect my rating ^-^
•  » » 12 months ago, # ^ |   0 I thought if you presented, no matter have you submit or no, It will affect your rating?
 » 12 months ago, # |   +3 The wording of some of the problems are hard to understand :(
 » 12 months ago, # | ← Rev. 3 →   -8 Bruh, I don't even feel like trying C cause pretty sure it's beyond my capabilities. Spoiler**GUESSFORCES**
 » 12 months ago, # | ← Rev. 2 →   +8 In the last 12 rounds held over the past 3-4 months, this is the only round where I had to spend ~10 minutes on both A and B just to be able to interpret properly what the question wants to say, never mind attempt to solve them. Statements were weirdly put, and maybe difficulty wasn't properly calibrated going by the meagre amount of submissions ~4k for div2 A even after an hour.
 » 12 months ago, # |   0 statement of div2 c is beyond my scope :')
 » 12 months ago, # |   0 one silly mistake costed 4 penalties on problem B :(
•  » » 12 months ago, # ^ |   0 Same :(
 » 12 months ago, # |   +398 Thanks a lot for clarifying this, before the round I was worried that my solution might also be tested on a negative number of test cases.
 » 12 months ago, # |   0 I am even not able to solve B this time, my brain exploded thinking of it :(
•  » » 12 months ago, # ^ |   +3 I am not even able to solve A :/
 » 12 months ago, # |   +155 A flaw in the rating system exposed brutally by this round: in div 1, those who do manage to solve A, but do so late and/or with penalty attempts, will see their rating fall significantly, while those who bail out completely on solving A will see no change to their rating.
•  » » 12 months ago, # ^ |   +73 that's why A should always be comparatively easy in both divisions, can you imagine less than 6k participation in div2 :(
•  » » » 12 months ago, # ^ |   +45 100% agree
 » 12 months ago, # |   0 Me on this contest be like:Click this
 » 12 months ago, # |   +27 Damn, That's the hardest A and C I've ever seen
•  » » 12 months ago, # ^ |   +3 For me B Totaly was pain in ass!
 » 12 months ago, # | ← Rev. 2 →   +1 HardForces :(UPD: Sad, Got FST on Div.2 B :(
 » 12 months ago, # |   0 Today I will get -101 rating change :(
 » 12 months ago, # |   +5 Its really good, but is so hard:(
 » 12 months ago, # |   +4 20000 registers 6000 contestants hmmm makes sense
 » 12 months ago, # | ← Rev. 2 →   +15 Didn't participate in a Div. 0 before, but today i did.
 » 12 months ago, # |   0 I Just guessed the solution for A after 1hr 20 mins of waste analysis and it worked :/ and then solved B under 30 mins , strange round
 » 12 months ago, # |   0 Damn, couldn't even solve A.This contest really was amazing!
 » 12 months ago, # |   +16 Is O(N^3) the intended solution for Div1 A or are you expecting O(N^2)?
•  » » 12 months ago, # ^ |   +3 $O(N^3)$.
•  » » » 12 months ago, # ^ | ← Rev. 2 →   +20 Amazing. I had an O(N^3) time with O(N^3) space initially which got MLE.Optimized it to O(N^3) time with O(N^2) memory and still got TLE. Tried optimizing it further and got WA. What a waste of 2 hours for me.
•  » » » » 12 months ago, # ^ |   +3 I solved it O(N^3 * 16) it gave me TLE but i break when i find solution with value greater than 16 and passed the pretests
•  » » » » 12 months ago, # ^ |   +25 Atleast you were writing codes. I was just crying for 2 hrs T_T
•  » » 12 months ago, # ^ |   0 Yes, my O(n^3) passed
•  » » 12 months ago, # ^ |   0 I think O(n^3)
 » 12 months ago, # |   +5 Are the CF div2 rounds are getting harder or me getting dumber?
•  » » 12 months ago, # ^ |   +19 You just happened to be in the wrong contest at the wrong time
•  » » 12 months ago, # ^ | ← Rev. 2 →   0 I am trying again on CF after spending a long time on Leetcode and DSA style problems. I am thinking to quit. It feels like a waste of time now but I am interested too lol. I don't know. I can spend this time on learning other skills like more backend. Will maybe just try to solve some harder problems for my range in the next few days.
 » 12 months ago, # |   +16 A contest with really high difficulty...
 » 12 months ago, # | ← Rev. 2 →   +3 So is there anyone tell me how to solve Div.1A/Div.2C. I got WA on pretest 18 :(
•  » » 12 months ago, # ^ |   +13 You can discover the property that the answer will not be more than 16. and just do it in brute force. it will be O(n * m * 16 * 16). Hint : try to use the number larger than 16, maybe 17.
•  » » » 12 months ago, # ^ |   0 I found that but failed to have further a further thought. Could you please tell me what the (16 * 16) is?
•  » » 12 months ago, # ^ |   0 You are just checking min answer for the 5x4 matrix. But there may be test cases like 8x8 size 01111110 10000001 10000001 10000001 10000001 10000001 10000001 11111110 In this case, the answer is 0. But according to your approach, the answer will come to more than zero.
 » 12 months ago, # |   +7 Any O(nlogn) solution for Div1C/Div2E? I only came up with sqrt decomposition XD
•  » » 12 months ago, # ^ |   0 wasn't time limit a little too tight for sqrt decomp? :(
•  » » » 12 months ago, # ^ |   +8 O(nsqrt(nlogn)) should work for n=100k.Example: 1540D - Inverse Inversions
•  » » » » 12 months ago, # ^ |   0 It has 5 sec limit. 1 sec limit kept me under the impression that nsqrt(n) is being punished here. anyways my fault :(
•  » » » » 12 months ago, # ^ |   0 Not in this task, not with 1 second time limit, IMO. 1.5 seconds would probably be enough.
•  » » » 12 months ago, # ^ |   +10 I passed system tests with $O(n\cdot \sqrt{n}\cdot \log{n})$.I used a fenwick tree for big values of $x[i]+y[i]$ instead of just maintaining when to add or subtract something.
•  » » » » 12 months ago, # ^ |   +10 Weak systests :/ But I see that this solution was already hacked after contest. My solution with FT was a bit too slow too.
•  » » 12 months ago, # ^ |   -84 I can only come up with $O(n\sqrt{n})$ solution too.
•  » » 12 months ago, # ^ |   +26 The TL may be a bit tight for O(N sqrt(N))
•  » » » 12 months ago, # ^ |   -126 In fact the constant of the solution is very small. I think even $O(n\sqrt{n}log_n)$ solution can also pass.
•  » » » » 12 months ago, # ^ |   +16 And O(n^5) can pass B with some optimization...
•  » » » » » 12 months ago, # ^ |   -130 $O(n^5)$ is the expected complexity.In fact you will find the constant is extreme small. A lot of state is useless.
•  » » » » » » 12 months ago, # ^ |   +36 That's really unexpected.. I thought author may have come up with something smarter to get $O(n^4)$ solution and set $n=100$ to cut $O(n^5)$ one..
•  » » » » » » » 12 months ago, # ^ |   -6 In fact there do have a solution works in $O(n^3log^2n)$, but it's too difficult for D1B.
•  » » » » » » 12 months ago, # ^ |   +146 Well, that's surprising. Why didn't you set $n \leq 50$? It would be much easier to understand that the solution is fast enough and it's worth coding. Was it that important to fail $O(n^6)$ solution?
•  » » » » » » » 12 months ago, # ^ |   +20 Um_nik played these games.
•  » » » » » » » 12 months ago, # ^ |   -65 Sorry but I didn't come up with any $O(n^6)$ solution.
•  » » » » » » » » 12 months ago, # ^ |   +107 Then my question remains: why $n \leq 100$? Only to make some people not implement a correct solution?
•  » » » » » » » » » 12 months ago, # ^ |   -8 "In fact the constant of the solution is very small." -interlude
•  » » » » » » » » » 12 months ago, # ^ |   +18 I understand that the solution is fast enough. It would still be with $n \leq 50$. And this fact would be clear before implementing.
•  » » » » » » » » » 12 months ago, # ^ |   +18 Just because the $O(n^5)$ solution can pass the task and it runs very fast. In fact you will find that the answer will be 0 if $m,k$ is larger than $\frac{n}{2}$.
•  » » » » » » » » » 12 months ago, # ^ |   +45 Not always, my $O(n^5)$ solution got TL.
•  » » » » » » 12 months ago, # ^ |   +38 Are u serious? Only because of this, I didn't code the solution and thought for an O(n^4) solution for almost 1.5 hrs. And the constant isn't that small, recursion won't pass easily without optimizations (returning wherever you can).
•  » » » » 12 months ago, # ^ |   +16 However my O(N sqrt(N)) solution didn't pass.
•  » » » 12 months ago, # ^ |   +5 I failed on both BC for TL
•  » » » » 12 months ago, # ^ |   0 I think your solution for C is slow because of "cin and cout" .
•  » » » » » 12 months ago, # ^ |   +3 I'm sorry but as a matter of fact,coder's submission has got another TL after replace cin/cout with faster IO streams. Isn't it obvious that this failure it due to implementing or Time Limit but not IO ways?
•  » » » » » 12 months ago, # ^ |   0 You are right..my bad
•  » » 12 months ago, # ^ |   0 O(nsqrt(n)), you can operate trains whose x+y>=sqrt(n) and whose x+y
 » 12 months ago, # |   +1 Perfect! Problems are challengable and of benifits. especially C, its great on both Thinking and Realization. All in all, great round.
 » 12 months ago, # |   +2 Ooh man! one of the toughest A in div 2!!!
•  » » 12 months ago, # ^ |   +5 maybe you can just look at sample test cases to find the pattern without thinking much or proving anything . At least that's what I did XD
•  » » » 12 months ago, # ^ |   0 I figured it out with PnC but still I would say it shouldn't have had been an A problem, maybe B or B1 if possible!
•  » » 12 months ago, # ^ |   0 The hardest ever
 » 12 months ago, # | ← Rev. 2 →   +46 So difficult.UPD : also fstforces.
 » 12 months ago, # |   +48 Why is the diameter of the graph given as <=k-1, why not <=k ?
•  » » 12 months ago, # ^ |   +32 It's < k-1, My guess is just to add more confusion.it's like let's make the contest very hard and confusing for the people who decided to participate at 3 AM as they already have a brain fog at this time :D
 » 12 months ago, # |   +6 Am I the only one who has a bad feeling about system tests?
 » 12 months ago, # |   +40 Well at least the judge queue will be short lol
 » 12 months ago, # | ← Rev. 3 →   +20 Please, if you will write a round again in the future think about the quality of statements twice before. Some things were not very clear, like in div2 B where you should have specified strictly less, as you did while the round was going. Also, talking about B, it wouldn't be a bad idea to also write in bold important words, not just numbers and variables(here I'm talking about the word CONNECTED); you can see in other past problems that this is sometimes done, especially for important conditions like the connectivity one in B(for example for saying that all the numbers in an array need to be different). One last thing about this problem would be the limits for k, like were those special cases of k <= 1 really that important to be placed in the tests, ore are they even in the tests?
•  » » 12 months ago, # ^ |   +9 I am really confused why the diameter of the graph must be strictly less than k−1. other than k, it causes unnecessary confusion.
 » 12 months ago, # |   0 hardforce
•  » » 12 months ago, # ^ |   0 don't worry, system_fail_forces also coming soon.
 » 12 months ago, # |   0 Couldnt solve anything. PS: Writing bruteforce and searching on OEIS is not "solving".
 » 12 months ago, # |   +11 I feel like today's contest is suitable for Div 1.5 and Div 0.5
 » 12 months ago, # |   +25 The round had nice problems (I read only div1A-C tho), though ridiculously hard. Could've easily been split into two separate div1 rounds with some easier, filler problems. I want to express my deepest sympathy to all div2 participants, who had those super tough problems given as C-E. Div1A should've been div2E IMO. To the authors, you did very good job with inventing interesting problems. Only that you underestimated their hardness and the overall contest turned out too hard.
•  » » 12 months ago, # ^ |   +14 That is the job of coordinators and testers but I don't know how can they all agree to serve all these problems together.
 » 12 months ago, # |   +1 Some solutions for Div2 B will fail for this test: n = 1, m = 0, k = 1
•  » » 12 months ago, # ^ |   0 is it NO ?
•  » » » 12 months ago, # ^ |   0 Yeah, some solutions didn't consider that the given diameter could be negative as well.
•  » » » » 12 months ago, # ^ |   0 In first sample he told u diameter of the graph is zero, its obv. not strictly less than 0
•  » » » » 12 months ago, # ^ |   +6 Although I honestly don't see why the authors would want to make k start from 0 and not 1.
•  » » » » 12 months ago, # ^ |   0 I felt,The definition of diameter is valid only when there are atleast 2 vertices. Is this assumption wrong?
•  » » » 12 months ago, # ^ |   0 Yes. The Answer will be NO
•  » » 12 months ago, # ^ | ← Rev. 3 →   0 YES
 » 12 months ago, # |   0 Honestly how to iterate on this map in c++ for Div1 C ? map[(xi + yi)][day % (xi + yi)][xi].push_back({st_j, end_j}) Plus is it really the solution ?
•  » » 12 months ago, # ^ |   -14 Just use map.insert.The solution is about O(n sqrt m)
 » 12 months ago, # |   -14 Maybe the pretest is too weak?
 » 12 months ago, # |   +6 system testing is super fast today!
 » 12 months ago, # |   +31 Despite so many testers spanning across different ratings, a pretty unbalanced round.
 » 12 months ago, # |   +3 Saw a pattern in the ans of A for n = 2 , half of the total-permutations and it actually worked for larger n
 » 12 months ago, # |   +10 You could bold "connected" in B :/
 » 12 months ago, # |   0 k = 0 on problem B. Damn you guys got me!!!
 » 12 months ago, # |   +7 Awesome difficulty Forces+FST Forces = Codeforces Round #745
 » 12 months ago, # |   0 system testing gotta go fast
 » 12 months ago, # |   +50 Did someone mistakenly swap B and C?
•  » » 12 months ago, # ^ |   -44 Before the contest I thought B is easier but it turns out that I was wrong.
 » 12 months ago, # |   +18 Who actually proved their solutions for A lol
•  » » 12 months ago, # ^ |   0 lol haha lmao . rofl . so funny .
•  » » 12 months ago, # ^ |   0 Not me lol
•  » » 12 months ago, # ^ |   +7 I thought of something like, if a permutation is valid, it's reverse is always invalid. So half would be valid and half invalid
•  » » 12 months ago, # ^ |   +9 Proof by symmetry is valid, and very simple.
•  » » 12 months ago, # ^ | ← Rev. 2 →   +8 I think you can make a one-to-one mapping from valid permutations to invalid ones; let's say you want to count permutations which have exactly $n$ indices such that $p[i]p[i+1]$ (note that we don't have any repeated numbers, so this condition is the exact opposite of the one we consider valid.) Let's assume $\Pi(i)$ is the count of all permutations that have $i$ consecutive increasing pairs, and $\neg\Pi(i)$ the same for decreasing pairs; we know a pair is either increasing or decreasing; so we have: $\Pi(i)=\neg\Pi(2n-1-i)$; so we have: $\neg\Pi(i)=\Pi(2n-1-i)$; therefore, we have $\sum_{i=0}^{n-1}\Pi(i)=\sum_{i=n}^{2n-i}\Pi(i)$; I think by now you can see why $\frac{n!}{2}$ works in this problem.
 » 12 months ago, # |   +4 What is test 4 for problem Div2B ?
•  » » 12 months ago, # ^ |   0 I'm crying
 » 12 months ago, # |   +3 I came up with a (n^2)m solution for Div1 A but it gave TLE. Is there a more efficient solution?
•  » » 12 months ago, # ^ |   0 I also have this exact complexity and it passed. I used only vectors and pairs, maybe you used some more complex data structures?
•  » » » 12 months ago, # ^ |   0 Hmm. Seems like the constraints were tight. I see many solutions giving TLE in the system test.
•  » » » » 12 months ago, # ^ |   0 Agreed, If the intended solution was O(n*n*m) it's just sad.
•  » » » » 12 months ago, # ^ |   0 yep. FST for $O(n^2 m)$130355128
•  » » » » » 12 months ago, # ^ |   0 update: constant optimization (using tuple instead of vector) passed with 139ms.130386010
 » 12 months ago, # |   0 But it actually feels good to see a tough round now and then.
 » 12 months ago, # |   0 How to solve Problem C ?
 » 12 months ago, # |   +8 hardforces + clumsyforces + (1.5 + 0.5 ) forces +system test fail forces + ....
•  » » 12 months ago, # ^ |   0 so, bad round. lol
 » 12 months ago, # |   0 It is so reassuring to see that the problems were hard for everyone! Nevertheless, they were also really engaging.
 » 12 months ago, # |   +16 important rule : don't join chinese contest LOL
 » 12 months ago, # |   -17 FSTForces :(
 » 12 months ago, # |   0 shitty contest
 » 12 months ago, # | ← Rev. 2 →   +3 I think the pretest in Div 2 B could have been a little better. Other than that, contest had really interesting problems, except C. I didn't like C, as the only solution that I thought of was implementation heavy, so I simply skipped it and moved ahead. E was good. Just missed the window for submission.
 » 12 months ago, # |   +3 Annoucement-cum-Hardforces
 » 12 months ago, # |   -12 everything in this contest was bad. even the problems.
 » 12 months ago, # |   +8 Is it possible to solve Div1B faster than $O(n^5)$?
•  » » 12 months ago, # ^ |   -67 With 2D FFT we can solve it in $O(n^3log^2n)$.
•  » » » 12 months ago, # ^ |   +8 I wondered if it's possible during the round. Nice :|
•  » » » 12 months ago, # ^ |   +35 How? Don't you need division for that?
•  » » » » 12 months ago, # ^ | ← Rev. 6 →   0 let dp[ k ][ i ][ j ] denote number of sequences of length i that have exactly j elements with k maximums (e.g. as in the number of maximums in the process defined in statement, e.g. the variable m). Then dp[ k ][ i ][ j ] is the sum over all i1 , i2, j1 , j2 , such that i1 + i2 == i-1 , j1 + j2 == j of ((i-1 choose i1) * dp[k-1][i1][j1] ) * dp[k-1][i2][j2]. This thing can be done with 2d FFT. E.g. we are fixing the position of the maximum element. Edit: right, the modulo isnt guaranteed to be prime, not too sure.
•  » » » » » 12 months ago, # ^ |   -12 No, its wrong.
•  » » » » » 12 months ago, # ^ | ← Rev. 4 →   0 1) it should be i1+i2=i-1, though that's pretty easy to handle 2) you forgot an (i-1 choose i1) factor, which requires dividing by i! before the convolution.
•  » » » » » » 12 months ago, # ^ |   0 ohh right
•  » » » » » » 12 months ago, # ^ | ← Rev. 2 →   0 I am not too sure i understand part 2). Do you mean calculation of (i-1 choose i1) mod p needs division ? If so, we can pre calculate that with pascal's triangle equality. Edit : I see, makes sense.
•  » » » » » » » 12 months ago, # ^ |   0 No; you can divide by i! before convolution so you just do m convolutions of size nk; unless I'm missing something, directly calculating (i-1 choose i1) requires you to do mn^2 convolutions of size k, which is slower than the complexity interlude mentioned.
 » 12 months ago, # |   0 How to solve A ?? Intution ??
•  » » 12 months ago, # ^ |   0 just apply bruteforce and and think what will be the answer when n =3,4,5
•  » » 12 months ago, # ^ |   0 Most people just deduced it from the given sample cases
•  » » 12 months ago, # ^ |   0 $ans=\begin{cases}1&n=1\\\prod\limits_{i=3}^{2n} i&n\geq 2\end{cases}$
•  » » » 12 months ago, # ^ |   0 why ?? i need intution not just the plain answer :_:
•  » » » » 12 months ago, # ^ |   0 You can find this rule by directly calculating the result when $n=3,4,5$
•  » » » » 12 months ago, # ^ | ← Rev. 2 →   +9 It's in fact $\frac{(2n)!}{2}$. For any permutation with $k(k>=n)$ such $i$, you can reverse the permutation and get another unique one with $2n-1-k(2n-1-k •  » » 12 months ago, # ^ | +11 the answer is (2*n)!/2, because each permutation whose the number of pi < pi+1 is no less than n corresponds to another permutation whose the number of pi < pi+1 is less than n after you reverse this permutation •  » » 12 months ago, # ^ | ← Rev. 3 → 0 its simply a combinatrics problem ans : (2*n-1)!*(n) approach: for n there will be 2*n numbers and we can have no more than n positions with pi>n; fo(i,1,2*n) { ans=(ans*i)%MOD; } ans=(ans*n)%MOD; cout<  » 12 months ago, # | 0 Tight ML in div2 C ;(  » 12 months ago, # | +57 Sometimes I miss the old times when this assumption at least sometimes was incorrect. •  » » 12 months ago, # ^ | 0 The data of this task is actually difficult to make, and I can't come up with all the wrong solution too. Maybe we could communicate more about this task. •  » » » 12 months ago, # ^ | +8 maroonrk what's incorrect in your code? •  » » » » 12 months ago, # ^ | +10 My solutions is like this:The essential part of this problem is to make in-degrees of all vertices except the root no less than 2. For each vertex$i$, we know to which vertex$j$we can span an edge. Valid$j$'s form a set of ranges if we order vertices by their distances from the root, and the number of ranges is at most the degree of the vertex$i$. My solution consumes O(number of ranges of the queried vertex) time, so it's a$O(NQ)\$ time algorithm. However, it passed somehow.
•  » » » » » 12 months ago, # ^ |   +9 Well, doesn't look like "we can't come up with all the wrong solution".
 » 12 months ago, # | ← Rev. 2 →   +18 I didn't enjoy this round very much,not only because I got a negative delta.(For div2) The problems are way harder than a typical Div2 round,Problem ABC are not very suitable for their place and Problem DEF are so hard that very few people can solve them during the 2h contest.Also the statement is kinda hard to understand(maybe it's my fault for not reading the statement that carefully but).You have to read it mutiple times to find certain words like "connected" or figure out what the problem is asking.Contests are used to seperate contestants into two parts:Those who are very good at Problem Solving and those who are not that good.They are NOT used to find out who are better at English reading and typing.upd: obviously the pretests are weak too but I blame myself
 » 12 months ago, # |   +2 one of the worst contests with worst pretests with hard problemsat all, you won the worst contest
•  » » 12 months ago, # ^ |   0 true!
•  » » 12 months ago, # ^ | ← Rev. 2 →   -20 I agree with you.
 » 12 months ago, # |   0 how to solve c?
 » 12 months ago, # |   +1 Every Time I think I am now good enough to become CM , My Rating Starts Falling Until it reaches Specialist Range
 » 12 months ago, # | ← Rev. 5 →   +59 Sincerely, this round is not a good round. I have taken nearly 40 contests, and this contest is the first contest that I downvote. 1, the problem are so unbalanced, for div1, so many people are not able to solve A, and for div2, difficulty gap between B and C are too large, so as C and D. 2, div2 B is the worst problem I have ever met. It is so tricky, and description are soooooooo uncomfortable. What is "less than k-1" mean? Could you just write less than k" or even better "no greater than k". "n=1" is meaningless because we cannot pick two points from n=1, and "k=0 or k=1" is even more meaningless because, who want to discuss a graph with negative k?3, div2 A are so unfriendly to newbies. 4, weak pretests5, Except for the balance and description, a good contest should be educated. Even I cannot solve the problem during the contest and get negative delta, it doesn't matter. We can learn something after the contest, and improve our ways of thinking or coding skills. But in this contest, I just want to say, what the hell of this ?? just a bunch of mind tricks or brain teaser? Can I learn something in the contest? In all, getting up at 5:30 AM and take part in such contest is the most wasteful thing I have ever done.
•  » » 12 months ago, # ^ |   +1 systest is pretty weak too? I uphacked myself with a case with obvious 32-bit integer overflow...130380959
 » 12 months ago, # |   +11 I'm wondering if the pretest is too weak. I see many people got FST on 1C
 » 12 months ago, # |   +9 In Div2 B, what is the expected answer for case: n = 1, m = 0 and any k. I guess according to definition of diameter given in the problem statement, diameter is not defined for n = 1 (we need at least 2 nodes to define diameter). I think the answer for this case is ambiguous and the problem should accept both "Yes" and "No" as the correct answers for the above-mentioned case.
•  » » 12 months ago, # ^ |   0 totally argee
•  » » 12 months ago, # ^ |   0 I think they made it quite clear the if n=1 the diameter is 0 by putting an example with n=1 and a note that says in this case the diameter is 0.
•  » » » 12 months ago, # ^ | ← Rev. 2 →   0 I agree but this thing should be mentioned in the problem statement itself, most people don't see the explanation of the sample test case.
 » 12 months ago, # |   +3 Why condition in div2 B use less than k-1 instead of less than just k? It just creates some confusions.
•  » » 12 months ago, # ^ |   0 something likecin>>k--;
 » 12 months ago, # |   0 what is test 61 in