Anti

"A DDoS-type string is a string of length 4 consisting of uppercase and lowercase English letters sa(...TRUNCATED)

atcoder

hard

"[{\"input\": \"DD??S\\n\", \"output\": \"676\\n\", \"testtype\": \"stdin\"}, {\"input\": \"????????(...TRUNCATED)

"[{\"input\": \"vgxgpuamkxkszhkbpphykinkezplvfjaqmopodotkrjzrimlvumuarenexcfycebeurgvjyospdhvuyfvtvn(...TRUNCATED)

{}

Labyrinth-1

You have a robot in a two-dimensional labyrinth which consists of *N*<=×<=*M* cells. Some pairs of cells adjacent by side are separated by a wall or a door. The labyrinth itself is separated from the outside with walls around it. Some labyrinth cells are the exits. In order to leave the labyrinth the robot should reach any exit. There are keys in some cells. Any key can open any door but after the door is opened the key stays in the lock. Thus every key can be used only once. There are no labyrinth cells that contain both a key and an exit. Also there can not be both a wall and a door between the pair of adjacent cells. Your need to write a program in *abc* language (see the language description below) that will lead the robot to one of the exits. Lets numerate the labyrinth rows from 0 to *N*<=-<=1 top to bottom and the columns – from 0 to *M*<=-<=1 left to right. In *abc* language the following primary commands are available: - move-DIR – move to the adjacent cell in the direction. *down* increases the number of the row by 1, *right* increases the number of the column by 1. In case there’s a wall or a closed door in this direction, nothing’s happening. - open-DIR – open the door between the current cell and the adjacent one in *DIR* direction. In case there isn’t any door in this direction or it’s already been opened or the robot doesn’t have a key, nothing’s happening.- take – take the key in the current cell. In case there isn’t any key or the robot has already picked it up, nothing’s happening. The robot is able to carry any number of keys.- terminate – terminate the program. This command is not obligatory to use. In case it’s absent the command is added at the end of the program automatically. Also, there are the following control commands in *abc* language: - for-N OPS end – repeat the sequence of the *OPS* commands *N* times, 0<=&lt;<=*N*<=≤<=100000. Each loop counter check counts as a command fulfilled by the robot. - if-ok OPS1 else OPS2 endif – carries out the sequence of the *OPS*1 commands, if the previous command of moving, taking the key or opening the door was successful, otherwise the sequence of the *OPS*2 commands is being carried out. Should there be no previous command run, the sequence *OPS*1 will be carried out. If-ok check counts as a command fulfilled by the robot. - break – stops the current *for* loop. - continue – finishes the current *for* loop iterations. Note that the control and the primary commands can be fit into each other arbitrarily. The robot will fulfill your commands sequentially until it exits the labyrinth, or it runs out of the commands, or the *terminate* command is run, or the quantity of the fulfilled commands exceeds the bound number 5·106. In *abc* language each command is a separate word and should be separated from other commands with at least one space symbol. You should write a program that prints the sequence of commands leading the robot out of the labyrinth. Of course, as you are a good programmer, you should optimize these sequence. The number of the non-space symbols in the sequence should not exceed 107. If you succeed in finding the way out of the labyrinth *i* you’ll be granted the number of points equal to: - *W**i* – labyrinth’s weight, some fixed constant. - *G**i* – number of robots moves. - *O**i* – number of fulfilled commands. Note that this number includes commands like *take* executed in the cells with no key, as well as opening commands applied to the already opened doors. - *L**i* – sequence length in symbols, excluding space symbols and line breaks. - *Q*<==<=10·*N*·*M*. In case your sequence doesn’t lead the robot to the exit you’ll be granted 0 points. Your programs result will be the sum of all *S**i*. You should maximize this total sum. All labyrinths will be known and available to you. You can download the archive with labyrinths by any of the given links, password to extract files is aimtechiscool: 1. [https://drive.google.com/file/d/1dkIBfW_Gy6c3FJtXjMXZPMsGKRyn3pzp](https://drive.google.com/file/d/1dkIBfW_Gy6c3FJtXjMXZPMsGKRyn3pzp) 1. [https://www.dropbox.com/s/77jrplnjgmviiwt/aimmaze.zip?dl=0](https://www.dropbox.com/s/77jrplnjgmviiwt/aimmaze.zip?dl=0) 1. [https://yadi.sk/d/JNXDLeH63RzaCi](https://yadi.sk/d/JNXDLeH63RzaCi) In order to make local testing of your programs more convenient, the program calculating your results (checker) and the labyrinth visualizer will be available. This program is written in *python*3 programming language, that’s why you’re going to need *python*3 interpreter, as well as *pillow* library, which you can install with the following command pip3 install pillow. *pip*3 is a utility program for *python*3 package (library) installation. It will be installed automatically with the *python*3 interpreter. Example command to run checker and visualizer: python3 aimmaze.py maze.in robot.abc --image maze.png. The checker can be run separately of visualization: python3 aimmaze.py maze.in robot.abc. Flag --output-log will let you see the information of robots each step: python3 aimmaze.py maze.in robot.abc --output-log. Note *python*3 can be installed as *python* on your computer. To adjust image settings, you can edit constants at the beginning of the program *aimmaze*.*py*.The first line contains integers *i*,<= *W*,<= *N*,<= *M*,<= *x*0,<= *y*0,<= *C*,<= *D*,<= *K*,<= *E*. - 1<=≤<=*i*<=≤<=14 – labyrinth’s number, which is needed for a checking program. - 1<=≤<=*W*<=≤<=1018 – labyrinth’s weight, which is needed for a checking program. - 2<=≤<=*N*,<=*M*<=≤<=1000 – labyrinth’s height and width. - 0<=≤<=*x*0<=≤<=*N*<=-<=1,<= 0<=≤<=*y*0<=≤<=*M*<=-<=1 – robot’s starting position (*x*0,<=*y*0). - 0<=≤<=*C*<=≤<=2·*NM* – number of walls. - 0<=≤<=*D*<=≤<=105 – number of doors. - 0<=≤<=*K*<=≤<=105 – number of keys. - 1<=≤<=*E*<=≤<=1000 – number of exits. The *x* coordinate corresponds to the row number, *y* – to the column number. (0,<=0) cell is located on the left-up corner, so that *down* direction increases the *x* coordinate, while *right* direction increases the *y* coordinate. Each of the next *C* lines contains 4 integers each *x*1,<=*y*1,<=*x*2,<=*y*2 – the coordinates of cells with a wall between them in a zero based indexing. It is guaranteed that |*x*1<=-<=*x*2|<=+<=|*y*1<=-<=*y*2|<==<=1,<= 0<=≤<=*x*1,<=*x*2<=≤<=*N*<=-<=1,<= 0<=≤<=*y*1,<=*y*2<=≤<=*M*<=-<=1. Also there are always walls around the labyrinth’s borders, which are not given in the labyrinths description. Each of the next *D* lines contains door description in the same format as walls description. It is guaranteed that doors and walls don’t overlap. Each of the next *K* rows contains a pair of integer which are the key coordinates in a zero based indexing. Each of the next *E* rows contains a pair of integer which are the exit coordinates in a zero based indexing. It is guaranteed that the robots starting position as well as keys and exits are located in pairwise different cells.Print a program in *abc* language which passes the given labyrinth. Commands have to be separated by at least one space symbol. You can use arbitrary formatting for the program.

codefo

Word Capitalization

Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word. Note, that during capitalization all the letters except the first one remains unchanged.A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.Output the given word after capitalization.

codeforces

A

281

2025-01-20T20:58:05.360400

easy

```python # Read the word from stdin word = input() # Capitalize the first letter of the word and make the rest lowercase capitalized_word = word.capitalize() # Print the capitalized word print(capitalized_word) ```

# Read the word from stdin word = input() # Capitalize the first letter of the word and make the rest lowercase capitalized_word = word.capitalize() # Print the capitalized word print(capitalized_word)

OK

diff --git a/astropy/modeling/separable.py b/astropy/modeling/separable.py --- a/astropy/modeling/separable.py +++ b/astropy/modeling/separable.py @@ -242,7 +242,7 @@ def _cstack(left, right): cright = _coord_matrix(right, 'right', noutp) else: cright = np.zeros((noutp, right.shape[1])) - cright[-right.shape[0]:, -right.shape[1]:] = 1 + cright[-right.shape[0]:, -right.shape[1]:] = right return np.hstack([cleft, cright])

diff --git a/astropy/modeling/tests/test_separable.py b/astropy/modeling/tests/test_separable.py --- a/astropy/modeling/tests/test_separable.py +++ b/astropy/modeling/tests/test_separable.py @@ -28,6 +28,13 @@ p1 = models.Polynomial1D(1, name='p1') +cm_4d_expected = (np.array([False, False, True, True]), + np.array([[True, True, False, False], + [True, True, False, False], + [False, False, True, False], + [False, False, False, True]])) + + compound_models = { 'cm1': (map3 & sh1 | rot & sh1 | sh1 & sh2 & sh1, (np.array([False, False, True]), @@ -52,7 +59,17 @@ 'cm7': (map2 | p2 & sh1, (np.array([False, True]), np.array([[True, False], [False, True]])) - ) + ), + 'cm8': (rot & (sh1 & sh2), cm_4d_expected), + 'cm9': (rot & sh1 & sh2, cm_4d_expected), + 'cm10': ((rot & sh1) & sh2, cm_4d_expected), + 'cm11': (rot & sh1 & (scl1 & scl2), + (np.array([False, False, True, True, True]), + np.array([[True, True, False, False, False], + [True, True, False, False, False], + [False, False, True, False, False], + [False, False, False, True, False], + [False, False, False, False, True]]))), }

Modeling's `separability_matrix` does not compute separability correctly for nested CompoundModels Consider the following model: ```python from astropy.modeling import models as m from astropy.modeling.separable import separability_matrix cm = m.Linear1D(10) & m.Linear1D(5) ``` It's separability matrix as you might expect is a diagonal: ```python >>> separability_matrix(cm) array([[ True, False], [False, True]]) ``` If I make the model more complex: ```python >>> separability_matrix(m.Pix2Sky_TAN() & m.Linear1D(10) & m.Linear1D(5)) array([[ True, True, False, False], [ True, True, False, False], [False, False, True, False], [False, False, False, True]]) ``` The output matrix is again, as expected, the outputs and inputs to the linear models are separable and independent of each other. If however, I nest these compound models: ```python >>> separability_matrix(m.Pix2Sky_TAN() & cm) array([[ True, True, False, False], [ True, True, False, False], [False, False, True, True], [False, False, True, True]]) ``` Suddenly the inputs and outputs are no longer separable? This feels like a bug to me, but I might be missing something?

2022-03-03T15:14:54Z

4.3

["astropy/modeling/tests/test_separable.py::test_separable[compound_model6-result6]", "astropy/modeling/tests/test_separable.py::test_separable[compound_model9-result9]"]

["astropy/modeling/tests/test_separable.py::test_coord_matrix", "astropy/modeling/tests/test_separable.py::test_cdot", "astropy/modeling/tests/test_separable.py::test_cstack", "astropy/modeling/tests/test_separable.py::test_arith_oper", "astropy/modeling/tests/test_separable.py::test_separable[compound_model0-result0]", "astropy/modeling/tests/test_separable.py::test_separable[compound_model1-result1]", "astropy/modeling/tests/test_separable.py::test_separable[compound_model2-result2]", "astropy/modeling/tests/test_separable.py::test_separable[compound_model3-result3]", "astropy/modeling/tests/test_separable.py::test_separable[compound_model4-result4]", "astropy/modeling/tests/test_separable.py::test_separable[compound_model5-result5]", "astropy/modeling/tests/test_separable.py::test_separable[compound_model7-result7]", "astropy/modeling/tests/test_separable.py::test_separable[compound_model8-result8]", "astropy/modeling/tests/test_separable.py::test_custom_model_separable"]

298ccb478e6bf092953bca67a3d29dc6c35f6752

15 min - 1 hour

2024-II-4

Let $x,y$ and $z$ be positive real numbers that satisfy the following system of equations: \[\log_2\left({x \over yz}\right) = {1 \over 2}\] \[\log_2\left({y \over xz}\right) = {1 \over 3}\] \[\log_2\left({z \over xy}\right) = {1 \over 4}\] Then the value of $\left|\log_2(x^4y^3z^2)\right|$ is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Denote $\log_2(x) = a$, $\log_2(y) = b$, and $\log_2(z) = c$. Then, we have: $a-b-c = \frac{1}{2}$, $-a+b-c = \frac{1}{3}$, $-a-b+c = \frac{1}{4}$. Now, we can solve to get $a = \frac{-7}{24}, b = \frac{-9}{24}, c = \frac{-5}{12}$. Plugging these values in, we obtain $|4a + 3b + 2c| = \frac{25}{8} \implies \boxed{033}$.

Convert the point $(0,3)$ in rectangular coordinates to polar coordinates. Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$

We have that $r = \sqrt{0^2 + 3^2} = 3.$ Also, if we draw the line connecting the origin and $(0,3),$ this line makes an angle of $\frac{\pi}{2}$ with the positive $x$-axis. [asy] unitsize(0.8 cm); draw((-0.5,0)--(3.5,0)); draw((0,-0.5)--(0,3.5)); draw(arc((0,0),3,0,90),red,Arrow(6)); dot((0,3), red); label("$(0,3)$", (0,3), W); dot((3,0), red); [/asy] Therefore, the polar coordinates are $\boxed{\left( 3, \frac{\pi}{2} \right)}.$

\left( 3, \frac{\pi}{2} \right)

Precalculus

2

test/precalculus/807.json

1

有些这样的“洋人”就站在大众之间，如同鹤立鸡群，毫不掩饰自己的优越感。他们排在非凡的甲菜盆后面，虽然人数寥寥无几，但却特别惹眼。

true

{ "span1_index": 6, "span1_text": "洋人", "span2_index": 35, "span2_text": "他们" }

0

下列关于资本结构理论的说法中，不正确的是____。

代理理论、权衡理论、有企业所得税条件下的MM理论，都认为企业价值与资本结构有关

按照优序融资理论的观点，考虑信息不对称和逆向选择的影响，管理者偏好首选留存收益筹资，然后是发行新股筹资，最后是债务筹资

权衡理论是对有企业所得税条件下的MM理论的扩展

代理理论是对权衡理论的扩展

中华文化

中医

伏兔穴所属的经脉是什么？

足阳明胃经

[ "https://zh.wikipedia.org/wiki/%E4%BC%8F%E5%85%94%