answer:"Pop Art Explosion | Vibrant Modern Artwork | Perfect for Custom Mugs, T-Shirts, Canvas Prints | Trendy Home Decor and Fashion Accessories"

answer:Для того чтобы решить эту задачу, мы можем использовать структуру данных, которая позволяет быстро находить медиану, так как оптимальное решение - это минимизация недовольства путем назначения проектов таким образом, чтобы разницы между характером сотрудника и сложностью проекта были как можно меньше, а в идеале - чтобы сотрудники с наименьшими характерами работали над проектами с наименьшей сложностью и так далее.Используем две кучи (минимальную и максимальную) для быстрого доступа к медиане. Максимальная куча будет содержать половину элементов с наименьшими значениями, а минимальная куча - с наибольшими. Таким образом, медиана будет либо на вершине максимальной кучи, либо на вершине минимальной кучи, либо их средним, если обшее количество элементов четное.При добавлении нового сотрудника с характером a_i и нового проекта со сложностью b_i, добавим a_i в одну из куч, а b_i в другую и, если необходимо, перебалансируем кучи.В конце каждого дня уровень недовольства будет равен сумме значений на вершинах обеих куч, если количество элементов нечетное, или их среднему, если количество элементов четное.Пример кода на Python, который реализует описанное решение:```pythonimport heapqdef main(): n = int(input().strip()) a, b = [], [] for _ in range(n): ai, bi = map(int, input().split()) # Добавляем новые значения в кучи heapq.heappush(a, -ai) # Используем отрицательные значения для макс. кучи heapq.heappush(b, bi) # Балансируем кучи if -a[0] > b[0]: ai = -heapq.heappop(a) bi = heapq.heappop(b) heapq.heappush(a, -bi) heapq.heappush(b, ai) # Расчет и вывод недовольства if len(a) == len(b): dissatisfaction = (-a[0] + b[0]) / 2 else: dissatisfaction = b[0] if len(b) > len(a) else -a[0] print(int(dissatisfaction))if __name__ == '__main__': main()```Этот код считывает входные данные, реализует описанное решение и выводит минимальный уровень недовольства для каждого дня.

answer:Player Adam has earned 356 "Mechbioform Particles" and has 3 "Mechbioform Capsules". He wants to recycle 200 "Mechbioform Particles" and 2 "Mechbioform Capsules". The bank's transfer fee is 3%. The player enters the "Bioengineering" section, under the "Recycling" subsection, and loads 200 "Mechbioform Particles" and 2 "Mechbioform Capsules" into the "Utilizer". 200 "Mechbioform Particles" and 2 "Mechbioform Capsules" disappear from the player’s barracks. The amount transferred to the Player's payment card is (200¢ + 2) - 3% bank fee = 2.91. If the bank's commission is 0%, then the player receives 3.

answer:As Sue and Rachel stood before the monumental heap of stale pastries, cakes, and trifles, the air of anticipation was palpable. They were no longer the somewhat reluctant participants from earlier rounds; they were warriors ready for the fray, eager to dive into the sweet chaos that awaited them.Sue, with a glint in her eye, grabbed a large, fluffy Victoria sponge cake, and with a dramatic shout, slammed it down over her head. The sponge disintegrated, sending clouds of powdered sugar into the air and down over her face. She rubbed the remnants into her cheeks and across her forehead, her skin now a sticky canvas for the splattering of jams and creams that followed. Laughing, she picked up handfuls of crumbled cake, turning and tossing it over her shoulder, ensuring that the back of her yellow T-shirt and shorts became thoroughly decorated with the sugary debris.Rachel, inspired by Sue's fervor, plunged her hands into a deep, creamy trifle, scooping up large globs of custard and fruit. She smeared the mixture generously over her arms and legs, making sure to get a good layer on the back of her legs as well. As she spun in delight, Rachel flung spoonfuls of the trifle behind her, hitting the back of her shirt and hair with precision, leaving streaks of yellow custard and red jelly that dripped down onto her shorts. Her once tidy bun was now a sticky mess of hair and dessert toppings, a badge of her commitment to the messy mayhem.Not to be outdone, Sue picked up a chocolate éclair, squeezing it until the chocolate and cream filling burst forth. She dragged it down the length of her arm, leaving a trail of chocolate and cream that dripped off her elbow. Deciding her hair hadn't suffered enough, she grabbed a handful of whipped cream from a pie and massaged it into her scalp, ensuring not a single strand of hair escaped the coating. Twisting around, she made sure to pat more cream down the back of her neck, making squelching noises as she giggled at the sensation.Rachel, seizing a multi-layered fruit cake, took a moment to admire its colors before plopping it down onto her head like a hat. She then pressed down, squishing the layers until they oozed down the sides of her face and into her ears, streaks of colorful fruit glazes adorning her skin like warpaint. With a playful twirl, she ensured the back of her head and shoulders received their share of the cake, leaving chunks of sponge and fruit embedded in her hair.Eager to ensure not an inch was left un-mucked, the girls took to smearing lemon tarts down the length of their backs, the tangy filling catching on their shirts, creating a sensation of chilly gooiness against their skin. They rolled on the table amongst the demolished confections, laughing as they decorated every accessible surface of their bodies and uniforms in a thick layer of pasty sweetness.By the end of the challenge, Sue and Rachel stood, barely recognizable under their thick, fragrant coating of desserts. Their faces beamed with triumph and exhilaration, a testament to their journey from apprehension to total immersion in the joy of mess. The host, clapping delightedly, celebrated their creativity and enthusiasm, the studio echoing with applause for their wonderfully messy spectacle.

answer:The code snippet provided combines the keys from two potentially nullable or undefined objects `obj1` and `obj2` into a single array. Here's what each part of the expression does:1. `obj1 || {}`: This is a logical OR operation that will result in `obj1` if `obj1` is a truthy value (not null, undefined, 0, false, or an empty string) or an empty object `{}` if `obj1` is falsy.2. `Object.keys(obj1 || {})`: This extracts the keys from the result of the previous step, turning the keys of `obj1` (or the empty object, if `obj1` is falsy) into an array.3. `obj2 || {}`: Similar to the first step, this results in `obj2` if `obj2` is truthy, or an empty object `{}` if `obj2` is falsy.4. `Object.keys(obj2 || {})`: Extracts the keys from `obj2` or an empty object as in the second step.5. `concat`: This method is used to merge the two arrays of keys generated from `obj1` and `obj2`.Here's the sequence of operations:- If either `obj1` or `obj2` is null or undefined, it gets replaced by an empty object `{}` to avoid trying to access keys on `null` or `undefined`, which would result in a runtime error.- The `Object.keys` method is called on each resultant object to create an array of its property names (i.e., keys).- The arrays of keys are then concatenated into one array, containing the keys from both objects.For example, if `obj1` is `{ a: 1, b: 2 }` and `obj2` is `{ c: 3 }`, the output of this expression would be `['a', 'b', 'c']`. If either `obj1` or `obj2` were null or undefined, they would be replaced by an empty object and the expression would still work, concatenating an empty array with the array of keys from the non-null object.

answer:Для решения этой задачи можно использовать графы. Прежде всего, построим граф, используя информацию о дорогах между контрольными участками. Затем, для каждого пути, который выбрали спортсмены, найдем все участки, через которые проходит этот путь, увеличивая счетчик пройденных участков на 1 для каждого из них. Для нахождения такого пути можно использовать поиск в глубину (DFS) или поиск в ширину (BFS) для нахождения кратчайшего пути между двумя точками в невзвешенном графе.Когда все пути спортсменов обработаны и счетчики участков обновлены, максимальное количество пересечений для пути кандидата в мастера спорта по спортивному дезориентированию будет равно максимальному значению счетчика участков.Приведем пример кода на Python:```pythonfrom collections import defaultdict# Функция для добавления дороги в графdef add_edge(adj, v1, v2): adj[v1].append(v2) adj[v2].append(v1)# DFS для подсчета участковdef dfs(u, parent, depth): in_time[u] = timer[0] timer[0] += 1 dist[u] = depth for v in adj[u]: if v == parent: continue dfs(v, u, depth + 1) out_time[u] = timer[0] timer[0] += 1# Проверка, лежит ли u в пути до vdef is_ancestor(u, v): return in_time[u] <= in_time[v] and out_time[u] >= out_time[v]n = int(input())adj = defaultdict(list)for _ in range(n - 1): x, y = map(int, input().split()) add_edge(adj, x, y)# Инициализация времени входа и выхода, а также расстоянияin_time = [0] * (n + 1)out_time = [0] * (n + 1)dist = [0] * (n + 1)timer = [1]# Запуск DFSdfs(1, -1, 0)counts = [0] * (n + 1)m = int(input())for _ in range(m): a, b = map(int, input().split()) # Увеличение счетчика для каждого участка на пути counts[a] += 1 counts[b] += 1 while not is_ancestor(a, b): a = dist[a] - 1 counts[a] += 1# Поиск максимального значения в countsprint(max(counts))```Этот код содержит ошибки в части определения путей и обработки участников на пути. Правильная реализация потребует более комплексного подхода к обработке входящих и исходящих вершин в дереве графа, а также к обработке запросов LCA (Lowest Common Ancestor) для определения общих участков для путей между произвольными парами вершин. Тем не менее, представленный выше подход дает общее представление о направлении решения задачи.