Prompt:
Văn nhận được n kẹo mút vào ngày sinh nhật của mình và anh muốn tặng một số kẹo cho Hoa. Anh biết rằng Hoa quan tâm đến số lần được nhận quà, vì vậy Văn muốn tặng Hoa kẹo nhiều lần nhất có thể. Tuy nhiên, Hoa nhớ số kẹo mình được nhận mỗi lần Văn tặng và Văn không muốn tặng Hoa số kẹo giống nhau ở 2 lần liên tiếp. Với n chiêc kẹo, Văn có thể tặng Hoa mấy lần?

Ví dụ

Với n = 3, đầu ra là happyPresents(n) = 2.
Đầu vào/Đầu ra

[giới hạn thời gian chạy] 0.5 seconds

[đầu vào] integer n

Số nguyên dương.

Điều kiện tiền đề:
0 ≤ n ≤ 109.

[đầu ra] integer

Số lần nhiều nhất mà Văn có thể tặng kẹo cho Hoa.

code python


Prompt: Сумма первых n членов арифметической прогрессии, разность которой отлична от нуля, равна половине суммы следующих n членов этой про- грессии. Найти отношение суммы первых 3n членов прогрессии к сумме ее первых n членов.


Prompt:
Se aplica un fuerza de 500 lb a un alambre de cobre con diámetro 0,1 pulg, con un esfuerzo de fluencia de 20 000
psi. ¿Se deformará plásticamente el alambre?


Prompt:
solve this in cpp

Given two positive integers n and k, and another array a of n

integers.

In one operation, you can select any subarray of size k
of a, then remove it from the array without changing the order of other elements. More formally, let (l,r) be an operation on subarray al,al+1,…,ar such that r−l+1=k, then performing this operation means replacing a with [a1,…,al−1,ar+1,…,an]

.

For example, if a=[1,2,3,4,5]
and we perform operation (3,5) on this array, it will become a=[1,2]. Moreover, operation (2,4) results in a=[1,5], and operation (1,3) results in a=[4,5]

.

You have to repeat the operation while the length of a
is greater than k (which means |a|>k). What is the largest possible median† of all remaining elements of the array a

after the process?

†
The median of an array of length n is the element whose index is ⌊(n+1)/2⌋ after we sort the elements in non-decreasing order. For example: median([2,1,5,4,3])=3, median([5])=5, and median([6,8,2,4])=4

.
Input

The first line contains a single integer t
(1≤t≤104

) — the number of test cases.

The first line of each test case contains two integers n
and k (1≤n,k≤5⋅105

).

The second line contains n
integers a1,a2,…,an (1≤ai≤109) — the array a

.

It is guaranteed that the sum of n
over all test cases does not exceed 5⋅105

.
Output

For each test case, print a single integer — the largest median possible after performing the operations.


Prompt: Дан список чисел. Разбить спиоск на пары так, чтобы сумма модулей разностей пар была минимальна. Если чисел нечетное количество - сохранить лишнее в отдельный массив


Prompt:
Consider the following recurrence relation and initial condition for a recursive algorithm

M(n) = 2 * M(n-1) + 1 // recurrence relation

M(1) = 1                 // initial condition

This is the first step of the backward substitution to get the time complexity of the algorithm. Select the correct result of the first substitution.

M(n) = 2* M(n-1) + 1
         =  ?

Group of answer choices

2 * M(n-2) + 1 + 1

22 * M(n-2) + 1 + 1

22 * M(n-2) + 2 + 1

None of these.

2 * M(n-2) + 2 + 1


Prompt: If neodymium magnets have 2 and 20 lbs pull force on steel, what would the relative attractive forces be when 1 cm away from the steel?


Prompt:
Three out of the last ten candidates wins a treasure hunt game. Previous record shows fraction of winners follows the Beta
(
20.0
,
b
)
(20.0,b) distribution with an average of
20.0
%
20.0%. Estimate the long-term fraction of winners of the treasure hunt game. Write your answer correct to two decimal places.


Prompt:
本题与问题1类似，都是通过构建多目标约束优化得出2024-2030年农作物最大利润，不同之处是引入了农作物预期销售量、亩产量、种植成本和销售价格等不确定因素，继而利用不确定因素构建新的约束条件。
7.1.1目标函数构建
由题意可知，此题仅考虑2024-2030的农作物每季总产量超出部分以2023年销售价格的50%降价出售的情况，得出如下目标函数
〖max  Z〗_1=∑_i▒∑_j▒〖min(〗 P_jt∙X_ijt，M_ij)∙Q_ijt+1/2 ∑_i▒∑_j▒〖max(〗 P_jt∙X_ijt-M_ij,0)∙Q_ijt-∑_i▒∑_j▒C_jt ∙X_ijt
7.1.2约束条件的构建
本题在问题1的基础上引入预期销售量、亩产量、种植成本和销售价格等波动因素，通过在问题1的约束条件上建立如下数学表达式
m_ijt=M_ijt∙〖(1+a_1)〗^t
m_ijt=M_ijt∙(1+a_1 )
q_ijt=Q_ijt∙〖(1+a_3)〗^t
c_ijt=C_ijt∙〖(1+a_4)〗^t
p_ijt=P_ijt∙(1+a_5 )^t   17≤j≤37
p_ijt=P_ijt∙(1+a_6 )^t   38≤j≤40
p_ijt=P_ijt∙(1+a_7 )^t   j=41
7.2蒙特卡罗模拟优化
蒙特卡罗模拟是一种通过随机抽样来模拟不确定性因素，并通过多次模拟来评估不同方案的风险和收益的方法，非常适合解决本问题。首先，先对本题的不确定性参数多次随机抽样，每次抽样后求解一个确定性的优化问题，从而获得一系列可能的种植方案和相应的经济效益。再从一系列种植方案中找出最大化利润的方案，完成结果表格2。
7.2.1不确定参数输入
利用蒙特卡罗法假设问题2中的不确定因素均服从正态分布，

其中，μ为2023年的销售量，σ为
7.2.2


Prompt:
Finish this function def dy_dw(x):
    """Implement the derivative of the function with respect to the parameter w.

    Arguments:
    x - a scalar value, the input to the function

    Returns:
    derivative - a scalar value, the derivative of the function with respect to w
    """
    derivative = None
    # YOUR CODE HERE
    return derivative


Prompt: What part of the distribution would [-1; +2] standard deviation range capture?


Prompt:
Dado um numero n, representa q quantidade de cubinhos do lado de um cubo, ou seja, a quantidade de cubos eh n^3, onde todas as faces de cada cubinho que esta visivel esta pintada de preto, diga:

quantidade de cubinhos com 0 faces pintadas
quantidade de cubinhos com 1 face pintada
quantidade de cubinhos com 2 faces pintadas
e quantidade com 3 faces pintadas.


Prompt: Какое усилие оказывает ветер с разной скоростью в секунду на стоящего человека с поднятыми вверх руками, ветер дует прямо в лицо. И какое усилие на человека лежащего в пляжном шезлонге? И усилие на человека сидящего на табурете? Выведи в форме таблицы. Колонки выведи с ветром от 5 до 15 метров вс екунду с шагом в 1 метр.


Prompt: compute the gradient of -logdet(I+AHBA)


Prompt: calculate how many seconds an Olympic marathon runner can expect to lose by having long hair (past shoulders) flowing in the wind


Prompt: how to estimate distance between different curves parametrizations


Prompt: in the identity 7 * 4 + 8 * 8 = 92, can you modify exactly one integer on the left hand side of the equation so that the right hand side becomes 106? Don't guess. First do the reasoning on what change do we need to make, then plan how to achieve that.


Prompt:
40 g of IMPURE calcium carbonate reacts with a 200 cm' of a dilute sulphuric acid with a concentration of 1,5 mol-dm-3. Al the calcium carbonate and sulphuric acid react completely leaving the impurities unreacted at the completion of the reaction.
CaCO3 (s) +H2SO4 (aq) →CaSO (s) +COz(g) + H20 (P)

8.3.1 Calculate the percentage purity of the calcium carbonate.
 To obtain the sulphuric acid solution of concentration 1,5 mol.dm-3 that reacted with the IMPURE calcium carbonate, 10 cm of a concentrated sulphuric acid solution of concentration 9 mol-dm-3 was added to water.
8.3.2 Calculate the volume of water required to dilute the concentrated sulphuric acid solution to a concentration of 1,5 mol-dm-3.


Prompt: 一瓶水一元钱，同时，每有两个空瓶可以换一瓶水。假设你有40元，最多可以喝几瓶水？


Prompt:
C2=atand(M2)；
PAR_angle=C2-C2';

    % 找到平行线
    parallel_lines = abs(PAR_angle) < 0.5;
M2是一个列向量，PAR_angle=C2-C2';得到一个方阵。帮我续写代码，通过找到parallel_lines 矩阵中非对角线上为1的元素，找到C2中哪几对元素的差满足条件。成对元素的索引，放入一个cell中。


Prompt:
Use the method of Instantaneous Center of Zero Velocity (IC) when applicable. For a limited range, disk A rolls without slipping on the floor. If disk A has the motion shown, determine the angular velocity and the angular acceleration of CD at the instant when CD is vertical, AB is horizontal, and BC forms a 20° angle with respect to the horizontal.

The diagram shows:
ωA = 0.1 rad/s
αA = 0.2 rad/s²
R = 4'
BC = 18'
CD = 3'
∠BCD = 20°


Prompt:
2023 年高教社杯全国大学生数学建模竞赛题目
（请先阅读“全国大学生数学建模竞赛论文格式规范”）
D 题 圈养湖羊的空间利用率
规模化的圈养养殖场通常根据牲畜的性别和生长阶段分群饲养，适应不同种类、不同阶段
的牲畜对空间的不同要求，以保障牲畜安全和健康；与此同时，也要尽量减少空间闲置所造成
的资源浪费。在实际运营中，还需要考虑市场上饲料价格和产品销售价格的波动以及气候、疾
病、种畜淘汰、更新等诸多复杂且关联的因素，但空间利用率是相对独立并影响养殖场经营效
益的重要问题。
湖羊是国家级绵羊保护品种，具有早期生长快、性成熟早、四季发情并且可以圈养等优良
特性。湖羊养殖场通常建有若干标准羊栏，每一标准羊栏所能容纳的羊只数量由羊的性别、大
小、生长阶段决定。
湖羊养殖的生产过程主要包括繁殖和育肥两大环节。人工授精技术要求高，因此湖羊繁殖
大多采用种公羊和基础母羊自然交配的方式。怀孕母羊分娩后给羔羊哺乳，羔羊断奶后独立喂
饲，育肥长成后出栏。自然交配时将若干基础母羊与一只种公羊关在一个羊栏中，自然交配期
约为 3 周，然后将种公羊移出。受孕母羊的孕期约为 5 个月，每胎通常产羔 2 只。母羊分娩后
哺乳期通常控制在 6 周左右，断奶后将羔羊移至育肥羊栏喂饲。一般情况下，羔羊断奶后经过
7 个月左右育肥就可以出栏。母羊停止哺乳后，经过约 3 周的空怀休整期，一般会很快发情，
可以再次配种。按上述周期，正常情况下，每只基础母羊每 2 年可生产 3 胎。在不考虑种公羊
配种能力差异的情况下，种公羊与基础母羊一般按不低于 1:50 的比例配置。种公羊和母羊在非
交配期原则上不关在同一栏中。
某湖羊养殖场设置标准羊栏，规格是：空怀休整期每栏基础母羊不超过 14 只；非交配期的
种公羊每栏不超过 4 只；自然交配期每栏 1 只种公羊及不超过 14 只基础母羊；怀孕期每栏不超
过 8 只待产母羊；分娩后的哺乳期，每栏不超过 6 只母羊及它们的羔羊；育肥期每栏不超过 14
只羔羊。原则上不同阶段的羊只不能同栏。
养殖场的经营管理者为保障效益，需要通过制定生产计划来优化养殖场的空间利用率。这
里的生产计划，主要是决定什么时间开始对多少可配种的基础母羊进行配种，控制羊只的繁育
期，进而调节对羊栏的需求量，以确保有足够多的羊栏，同时尽量减少羊栏闲置。当羊栏不够
时，可以租用其他场地。
请建立数学模型讨论并解决以下问题：
问题 1 不考虑不确定因素和种羊的淘汰更新，假定自然交配期 20 天，母羊都能受孕，孕
期 149 天，每胎产羔 2 只，哺乳期 40 天，羔羊育肥期 210 天，母羊空怀休整期 20 天。该湖羊
养殖场现有 112 个标准羊栏，在实现连续生产的条件下，试确定养殖场种公羊与基础母羊的合
理数量，并估算年化出栏羊只数量的范围。若该养殖场希望每年出栏不少于 1500 只羊，试估算
现有标准羊栏数量的缺口。
问题 2 在问题 1 的基础上，对 112 个标准羊栏给出具体的生产计划（包括种公羊与基础
母羊的配种时机和数量、羊栏的使用方案、年化出栏羊只数量等），使得年化出栏羊只数量最
大。
问题 3 问题 1 和问题 2 中用到的数据都没有考虑不确定性，一旦决定了什么时间开始对
多少可配种的基础母羊进行配种，后续对羊栏的安排和需求也就随之确定。例如，用 3 个羊栏
给 42 只母羊进行配种，孕期需要 6 个羊栏，哺乳期需要 7 个羊栏给怀孕母羊分娩和哺乳，哺乳
期结束就需要给 84 只断奶羔羊和 42 只母羊共安排 9 个羊栏进行育肥和休整。但实际情况并非
如此，配种成功率、分娩羔羊的数目和死亡率等都有不确定性，哺乳时间也可以调控，这些都
会影响空间需求。
现根据经验作以下考虑：
(1) 母羊通过自然交配受孕率为 85%，交配期结束后 30 天可识别出是否成功受孕；
(2) 在自然交配的 20 天中受孕母羊的受孕时间并不确知，而孕期会在 147-150 天内波动，
这些因素将影响到预产期范围；
(3) 怀孕母羊分娩时一般每胎产羔 2 只，少部分每胎产羔 1 只或 3 只及以上，目前尚没有
实用手段控制或提前得知产羔数。羔羊出生时，有夭折的可能，多羔死亡率高于正常。通常可
以按平均每胎产羔 2.2 只、羔羊平均死亡率 3%估算。
(4) 母羊哺乳期过短不利于羔羊后期的生长，通常是羔羊体重达到一定标准后断奶；而哺乳
期过长，母羊的身体消耗就越大，早点断奶，有利于早恢复、早发情配种。一种经验做法是将
哺乳期控制在 35-45 天内，以 40 天为基准，哺乳期每减少 1 天，羔羊的育肥期增加 2 天；哺乳
期每增加 1 天，羔羊的育肥期减少 2 天。除此之外，母羊的空怀休整期可在不少于 18 天的前提
下灵活调控。
此外，如有必要，允许分娩日期相差不超过 7 天的哺乳期母羊及所产羔羊同栏，允许断奶
日期相差不超过 7 天的育肥期羔羊同栏，允许断奶日期相差不超过 7 天的休整期母羊同栏。为
简化问题，不考虑母羊流产、死亡以及羔羊在哺乳期或育肥期夭折和个体发育快慢等情况。
在以上不确定性的考虑下，生产计划的制定与问题 1 和问题 2 将有较大的不同：一旦作出
了“什么时间开始对多少可配种的基础母羊进行配种”的决定，后续羊栏的需求和安排不再是
随之确定的，而是每一步都会出现若干种可能的情况需要作相应的并遵从基本规则的安排处理，
但无法改变或调整上一步。因此，某种意义上，本问题要讨论研究的生产计划将是一个应对多
种可能情况的“预案集”。
请综合考虑可行性和年化出栏羊只数量，制定具体的生产计划，使得整体方案的期望损失
最小。其中整体方案的损失由羊栏使用情况决定，当羊栏空置时，每栏每天的损失为 1；当羊栏
数量不够时，所缺的羊栏每栏每天的损失（即租用费）为 3。


Prompt:
Задача 1
Портфель состоит из 2 акций – А и В с корреляцией –1. Доходности активов соответственно А – 28 % и В – 20 %. Стандартное отклонение А – 20 %, В – 15 %. Определите удельные веса активов А и В, чтобы риск портфеля был равен 0, и найдите доходность сформированного безрискового портфеля.


Prompt:
1. What is the smallest integer on the number line such that its square is between 15 and 30?
2. Are there primes whose digits sum to 9?


Prompt: Привет, у тебя есть три переменные которые содержать числа 2, 10, 3, как из них сделать 36, если ты можешь ставить между ними только +, *, и по разному закрывать скобками, также нельзя двигать переменные a b c, не может быть a c b


Prompt:
Deskripsi
Diberikan sebuah grid yang tersusun oleh
N
N baris dan
M
M kolom. Baris-barisnya dinomori dari
1
1 hingga
N
N dari atas ke bawah. Kolom-kolomnya dinomori dari
1
1 hingga
M
M dari kiri ke kanan. Awal mulanya semua petak berwarna putih.

Terdapat barisan
C
C dan
A
A yang masing-masing terdiri dari
M
M bilangan. Tidak hanya itu, terdapat juga barisan
R
R yang terdiri dari
N
N bilangan.

Anda, sebagai asisten Pak Chanek, diminta untuk mewarnai seluruh petak pada grid tersebut menjadi merah dengan beberapa kali operasi. Setiap operasi dapat berupa salah satu dari tiga jenis operasi berikut:

Operasi tipe
1
1, yaitu mewarnai seluruh petak pada baris
i
i menjadi merah dengan biaya
R
i
R
i

 .
Operasi tipe
2
2, yaitu mewarnai seluruh petak pada kolom
i
i menjadi merah dengan biaya
C
i
C
i

 .
Operasi tipe
3
3, yaitu mewarnai salah satu petak yang berada di kolom
i
i menjadi merah dengan biaya
A
i
A
i

 .
Catat bahwa sebuah petak dapat diwarnai lebih dari sekali.

Berapakah total biaya minimum untuk mewarnai seluruh petak putih menjadi warna merah?

Batasan
1
≤
N
,
M
≤
200
000
1≤N,M≤200000
1
≤
R
i
,
C
i
,
A
i
≤
1
0
6
1≤R
i

 ,C
i

 ,A
i

 ≤10
6

Masukan
N M
R1 R2 R3 … RN
C1 C2 C3 … CM
A1 A2 A3 … AM
Keluaran
Sebuah bilangan bulat yang menyatakan total biaya minimum untuk mewarnai seluruh petak menjadi berwarna merah.

Contoh Masukan 1
4 6
100 4 100 100
1 1 100 100 100 100
1 2 3 4 5 6
Contoh Keluaran 1
60
Penjelasan Contoh 1
Berikut salah satu cara pewarnaan yang optimal:

Lakukan operasi tipe
2
2 pada kolom
1
1 dan kolom
2
2.
Lakukan operasi tipe
1
1 pada baris
2
2.
Sisanya yang belum terwarnai, lakukan operasi tipe
3
3.
Contoh Masukan 2
4 6
1000 1000 1000 1000
1000 1000 1000 1000 1000 1000
1 5 9 12 15 21
Contoh Keluaran 2
252


Prompt: A ball is dropped from a height of 100 meters. It bounces back to 50% of its previous height after each bounce. How far does the ball travel before it comes to rest?


Prompt:
2=14
3=39
4=84
5=155
6=258
9=?


Prompt:
import numpy as np
import pandas as pd
from sklearn.cluster import KMeans
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib.colors import ListedColormap
from scipy.interpolate import griddata
from tqdm import tqdm

# 读取Excel文件
df = pd.read_excel('data.xlsx', index_col=0)
# 获取x轴和y轴数据
x = df.columns.astype(float) * 1852
y = df.index.astype(float) * 1852

# 获取z轴数据（二维数组）
z = -df.values

# 计算x和y轴上的间距
dx = x[1] - x[0]
dy = y[1] - y[0]

# 计算z轴上的梯度
gradient_x, gradient_y = np.gradient(z, dx, dy)

# 计算每个点的倾斜角
slope = (np.arctan(np.sqrt(gradient_x**2 + gradient_y**2)) * 180 / np.pi)

# 使用KMeans聚类算法对倾斜角进行聚类
n_clusters = 5  # 设置聚类数量
kmeans = KMeans(n_clusters=n_clusters, random_state=0)
kmeans = kmeans.fit(slope.reshape(-1, 1))
labels = kmeans.labels_.reshape(slope.shape)

import numpy as np
import pandas as pd
from sklearn.cluster import KMeans
import matplotlib.pyplot as plt
from scipy.interpolate import griddata
from tqdm import tqdm

def calculate_beam_width(depth, opening_angle):
    return 2 * np.abs(depth) * np.tan(np.radians(opening_angle / 2))

def calculate_overlap_rate(d, beam_width):
    return max(0, (2 * beam_width - d) / beam_width)

def optimize_survey_lines(x, y, z, slope, labels, min_depth):
    opening_angle = 120
    desired_overlap = 0.1
    min_line_spacing = 50

    # Create 2D meshgrid
    xi, yi = np.meshgrid(x, y)

    x_min, x_max = np.min(x), np.max(x)
    y_min, y_max = np.min(y), np.max(y)

    survey_lines = []
    total_length = 0
    total_area = (x_max - x_min) * (y_max - y_min)
    covered_area = 0
    excessive_overlap_area = 0

    for cluster in tqdm(range(np.max(labels) + 1), desc="Processing clusters"):
        mask = labels == cluster
        cluster_x = xi[mask]
        cluster_y = yi[mask]
        cluster_z = z[mask]
        cluster_slope = slope[mask]

        if len(cluster_z) == 0:
            continue

        avg_slope = np.mean(cluster_slope)
        adjusted_opening_angle = opening_angle - avg_slope

        avg_depth = np.mean(np.abs(cluster_z))
        beam_width = calculate_beam_width(avg_depth, adjusted_opening_angle)
        line_spacing = beam_width * (1 - desired_overlap)
        line_spacing = max(line_spacing, min_line_spacing)

        # Generate lines parallel to the longer side of the cluster
        if np.ptp(cluster_x) > np.ptp(cluster_y):
            for y_pos in np.arange(np.min(cluster_y), np.max(cluster_y), line_spacing):
                start = np.array([np.min(cluster_x), y_pos])
                end = np.array([np.max(cluster_x), y_pos])
                survey_lines.append((start, end))
        else:
            for x_pos in np.arange(np.min(cluster_x), np.max(cluster_x), line_spacing):
                start = np.array([x_pos, np.min(cluster_y)])
                end = np.array([x_pos, np.max(cluster_y)])
                survey_lines.append((start, end))

    # Calculate total length and overlap
    for i, line in enumerate(survey_lines):
        start, end = line
        line_length = np.linalg.norm(end - start)
        total_length += line_length
        line_covered_area = line_length * beam_width
        covered_area += line_covered_area

        if i > 0:
            prev_start, prev_end = survey_lines[i-1]
            overlap = calculate_overlap_rate(np.linalg.norm(start - prev_start), beam_width)
            if overlap > 0.2:
                excessive_overlap_area += line_covered_area * (overlap - 0.2)

    uncovered_area = max(0, total_area - covered_area)
    uncovered_percentage = (uncovered_area / total_area) * 100
    excessive_overlap_percentage = (excessive_overlap_area / total_area) * 100

    return survey_lines, total_length, uncovered_percentage, excessive_overlap_percentage, adjusted_opening_angle

# Run optimization algorithm
print("Starting survey line optimization...")
survey_lines, total_length, uncovered_percentage, excessive_overlap_percentage, adjusted_opening_angle = optimize_survey_lines(x, y, z, slope, labels, min_depth=-10)

print(f"\nTotal survey line length: {total_length:.2f} meters")
print(f"Uncovered area percentage: {uncovered_percentage:.2f}%")
print(f"Excessive overlap area percentage: {excessive_overlap_percentage:.2f}%")
print(f"Adjusted opening angle: {adjusted_opening_angle:.2f} degrees")

# Plot survey lines and bathymetry
print("Plotting results...")
plt.figure(figsize=(12, 10))
plt.contourf(x, y, z, cmap='terrain')
for line in survey_lines:
    plt.plot([line[0][0], line[1][0]], [line[0][1], line[1][1]], 'r-')
plt.colorbar(label='Depth (m)')
plt.title('Survey Lines and Bathymetry')
plt.xlabel('X (m)')
plt.ylabel('Y (m)')
plt.gca().set_aspect('equal', adjustable='box')
plt.show()

Total survey line length: 1140128.24 meters
Uncovered area percentage: 0.00%
Excessive overlap area percentage: 170.15%
Adjusted opening angle: 119.67 degrees
Plotting results...

要求为计算如下指标：(1) 测线的总长度；(2) 漏测海区占总待测海域面积的百分比；(3) 在重叠区域中，
重叠率超过 20% 部分的总长度。上述代码的结果不符合要求，请你改进。


Prompt: Find the charge of a peptide chain Arg-Ile-Phe-Leu-Asp-Arg-Glu-His at pH 1 and 14.


Prompt:
You have a tank with e capacity of Cities that can be bought and sold by N peope The рещ people are standing in a queeze served sequentially the order of anay A

Some of then waarisella lecof all and some of then want to buy a litre of oil and A destribes tive Here Ajij-1 denotes that the person denotes that the person wants to buy a litre of m wants to sell litre of oil and A=-1

When a pesos wants to sell a be of od but the tank is fall they cannot sell it and become spnet, Smily, when a person wants to buy a litra ef of but the tans is empty, they cannot buy and sесечна цpset. Botiv these cases cause disturbances.

You can musmize the afsturbance by ing the tans initially with a certain X litres of oil

Find the minimum initial amount of oil X that results in the least number of disturbances.

Input Format

The first time contains an enge. I denoting the number of niemants in A.

The rest the credans an vergel: C, daneting the capacity of the tank

Each line of the Nabeequent lines (were si Ni.contains an integer derling Ali

Constraints


Prompt: how can i dilute 500ml of boiling water to have a 95 celsius degree water


Prompt: 一个n级台阶，可以走一步也可以走两步，生成一个java程序，计算出一共有多少种方法，并把方法列出来


Prompt: $\overline{z}$ を $z$ の複素共役として表します。$z_1$ と $z_2$ を、$\overline{z_1}z_2 + z_1\overline{z_2} = 32$ かつ $|z_1\overline{z_1}| \cdot |z_2\overline{z_2}| = 300$ となるような複素数とします。$z_1$ と $z_2$ が原点 $O$ の複素平面上の点 $A$ と $B$ を表す場合、$\triangle AOB$ の面積はいくらでしょうか。


Prompt: draft a python code that generate x,y cordinate from int 49 using d.G=x,y ecc calculation in base16 hexadecimal format


Prompt:
% 清空工作区和关闭所有图形窗口
clear; close all; clc;

% 定义全局变量
global DH_params target_pos joint_inertia joint_avg_speed;

% 设置DH参数 [a, alpha, d, theta_min, theta_max]
DH_params = [
    300, -90, 600, -160, 160;
    1200, 0,   0, -150,  15;
    300, -90,  0, -200,  80;
    0,   -90, 1200, -180, 180;
    0,   -90,  0, -120, 120;
    0,    0,   0, -180, 180
];

target_pos = [1500; 1200; 200];  % 目标位置
joint_inertia = [0.5, 0.3, 0.4, 0.6, 0.2, 0.4];  % 关节转动惯量
joint_avg_speed = [2.0, 1.5, 1.0, 2.5, 3.0, 2.0];  % 关节平均角速度

% 设置多目标遗传算法选项
opts = optimoptions('gamultiobj', 'PopulationSize', 1000, 'MaxGenerations', 100, ...
    'PlotFcn', @gaplotpareto, 'Display', 'iter');

% 多启动法参数
numStarts = 5; % 启动次数
bestSolutions = cell(numStarts, 1);
bestFvals = cell(numStarts, 1);

for start = 1:numStarts
    % 运行多目标优化
    [x, fval] = gamultiobj(@objective_function, 6, [], [], [], [], ...
        deg2rad(DH_params(:,4)), deg2rad(DH_params(:,5)), @constraint_function, opts);

    % 存储结果
    bestSolutions{start} = x;
    bestFvals{start} = fval;

    % 可视化每次运行的结果
    visualize_results(fval);
    disp_results(x, fval);
end

% 比较所有结果，找到全局最优解
globalBestFval = [inf, inf];
globalBestSolution = [];

for i = 1:numStarts
    % 找到最小误差的解
    [minError, idx] = min(bestFvals{i}(:,1));
    if minError < globalBestFval(1) || (minError == globalBestFval(1) && bestFvals{i}(idx,2) < globalBestFval(2))
        globalBestFval = bestFvals{i}(idx,:);
        globalBestSolution = bestSolutions{i}(idx,:);
    end
end

% 显示全局最优解
disp('全局最优解 - 关节角度 (度):');
disp(array2table(rad2deg(globalBestSolution), 'VariableNames', {'Joint1', 'Joint2', 'Joint3', 'Joint4', 'Joint5', 'Joint6'}));
disp('全局最优解 - 目标函数值:');
disp(array2table(globalBestFval, 'VariableNames', {'EndEffectorError_mm_', 'EnergyConsumption_J_'}));

% 目标函数：最小化末端误差和能耗
function f = objective_function(theta)
    global target_pos joint_inertia joint_avg_speed;
    [end_pos, ~] = forward_kinematics(theta);
    f(1) = norm(end_pos - target_pos);  % 末端误差

    mass = 5; g = 9.81;  % 质量和重力加速度
    gravitational_energy = sum(mass * g * forward_kinematics(theta));  % 重力势能

    f(2) = sum(joint_inertia .* (joint_avg_speed .* theta).^2) + gravitational_energy;  % 总能耗
end

% 约束函数：限制末端误差不超过200mm
function [c, ceq] = constraint_function(theta)
    global target_pos;
    [end_pos, ~] = forward_kinematics(theta);
    c = norm(end_pos - target_pos) - 200;
    ceq = [];
end

% 正向运动学计算
function [end_pos, T] = forward_kinematics(theta)
    global DH_params;
    T = eye(4);
    for i = 1:6
        T = T * dh_matrix(DH_params(i,:), theta(i));
    end
    end_pos = T(1:3,4);
end

% DH矩阵计算
function T = dh_matrix(params, theta)
    a = params(1); alpha = deg2rad(params(2)); d = params(3);
    T = [
        cos(theta), -sin(theta)*cos(alpha),  sin(theta)*sin(alpha), a*cos(theta);
        sin(theta),  cos(theta)*cos(alpha), -cos(theta)*sin(alpha), a*sin(theta);
        0,           sin(alpha),             cos(alpha),            d;
        0,           0,                      0,                     1
    ];
end

% 更新 visualize_results 和 disp_results 函数
function visualize_results(fval)
    figure('Name', 'Pareto Fronts', 'NumberTitle', 'off');
    scatter(fval(:,1), fval(:,2), 'filled');
    xlabel('End Effector Error (mm)');
    ylabel('Energy Consumption (J)');
    title('Pareto Fronts');
    grid on;
end

function disp_results(x, fval)
    % 显示最高能耗点，最低误差点，以及均衡点相应的角度及数值
    [minError, minErrorIdx] = min(fval(:,1));
    [maxEnergy, maxEnergyIdx] = max(fval(:,2));

    disp('最高能耗点 - 关节角度 (度):');
    disp(array2table(rad2deg(x(maxEnergyIdx,:)), 'VariableNames', {'Joint1', 'Joint2', 'Joint3', 'Joint4', 'Joint5', 'Joint6'}));
    disp('最高能耗点 - 目标函数值:');
    disp(array2table([minError, fval(maxEnergyIdx,2)], 'VariableNames', {'EndEffectorError_mm_', 'EnergyConsumption_J_'}));

    disp('最低误差点 - 关节角度 (度):');
    disp(array2table(rad2deg(x(minErrorIdx,:)), 'VariableNames', {'Joint1', 'Joint2', 'Joint3', 'Joint4', 'Joint5', 'Joint6'}));
    disp('最低误差点 - 目标函数值:');
    disp(array2table([minError, fval(maxEnergyIdx,2)], 'VariableNames', {'EndEffectorError_mm_', 'EnergyConsumption_J_'}));

    % 找到能量和误差均衡的点
    balancedIdx = find(fval(:,1) <= 1.1 * minError
修改代码去bug


Prompt: 黒玉3個、赤玉4個、白玉5個が入っている袋から玉を1個ずつ取り出し、取り出した玉を順に横一列に12個すべて並べる。ただし、袋から個々の玉が取り出される確率は等しいものとする。どの赤玉も隣り合わない確率pを求めよ。


Prompt: alright chat, master of astrophysics. since the "temperature" of a black hole is inversely proportional to his mass, what would be the "temperature" of a black hole with 112.2 billion suns? would the expanding universe become colder than this temperature allowing the hole to evaporate? how long would it take?


Prompt:
请帮忙估算下面电器每日每周每月的耗电量以及电费，电费1度0.65元：
电饭煲500w, 一天使用2次，一次使用25分钟；
热水器1500w，一天使用1次，一次使用5分钟；
电磁炉1000w，一天使用1次，一次使用30分钟；
笔记本150瓦，每天使用12小时；
此外还有苹果13手机一台，小米3路由器一台，请合理估算


Prompt:
from Crypto.Util.number import getPrime, bytes_to_long
from random import getrandbits
from decimal import Decimal, getcontext
from secret import FLAG

N = 3
MSIZE = 64
PSIZE = 128

getcontext().prec = 2024

def chunk(inp: bytes, n: int) -> list[bytes]:
    return [inp[i:i + n] for i in range(0, len(inp), n)]

def generate() -> list[int]:
    return sorted(getPrime(PSIZE) for _ in range(N))

def otp(data: int) -> list[int]:
    key = [getrandbits(MSIZE) for _ in range(N - 1)]

    # Apply multiple times for extra security! ;)
    for k in key:
        data ^= k

    return key + [data]

def enc(data: int, key: list[int]) -> Decimal:
    return sum(a * Decimal(p).sqrt() for a, p in zip(otp(data), key))

def encrypt(plaintext: bytes, key: list[int]) -> Decimal:
    out = []
    for pt in chunk(plaintext, MSIZE // 8):
        out.append(enc(bytes_to_long(pt), key))

    return out

def main() -> None:
    key = generate()
    ct = encrypt(FLAG, key)

    print(ct)

if __name__ == "__main__":
    main()
otput: [Decimal('613865483278018068252699664344014709603.51690674693116605142775061636120046870769083140402021865687508441855022441399605307934286489840179576316516207824661811160477759304066064291205730622145206447227404607082151753919361349868545307462428210932437992623119255718054243534096986413211959304171135887211540587436459888779324151143025499619973135708235491605216411394962317959813656209629567607308778163987166821943423002154894970121188451784354115735408809579052563235078363998602200794840158081619720701643652284259066561050115602486569569690456649269369511972329349611423116985630502303784458447333880791258687570760036481989066910025496948546865951981177430643941260666483918647746947170525925083317666875914436471646901318749705595856856883346891726769074746650987330068104292614480104709810601198864073890537548001433033723615776224146304045036932985711688197452123005099236554526727107165691391180111605054039864164377438525310672107170437281478322426514683178129245151967154583110424925964982504343607726182942419964058684010006450471850408468724523024702559055514056791838218373955386876710426728739081458827252802226359687030921900842002785720726229892098824084346436742345912593840934579190526250607721076769764172513184565791506300616867803697312829413891917781663107526096475492046133891964020871514785931302212911910422061349227840920633130097807344731335384126657969326959305179863489255473137828869343169464391631468853227921632272791516165605848151467308601405844760824881338357444888188035204000407673339331919346899410420054651077086850532999310419256243385542853597931278472451318309939063364567112084455650629429599265528219435289213175349567467961574398181095962343974433190931067322110878499680729907288265465620467270493396020598011075090015918395921964211257921275263305336676548841001128867189130080538627211844161021733430009452985356108168888790283056201446152044771750329866340679923824523137794621308627512829896733426062792367679307468573465298446962647695970868442917945963296248995830871462803'), Decimal('245036407881857959287329830175942706463.41676802782442994887934627742200369945701412601647195354905097451178257098540606801565385675107318120460650888047274089786443222032370562593772359240722830861471549006005870906901656856743873456596259047367300567575908527230366385292440919562964298317812232800102576149495809413427952810634193186781922614214268017709482845572771078220390794231892763512491653218636267112852524278614145782738125962122310806946368712582799445479001874840170209715004731276988049273955290890839588161009103614134479683822558461311134193156291806539774288015592213012944114677791107334704048050248067790961044976635609645113466054839476916586971455181287563116833537875679872852426596500045916084590678852453213603747411621009244578584574321606967288315507350789499360210767971560504476491167021056575773203913674206412928683729647182751865762672106362511792516860263344700590671316013798899626125150168033775577935989491544189330151349377248071416776567110915867894841003988715963213282301539772548019203054604324657682585086974883293993693364621477765424253486620846417121914627069652179323747638053294808478946841684842391124370675090921956558456987438962361101647423473932215595029566302891192438202103166531605890363352381011015329475002801933772777494625189852526005526409489415563470356698119177369784245367090617016832725703245455210343971981483284742666226075984764682029024194412093587469973865623264825887838718407665448451327538641035583414835042280186524871430174844859199371418365330179156143049238655596474961217964906635309020982967098088279871772621354935094165104148277851308675528401723829872589324051699054275878133606165491429811386740566384976019587172015296076695877847670182335353621529186538640618471230538158759335554894682026973199860563823746052413212330496106907455577991532192244914548792485371507973770044500533179391911148716837049004674981572325374669719464763272448641196278444851808838959425053174592256179039838611577554410785172685506776747109456418706054407235309341'), Decimal('137345167246666152708695783992412451677.21558050568623481295196376367975242130555102201515580316330966697148890411534221923225952905285897183010922406175094262069208074739275097805536185634560982617648832780575658021059502299936366800046975825224832714908010519763833830348421655509168881358855533351044140363192436142763626456404652383592376546952793513967937938636543810723145564947948780573116974908526774605955951469943187865644596524139107481653386681950240101520692953876806620084423038367388560905871875836530739404184303742718262418117805718036122357051928716596276399762849629107495322882675719409051073299089695124393278836941013272789785477953131515760862594743832917404252472431732912434480355250761546326425334589230931109133966112698705631064298540629644087174339495451233415165784208154862632994746999100506080852291240231466075807689137982406781289386109243195384399444517321654449447540787541127685794368775611662723601685545144635953407926414192981817325925467123186823453700610674338813375219959042008201960857405988222431762449068615565770592161380029925528004633925411749903272512805896758350376028759094075163759541252996218647962921737695225607000672827355712319484834139082006323871120783473891599300315829557138141744616522321981481762618665822002496867084381222827633659627360022043393386819320488647893458122132426243653269827396864109912568570075611460793016682214280860824555257535036669613199445516497725227349649115460430628789851678786053832488469704975290639037623666238775508475503615248256142687325810589424170821237702234472583990907852983428072024771329547328770035145579513303770469872758505372515800860236159229593770017270383301142993264982903473737872532789578999358701823743117676045641159467315897055924399632998579024195194906804156820621748747515175782908138772828776041903750590418332731770693214602115172315637279313855936478443754676804416808777432960976763034716998360016818363826448918529745873789717604991019505220078545941396511331142324037841720139678985890298577830351331'), Decimal('248598813941244564314190366976956238846.21287154177193010973910605580779261210365984174749796180419923931211513873961571752231238459134695223104083276961569452018392982079324317136535206452261881826873414602629594471707660157457896214988591398626491707261264840694400878016171848489165321633160306766022332706508213328119352706170273262931280561423851178335709726505935522918272264480451898981178517451201483151591554008657960692345350455891365321593940275460890295306738282405071381298895783171740603832308867778448785711118127825021180253182388515151158156028351993831870085534975634188825697876813579914222802552156594566015228528827397850794636185597567790199750813063356917626291927109597628605896564415404785475255907843090500284371578116809505806089285850503289523274957689703557226650819188753255887580294403919233020671057076485044952433816771965385673494711771474126428651980119604119559562596188714862777544919355156232243652718517709528203262205358442941280815388198987743628689516005444663613668047007886122931571139930784688640045996582570962554713967405406762001906063378302125401898155550603489814135391137430186059041384496922239998747587303366147421038091369722510204804734395780572006815615350115111367222110517089689188974108653902029376056551397900019653419884380778042660817204485470749571339982927039079734483080043934563757326484950182811230031559346385457298240850154178951900739178353564235216322232791700369486112525607172179782914690610012323349746725297341069764127541218762893367556754429312136345943483494236203659704464226488931565438262759788872114208783818792121079840366002113316654063450919686404664767226446116187197107156555434824485171863955895656710114494285915504850382199832564322800850387423097212704670656698969249945269974571051395109564672354772654744585684053379753599154369914818372119637829894516195738013893515476266773802929361065238109305019468285763072333229080155510459572404321239788374283507107268753407976278020430629150787678790300001623259645896920037795093793915002')]
найди флаг


Prompt:
 Петя, Вася и Миша сдают экзамен, чтобы
поступить на курсы по машинному обучению.
Вероятность того, что Петя сдаст экзамен -0,9
(он усердно готовился), Вася - 0,1 (он вообще
не готовился), Миша - 0,5 (как повезет!). После
окончания экзамена объявили, что, к
сожалению, экзамен сдал только один. Какова
вероятность, что это был Вася?
Формат ответа: натуральное число или
несократимая дробь вида plq, где р - целое
число, q - натуральное число.


Prompt:
Sử dụng kiến thức kinh tế học chuyên môn và critical skills để giải bài tập sau. Bên cạnh đó, hãy giải thích để giúp tôi hiểu rõ hơn. "4. Consider the Cobb–Douglas production function:
Yt = AKt 1/3Lt 2/3
where Yt is the amount of output, Kt is the amount of capital, Lt is the amount of labor, and A is a
parameter that measures the state of technology. Saving in the economy is 20% (s=0.2) a year
a. Does this production function exhibit constant return to scale?
b. For this production function, the marginal product of labor (MPL) and the marginal product of
capital are:
MPL = (2/3) A(K/L)1/3
MPK = (1/3) A(L/K)2/3
Verify that these marginal products are diminishing.
c. The labor and capital markets are competitive, so labor and capital are paid, in real term, their marginal products. Find the labor’s share and capital’s share of output.
d. Find the production function for output per labor (yt = Yt/Lt)
e. If the amount of capital depreciates 3% (d = 0.03) a year and labor grows 7% (n = 0.07) a year.
Find the amount of investment needed to keep the capital-labor ratio (kt = Kt/Lt) stays the same over time.
f. If A = 1, find the steady-state levels of the capital-labor ratio and output per worker."


Prompt:
Suppose I tell you that the information provided by the character being yellow (or not) about
whether the character’s name starts with ‘H’ is 0.0015 bits. How can you now compute the
conditional mutual information from the variable “sitting in something” to the variable “name
starts with ‘H’”, conditioned on “colour is yellow?”, without referring back to the probabilities
here? (2 marks)


Prompt: 你有四个装药丸的罐子，每个药丸都有一定的重量，被污染的药丸是没被污染的药丸的重量+1。只称量一次，如何判断哪个罐子的药被污染了？


Prompt: the smallest number which when divided by 20,25,35,40 leaves the remainder 6 .when divided by 14,19,23,34 respectively the difference between the divisor and the corresponding remainder is 6. what is the number


Prompt: I show you two candy papers, one of them is empty one of them has a candy in it, I ask you to pick one, when you pick one i show you that you have the one with the candy inside, now you have an option to either switch or stay with the choice you made, which option has the best outcome?


Prompt:
## Problem statement
There is a grid of size $N \times N$, and each square has $-1$ or $1$ written on it. You want to start from the top left square $(1, 1)$ and move to the bottom right square $(N, N)$. However, you must follow the following rules to move.

* The squares you can move to from square $(i, j)$ are $(i+1, j)$, $(i, j+1)$, and $(i+1, j+1)$.
* You must move so that the product of the numbers written on the squares you pass through when moving is $-1$. In other words, you must pass through squares with an odd number of $-1$s written on them.

You must determine whether there is a path that you can move through according to the above rules, and if so, print one of them.

## Constraints

* $2 \leq N \leq 1000$
* Each square in the grid has $-1$ or $1$ written on it.

## Input

Input is given from standard input in the following format.

```
N
a_{1, 1} a_{1, 2} ... a_{1, N}
a_{2, 1} a_{2, 2} ... a_{2, N}
:
a_{N, 1} a_{N, 2} ... a_{N, N}
```

* $a_{i, j}$ represents the number written in the $i$th row and $j$th column of the grid.

## Output

If a path exists, output one of them in the following format.

```
YES
(1, 1)
(x_2, y_2)
:
(x_k, y_k)
(N, N)
```

* $(x_i, y_i)$ represents the $i$th square on the path.
* If no path exists, output `NO`.


Prompt:
from pulp import *
import random

def generate_coverage(strategy_id):
    """
    根据策略ID生成覆盖区域。
    :param strategy_id: 策略的类型ID
    :return: 覆盖区域的坐标偏移列表
    """
    if  strategy_id == 1:
        return [(i, j) for i in range(-1, 2) for j in range(-1, 2)] # 3x3
    elif strategy_id == 2:
        return [(i, j) for i in range(-2, 2) for j in range(-1, 2)] # 4x3
    elif strategy_id == 3:
        return [(i, j) for i in range(-2, 3) for j in range(-2, 3)] # 5x5
    elif strategy_id == 4:
        return [(0, j) for j in range(-6, 7)] # 全列覆盖
    elif strategy_id == 5:
        return [(i, 0) for i in range(-8, 9)] # 全行覆盖
    elif strategy_id == 6:
        return [(0, 0), (0, 1), (0, -1), (1, 0), (-1, 0)] # 小十字
    elif strategy_id == 7:
        return [(i, j) for j in range(-1, 2) for i in range(-8, 9)] # 三行覆盖
    else:
        raise ValueError("未知的策略ID")
def generate_cross_coverage(rows, cols):
    """
    根据给定的行数和列数生成十字覆盖范围。
    :param rows: 十字的行数
    :param cols: 十字的列数
    :return: 覆盖区域的坐标偏移列表
    """
    coverage = []
    # 纵向覆盖
    for i in range(-rows, rows + 1):
        coverage.append((i, 0))
    # 横向覆盖
    for j in range(-cols, cols + 1):
        coverage.append((0, j))
    return coverage

def add_strategy(strategies, strategy_id, cost, rows=None, cols=None):
    """
    添加策略到策略字典中，并动态生成唯一的策略ID和覆盖范围。
    :param strategies: 策略字典
    :param strategy_id: 策略的类型ID
    :param cost: 策略的成本
    :param rows: 十字的行数（仅用于十字策略）
    :param cols: 十字的列数（仅用于十字策略）
    """
    global strategy_count
    strategy_count += 1 # 增加计数器

    # 根据策略类型ID生成覆盖范围
    if strategy_id == 8:
        coverage = generate_cross_coverage(rows, cols)
    elif strategy_id == 9:  # 复制策略
        coverage = []  # 复制策略的覆盖范围将在求解过程中动态确定
    else:
        coverage = generate_coverage(strategy_id)

    # 添加策略到字典，使用唯一ID
    strategies[f"{strategy_count}"] = {"coverage": coverage, "cost": cost}
def solve_special_card_problem(points_to_cover, obstacles, consider_copy_strategy):
    # 定义问题
    prob = LpProblem("Map Coverage Problem", LpMinimize)

    # 定义常量
    MAP_WIDTH = 9
    MAP_HEIGHT = 7

    strategies = {}
    global strategy_count
    strategy_count = 0

    # 添加策略
    add_strategy(strategies, 3, 275)  # 5x5
    add_strategy(strategies, 1, 275)  # 5x5
    add_strategy(strategies, 2, 2275)  # 5x5
    add_strategy(strategies, 4, 275)  # 5x5
    add_strategy(strategies, 6, 275)  # 5x5
    add_strategy(strategies, 2, 275)  # 5x5
    add_strategy(strategies, 2, 275)  # 5x5
    add_strategy(strategies, 3, 2375)  # 5x5
    add_strategy(strategies, 4, 275)  # 5x5
    add_strategy(strategies, 5, 275)  # 5x5
    add_strategy(strategies, 7, 25)  # 5x5
    add_strategy(strategies, 8, 325,2,2)  # 5x5
    add_strategy(strategies, 8, 275,3,5)  # 5x5
    add_strategy(strategies, 8, 2735,2,4)  # 5x5
    add_strategy(strategies, 9, 325)  # 复制策略
    add_strategy(strategies, 3, 325)  # 复制策略
    add_strategy(strategies, 2, 13)  # 复制策略
    add_strategy(strategies, 2, 12)  # 复制策略
    add_strategy(strategies, 4, 8)  # 复制策略
    add_strategy(strategies, 6, 6)  # 复制策略
    add_strategy(strategies, 7, 10)  # 复制策略

    # 创建决策变量
    x = LpVariable.dicts("strategy",
                         [(i, j, s) for i in range(1, MAP_WIDTH + 1)
                          for j in range(1, MAP_HEIGHT + 1)
                          for s in strategies.keys()],
                         cat='Binary')

    # 为复制策略创建新的决策变量
    copy_coverage = {}
    if consider_copy_strategy:
        copy_coverage = LpVariable.dicts("copy_coverage",
                                         [(i, j, s, str(strategy_count)) for i in range(1, MAP_WIDTH + 1)
                                          for j in range(1, MAP_HEIGHT + 1)
                                          for s in strategies.keys() if s != str(strategy_count)],
                                         cat='Binary')

    # 目标函数
    prob += lpSum([strategies[s]["cost"] * x[i, j, s] for i in range(1, MAP_WIDTH + 1)
                   for j in range(1, MAP_HEIGHT + 1)
                   for s in strategies.keys()])

    # 约束条件
    # 1. 每个待处理点位至少被覆盖一次
    for point in points_to_cover:
        if point not in obstacles:  # 排除既是障碍物又是待覆盖点的情况
            i, j = map(int, point.split('-'))
            coverage_constraint = lpSum([x[i_s, j_s, s] for i_s in range(1, MAP_WIDTH + 1)
                                         for j_s in range(1, MAP_HEIGHT + 1)
                                         for s in strategies.keys() if s != str(strategy_count)
                                         if (i - i_s, j - j_s) in strategies[s]["coverage"]])

            if consider_copy_strategy:
                coverage_constraint += lpSum([copy_coverage[i_s, j_s, s, str(strategy_count)]
                                              for i_s in range(1, MAP_WIDTH + 1)
                                              for j_s in range(1, MAP_HEIGHT + 1)
                                              for s in strategies.keys() if s != str(strategy_count)
                                              if (i - i_s, j - j_s) in strategies[s]["coverage"]])

            prob += coverage_constraint >= 1

    # 2. 不能在障碍上放置策略
    for obstacle in obstacles:
        i, j = map(int, obstacle.split('-'))
        for s in strategies.keys():
            prob += x[i, j, s] == 0

    # 3. 每个策略只能被放置一次
    for s in strategies.keys():
        prob += lpSum([x[i, j, s] for i in range(1, MAP_WIDTH + 1)
                       for j in range(1, MAP_HEIGHT + 1)]) <= 1

        if consider_copy_strategy:
            # 确保复制策略只复制一个其他策略
            for i in range(1, MAP_WIDTH + 1):
                for j in range(1, MAP_HEIGHT + 1):
                    prob += lpSum([copy_coverage[i, j, s, str(strategy_count)]
                                   for s in strategies.keys() if s != str(strategy_count)]) == x[i, j, str(strategy_count)]

            # 确保只有在原策略被使用时才能复制
            for s in strategies.keys():
                if s != str(strategy_count):
                    prob += lpSum([copy_coverage[i, j, s, str(strategy_count)]
                                   for i in range(1, MAP_WIDTH + 1)
                                   for j in range(1, MAP_HEIGHT + 1)]) <= lpSum([x[i_s, j_s, s]
                                                                                for i_s in range(1, MAP_WIDTH + 1)
                                                                                for j_s in range(1, MAP_HEIGHT + 1)])

    # 求解问题
    prob.solve()

# 输出结果
    print("Status:", LpStatus[prob.status])
    if LpStatus[prob.status] == "Optimal":  # 有解
        print("火苗花费 =", value(prob.objective))

        # 存储复制策略的信息
        copy_strategy_info = None

        # 输出第一组完整的解
        for s in strategies.keys():
            if s != str(strategy_count):  # 不是复制策略
                for i in range(1, MAP_WIDTH + 1):
                    for j in range(1, MAP_HEIGHT + 1):
                        if value(x[i, j, s]) > 0.99:
                            print(f"对策卡 {s} 放置于 ({i},{j})")

                            # 检查是否有复制这个策略的复制策略
                            for i_copy in range(1, MAP_WIDTH + 1):
                                for j_copy in range(1, MAP_HEIGHT + 1):
                                    if value(x[i_copy, j_copy, str(strategy_count)]) > 0.99 and \
                                            value(copy_coverage[i_copy, j_copy, s, str(strategy_count)]) > 0.99:
                                        print(
                                            f"对策卡 {strategy_count} (复制了 {s}) 放置于 ({i_copy},{j_copy})")
                            break  # 找到策略放置位置后跳出内层循环
                    if value(x[i, j, s]) > 0.99:
                        break  # 找到策略放置位置后跳出外层循环

    else:
        print("问题无解")
# 定义待处理点位列表
points_to_cover = ["9-5", "3-2", "9-2", "9-1","9-4","9-3","9-7","8-3","1-1","1-1"] # 添加所有待处理点位

# 定义障碍列表
obstacles = ["1-1", "2-3", "9-6"] # 添加所有障碍点位

# 调用求解函数
consider_copy_strategy = True # 设置为True以考虑复制策略，设置为False以忽略

solve_special_card_problem(points_to_cover, obstacles, consider_copy_strategy)


Prompt: A piano is being raised to the third floor of a building using a rope and pulley. The piano is 10.9 m above the ground, moving upwards at 1.5 m s-1, when the rope snaps. Calculate how much time (in s) elapses before the piano hits the ground. Give your answer to 1 d.p.


Prompt:
При каких a корни уравнения x2 + x + a = 0 больше a?

Вопрос 3Выберите один ответ:

a >- 2

a≥-2

а< -2

a∈(0;2)∪(2;+∞)


Prompt: Please find the paramters a,b, and c for the following Ansatz of an equation f(x)=abs(a+x)+bx+c that should govern the following behaviour: linear from x=1 to x=-inf e.g. f(0)=0, f(-1)=-1 etc and f(2)=4 with a kink at f(1) that has the value of 1. Please explain every step of your solution and test your final results on at least 3 points by calculating the f(x) with your determined parameters a,b,c.


Prompt:
Где ошибка в коде
среднее квадратичное чисел a и b
num4 = pow(a + b, 2) / 2
print(sqrt(num4))


Prompt:
В горизонтально расположенной трубке, запаянной с одного конца, столбиком ртути длиной 12
 см заперт слой воздуха. Давление воздуха снаружи трубки равно 750
 мм рт. ст. Определите давление воздуха, запертого внутри трубки. Ответ дайте в мм рт. ст., округлив до целого числа.


Prompt:
Suppose your classmate Rachel is faced with the following decision: they either receive $6,000 with certainty, or a 50-50 chance of receiving $3,000 or $9,000. Suppose Rachel prefers the certain $6,000.
True or False: Based on their preferences, Rachel is not a risk averse person.

	True
	False
Suppose there is a disease certain to impact exactly 2% of a population and that each member of the population is equally likely become infected. Treatment for individuals who are infected costs $36,000. Assume in this scenario that the necessary treatment for the disease is the only healthcare cost faced by this population.
The expected cost of healthcare is
$________
. If members of the population (prefer or do not prefer)    to pay the expected cost of healthcare with certainty, rather than take on the 2% risk of having to pay the full cost of the treatment, they are risk averse.


Prompt: 你能提供在广告历史累计效应下企业在n个企业的行业竞争中的广告决策的数理模型吗


Prompt:
Data Science - Price Optimization Task You are provided with synthetic data from a pricing experiment conducted on embedded travel insurance within an OTA (Online Travel Agency) funnel for flights. Each time a customer proceeds to checkout a flight, an insurance quote is generated. The quotes dataset includes flight attributes and pricing details for each quote as described below: row_id country_of_origin    country_of_destination lead_time    trip_duration  ticket_price  number_of_passengers  return_trip   base_retail_premium   split  p    conversion   retail_premium modifier 1    Canada India  133   17   1572.96 3    TRUE  157.30 tr   0.2   1    173.03 10 2    Spain  Brazil 62   16   1751.35 1    TRUE  175.14 tr   0.2   0    192.65 10 3    USA   Japan  4    7    1961.71 4    FALSE  196.17 tr   0.2   0    235.41 20 4    USA   Australia    66   27   719.63 3    TRUE  71.96  tr   0.2   0    64.77  -10 5    France Australia    175   6    1932.60 1    FALSE  193.26 tr   0.2   0    173.93 -10 row_id column is a unique quote identifier. country_of_origin column indicates country from which journey starts. country_of_destination column indicates country where journey ends. lead_time column represents number of days between booking and departure. trip_duration column shows duration of trip in days. ticket_price column lists price of flight ticket. number_of_passengers column shows how many passengers are included in quote. return_trip column is a boolean indicating whether trip is a round trip. base_retail_premium column shows base price of travel insurance before any modifications. split column indicates whether data is part of training set ('tr') or test set ('te'). Note that the outcomes for the test set are not available - it is here so that your submission can be evaluated. p column represents the sizes of experiment groups for different modifiers. In this case they are equal at 20% of quotes. The modifier column represents a random modification to the base price based on a hashed customer ID. The retail_premium is calculated as base_retail_premium * (1 + modifier/100). If the insurance policy is purchased, the conversion field is set to 1, and the total retail premium received is retail_premium * conversion. Your task is to analyze the "training" data and construct a pricing rule that maps flight attributes to modifiers. This rule should aim to maximize the average converted premium per quote when deployed. Once you have a candidate rule, we will evaluate it on a hold out sample. To do this you should assign your proposed modifier to each row where split equals "te" (the test data). The submission will be evaluated by how much of total available lift over the default submission of always assigning modifier 0 has been captured: The scoring formula is given by: Score = 𝑉 ( 𝜋 optimal ) − 𝑉 ( 𝜋 proposed ) 𝑉 ( 𝜋 proposed ) − 𝑉 ( 𝜋 base ) Score= V(π proposed  )−V(π base  ) V(π optimal  )−V(π proposed  )  where: $ V(\pi_{\text{optimal}}) $ is the value of the optimal policy, representing the maximum possible value that could be achieved (remember this is a simulation). $ V(\pi_{\text{proposed}}) $ is the value of the proposed policy, representing the value achieved by the policy you are testing. $ V(\pi_{\text{base}}) $ is the value of the baseline policy, where no modifications are made to the pricing. The score measures the relative improvement of the proposed policy over the baseline, normalized by the maximum possible improvement (from baseline to optimal). Output your results in a CSV file named submission.csv with the following format: row_id proposed_modifier 499386 0 499387 0 499388 0 499389 -20 499390 -20 499391 -20 499392 -20 The CSV file should only include rows from the test split. Outline how you will model this task - what will be you dependent variable, what modelling framework is approriate (list at list 3). Give formulas for the optimisation and validation objectives.

Make sure to list your proposed solution step by step with formulas


Prompt: Mikä päivä on 1311861118074000


Prompt:
How do Box-Cox transformation in this data ?

"library(ExpDes.pt)
library(readxl)
library(car)
library(easyanova)
library(openxlsx)
library(dplyr)
library(tidyverse)
###                                 #####
### Carregando o conjunto de dados #####JUNCEA#########
###JUNCEA#####
data <- read_excel("../data/juncea.xlsx")
data$Tratamento = as.factor(data$Tratamento)
data$Rep = as.factor(data$Rep)
names(data)

##############################################
#############################################
names(data)
data0 = data[c(1,3:22)]
data0 <-ea1(data = data0, design=1, alpha=0.10, list = T)

data0"


Prompt: Assume I have a shape that looks like an ice cream cone (2d) consisting of a semi circle on top of a equilateral triangle. Assume that the triangle has a side length of 6. The density of the semicircle is p and the density of the triangle is 3p. Assume that the bottom coordinate of the triangle is at (0,0) and the left point is at (-3, 3 * sqrt(3)) while the right point of the triangle is at (3, 3* sqrt(3)). Assume that the top point of the semicircle is at (0, 3 + 3 * sqrt(3)). Also assume that the left and right points of the triangle correspond to the ends of the semi circle respectively. I need to find the center of mass of this shape in coordinate form. Since the shape is symetrical, assume that the x coordinate is 0. Find the y coordinate for this shape.


Prompt: Calculus application in Business (provide a sample problem with solution)


Prompt:
推荐下720*720p60 的porch值，填到下面参数中THW =
TVW =
AHW = 720
AVW = 720
HBP =
VBP =
HFP =
VFP =
HSW =
VSW =
HSP = 0
VSP = 0


Prompt:
dùng ngôn ngữ lập trình C Thuật toán Euclide mở rộng:
(1) Nếu b = 0 thì đặt d - a,x - 1,y - 0 và return (d, x,y)
(2) Đặt
X2 - 1, X1 - 0, У2 - 0, У1 - 1
(3) While b > 0 do
3.1. q- [a/b] , r-a- qb,
X - X2 - qX1 .
у - Уг - ЧУ1
3.2. a -b,b - r,x2 + X1
Х1 -Х,Уг - Ул,У1 - у
(4) Đặt d - a,x - x2, y - Y2 và return (d, x, y)


Prompt:
A loan of Rp11 million. The repayment is made with an installment of Rp1.1 million per month, starting next month for 12 installments. What is the effective annual rate?
Question 20Answer

a.
41.29%

b.
10%

c.
5%

d.
35.07%

e.
2.92%


Prompt: 1, 2, 5, 13, 35, 93, 260 ... try to determine the most likely next number from the following options 620, 717, 812


Prompt:
1.2.2 Radial Direction

Fr = mar (4)
The radial forces are tension and the radial component of gravity:

T − mg cos θ = mL  ̇θ
2

(5)

This equation determines the tension in the string.


Prompt:
Вычислить несобственный интеграл или показать, что он расходится:

\int_{0}^{\infty } \cos(x)e^{sin(x)}


Prompt:
修改并优化“针对自发上报的不良反应数据存在的数据不完整、错误上报、因果关系确定性不足以及单一信号挖掘方法的局限性等问题，设计一种集成多种统计学方法的不良反应信号挖掘模型，提高信号真实性，减少偏倚，优化不良反应信号质量，为上市后药物提供更为专业的预警支持和风险评估，并为药物分子不良反应预测模型的开发和升级提供数据支持。具体采用的信号监测方法包括：
（1）报告比数比法
报告比数比法(ROR)是比例失衡的一种测量方法，用于鉴别被报告药品与不良反应之间的潜在联系。通过比较目标药品与目标事件组合出现的频率与背景频率的差异，对异常信号进行诊断。假设2×2的药品与不良反应的四格表如下：
表2.1 药品与不良反应的四格表
药品	目标药品不良反应	其他药品不良反应	总计
目标药品	a	b	a+b
其他药品	c	d	c+d
总计	a+c	b+d	a+b+c+d
其中a,b,c,d分别各类别出现频数。ROR计算公式如下：

其95%的置信区间为：

在一个药品不良反应事件报告的数据库中，当目标药品与目标事件的组合出现的频率明显高于整个数据库的背景频率，报告比数比的数值越大。若数值达到设定阈值，则认为目标药品所导致的目标药品不良反应是一个值得注意的信号。常用的显著有意义的比例失衡的阈值是95%的置信区间下限大于1，要求样本个数等于或大于3。
（2）比例报告比值法
比例报告比值法(PRR)的基本原理与报告比数比值相同，比例报告比值越大，则目标药品与目标事件的组合出现的频率明显高于整个数据库的背景频率。计算公式为：

其95%的置区间为：

常用的显著有意义的比例失衡的阈值是95%置信区间的下限大于1，对样本量未提出特殊要求。
（3）贝叶斯可信传播神经网络法(BCPNN)
贝叶斯可信传播神经网络法反映一种关联程度，通过目标药品不良反应发生的先验概率和后验概率的比值的对数值来表示，即信息分数(IC)。计算公式为：

当信息分数的置信区间下限大于0时，说明目标药品与目标不良反应存在关联，即提示一个信号的存在。
（4）Yule’s Q方法
Yule’s Q方法也为目前所使用的一个衡量比例失衡的方法，其计算公式为：

（5）卡方检验方法
使用统计假设检验方法检验目标药品与目标不良反应是否存在关系，计算公式为：

使用p值度量，p值小于等于0.05，则提示一个信号的存在。
结合多种信号挖掘方法，根据专家经验或领域知识进行规则设定，根据历史数据或其他信息动态调整阈值。整体研究方案如下：
方案实施步骤如下：
步骤1：药品上市后阶段，对目标药品的不良反应事件报告进行收集，处理缺失值和异常值，确保数据质量。
步骤2：计算背景事件频率，目标药品与目标事件组合出现的频率，作为信号挖掘的基础数据；联合不良反应信号监测方法，包括ROR、PRR、RRR、BCPNN、Yule’s Q方法和卡方检验，计算反映目标药品与目标事件组合出现的频率与背景频率差异的统计量，根据药厂和监管机构的实际需求，建立合理的阈值。
步骤3：将不同方法的输出指标作为模型的输入特征，选择多种机器学习模型，包括逻辑回归、支持向量机、随机森林、梯度提升树等，对特征进行训练，学习不同指标之间的关系，预测不良反应信号的可能性。
步骤4：基于投票学习方法，融合各机器学习模型的预测结果，输出最终的不良反应信号。对模型进行优化，确保其鲁棒性与预测准确性。通过历史数据验证模型的有效性，调整模型参数与阈值。最终模型将用于实际药品上市后的不良反应监测，提供预警支持与风险评估。
通过集成多种信号挖掘方法并结合机器学习技术，提升不良反应信号的识别准确性与可靠性，对上市后药品安全性监测提供技术支持，助力药物分子不良反应预测模型的开发与升级。”


Prompt:
R语言编程：( 1) 首先，用Gaussian Copula、Student’s t Copula 、Gumbel Copula、Frank Copula 、Clayton Copula  函数对操作风险单元的相依结构进行拟合，估计其 Copula 函数的参数，并根据检验结果，确定一个拟合最佳的 Copula 函数C( u，v) ，其中，u = G1 ( x1 ) 、v = G2 ( x2 ) ，这里 u、v 均服从( 0，1) 均匀分布
( 2) 生成两个独立的服从( 0，1) 均匀分布的随机数 u 和 w，u 为第一个需要模拟的随机数。由Copula 理论知 Cu ( v) 、Cv( u) 均服从均匀分布，因此
令 Cu ( v) = w，通过 Cu ( v) 的逆函数即可计算得到v = C（-1）u ( w) ;
( 3) 根 据 风 险 单 元 的 边 际 分 布 G1 ( x1 ) 和G2 ( x2 ) ，计算与 u 和 v 相对应风险单元的操作风险损失 x1 = G 1（-1） ( u) ，x2 = G2（-1） ( v) ; 但由于 G1 ( x1 ) 和G2 ( x2 ) 没有具体的解析形式，我们利用单风险单元求 VaＲ 的思路，将模拟生成的不同操作风险单元损失L1(1)……L1（N),L2(1)……L2（N）按从大到小进行排序，令:x1 = G1（－ 1）( u) = min {L1(t)，t/N>=u}; x2 = G 2(-1)( v) = min {L2(t)，t/N>=v}
( 4) 计算两个风险单元操作风险: L = x1 + x2，由此得到未来银行内部整体操作风险损失的一个可能情况;
( 5) 将步骤( 3) 和步骤( 4) 重复多次，如 M 次，那么能模拟得到考虑风险单元相依结构下的银行未来所面临的内部整体操作风险的 M 种可能情
景，记为{ L1，L2，…，LM } 由此可以得到内部整体操作风险的损失分布 G( x) ;
( 6) 将 M 个整体操作风险可能损失按从小到大的顺序排序L1……LM， 则置信水平为 α 下的为 VaＲα :
VaＲα = min{L(t):t/M>=α}其中1<=t<=m*N


Prompt: 求解以下物理问题：汽车以 20m/s 的速度在平直公路上行驶，急刹车时的加速度大小为 5m/s^2 ，则自驾驶员急踩刹车开始，经过几秒后完全刹停？经过 5 秒后的位移大小为多少米？


Prompt: If A + B = B*C, B + C = C*A, and C + A = A*B; Find ratio of A:B:C


Prompt: Jen enters a lottery by picking $4$ distinct numbers from $S=\{1,2,3,\cdots,9,10\}.$ $4$ numbers are randomly chosen from $S.$ She wins a prize if at least two of her numbers were $2$ of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of her winning the grand prize given that she won a prize is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.


Prompt:
(((((((((((((((((((((((((((((((((((((0.5x(1+1))x((0.5x(1+1))+(0.5x(1+1))x((0.5x(1+1))+(0.5x(1+1)+((0.5x(1+1))x((0.5x(1+1))+((0.5x(1+1)+((0.5x(0.5x(1+1)))+(((0.5x(1+1))x((0.5x(1+1))+(0.5x(1+1)x((0.5x(1+1))+(0.5x(1+1))+(((0.5x(2+2))x2))+0.5) / ((((((0.25x(((0.5x(1+1)) / (1+1))) x (1+1))) x (((0.5x(1+1))) x (1+1))))x((1+1)x(1+1)))x(((((((0.25x(((0.5x(1+1)) / (1+1))) x (1+1))) x (((0.5x(1+1))))))))))))))))))))))))))))))))))-6.5)))-(((10+((0.5x4))))) = y

Find y


Prompt:
1.预测：使用当前的权重向量$\beta_{n-1}$来预测新数据点的目标值$\hat{y}_n=t3_n^{\prime}\cdot\beta_{n-1}$。
2.计算误差：计算预测误差$e_n=y_n-\hat{y}_n$。
3.更新协方差矩阵：使用下面的公式更新协方差矩阵$P_n:$
$$P_n=\dfrac{1}{\lambda}(P_{n-1}-\dfrac{P_{n-1}\cdot t3_n\cdot t3_n'\cdot P_{n-1}}{\lambda+t3_n'\cdot P_{n-1}\cdot t3_n})$$
其中$\lambda$是正则化参数。

4.更新权重向量：使用下面的公式更新权重向量$\beta_n:$
$$\beta_n=\beta_{n-1}+P_n\cdot t3_n\cdot e_n$$


Prompt:
hivesql,已知每个月的交易次数,求近3月月交易次数变异系数
近6月月交易次数变异系数
近12月月交易次数变异系数


Prompt: E. coli cells are rod-shaped, 2 micrometers long and 0.8 micrometers in diameter. One E. coli cell contains 15,000 spherical ribosomes with a diameter of 18 nm. What percentage of the cell volume do the ribosomes occupy?


Prompt: Prove that  {x:∑∞n=1n4|xn|2<∞} is a  closed subspace of l^2


Prompt: Antiderivar la siguiente función f(x)=(x+1)(2x-1)


Prompt:
assume a current value of property is 600,000 account for 3% inflation every year and deduct these annual taxes, and give me the value of the initial investment each year
taxes:
2024: $10,000
2025: $10,350
2026:$10,700
2027: $11,050
2028:$11,400
2029:$11,750
2030:$12,100
2031: $12,450
2032: $12,800
2033: $13,150
2034: $13,500
2035:$13,850
2036: $14,200
2037: $14,550
2038: $14,900
2039:$15,250
2040:$15,600
2041: $15,950
2042:$16,300
2043: $16,650


Prompt: Using the binary operation defined as (a,b) = a^2 + b^2, how can we transform the expression a^2 + b^2 + c^2 such that only this operation that i defined and no other mathematical operations(+,-,*,/,exp.,etc) except for the square root (√) are present in the final expression? Please provide a step-by-step deductive reasoning process to arrive at the answer.


Prompt: 4 + 3 = 21 ，2 + 5 = 35， 7 + 4 = ??


Prompt: распредели 53 файла по номерам между четырьмя поставщиками, чтобы каждый файл был отправлен троим поставщикам


Prompt: Verdantix finds that the overall climate financial data and analytics market was worth $468 million in 2022 and will grow to more than $1.3 billion by 2028, at a CAGR of 19%. What was the market worth in 2023 and in 2024?


Prompt: Simplify (p↔q)∧(r↔q)


Prompt: 已知二次型f（x1,x2,x3）= 3x1^2+4x2^2+3X3^2+2x1x3；（1）求正交变换x=Qy将f（x1,x2,x3）化为标准形


Prompt: Assume 50% women have 1 child and 50% women have 3 children, meaning the average fertility rate is 2. What is the average number of siblings?


Prompt:
Jonain tulevaisuuden aikana, kun kotona pysymisen määräykset on poistettu, aviopari, Lauri ja Maija, suuntaavat baariin juhlimaan uutta vapauttaan.

He löytävät sieltä neljä muuta pariskuntaa, joilla oli sama idea.

Innoissaan sosiaalisesta kontaktista, jokainen henkilö viidestä pariskunnasta kopauttaa kyynärpäitään (uusi tervehdys) innokkaasti jokaisen henkilön kanssa, jota he eivät vielä tunne.

Käy ilmi, että monet ihmiset tunsivat toisensa ennestään, joten kun Maija kysyy kaikilta, montako kyynärpäätä he kopauttivat, hän saa hämmästyttävästi yhdeksän eri vastausta!

Kysymys: Montako kyynärpäätä Lauri kopautti?


Prompt:
# RATING: 3100
# TAGS: greedy,implementation,sortings
# LANGUAGE IS python3
# CORRECT SOLUTION
# In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game.
# After days of struggling, you finally find the binary string that encodes the gacha system of the game.
# In order to decode it, you must first solve the following problem.
#
# You are given a binary string s of length n.
# For each pair of integers (l, r) (1 ≤ l ≤ r ≤ n), count the number of pairs (x, y) (l ≤ x ≤ y ≤ r) such that the amount of 0 equals the amount of 1 in the substring s_x,s_x+1,...,s_y.
#
# Output the sum of counts over all possible (l, r) modulo 10^9 + 7.
#
# Input
#
# The first line contains t (1 ≤ t ≤ 1000) — the number of test cases.
#
# Each test case contains a binary string s (1 ≤ |s| ≤ 2 ⋅ 10^5).
# It is guaranteed s only contains characters 0 and 1.
#
# It is guaranteed the sum of |s| over all test cases does not exceed 2 ⋅ 10^5.
#
# Output
#
# For each test case, output an integer, the answer modulo 10^9+7.
#
#
# Example
#
# Input
#
#
# 4
# 0000
# 01010101
# 1100111001
# 11000000111
#
#
# Output
#
#
# 0
# 130
# 147
# 70


Prompt: Привет! Напиши формулу для нормированной диаграммы направленности трехмерной фазированной решетки.


Prompt: 一根木棍折成3段，围成正方形的概率是多少？


Prompt: Given one array which depicts a line and a list of arrays which depict a line each, how can we find the array which has the best continuation to the given line? The arrays contain the xy coordinates. Please provide the solution in python.


Prompt:
La suma de las longitudes de tres palos es de 12 cm.
El palo más largo mide 4 cm más que el palo más corto.
Todos los palos miden como mínimo 2 cm.
Cuál es la longitud de cada palo?.


Prompt: "The number of Library's HVAC Crew equals 4 more than Northshore Region's Project. The number of Apartment Project's Concrete Crew equals 3 times Bayview Region's Office Building. The number of Northshore Region's Library equals the sum of Bayview Region's Apartment Project, Brookside Region's Project, and Northshore Region's Apartment Project. The number of Bayview Region's Office Building equals 3. The number of Northshore Region's Office Building equals the difference of Bayview Region's Office Building and Brookside Region's Project. The number of Apartment Project's HVAC Crew equals Brookside Region's Office Building. The number of Bayview Region's Apartment Project equals 0 more than the sum of Bayview Region's Office Building and Apartment Project's Concrete Crew. The number of Office Building's HVAC Crew equals 1 times the difference of Bayview Region's Apartment Project and Bayview Region's Office Building. The number of Brookside Region's Office Building equals the difference of Office Building's HVAC Crew and Apartment Project's Concrete Crew. The number of Apartment Project's Welding Crew equals 6. The number of Library's Concrete Crew equals the sum of Northshore Region's Apartment Project, Library's Welding Crew, and Brookside Region's Library. The number of Northshore Region's Apartment Project equals 1 times the sum of Brookside Region's Library, Bayview Region's Office Building, and Office Building's HVAC Crew. The number of Brookside Region's Library equals the sum of Apartment Project's Concrete Crew and Office Building's HVAC Crew. The number of Library's Welding Crew equals 2 times the sum of Northshore Region's Office Building, Northshore Region's Project, and Bayview Region's Office Building. The number of Office Building's Welding Crew equals Apartment Project's Welding Crew. The number of Office Building's Concrete Crew equals the difference of Office Building's Welding Crew and Apartment Project's Welding Crew. Determine the quantity of HVAC Crew in Library."


Prompt: You have two jars, 50 red marbles, and 50 blue marbles. You need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. When picking, you’ll first randomly pick a jar, and then randomly pick a marble out of that jar. You can arrange the marbles however you like, but each marble must be in a jar.


Prompt: Is ArcSinh[Cosh[41] equal to 41?


Prompt: Imagine I have a list of n numbers. I want to group these numbers into k clusters such that the mean of the numbers within each cluster is kind of the same, measured by the variance. Give a solution using convex programming using cvxpy


Prompt:
import math
import numpy as np
import pandas as pd

# 常数定义
PI = math.pi
R = 55 / (2 * PI)  # 螺线半径
PITCH = 55  # 螺线螺距
V = 1  # 龙头前把手的行进速度
TOTAL_TIME = 300  # 总时间
dt = 0.01  # 时间步长
L = 223  # 板凳数量

# 初始化数组
t = np.arange(0, TOTAL_TIME, dt)
x = np.zeros((L, len(t)))
y = np.zeros((L, len(t)))
vx = np.zeros((L, len(t)))
vy = np.zeros((L, len(t)))

# 计算龙头前把手的位置和速度
for i in range(len(t)):
    x[0, i] = R * math.cos(2 * PI * t[i] / PITCH)
    y[0, i] = R * math.sin(2 * PI * t[i] / PITCH)
    vx[0, i] = -V * math.sin(2 * PI * t[i] / PITCH)
    vy[0, i] = V * math.cos(2 * PI * t[i] / PITCH)

# 计算龙身和龙尾的位置和速度
for j in range(1, L):
    for i in range(len(t)):
        x[j, i] = x[j-1, i] - 220 / 1000 * math.cos(2 * PI * t[i] / PITCH)
        y[j, i] = y[j-1, i] - 220 / 1000 * math.sin(2 * PI * t[i] / PITCH)
        vx[j, i] = vx[j-1, i] + V * math.sin(2 * PI * t[i] / PITCH)
        vy[j, i] = vy[j-1, i] - V * math.cos(2 * PI * t[i] / PITCH)

# 计算每节龙身的切线速度
def calculate_tangent_velocity(x, y, dt):
    dx_dt = np.gradient(x, dt, axis=1)
    dy_dt = np.gradient(y, dt, axis=1)
    return np.sqrt(dx_dt**2 + dy_dt**2)

# 计算切线速度
tangent_velocity = calculate_tangent_velocity(x, y, dt)

# 确定舞龙队盘入的终止时刻
min_distance = float('inf')
end_time = 0
for i in range(len(t)):
    distance = math.sqrt((x[-1, i] - x[0, i])**2 + (y[-1, i] - y[0, i])**2)
    if distance < min_distance:
        min_distance = distance
        end_time = t[i]

# 计算终止时刻的位置和速度
end_x = x[:, int(end_time / dt)]
end_y = y[:, int(end_time / dt)]
end_speed = tangent_velocity[:, int(end_time / dt)]

# 保存结果到文件
result2 = pd.DataFrame({
    'x': end_x,
    'y': end_y,
    'speed': end_speed  # 使用切线速度
})
result2.to_excel('result2.xlsx', index=False)

# 打印终止时刻的位置和速度
print("终止时刻：", end_time)
for i in [0, 1, 51, 101, 151, 201, -1]:
    print(f"第 {i} 节龙身前把手位置：", (end_x[i], end_y[i]))
    print(f"第 {i} 节龙身前总速度：", end_speed[i])
解释以上代码 解释如何建模的


Prompt: 一个西瓜进价50元, 卖了70元, 老板收了100元假币, 问亏了多少钱? 列出详细推理步骤.


Prompt:
I want to make a cheese based keto naan/flatbread. I put 272g of shredded cheese (3/4 mozarella, 1/4 parmesan) in a microwave for it to melt. In context of baker's percentages, what would be the water content of this in the dough?

With the right hydration value for naan, suggest the right amounts of flour to add (for flour blend you may use any of almond flour, golden flaxseed meal, vital wheat gluten, bamboo fiber, lupin flour, coconut flour)

I would also reckon 1tsp of baking powder would be a good idea, also some onion/garlic powder, nutritional yeast for taste
potentially xanthan gum/glucomannan depending on whether vital wheat gluten or egg is used and the binding properties of the dough.
Aim for a chewy, airy, more firm/crunchy texture rather than saggy and moist


Prompt: Phương trình chuyển động của một vành tròn bán kính r lăn không trượt trên mặt phẳng với vận tốc v không đổi


Prompt: If you take two random people who are both 100% Ashkenazi, what degree of "cousin" will they be on average? How does this change if one of them is 50% Ashkenazi and the other is 25% Ashkenazi?


Prompt:
If 'a' implies "+", 'b' implies "-", 'c' implies "x" and 'd' implies "/", insert proper letter between the figures in the following equation. 40_20_30_6=55
*
1 point
a,b,c
b,c,d
a,b,d
d,b,a


Prompt:
G
S
Question
cannot review this problem again. You can use print to debug your code. The print may not work in case of syntax/runtime error. The version of Python being used is 3.4.3
In a science research lab, combining two nuclear chemicals produces a maximum energy that is the product of the energy of the two chemicals. The energy values of the chemicals can be negative or positive. The scientist wishes to calculate the sum of the energies of the two chemicals which produces maximum energy on reaction.
Write an algorithm to find the sum of energy of the two chemicals which produces maximum energy on reaction.
Input
The first line of the input consists of an integer - numOfChem, representing the number of chemicals (N).
The second line consists of N space-separated integers - enero, ener...
ener.1 representing the energies of the chemicals.
Output
Print an integer representing the sum of energy of the two chemicals which produces maximum energy on reaction.
Constraints
0s numOfChem≤ 106
-106 eners 106
0si<numOfChem
Example
Input:
7
9-38-6-7810
41:00
1
2
3
4
5
def maxEnergy(ener):
6
#Write your code here
7
8
return
9
10
def main():
11
#input for ener
12
ener = []
13
ener size int(input())
14
ener = list(map(int,input().split()))
15
16
17
result maxEnergy (ener)
18
print(result)
19 20 if name =="main":
21 main()

Write a python code


Prompt:
根据下面一段描述，编写出matlab代码
平面上有N的点，这些点均为稳态点，然后需要根据这些稳态点进行稳态点调度。待估计点是空间上的任意一点，需要估计出这个待估计对应的稳态。

（1） 考虑将第一个稳态点和最后一个稳态点连接，形成一个简化的基线向量.
（2）待估计点向该基线向量作垂线两线交点对应的稳态点即是目标稳态点。（由向量知识可知，从待估计点向该基线向量作垂线寻找交点实际上就是构建从第一个稳态点到待估计点的向量，该向量与基线向量做内积即可得到交点。）


Prompt: connaissant la mesure I1=52 dB du bruit exprimé en décibel d'une source sonore située à 7 mètres de distance, est-il possible alors de connaitre la mesure I2 de ce même bruit mais cette fois-ci à 1 mètre de la même source sonore ?


Prompt: find the last non-zero digit of 50!


Prompt: Diketahui x dan y merupakan dua bilangan asli berbeda yang memenuhi 2x + 3y = 23. Pernyataan yang benar di bawah ini adalah .... Pilih kategori yang benar untuk setiap pilihan. Kategori Benar Salah Nilai xy selalu lebih besar dari 10 Pilih jawabanmu x +Y Nilai - selalu lebih besar dari 3 2 Pilih jawabanmu Kelipatan persekutuan terkecil dari x dan y selalu lebih besar dari 8 Pilih jawabanmu Jawab


Prompt: Obtain a quadratic sequence whose first, second, third and fifth terms form a geometric sequence.


Prompt: 全国大学生数学竞赛非数学类试题


Prompt:
làm bài tập này theo nguyên lí thống kê biết là phân  phối chuẩn: "A class's math test score has a normal distribution with X= 5.6 and
S=1.41.
1/ In what range will 68% of students' test scores fall?
2/ How many percentages (%) of students have scores below 2.78? 3/ If there is no hypothesis that student scores are normally distributed, at least how many students have scores in the range between 2.78 and 8.42?
"


Prompt:
Use the cross product to find the sine of the angle between the vectors u = (2, 3, 6) and v = (2, 3, 6).
—


Prompt:
A. Closest Point
time limit per test2 seconds
memory limit per test512 megabytes
Consider a set of points on a line. The distance between two points i
 and j
 is |i−j|
.

The point i
 from the set is the closest to the point j
 from the set, if there is no other point k
 in the set such that the distance from j
 to k
 is strictly less than the distance from j
 to i
. In other words, all other points from the set have distance to j
 greater or equal to |i−j|
.

For example, consider a set of points {1,3,5,8}
:

for the point 1
, the closest point is 3
 (other points have distance greater than |1−3|=2
);
for the point 3
, there are two closest points: 1
 and 5
;
for the point 5
, the closest point is 3
 (but not 8
, since its distance is greater than |3−5|
);
for the point 8
, the closest point is 5
.
You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input
The first line contains one integer t
 (1≤t≤1000
) — the number of test cases.

Each test case consists of two lines:

the first line contains one integer n
 (2≤n≤40
) — the number of points in the set;
the second line contains n
 integers x1,x2,…,xn
 (1≤x1<x2<⋯<xn≤100
) — the points from the set.
Output
For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Example
InputCopy
3
2
3 8
2
5 6
6
1 2 3 4 5 10
OutputCopy
YES
NO
NO
Note
In the first example, the point 7
 will be the closest to both 3
 and 8
.

In the second example, it is impossible to add an integer point so that it becomes the closest to both 5
 and 6
, and is different from both of them.
c++ code


Prompt:
定义两个数 x,y 的 k-xor 为 x,y 在正整数 k(k≥2) 进制意义下的不进位加法。现在 cats 有两个整数
a,b，cats 算出了它们在某个进制 k 下的 k-xor 为 c。但在 cats 计算出 c 后，cats 忘记了 k 的值。你能帮 cats 算出所有大于等于
2 的可能是 k的不同正整数个数吗？如果有无穷多个满足条件的 k，输出−1。
注：两个数在 k进制下的不进位加法为，将两个数分别写出它们的 k 进制表示，并将两个数对应的位分别相加，然后将每一位相加得到的结果分别对 k 取模，将结果看做一个新的
k 进制数，这个结果即为两个数 k 进制下不进位加法的结果。例如 16=(121) 3和 8=(022) 3在 3 进制下的不进位加法的结果即为 (110) 3=12。
Input
第一行包含一个整数
T(1≤T≤100)，表示一共有 T 组测试数据。每组测试数据包含一行三个整数 a,b,c(0≤a,b,c≤1000000009)，表示参与 k-xor 运算的两个数和运算结果。
Output
对于每组测试数据，输出一个整数，表示所有大于等于 2 的可能是 k 的不同正整数个数。如果有无穷多个满足条件的 k，输出 −1。
Sample Input
5
3 5 6
16 8 12
0 0 0
21 21 0
123 456 789
Sample Output
1
2
-1
3
0
Time Limit (Java / Others)
4000 / 2000 MS
Memory Limit (Java / Others)
524288 / 524288 K
Ratio (Accepted / Submitted)
64.21% (244/380)
c++简单的ACM代码写，并且符合过题条件，加思路


Prompt:
C. Even Positions
time limit per test2 seconds
memory limit per test256 megabytes
Monocarp had a regular bracket sequence s
 of length n
 (n
 is even). He even came up with his own way to calculate its cost.

He knows that in a regular bracket sequence (RBS), each opening bracket is paired up with the corresponding closing bracket. So he decided to calculate the cost of RBS as the sum of distances between pairs of corresponding bracket pairs.

For example, let's look at RBS (())(). It has three pairs of brackets:

(__)__: the distance between brackets at position 1
 and at 4
 is 4−1=3
;
_()___: the distance is 3−2=1
;
____(): the distance is 6−5=1
.
So the cost of (())() is 3+1+1=5
.
Unfortunately, due to data corruption, Monocarp lost all characters on odd positions s1,s3,…,sn−1
. Only characters on even positions (s2,s4,…,sn
) remain. For example, (())() turned to _(_)_).

Monocarp wants to restore his RBS by placing brackets on the odd positions. But since the restored RBS may not be unique, he wants to choose one with minimum cost. It's too hard to do for Monocarp alone, so can you help him?

Reminder: A regular bracket sequence is a string consisting of only brackets, such that this sequence, when inserted 1-s and +-s, gives a valid mathematical expression. For example, (), (()) or (()())() are RBS, while ), ()( or ())(() are not.

Input
The first line contains a single integer t
 (1≤t≤5000
) — the number of test cases. Next t
 cases follow.

The first line of each test case contains a single integer n
 (2≤n≤2⋅105
; n
 is even) — the length of string s
.

The second line of each test case contains a string s
 of length n
, where all characters on the odd positions are '_' and all characters on the even positions are either '(' or ')'.

Additional constraints:

s
 can be restored to at least one regular bracket sequence;
the total sum of n
 over all test cases doesn't exceed 2⋅105
.
Output
For each test case, print one integer — the minimum cost of the regular bracket sequence that can be obtained from s
 by replacing '_'-s with brackets.

Example
InputCopy
4
6
_(_)_)
2
_)
8
_)_)_)_)
8
_(_)_(_)
OutputCopy
5
1
4
8
Note
In the first test case, it's optimal to make s
 equal to (())(). The cost of s
 will be equal to 3+1+1=5
.

In the second test case, the only option is to make s
 equal to () with cost 1
.

In the third test case, the only possible RBS is ()()()() with cost 1+1+1+1=4
.

In the fourth test case, it's optimal to make s
 equal to (())(()) with cost 3+1+3+1=8
.
c++ optimized code


Prompt: The base of a right prism is a rhombus with a side of 20. The perimeter of one of the diagonal sections of the prism is 58. Determine the volume of the prism if its height is 5.


Prompt: Q2. A Class of 45 students is randomly split into 3 practice groups of the same size. A pair of friends in the class really like to work together and hope they won’t end up in diﬀerent groups. What are the chances they’re assigned to the same group?


Prompt: If i increased size by 25% how much more volume do i get


Prompt: чему равна напряжённость равномерно заряженного объёмной плотностью шара


Prompt:
1. (a) Compare the rates of spontaneous and stimulated emission at room temperature
(T=300 K)
for an atomic transition where the frequency associated with the transition is about
3×10
10
  Hz
, which is in the microwave region
(KB=1.38×10
23
  J/K)
. (b) What will be the wavelength of the line spectrum resulting from the transition of an electron from an energy level of
40×10
−20
  J
to a level of
15×10
−20
  J


Prompt:
Marks of tenth class students of a school follow the normal distribution with an unknown mean
μ
μ and variance
36.
36. Marks of 10 students of the tenth class are 24, 47, 43, 81, 94, 80, 60, 35, 48, 71. Find the Bayesian estimate (posterior mean) of
μ
μ assuming the Normal
(
45
,
25
)
(45,25) prior distribution. Write your answer correct to two decimal places.


Prompt:
 Ông Bách vay ngân hàng 500 triệu đồng trong 5 năm, lãi gộp vốn 3 tháng 1 lần.
Khi đáo hạn ông Bách phải trả cho ngân hàng cả gốc và lãi là 840 triệu đồng. Xác
định lãi suất cho vay


Prompt:
вырази из системы уравнений неизвестные переменные: x_A, x_N, x_C, y_A, y_N, y_C, R
((x_A - x_N)^2 + (y_A - y_N)^2 = R^2)
((x_A - x_M)^2 + (y_A - y_M)^2 = AM^2)
((x_C - x_N)^2 + (y_C - y_N)^2 = R^2)
((x_C - x_K)^2 + (y_C - y_K)^2 = CK^2)
((x_B - x_N)^2 + (y_B - y_N)^2 = R^2)
(\frac{x_A - x_M}{x_N - x_M} = \frac{y_A - y_M}{y_N - y_M})
(\frac{x_C - x_K}{x_N - x_K} = \frac{y_C - y_K}{y_N - y_K})


Prompt:
51 840
ДАНО:
А) Кредит (ТК) = 51 840 BYN;
Срок кредита = 120 мес.;
Б) Схема погашения кредита: дифференцированная, точно в срок. Расчетный срок кредита (М) = 120 мес.;
В) Льготный период (Л)=36 мес.;
Г) Ставка льготного периода  (КСДЛ)=0,072;
Д) Ставка кредита  (ГСДЮ) =0,125;
Е)  Выплата займа по данным кредитора
(СумК@) = 77209,58 BYN;
Ж) Эталон итоговой  выплаты (ЭДв) =   77 469,48  BYN.

НАЙТИ:
1) МКСДЛ=КСДЛ/12;

2) МЯ=ГСДЮ/12;

3) МДТК=ТК/М;

4а)   ЛГ= Л-1;

4б)  ЛПТ=МДТК*ЛГ;

4в) СумТКЛ=2ТК- ЛПТ

5) ПолТКЛ = СумТКЛ / 2;
6) Среднемесячная плата льготных процентов
(Точ%П=ПолТКЛ*МКСДЛ);
7) Плата льготных процентов за срок льготы (ЛТоч%Л=Точ%П*Л);
8) ВТЛ=МДТК*Л;

9) СумТЖ=ТК - ВТЛ+МДТК

10) ПолТЖ=СумТЖ/2;
11) Пол%ТЖ=ПолТЖ*МЯ;
12) Ж = М - Л;
13) Точ%Ж=Пол%ТЖ*Ж;
14) Выплата займа по расчету нейросети
(ТСВТЦ=ТК+ЛТоч%Л+Точ%Ж);
 15) Разница  выплаты займа по данным кредитора и бота (РПРТЦ = СумК@ -ТСВТЦ);
16) Переплата по кредиту (выплата процентов)
ПВП=ТСВТЦ -  ТК;
17) Относительная переплата по кредиту,%;
ППК=100(ТСВТЦ/ТК -1);
 18) Индекс совпадения выплаты займа по данным бота с эталоном (ИнБЭ = ТСВТЦ / ЭДв);
19) Сделай выводы по пунктам
  15,  17 и 18.


Prompt: You are playing Russian roulette with a six-shooter revolver. Your opponent puts in five bullets, spins the chambers and fires at himself, but no bullet comes out. He gives you the choice of whether or not he should spin the chambers again before firing at you. Should he spin again?


Prompt:
(15 pts) Consider a causal linear time-invariant system represented by the following frequency response:

H(jω) = 1 / (4jω + 3)

(a) (5 pts) Find the impulse response of the system.

(b) (10 pts) Find the input x(t), when we observe the following output:

y(t) = (e^(-5t) - e^(-10t))u(t). work it out step by step. be wise and correct


Prompt: I have 4 camels that can carry 23kg each. How many 10kg bricks can I carry? Please write how I should distribute the bricks. Check your answer to see if none of the camel carry too much weight


Prompt:
Example) There are 10 people, and 2 people are playing a 1v1 game. Each player has to play once, and if they win, they get 1 point, and if they lose, they get the same number of points. There are no ties.

Problem 1) What is the total number of games played by the ten players?

Problem 2) There is at most one person with a score of 9 points. Why is that?

Problem 3) How many people have a score of 8 or more?


Prompt: посчитай мне в виде примера показатели рентабельности и маржинальностии для топливного бизнеса


Prompt:
 Una marca de refresco lanza una promoción en la
cual cambian cuatro botellas vacías del refresco por
una botella llena de refresco. Si una persona tiene
inicialmente 19 botellas vacías, entonces la mayor
cantidad de botellas llenas que puede obtener de
esta promoción es
A. 6
B. 7
C. 4
D. 5


Prompt:
Hide Transcribed Text

A steady-state system for producing power consist of a pump, heat exchanger and a turbine. Water at 1.0 bar and
20
∘
 C
(state 1) enters the adiabatic pump and leaves at 10 bar (state 2). The pump draws
110 kW
of power, and the mass flow rate of water is
45 kg/s
. The water leaving the pump enters a heat exchanger and heated at constant pressure to
400
∘
 C
(state 3 ) using exhaust gases (Cp of gases
=1.1 kJ/kgK
) that enters at
500
∘
 C
and exits at
182
∘
 C
. The steam is adiabatically expanded in a turbine having an isentropic efficiency of 0.71 . The turbine exhausts (state 4 ) to the surroundings at 1.0 bar. a.) What is the rate at which heat must be supplied from the heat source to the water to bring it to
400
∘
 C
? b.) Determine the power produced by the turbine c.) What is total rate of entropy production resulting from this power production process? d.) What is thermal efficiency of this power production process? e.) If the pump and turbine are reversible, would the efficiency of this system equal the maximum possible efficiency? (Answer Yes or NO and provide a short explanation.)


Prompt: A complex circuit consists of a 24-volt battery, a 3-ohm resistor (R1), a 6-ohm resistor (R2) connected in parallel, and an additional 5-ohm resistor (R3) connected in series with the parallel combination. What is the total power dissipated by the entire circuit?


Prompt:
.设无向树有8片树叶,1个度为4的分支点,其余的分支点的度为3,则树的结点
数为


Prompt:
Diethyl phthalate (DEP) là chất lòng không màu, không có múi đặc biệt nhưng có vị đắng, khô chịu. DEP thường được sử dụng như một chất làm mềm trong sản xuất các sản phẩm như nhưa PVC (polyvinyl chloride), cao su. Quá trình sản xuất DẸP được thực hiện theo sơ đồ bên. Tính khối lượng DEP có thể được sản xuất từ 10 tän anhidrit phtalic He

+CHOH/xt H₂SO₄ đặc

H-90%

0

Dietyl phtalat

Anhidrit


Prompt:
int exp(int n)
{
    if (n <= 1) return 1;
    return exp(n - 1) + exp(n - 1) + exp(n - 1);
} - а здесь сложность O(3^N)?


Prompt:
Solve the following problem:

These measurements that are part of a much longer (unknown) stream sequence that happens at equal intervals. There are 10 measurements presented. y = [ -0.1, 0.1, 1.0, 1.0, 1.1, 0.1, 0.0, 0.1, 0.1, 0.0] Please interpolate these values into a different time base that has  7 points across the same interval.


Prompt: The coil of a moving iron voltmeter has a resistance of 20Ω and an inductance of 0.3H. It is connected in series with a swamping resistance of 2000Ω. The capacitor connected in shunt with the swamping resistance for making the meter read correctly at dc as well 50Hz ac will have a value ofofq


Prompt: If I have a 0.20mg/ml solution (20mg in 100ml VF) and I want to perform two dilutions to make a final solution which has a concentration of 0.05% of 0.20mg/ml, tell me how I would do that.


Prompt: 11) Find the savings plan balance after 10 months with an APR of 3% and monthly payments of $390. 11) A) $3944.17 B) $4758.00 C) $4044.30 D) $3352.54


Prompt: Как решить обратную задачу кинематики методом обратных преобразований в python?


Prompt: 假设有一个半球半径为r，从圆面出对它施加压力，假设它在被压的过程中体积都变成二维平面，求下压过程中的底面映射的面积变化


Prompt:
You have decided to apply for a loan. The loan could be a loan for a vehicle, a home, or to start a new business. The loan officer requests that you complete a balance sheet, and an income statement to support your loan request.

With $600 in checking, $1200 in savings, $4000 in stocks, $140,000 in a 401(k) retirement account, $400,000 house with a remaining $250,000 mortgage, a jeep cherokee vehicle worth $11,400 that has an outstanding loan of $9,200, $17200 in credit card debt, $34000 in student loan debt, create a personal balance sheet.

With gross pay of $11680, $2000 in federal income tax, $780 in social security tax, $180 in medicare tax, $490 in state income tax, $380 in health care premiums, $700 401(k) payment, self-employment income of $440, dividend income of $10, mortgage payment of $1540, $500 credit card payment, $260 student loan payment, $390 car payment, $40 gym payment, $160 cell phone service payment, $80 internet service payment, $120 streaming services payments, $1,500 in tuition payment, $220 doctor payment visit, $10 prescription drug payment, $480 was spent on dining out, $300 was spent on gas, $710 was spent on groceries, $340 was spent on entertainment, $180 was spent on car insurance, $140 was spent on electricity, $360 was spent on property taxes, $130 was spent on homeowners insurance, create a personal income statement.

After completing these statements, compute your current debt ratio, your liquidity ratio, and your debt service ratio prior to receiving the loan. Include your computations for these ratios. After completing these statements, do you think you would qualify for the loan based only on your financial statements? What actions could you take to improve your financial statements (please use realistic actions within your power to implement).

In your conclusion, create three positively correlated actions that you could take to strengthen your financial position over a three year timeline.


Prompt: A bee flies at 15 feet per second directly to a flowerbed from its hive. The bee stays at the flowerbed for 15 minutes, and then flies directly back to the hive at 9 feet per second. It is away from the hive for a total of 19 minutes.


Prompt:
A 50,000 kg airplane initially flying at a speed of 60.0 m/s accelerates at 5.0 m
/s2 for 600 meters. What is its velocity after this acceleration? What is the net
force that caused this acceleration?


Prompt:
 复制Markdown  展开
题目背景
神秘数字
36
36 是位于
35
35 和
37
37 之间的神秘自然数，是偶数也是
6
6 的平方。

毕达哥拉斯学派认为
36
36 是前四个奇数
1
,
3
,
5
,
7
1,3,5,7 与前四个偶数
2
,
4
,
6
,
8
2,4,6,8 的和，是最神圣的数字。

中国民间有三十六计，《西游记》中猪八戒会三十六变，《水浒传》中的三十六天罡星。

题目描述
小贝有一个大小为
�
n 的正整数数组
�
=
{
�
1
,
�
2
,
.
.
.
,
�
�
}
A={a
1

 ,a
2

 ,...,a
n

 }。小贝有一项特殊能力，他可以从数组
�
A 中任意选择两个整数
�
�
,
�
�
a
x

 ,a
y

 ，并将两个数拼接至一起形成新的整数
�
b。

例如
�
=
{
10
,
36
,
45
,
72
}
A={10,36,45,72}，小贝可以选择
10
10 和
45
45 拼接构成整数
1045
1045。

由于该能力消耗非常大，所以小贝仅能使用一次该能力。现在小贝想知道，有多少种方案构成的
�
b 是
36
36 的倍数？

输入格式
输入的第一行，包含一个整数
�
n，表示数组
�
A 的大小。

输入的第二行，包含
�
n 个正整数，其中第
�
i 个整数表示
�
�
a
i

 。

输出格式
输出仅一行，包含一个整数表示答案。

输入输出样例
输入 #1复制
8
1 2 3 4 5 6 7 8
输出 #1复制
2
输入 #2复制
4
3 3 6 6
输出 #2复制
4
输入 #3复制
2
36 36
输出 #3复制
2
说明/提示
1
≤
�
≤
1
0
5
1≤n≤10
5

1
≤
�
�
≤
1
0
18
1≤a
i

 ≤10
18怎么在时间复杂度的要求下完成这个题


Prompt:
Zkus vyřešit tento příklad. Postupuj krok za krokem. Aproximuj co nejpřesnější řešení.
2^x + x^3  = 930.498025157860930399


Prompt:
Berechne die Aufteilung neu so dass,
- keine der Zahlen kleiner als 10 ist
- alle Zahlen zusammen 100 ergeben
- jede Zahl eine ganze Zahlen ist:
23.47%
17.63%
11.30%
10.73%
9.89%
8.29%


Prompt:
Here is the code, that creates naive base for the precipitation levels based on the provided elevation map (a global earth map):
latitudes = torch.linspace(-90, 90, elevation.shape[0]).to(elevation.device)
climate_factor = torch.cos(torch.deg2rad(latitudes)[:, None]).repeat(1, elevation.shape[1])

However the torch.cos is not actually good function to do so. It reaches 0 at poles. It is known that the base  annual precipitation at equators is 2000mm and at poles ~160mm. This mean that the torch.cos have to be replaced with a little bit more sophisticated function. What can you offer?


Prompt: A child reads twice as many books each day as the day before. If this child finished the book on day 30, when did he read a quarter of the book?


Prompt:
Describe theory and math behind RLHF reinforcement learning from human feedback. Write loss equation with all details in Latex

\section{Preliminaries}\label{section:prelims}

We review the RLHF pipeline in \citeauthor{ziegler2020finetuning} (and later \citep{stiennon2022learning, bai2022training, ouyang2022training}). It usually includes three phases: 1) supervised fine-tuning (SFT); 2) preference sampling and reward learning and 3) RL optimization.

\textbf{SFT}: RLHF typically begins by fine-tuning a pre-trained LM with supervised learning on high-quality data for the downstream task(s) of interest (dialogue, summarization, etc.), to obtain a model $\pisft$.

\textbf{Reward Modelling Phase}: In the second phase the SFT model is prompted with prompts $x$ to produce pairs of answers $(y_1, y_2)\sim \pisft(y \mid x)$. These are then presented to human labelers who express preferences for one answer, denoted as $y_w\succ y_l \mid x$ where $y_w$ and $y_l$ denotes the preferred and dispreferred completion amongst $(y_1, y_2)$ respectively. The preferences are assumed to be generated by some latent reward model $r^*(y, x)$, which we do not have access to. There are a number of approaches used to model preferences, the Bradley-Terry (BT) \cite{bradley1952rankanalysis} model being a popular choice (although more general Plackett-Luce ranking models \citep{plackett1975analysis, luce2012individual} are also compatible with the framework if we have access to several ranked answers). The BT model stipulates that the human preference distribution $p^*$ can be written as:
\begin{equation}\label{eq:bradley-terry}
    p^*(y_1\succ y_2 \mid x)=\frac{\exp\left(r^*(x, y_1)\right)}{\exp\left(r^*(x, y_1)\right) + \exp\left(r^*(x, y_2)\right)}.
\end{equation}
Assuming access to a static dataset of comparisons $\mathcal{D}=\bigl\{x^{(i)}, y_w^{(i)}, y_l^{(i)}\bigr\}_{i=1}^N$ sampled from $p^*$, we can parametrize a reward model $r_{\phi}(x, y)$ and estimate the parameters via maximum likelihood. Framing the problem as a binary classification we have the negative log-likelihood loss:
\begin{equation}\label{eq:reward_model}
    \mathcal{L}_R(r_{\phi}, \mathcal{D}) = -\mathbb{E}_{(x, y_w, y_l)\sim \mathcal{D}}\bigl[\log \sigma(r_{\phi}(x, y_w)- r_{\phi}(x, y_l))\bigr]
\end{equation}
where $\sigma$ is the logistic function. In the context of LMs, the network $r_{\phi}(x, y)$ is often initialized from the SFT model $\pisft(y \mid x)$ with the addition of a linear layer on top of the final transformer layer that produces a single scalar prediction for the reward value \cite{ziegler2020finetuning}. To ensure a reward function with lower variance, prior works normalize the rewards, such that  $\mathbb{E}_{x,y\sim \mathcal{D}}\left[r_\phi(x, y)\right] = 0$ for all $x$.

\textbf{RL Fine-Tuning Phase}: During the RL phase, the learned reward function is used to provide feedback to the language model. Following prior works~\citep{jaques2017sequence, jaques2020human}, the optimization is formulated as
\begin{equation}\label{eq:RL}
\max_{\pi_{\theta}}  \mathbb{E}_{x\sim \mathcal{D}, y\sim \pi_{\theta}(y \mid x)}\bigl[r_{\phi}(x, y)\bigr] - \beta\mathbb{D}_{\textrm{KL}}\bigl[\pi_{\theta}(y\mid x)\mid \mid \piref(y\mid x)\bigr],
\end{equation}
where $\beta$ is a parameter controlling the deviation from the base reference policy $\piref$, namely the initial SFT model $\pisft$.
In practice, the language model policy $\pi_\theta$ is also initialized to $\pisft$. The added constraint is important, as it prevents the model from deviating too far from the distribution on which the reward model is accurate, as well as maintaining the generation diversity and preventing mode-collapse to single high-reward answers. Due to the discrete nature of language generation, this objective is not differentiable and is typically optimized with reinforcement learning. The standard approach \citep{ziegler2020finetuning, stiennon2022learning, bai2022training, ouyang2022training} has been to construct the reward function ${r(x, y) = r_{\phi}(x, y) -\beta (\log \pi_{\theta}(y\mid x) - \log \piref(y\mid x))}$, and maximize using PPO \cite{schulman2017proximal}.


Prompt: Please invert the matrix [[9, 3, 4], [1, 7, 2], [4, 4, 5]].


Prompt:
95% confidence interval of the index price in the next 115 days:
95% confidence interval for ln (STS0) in the next 115 days:
STS0 = LN((𝜇 - 0.5𝜎2)T; 𝜎2T) = LN(0.223016*115252; 18%^2*115252)
=> STS0 = LN(0.1018, 0.0148)
⇔ ln(STS0) = N(0.1018, 0.0148)
95% CI for ln (STS0) = 0.1018  1.96 x 0.0148
				= (-0.1366; 0.3402)

95% confidence interval for the stock price in the next 115 days:
	= S0 * exp (-0.1366; 0.3402)
	= 5584.64 * exp (-0.1366; 0.3402)
	= ($4871.59, $7847.7) (*)

The probability that future index will be lower than 5584.64:
P( z < ln(5584.645584.64)-0.10180.0148  ) = P( z < -0.84) = 1 - 0.7995 = 0.2005 (**)

Firstly, as we can see that a trader believes a volatility approximately equal to 18%, which is higher than the historical volatility we have calculated as 11.16%. This information suggests that a trader believes that there will be a large move in the stock price. The strategy suitable for a large move in stock can be either straddles, strangles or reverse butterfly spread. We consider a high volatility of a trader, and calculate the 95% confidence interval of the stock price in the next 115 days depending on the volatility of 18%. The result shows that the stock price is expected to be volatile in a range from $4871.59 to $7847.70 (*) and the probability of the index to be lower than the current price is notably low, which is just 20.05% (**). When the index is highly estimated to increase rather than decrease, investors can make profit with a bull spread. Therefore, our group would like to suggest 3 trading strategies, which are long straddles, long strangles and call bull spread:

Long Straddles:

Buy a call option with a strike price K=$5600
Buy a put option with a strike price K=$5600

A long straddle is a neutral options strategy that involves buying both a put option and a call option for the underlying security with the same strike price and the same expiration date. It enables traders to benefit regardless of the market's direction under the circumstances of large volatility. If the index increases, there is a call to secure the trader’s position and make profit, otherwise if the index decreases, there is a put to take advantage. Moreover, it is important to choose the strike price to bring a profit. If the index just moves slightly and around the strike price, a trader will lose most awfully. Therefore, we need to choose a strike price far from the expected value of the future index. Based on the strong possibility of the index price rising, and the expected index price of $6228.77 (***), we prefer to choose the strike price in a low range from $5500 to $5700 to gain profit if there is a growth in index price. To make it more satisfying and easier to choose a strike price in this range, we would like to pick the average strike price of the range, which is $5600.

Expected stock price in the next 115 days:
Annual drift = 𝜇 - 0.5𝜎2 = 0.223016
=> 𝜇 = 0.223016 + 0.5𝜎2 = 0.223016 + 0.5*0.182 = 23.92%
Expected stock price
(E(ST)) = S0e 𝜇T = 5584.64 * e 0.2392*(115/252) = $6228.77 (***)

There are two break-even points for the strike price K=5600:
Strike price plus total premium (St>K): $5600 + $334.9183 = $5934.9183
Strike price minus total premium (St<K): $5600 - $334.9183 = $5265.0817

If the future index price higher than the strike price, the break-even point is $5934.9183, while the 10% confidence interval (the interval that future index price most likely to approach) we calculate is from $6092.27 to $6277.80 (****), which indicates that a trader is certain to gain profit if the index price goes up. If the price goes down, a trader will benefit if he is right (from large movements), otherwise, small movements result in a loss equal to the total premium paid.

10% CI of stock price:
10% CI for ln(STS0) = 0.1018  0.125 x 0.0148
				= (0.087, 0.117)
10% confidence interval for the stock price in the next 115 days:
	= S0 * exp (0.087, 0.117)
	= 5584.64 * exp (0.087, 0.117)
	= ($6092.27, $6277.80) (****)

Long Strangles:

Buy a call option with a strike price K2=$6000
Buy a put option with a lower strike price K1=$5550

The range of the future index is considerably big, it can either decrease to $4871.59 or increase to $7847.70. Therefore, to minimise the risk, our group would like to choose the strike price near the current price, which is $5584.64. Moreover, because the index is expected to experience growth in the future, we would like to choose the strike price for the call option below the expected value of the index, which is $6228.77, to gain profit if the index increased dramatically. So we would set a strike price for the call option at K2=$6000. In terms of the put option, we choose a strike price as $5550 because the probability of the future index goes down is low, which is only 20.05% (**). If the price decreases obviously, a trader can maximise his or her profit by selling at a strike price that is closest to the current price.

Let’s consider the break-even points to see the possibility of this strategy:
Higher strike price plus total premium (St>K2): $6000+$169.4217=$6169.42
Lower strike price minus total premium (St<K1): $5550-$169.4217=$5380.58

This strategy benefits from large movements in the underlying stock price, either up or down, but requires a more significant price movement than a straddle to be profitable.
Call Bull Spread

Long a call option at K1=$6000
Short a call option at a higher K2=$6400

A bull spread using call options is an ideal strategy when a trader anticipates a moderate rise in the stock index. By buying a call option at a lower strike price 𝐾1=$6000 and shorting a call option at a higher strike price 𝐾2=$6400, the trader can capitalise on the expected increase to $6228.77. The choice of 𝐾2=$6400 ensures that the call option at this strike price is less likely to be exercised, allowing the trader to gain the initial premium from selling the call. Meanwhile, the call option at 𝐾1=$6000 will likely be exercised if the index approaches the expected value, providing a profit. This strategy limits potential losses to the net premium paid while offering the opportunity to benefit from the anticipated upward movement in the index, aligning with the trader's market outlook.


Prompt: CÁCH DÙNG CHI TIẾT BÀN TÍNH ABACUS VỚI CÁC PHÉP TÍNH CĂN


Prompt: Tracy purchased a car for $19,500. She is financing the purchase at an 11% annual interest rate, compounded monthly for 3 years. What is the payment that Tracy is required to make at the end of each month?


Prompt: Моя зарплата 100 тысяч рублей, у жены 70 тысяч, рассчитай график как можно быстрее выплатить кредит за машину суммой 705 тысяч рублей с процентной ставкой 21%. Учитывай траты на жизнь и тд