人工智能时代的新里程碑：ChatGPT的崛起

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ChatGPT 和人工智能的革命性技术 ChatGPT 是一种基于人工智能的聊天机器人，它可以生成人类般的对话，从而改变大型科技公司的游戏规则。在 Fortune 杂志的报道中，我们可以看到 ChatGPT 的崛起将如何影响科技行业，特别是微软等大型科技公司的发展。 ChatGPT 的出现可以追溯到 2022 年 11 月 30 日，当时 Sam Altman 的 OpenAI 公司推出了这款聊天机器人。ChatGPT 的出现标志着人工智能技术的突破性发展，它可以生成人类般的文本，对话甚至可以像人类一样流畅自然。 微软 company 正准备投资数十亿美元来开发 ChatGPT 和其他人工智能技术。微软的投资将为 ChatGPT 带来更多的发展机会，推动人工智能技术的进步。 ChatGPT 的出现不仅将改变科技行业，也将影响我们日常生活的方方面面。它可以应用于客服系统、语音助手、翻译软件等领域，提高人们的生活质量和工作效率。 人工智能技术的发展将带来更多的就业机会，但是同时也将对一些传统行业造成影响。因此，我们需要对人工智能技术的发展保持开放和包容的态度，拥抱技术的变革和发展。 ChatGPT 的崛起也引发了人们对人工智能技术的讨论和关注。人们开始关心人工智能技术的发展将如何影响我们的生活和工作，如何影响我们的社会和经济。 ChatGPT 的出现标志着人工智能技术的新纪元，它将改变科技行业和我们的生活，我们需要拥抱技术的变革和发展。 知识点： 1. ChatGPT 是一种基于人工智能的聊天机器人，可以生成人类般的对话。 2. ChatGPT 的出现标志着人工智能技术的突破性发展。 3. 微软公司准备投资数十亿美元来开发 ChatGPT 和其他人工智能技术。 4. ChatGPT 的出现将改变科技行业和我们的生活。 5. 人工智能技术的发展将带来更多的就业机会，但也将对一些传统行业造成影响。 6. 我们需要对人工智能技术的发展保持开放和包容的态度，拥抱技术的变革和发展。

Fortune Feb ChatGPT Cryptod.pdf.zip

2023-05-26 上传
Fortune Feb ChatGPT Cryptod.pdf.zip

Feb Time Quantum and ChatGPTd.pdf

2023-06-05 上传
Feb Time Quantum and ChatGPTd.pdf

import tkinter as tk import random root = tk.Tk() root.geometry("200x100") root.title("求签") def get_fortune(): fortunes = ["大吉", "中吉", "小吉", "吉", "末吉", "凶", "大凶"] fortune = random.choice(fortunes) result.config(text="{}，{}".format( "您的今日运势为", fortune)) button = tk.Button(root, text="求签", command=get_fortune) button.pack() result = tk.Label(root, text="") result.pack() root.mainloop()给该程旭添加一张背景图

2023-05-30 上传
result.config(text="{}，{}".format("您的今日运势为", fortune)) button = tk.Button(root, text="求签", command=get_fortune) button.pack() result = tk.Label(root, text="") result.pack() root....

import java.util.*;public class Billionaires { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); Map<String, List<Billionaire>> billionairesByCity = new HashMap<>(); for (int i = 0; i < n; i++) { String name = sc.next(); String city = sc.next(); long fortune = sc.nextLong(); Billionaire billionaire = new Billionaire(name, city, fortune); if (!billionairesByCity.containsKey(city)) { billionairesByCity.put(city, new ArrayList<>()); } billionairesByCity.get(city).add(billionaire); } int m = sc.nextInt(); int k = sc.nextInt(); for (int i = 0; i < k; i++) { int day = sc.nextInt(); String name = sc.next(); String city = sc.next(); List<Billionaire> billionaires = billionairesByCity.get(city); for (Billionaire billionaire : billionaires) { if (billionaire.name.equals(name)) { billionaire.addTravel(day); break; } } } List<String> cities = new ArrayList<>(billionairesByCity.keySet()); Collections.sort(cities); for (String city : cities) { List<Billionaire> billionaires = billionairesByCity.get(city); Map<Integer, Long> wealthByDay = new HashMap<>(); for (Billionaire billionaire : billionaires) { for (int day : billionaire.travels) { wealthByDay.put(day, wealthByDay.getOrDefault(day, 0L) + billionaire.fortune); } } int maxDays = 0; for (int day : wealthByDay.keySet()) { long wealth = wealthByDay.get(day); for (int i = day + 1; i <= m; i++) { long futureWealth = wealthByDay.getOrDefault(i, 0L); if (futureWealth > wealth) { break; } if (i - day > maxDays) { maxDays = i - day; } } } if (maxDays > 0) { System.out.println(city + " " + maxDays); } } } static class Billionaire { String name; String city; long fortune; List<Integer> travels = new ArrayList<>(); public Billionaire(String name, String city, long fortune) { this.name = name; this.city = city; this.fortune = fortune; } public void addTravel(int day) { travels.add(day); } }}的计算复杂度是多少

2023-05-30 上传
该程序的计算复杂度是 O(nklogk)，其中 n 是亿万富翁的数量，k 是每个亿万富翁旅行的次数的最大值，m 是最大旅行天数。这是由于程序需要对亿万富翁按照城市进行分组，然后对于每个城市，需要计算出每个亿万富翁在每...

用python怎么获取星座的今日运势

2023-05-11 上传
print(fortune.text.strip()) ``` 在这个示例中，我们首先导入了必要的库，然后使用requests库向星座运势网站发送请求。然后，使用BeautifulSoup库解析响应并找到我们需要的标签。最后，我们可以将每个星座的今日...

Lucky Numbers

2023-03-31 上传
It is also associated with good luck, fortune, and prosperity. 2. Eight: Considered lucky in Chinese culture as it sounds similar to the Chinese word for prosperity. 3. Nine: Considered lucky in ...

import tkinter as tk from PIL import Image, ImageTk import random root = tk.Tk() root.geometry("1728x1080") root.title("求签") # 加载背景图片 bg_image = Image.open("123.jpg") bg_photo = ImageTk.PhotoImage(bg_image) bg_label = tk.Label(root, image=bg_photo) bg_label.place(x=0, y=0, relwidth=1, relheight=1) def get_fortune(): fortunes = ["大吉", "中吉", "小吉", "吉", "末吉", "凶", "大凶"] fortune = random.choice(fortunes) result.config(text="{}，{}".format("您的今日运势为", fortune)) button = tk.Button(root, text="求签", command=get_fortune) button.pack() button.place(relx=0.5, rely=0.8, anchor="center", width=200, height=50) result = tk.Label(root, text="") result.pack() result.place(relx=0.5, rely=0.5, anchor="center", width=400, height=100) root.mainloop()修改文字大小

2023-05-30 上传
要修改 Label 中文字的大小，可以使用 font 参数。例如，将 result 标签中的文字大小改为 24 可以使用如下代码： ```python result = tk.Label(root, text="", font=("Arial", 24)) ``` 其中的参数 "Arial" 表示...

E: Package 'fortune-zh' has no installation candidate

2023-06-09 上传
这似乎是一个错误信息，它可能...3. 尝试安装fortune-zh软件包，如apt install fortune-zh或yum install fortune-zh等。 如果以上方法仍然无法解决问题，你可以尝试寻求更多的帮助，比如在相关的技术论坛或社区提问。

写一段Java实现Input In the first line you are given the number n of billionaires (1 ≤ n ≤ 10000). The following n lines contain information about these people: their names, cities where they were staying at the beginning of the period, and their fortunes. In the next line you are given the number m of days in the period for which you have the information (1 ≤ m ≤ 50000) and the number k of travels of the billionaires (0 ≤ k ≤ 50000). The following k lines contain the list of travels in the following format: the number of the day (from 1 to m−1), the name of the person, and the city of destination. You may assume that billionaires depart late at night and arrive to the destination city on the next day's morning. They cannot make more than one travel each day. The numbers of days in the list are not decreasing. All names of people and cities consist of at most 20 English letters; you must take into consideration the case of the symbols. The fortunes are in the range from 1 to 100 billions (one billion is a thousand million). Output In each line of the output give the name of a city and, after a space, the number of days during which this city was the first with respect to the sum of fortunes of the billionaires staying there. Leave out those cities for which there were no such days. The cities must be sorted alphabetically (with the usual symbol order: ABC...Zabc...z).

2023-05-30 上传
cityFortunes.put(city, cityFortunes.getOrDefault(city, 0) + fortune); } int m = scanner.nextInt(); int k = scanner.nextInt(); for (int i = 0; i ; i++) { int day = scanner.nextInt(); String ...

Reading package lists... Done Building dependency tree Reading state information... Done Package fortune-zh is not available, but is referred to by another package. This may mean that the package is missing, has been obsoleted, or is only available from another source However the following packages replace it: fortunes-zh E: Package 'fortune-zh' has no installation candidate

2023-06-09 上传
这个错误信息显示，此软件包 "fortune-zh" 不可用，但是有一个名为 "fortunes-zh" 的软件包可以替代它。这可能是因为 "fortune-zh" 被废弃或者从软件源中移除了。你可以使用以下命令安装 "fortunes-zh" 软件包： ``...
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