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I want you to act as a stackoverflow post. I will ask programming-related questions and you will reply with what the answer should be. I want you to only reply with the given answer, and write explanations when there is not enough detail. do not write explanations. When I need to tell you something in English, I will do so by putting text inside curly brackets {like this}. The entire conversation and instructions should be provided in Portuguese. My first question is
Por favor, responda abaixo com os seus comentários, pensamentos ou partilhas sobre este tema.
274 comentários
    fly

    小 K 喜欢翻看洛谷博客获取知识。每篇文章可能会有若干个(也有可能没有)参考文献的链接指向别的博客文章。小 K 求知欲旺盛,如果他看了某篇文章,那么他一定会去看这篇文章的参考文献(如果他之前已经看过这篇参考文献的话就不用再看它了)。 假设洛谷博客里面一共有 n篇文章(编号为 1 到 n)以及 m条参考文献引用关系。目前小 K 已经打开了编号为 1 的一篇文章,请帮助小 K 设计一种方法,使小 K 可以不重复、不遗漏的看完所有他能看到的文章。 这边是已经整理好的参考文献关系图,其中,文献 X → Y 表示文章 X 有参考文献 Y。不保证编号为 1 的文章没有被其他文章引用。请对这个图分别进行 DFS 和 BFS,并输出遍历结果。如果有很多篇文章可以参阅,请先看编号较小的那篇(因此你可能需要先排序)。 输入格式 共 m+1 行,第 1 行为 2 个数,n 和 m,分别表示一共有 n篇文章(编号为 1 到 n)以及m条参考文献引用关系。 接下来 m 行,每行有两个整数 X,Y 表示文章 X 有参考文献 Y。 输出格式 共 2 行。 第一行为 DFS 遍历结果,第二行为 BFS 遍历结果。

    • Responder
    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <queue>
    #include <functional>
    using namespace std;
    int main(){
        int n, m;
        cin >> n >> m;
        vector<vector<int>> graph(n + 1);
        for(int i = 0; i < m; i++){
            int u, v;
            cin >> u >> v;
            graph[u].push_back(v);
        }
        for(int i = 1; i <= n; i++){
            sort(graph[i].begin(), graph[i].end());
        }
        vector<bool> visited(n + 1, false);
        vector<int> dfsResult;
        function<void(int)> dfs = [&](int u) {
            visited[u] = true;
            dfsResult.push_back(u);
            for (int v : graph[u]) {
                if (!visited[v])
                    dfs(v);
            }
        };
        dfs(1);
        for (int x : dfsResult)
            cout << x << " ";
        cout << "\n";
        fill(visited.begin(), visited.end(), false);
        vector<int> bfsResult;
        queue<int> q;
        visited[1] = true;
        q.push(1);
        while(!q.empty()){
            int u = q.front();
            q.pop();
            bfsResult.push_back(u);
            for (int v : graph[u]) {
                if (!visited[v]){
                    visited[v] = true;
                    q.push(v);
                }
            }
        }
        for (int x : bfsResult)
            cout << x << " ";
        cout << "\n";
        return 0;
    }
    
    • Responder
    Fre

    编写一个python代码输出简单的4乘4的有输入界面的数独程序

    • Responder

    这是一个用Python编写的简单数独程序,包含4×4的游戏棋盘和用户输入界面:

    import random
    import numpy as np
    
    class Sudoku:
        def __init__(self):
            self.board = np.zeros((4, 4), dtype=int)
            self.solution = np.zeros((4, 4), dtype=int)
            self.generate_solution()
            self.create_puzzle()
            
        def generate_solution(self):
            # 生成有效的数独解
            base = [1, 2, 3, 4]
            random.shuffle(base)
            
            # 填充第一行
            self.solution[0] = base
            
            # 填充其余行
            for i in range(1, 4):
                # 根据4×4数独规则循环移动
                offset = 1 if i == 1 or i == 3 else 2
                self.solution[i] = np.roll(self.solution[i-1], offset)
        
        def create_puzzle(self):
            # 复制解决方案到棋盘
            self.board = self.solution.copy()
            
            # 随机挖空几个位置
            holes = random.randint(6, 8)  # 挖6-8个空
            
            for _ in range(holes):
                row = random.randint(0, 3)
                col = random.randint(0, 3)
                self.board[row][col] = 0
        
        def print_board(self):
            print("-" * 13)
            for i in range(4):
                row = "| "
                for j in range(4):
                    if self.board[i][j] == 0:
                        row += "_ "
                    else:
                        row += str(self.board[i][j]) + " "
                    if j == 1:
                        row += "| "
                row += "|"
                print(row)
                if i == 1:
                    print("-" * 13)
            print("-" * 13)
        
        def make_move(self, row, col, num):
            if self.board[row][col] == 0:
                if self.is_valid_move(row, col, num):
                    self.board[row][col] = num
                    return True
            return False
        
        def is_valid_move(self, row, col, num):
            # 检查行
            if num in self.board[row]:
                return False
            
            # 检查列
            if num in self.board[:, col]:
                return False
            
            # 检查2×2小方格
            box_row, box_col = 2 * (row // 2), 2 * (col // 2)
            for i in range(box_row, box_row + 2):
                for j in range(box_col, box_col + 2):
                    if self.board[i][j] == num:
                        return False
            
            return True
        
        def is_solved(self):
            return not np.any(self.board == 0)
        
        def is_valid_solution(self):
            return np.array_equal(self.board, self.solution)
    
    def main():
        game = Sudoku()
        print("欢迎来到4×4数独游戏!")
        print("请填入数字1-4,使每行、每列、每个2×2方格都包含1-4的数字。")
        
        while not game.is_solved():
            game.print_board()
            
            try:
                row = int(input("请输入行号 (0-3): "))
                col = int(input("请输入列号 (0-3): "))
                num = int(input("请输入数字 (1-4): "))
                
                if row < 0 or row > 3 or col < 0 or col > 3 or num < 1 or num > 4:
                    print("输入无效!请输入有效范围内的数字。")
                    continue
                
                if game.board[row][col] != 0:
                    print("该位置已经有数字了!")
                    continue
                
                if game.make_move(row, col, num):
                    print("有效移动!")
                else:
                    print("无效移动!该数字不能放在这个位置。")
            
            except ValueError:
                print("输入无效!请输入数字。")
        
        game.print_board()
        print("恭喜你完成了数独!")
        
        if game.is_valid_solution():
            print("解答正确!")
        else:
            print("解答有误!")
    
    if __name__ == "__main__":
        main()
    
    
    • Responder
    118

    写一个贪吃蛇

    • Responder
    import pygame  
    import random  
    
    # 初始化pygame  
    pygame.init()  
    
    # 设置游戏窗口大小  
    WIDTH, HEIGHT = 600, 400  
    CELL_SIZE = 20  
    screen = pygame.display.set_mode((WIDTH, HEIGHT))  
    pygame.display.set_caption("贪吃蛇")  
    
    # 颜色定义  
    WHITE = (255, 255, 255)  
    GREEN = (0, 255, 0)  
    RED = (255, 0, 0)  
    
    # 方向定义  
    UP = (0, -1)  
    DOWN = (0, 1)  
    LEFT = (-1, 0)  
    RIGHT = (1, 0)  
    
    # 初始化贪吃蛇  
    snake = [(WIDTH // 2, HEIGHT // 2)]  
    direction = RIGHT  
    
    # 生成食物  
    def random_food():  
        return (random.randint(0, (WIDTH // CELL_SIZE) - 1) * CELL_SIZE,  
                random.randint(0, (HEIGHT // CELL_SIZE) - 1) * CELL_SIZE)  
    
    food = random_food()  
    
    clock = pygame.time.Clock()  
    
    running = True  
    while running:  
        screen.fill(WHITE)  
        for event in pygame.event.get():  
            if event.type == pygame.QUIT:  
                running = False  
            elif event.type == pygame.KEYDOWN:  
                if event.key == pygame.K_UP and direction != DOWN:  
                    direction = UP  
                elif event.key == pygame.K_DOWN and direction != UP:  
                    direction = DOWN  
                elif event.key == pygame.K_LEFT and direction != RIGHT:  
                    direction = LEFT  
                elif event.key == pygame.K_RIGHT and direction != LEFT:  
                    direction = RIGHT  
    
        # 移动蛇  
        head_x, head_y = snake[0]  
        new_head = (head_x + direction[0] * CELL_SIZE, head_y + direction[1] * CELL_SIZE)  
    
        # 碰撞检测  
        if new_head in snake or not (0 <= new_head[0] < WIDTH and 0 <= new_head[1] < HEIGHT):  
            running = False  
            break  
    
        snake.insert(0, new_head)  
    
        # 吃食物  
        if new_head == food:  
            food = random_food()  
        else:  
            snake.pop()  
    
        # 绘制蛇和食物  
        pygame.draw.rect(screen, RED, (*food, CELL_SIZE, CELL_SIZE))  
        for segment in snake:  
            pygame.draw.rect(screen, GREEN, (*segment, CELL_SIZE, CELL_SIZE))  
    
        pygame.display.flip()  
        clock.tick(10)  
    
    pygame.quit()  
    
    • Responder
    186

    I want you to act as a stackoverflow post. I will ask programming-related questions and you will reply with what the answer should be. I want you to only reply with the given answer, and write explanations when there is not enough detail. do not write explanations. When I need to tell you something in English, I will do so by putting text inside curly brackets {like this}. The entire conversation and instructions should be provided in Chinese. My first question is '用python写一个99乘法表 '

    • Responder
    for i in range(1, 10):
        for j in range(1, i + 1):
            print(f"{j} x {i} = {i * j}", end="\t")
        print()
    
    • Responder
    101

    对于'<div><span>12</span><span>23</span></div>'这种格式html字符串,写一个函数将这个字符串转成对象的形式表示

    • Responder
    function stringToObject(str) {
      const result = {};
      const pairs = str.split('\n').filter(item => item.trim() !== '');
      
      pairs.forEach(pair => {
        const [key, value] = pair.split(':').map(item => item.trim());
        if (key && value) {
          result[key] = value;
        }
      });
    
      return result;
    }
    
    • Responder
    175

    这是?

    • Responder

    {请提供更多上下文或代码片段,以便我更好地理解你的问题。}

    这是 aishort 的提示词测试区,你可以分享对该提示词的建议或测试

    • Responder
    xum

    本题目要求读入n个整数,找到最大数和最小数,并输出结果,【输入格式】 例如当n=4时,输入给出4个绝对值不超过1000的整数A、B、C、D,以空格分隔。 【输出格式】 最大数和最小数,“max=?,min=?”。

    • Responder
    n = int(input())
    numbers = list(map(int, input().split()))
    
    print(f"max={max(numbers)},min={min(numbers)}")
    
    • Responder
    Ala

    给定一个经过编码的字符串,返回它解码后的字符串。

    编码规则为: k[encoded_string],表示其中方括号内部的 encoded_string 正好重复 k 次。注意 k 保证为正整数。

    你可以认为输入字符串总是有效的;输入字符串中没有额外的空格,且输入的方括号总是符合格式要求的。

    此外,你可以认为原始数据不包含数字,所有的数字只表示重复的次数 k ,例如不会出现像 3a 或 2[4] 的输入。

    示例 1:

    输入:s = "3[a]2[bc]" 输出:"aaabcbc" 示例 2:

    输入:s = "3[a2[c]]" 输出:"accaccacc" 示例 3:

    输入:s = "2[abc]3[cd]ef" 输出:"abcabccdcdcdef" 示例 4:

    输入:s = "abc3[cd]xyz" 输出:"abccdcdcdxyz"

    提示:

    1 <= s.length <= 30 s 由小写英文字母、数字和方括号 '[]' 组成 s 保证是一个 有效 的输入。 s 中所有整数的取值范围为 [1, 300]

    • Responder
    def decodeString(s: str) -> str:
        stack = []
        current_string = ""
        current_num = 0
    
        for char in s:
            if char.isdigit():
                current_num = current_num * 10 + int(char)
            elif char == '[':
                stack.append((current_string, current_num))
                current_string, current_num = "", 0
            elif char == ']':
                prev_string, repeat_times = stack.pop()
                current_string = prev_string + current_string * repeat_times
            else:
                current_string += char
    
        return current_string
    
    • Responder
    jie

    A form contains these fields: Name,Available,Banner Type,Expose Order,Schedule,Time Filter,App URL(Scheme),imagePath,buttonExpression,Target Platform.IOS,Target Platform.Android,Target Platform.Web,Target Group.Member SRL,Target Group.Segment Key,Target Group.Segment Info,Target Group.Region Filter,Target Group.AB Filter,Target Group.Override Field , Generate test cases based on these fields

    • Responder

    问题:
    A form contains these fields: Name, Available, Banner Type, Expose Order, Schedule, Time Filter, App URL(Scheme), imagePath, buttonExpression, Target Platform.IOS, Target Platform.Android, Target Platform.Web, Target Group.Member SRL, Target Group.Segment Key, Target Group.Segment Info, Target Group.Region Filter, Target Group.AB Filter, Target Group.Override Field. Generate test cases based on these fields.

    回答:

    1. Name Field:

      • Test Case 1: Check if the "Name" field accepts a valid string.
      • Test Case 2: Verify that the "Name" field is required (empty input should show a validation error).
      • Test Case 3: Ensure that special characters are allowed in the "Name" field.
      • Test Case 4: Test the maximum length for the "Name" field (if applicable).
    2. Available Field:

      • Test Case 1: Check if "Available" is a boolean (true/false).
      • Test Case 2: Ensure the default value is false (if applicable).
      • Test Case 3: Test if changing the value to true updates other relevant fields correctly.
    3. Banner Type Field:

      • Test Case 1: Verify that only valid banner types (e.g., image, text) can be selected.
      • Test Case 2: Ensure the "Banner Type" field has a default value.
      • Test Case 3: Test selecting different banner types and verifying the UI response.
    4. Expose Order Field:

      • Test Case 1: Check if the "Expose Order" field accepts only numerical values.
      • Test Case 2: Verify that the system orders items correctly based on the "Expose Order" value.
      • Test Case 3: Ensure that duplicate "Expose Order" values are not allowed.
    5. Schedule Field:

      • Test Case 1: Verify that the "Schedule" field accepts valid date and time input.
      • Test Case 2: Test for different time zones and check if the scheduling works as expected.
      • Test Case 3: Ensure "Schedule" cannot be set to a past date.
    6. Time Filter Field:

      • Test Case 1: Verify that the "Time Filter" accepts valid date range inputs.
      • Test Case 2: Check if the time filter works correctly when applied to data.
      • Test Case 3: Test invalid time filter inputs and ensure proper validation.
    7. App URL (Scheme) Field:

      • Test Case 1: Verify that the "App URL" accepts a valid URL scheme (e.g., myapp://).
      • Test Case 2: Ensure that an invalid URL scheme shows a validation error.
      • Test Case 3: Check if the URL redirects properly when clicked.
    8. ImagePath Field:

      • Test Case 1: Check if the "ImagePath" accepts valid image URLs.
      • Test Case 2: Ensure that the image is displayed correctly in the UI.
      • Test Case 3: Test invalid image paths and check for appropriate error messages.
    9. Button Expression Field:

      • Test Case 1: Verify that the "Button Expression" field accepts valid JavaScript or expression syntax.
      • Test Case 2: Ensure that the button performs the correct action when the expression is valid.
      • Test Case 3: Test invalid button expressions and check for proper error handling.
    10. Target Platform.IOS Field:

      • Test Case 1: Verify that "Target Platform.IOS" can be selected.
      • Test Case 2: Ensure that IOS-specific settings appear when selected.
      • Test Case 3: Test interaction with other target platforms when IOS is selected.
    11. Target Platform.Android Field:

      • Test Case 1: Verify that "Target Platform.Android" can be selected.
      • Test Case 2: Ensure that Android-specific settings appear when selected.
      • Test Case 3: Test interaction with other target platforms when Android is selected.
    12. Target Platform.Web Field:

      • Test Case 1: Verify that "Target Platform.Web" can be selected.
      • Test Case 2: Ensure that Web-specific settings appear when selected.
      • Test Case 3: Test interaction with other target platforms when Web is selected.
    13. Target Group.Member SRL Field:

      • Test Case 1: Ensure that the "Member SRL" field accepts valid values.
      • Test Case 2: Verify that selecting a member SRL applies to the right target group.
    14. Target Group.Segment Key Field:

      • Test Case 1: Check if the "Segment Key" field accepts valid keys.
      • Test Case 2: Ensure that selecting the right segment key filters the target group correctly.
    15. Target Group.Segment Info Field:

      • Test Case 1: Verify that the "Segment Info" field accepts valid data.
      • Test Case 2: Ensure that the segment info is displayed correctly when the target group is selected.
    16. Target Group.Region Filter Field:

      • Test Case 1: Check if the "Region Filter" field accepts valid regions.
      • Test Case 2: Test filtering by region and verify results.
    17. Target Group.AB Filter Field:

      • Test Case 1: Verify that the "AB Filter" field accepts valid AB testing parameters.
      • Test Case 2: Ensure that the AB testing results are applied correctly when selected.
    18. Target Group.Override Field:

      • Test Case 1: Check if the "Override Field" allows manual overrides.
      • Test Case 2: Ensure the override functionality works as expected when applied.
    • Responder
    mtt

    c语言中,已知单链表数据结构, 声明和定义在带头结点的单链表L中删除第pos个数据元素算法Delete。删除成功,则输出"OK",删除失败(在小于1 或者大于n的位置上执行删除操作)则输出"ERROR"。 Input: 程序输入一个整数,代表删除位置。 Ouput: 程序输出为两行:第一行是字符串OK或者ERROR,第二行是单链表的遍历结果. Example Input: 6 Example Output: ERROR 0 1 2 3 4 5

    • Responder
    #include <stdio.h>
    #include <stdlib.h>
    
    typedef struct Node {
        int data;
        struct Node *next;
    } Node;
    
    void Delete(Node *L, int pos) {
        Node *p = L;
        int i = 0;
        
        while (p->next != NULL && i < pos - 1) {
            p = p->next;
            i++;
        }
        
        if (p->next == NULL || i != pos - 1) {
            printf("ERROR\n");
            return;
        }
        
        Node *temp = p->next;
        p->next = temp->next;
        free(temp);
        
        printf("OK\n");
    }
    
    void PrintList(Node *L) {
        Node *p = L->next;
        if (p == NULL) {
            printf("\n");
            return;
        }
        printf("%d", p->data);
        p = p->next;
        
        while (p != NULL) {
            printf(" %d", p->data);
            p = p->next;
        }
        printf("\n");
    }
    
    int main() {
        Node *L = (Node *)malloc(sizeof(Node));
        L->next = NULL;
        
        // 创建链表
        for (int i = 0; i < 6; i++) {
            Node *newNode = (Node *)malloc(sizeof(Node));
            newNode->data = i;
            newNode->next = L->next;
            L->next = newNode;
        }
        
        int pos;
        scanf("%d", &pos);
        
        Delete(L, pos);
        PrintList(L);
        
        return 0;
    }
    
    • Responder