输入一个英文句子 This is a chair, that is a desk.按字符串大小输出单词，重复单词只能出现一次

作为人工智能语言模型，我没有亲身经验，但我可以根据数据和常识给出一些建议。

一般来说，实习面试相对较容易，因为公司对实习生的要求相对低一些，更注重实习生的潜力和学习能力，而非实际工作经验。此外，实习生的薪水和福利也相对较低，公司更容易承担风险。

相比之下，校招面试可能会更具挑战性。首先，校招面试通常是针对毕业生，公司对其要求更高，需要具备一定的专业知识和实践经验。其次，校招竞争激烈，公司会收到大量简历，而招聘人员需要筛选出最合适的候选人，这需要一定的时间和精力。

综上所述，实习面试相对容易，但也需要认真准备和展现自己的潜力和学习能力；而校招面试则需要更多的专业知识和实践经验，并且需要在众多候选人中脱颖而出。

相关问题

The Sleeping Teaching Assistant A university computer science department has a teaching assistant (TA) who helps undergraduate students with their programming assignments during regular office hours. The TA’s office is rather small and has room for only one desk with a chair and computer. There are three chairs in the hallway outside the office where students can sit and wait if the TA is currently helping another student. When there are no students who need help during office hours, the TA sits at the desk and takes a nap. If a student arrives during office hours and finds the TA sleeping, the student must awaken the TA to ask for help. If a student arrives and finds the TA currently helping another student, the student sits on one of the chairs in the hallway and waits. If no chairs are available, the student will come back at a later time. Using POSIX threads, mutex locks, and/or semaphores, implement a solution that coordinates the activities of the TA and the students. Details for this assignment are provided below. Using Pthreads, begin by creating N students. Each will run as a separate thread. The TA will run as a separate thread as well. Student threads will alternate between programming for a period of time and seeking help from the TA. If the TA is available, they will obtain help. Otherwise, they will either sit in a chair in the hallway or, if no chairs are available, will resume programming and will seek help at a later time. If a student arrives and notices that the TA is sleeping, the student must notify the TA using a semaphore. When the TA finishes helping a student, the TA must check to see if there are students waiting for help in the hallway. If so, the TA must help each of these students in turn. If no students are present, the TA may return to napping. Perhaps the best option for simulating students programming—as well as the TA providing help to a student—is to have the appropriate threads sleep for a random period of time using the sleep() API:

This is a programming assignment that requires the use of POSIX threads, mutex locks, and/or semaphores to coordinate the activities of the TA and the students. Here is one possible solution:

1. Create a mutex lock and two semaphores: one for the TA and one for the students waiting in the hallway.
2. Create N student threads and one TA thread.
3. Each student thread should loop indefinitely, alternating between programming and seeking help from the TA.
4. When a student needs help, they should try to acquire the mutex lock. If the TA is sleeping, the student should signal the TA semaphore and wait on the student semaphore. If the TA is helping another student, the student should wait on the student semaphore.
5. When the TA wakes up, they should try to acquire the mutex lock. If there are students waiting in the hallway, the TA should signal the student semaphore N times to wake up the students. The TA should then help each student in turn, releasing the mutex lock after each one is helped.
6. If there are no students waiting, the TA should release the mutex lock and go back to sleep.

Here is some sample code to implement this solution:

#include <pthread.h>
#include <semaphore.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>

#define N 10 // number of students
#define CHAIRS 3 // number of chairs in hallway

pthread_t students[N], ta;
pthread_mutex_t mutex;
sem_t student_sem, ta_sem;

int waiting = 0;

void *student(void *arg) {
    int id = *(int*)arg;
    while (1) {
        // program for a random amount of time
        sleep(rand() % 10 + 1);
        printf("Student %d needs help\n", id);
        pthread_mutex_lock(&mutex);
        if (waiting < CHAIRS) {
            // there is a free chair in the hallway
            waiting++;
            printf("Student %d waiting in hallway (%d/%d)\n", id, waiting, CHAIRS);
            pthread_mutex_unlock(&mutex);
            sem_wait(&student_sem);
            waiting--;
        } else {
            // no free chairs in the hallway
            printf("Student %d will come back later\n", id);
            pthread_mutex_unlock(&mutex);
        }
        // get help from TA
        printf("Student %d getting help from TA\n", id);
        // help for a random amount of time
        sleep(rand() % 5 + 1);
    }
}

void *ta_helper(void *arg) {
    while (1) {
        // sleep until a student wakes us up
        sem_wait(&ta_sem);
        while (1) {
            pthread_mutex_lock(&mutex);
            if (waiting == 0) {
                // no students waiting, go back to sleep
                pthread_mutex_unlock(&mutex);
                break;
            }
            // help the next student in line
            waiting--;
            sem_post(&student_sem);
            printf("TA helping a student (%d/%d)\n", waiting, CHAIRS);
            pthread_mutex_unlock(&mutex);
            // help for a random amount of time
            sleep(rand() % 5 + 1);
        }
    }
}

int main() {
    // initialize mutex lock and semaphores
    pthread_mutex_init(&mutex, NULL);
    sem_init(&student_sem, 0, 0);
    sem_init(&ta_sem, 0, 0);
    // create student threads
    int student_ids[N];
    for (int i = 0; i < N; i++) {
        student_ids[i] = i;
        pthread_create(&students[i], NULL, student, &student_ids[i]);
    }
    // create TA thread
    pthread_create(&ta, NULL, ta_helper, NULL);
    // wait for threads to finish
    for (int i = 0; i < N; i++) {
        pthread_join(students[i], NULL);
    }
    pthread_join(ta, NULL);
    // clean up mutex lock and semaphores
    pthread_mutex_destroy(&mutex);
    sem_destroy(&student_sem);
    sem_destroy(&ta_sem);
    return 0;
}

Note that this is just one possible solution and there may be other ways to implement the same behavior using different synchronization primitives.

用Java实现模拟翻译器 用户输入英文，控制台输出对应中文。 1.要求使用Map存储英文字典，只要求10个单词即可。 2.要求可以循环使用，输入x退出程序。 3.如果查不到，提示字典中不包含该单词。

好的，我可以帮您解答这个问题。

首先，我们需要创建一个 Map 来存储英文字典。这里我们只存储 10 个单词，可以使用 HashMap 来实现。代码如下：

Map<String, String> dictionary = new HashMap<>();
dictionary.put("hello", "你好");
dictionary.put("world", "世界");
dictionary.put("java", "Java语言");
dictionary.put("computer", "计算机");
dictionary.put("book", "书籍");
dictionary.put("table", "桌子");
dictionary.put("chair", "椅子");
dictionary.put("apple", "苹果");
dictionary.put("orange", "橙子");
dictionary.put("banana", "香蕉");

接下来，我们需要使用 Scanner 来读取用户的输入，并在 Map 中查找对应的中文释义。如果找到了，就输出对应的中文释义；如果没有找到，就输出提示信息。代码如下：

Scanner scanner = new Scanner(System.in);
while (true) {
    System.out.print("请输入英文单词：");
    String word = scanner.nextLine().toLowerCase();
    if (word.equals("x")) {
        break;
    }
    String meaning = dictionary.get(word);
    if (meaning != null) {
        System.out.println("中文释义为：" + meaning);
    } else {
        System.out.println("字典中不包含该单词。");
    }
}

在这个程序中，我们使用了一个死循环来不断读取用户的输入。如果用户输入的单词为 x，则退出循环；否则，我们将输入的单词转换为小写字母，并在 Map 中查找对应的中文释义。如果找到了，就输出中文释义；否则，就输出提示信息。

完整代码如下：