int selected = scanner.nextInt();

这段代码通过调用Scanner的nextInt()方法从标准输入流中读取一个整数，并将其赋值给变量selected。nextInt()方法将读取下一个整数，如果输入的内容不是整数，则会抛出InputMismatchException异常。因此，在实际应用中，我们通常需要对nextInt()方法的返回值进行检查，以确保输入的内容是符合要求的。例如：

Scanner scanner = new Scanner(System.in);
System.out.println("请输入一个整数：");
int num;
while (true) {
    try {
        num = scanner.nextInt();
        break;
    } catch (InputMismatchException e) {
        System.out.println("输入的内容不是整数，请重新输入：");
        scanner.nextLine();
    }
}
System.out.println("您输入的整数是：" + num);

这段代码通过使用try-catch语句捕获InputMismatchException异常，如果捕获到异常，则说明输入的内容不是整数，需要提示用户重新输入。如果输入的内容是整数，则跳出循环，并将读取的整数赋值给变量num。

相关问题

public static void searchStudent(Scanner scanner) { System.out.println("------ 查询学生信息 ------"); System.out.println("1. 按学号查询学生信息"); System.out.println("2. 按姓名查询学生信息"); System.out.println("3. 查询某个班级的学生信息"); System.out.println("4. 查询某个学院的学生信息"); System.out.print("请选择操作："); int choice = scanner.nextInt(); switch (choice) { case 1: searchStudentById(scanner); break; case 2: searchStudentByName(scanner); break; case 3: searchStudentByClass(scanner); break; case 4: searchStudentByCollege(scanner); break; default: System.out.println("无效的选项，请重新选择！"); break; } }帮我写完

Sure, here's the complete code:

public static void searchStudent(Scanner scanner) {
    System.out.println("------ 查询学生信息 ------");
    System.out.println("1. 按学号查询学生信息");
    System.out.println("2. 按姓名查询学生信息");
    System.out.println("3. 查询某个班级的学生信息");
    System.out.println("4. 查询某个学院的学生信息");
    System.out.print("请选择操作：");
    int choice = scanner.nextInt();
    switch (choice) {
        case 1:
            searchStudentById(scanner);
            break;
        case 2:
            searchStudentByName(scanner);
            break;
        case 3:
            searchStudentByClass(scanner);
            break;
        case 4:
            searchStudentByCollege(scanner);
            break;
        default:
            System.out.println("无效的选项，请重新选择！");
            break;
    }
}

public static void searchStudentById(Scanner scanner) {
    System.out.print("请输入学生的学号：");
    int id = scanner.nextInt();
    // TODO: 根据学号查询学生信息并输出
}

public static void searchStudentByName(Scanner scanner) {
    System.out.print("请输入学生的姓名：");
    String name = scanner.next();
    // TODO: 根据姓名查询学生信息并输出
}

public static void searchStudentByClass(Scanner scanner) {
    System.out.print("请输入班级名称：");
    String className = scanner.next();
    // TODO: 根据班级名称查询学生信息并输出
}

public static void searchStudentByCollege(Scanner scanner) {
    System.out.print("请输入学院名称：");
    String collegeName = scanner.next();
    // TODO: 根据学院名称查询学生信息并输出
}

You can fill in the TODO parts with your implementation code to search for student information based on the selected option.

John wantsto have a journey around Ireland. He has many items that he likes to carry with him, but although he has three suitcases, unfortunately, he cannot accommodate all his itemsin the suitcases. Thus, he needs to decide which items to pick. He knows the size and worth of each item. Therefore, he likes to pick the items so that the total values of the selected items be maximum. There are several items that he cannot take part of them, like his violin. But some of them, he can take part of them like sugar. You need to write a program in Java to help John select his items and tell John that each item should be put in which suitcase. The input of the program is the number of suitcases that John has and also the capacity of each of them. In addition, the total number of the items, the size and worth of each, and whether John can take a part of the item.

好的，这是一个背包问题（Knapsack Problem），我们可以使用动态规划算法来解决。首先，我们需要定义以下几个变量：

- `n`：旅行者带的物品总数
- `m`：旅行者拥有的行李箱数量
- `w[i]`：第 `i` 个物品的体积
- `v[i]`：第 `i` 个物品的价值
- `t[i]`：第 `i` 个物品是否可以分开带（是则为 1，否则为 0）
- `c[j]`：第 `j` 个行李箱的容量
- `f[i][j][k]`：前 `i` 个物品，放置在前 `j` 个行李箱中，第 `k` 个行李箱还剩余的容量时的最大价值

那么，我们可以得到状态转移方程：

- 当第 `i` 个物品不能分开带时，有：

$$
f[i][j][k] = \max \{f[i-1][j][k], f[i-1][j-1][c[j]] + v[i]\}
$$

- 当第 `i` 个物品能分开带时，有：

$$
f[i][j][k] = \max \{f[i-1][j][k], f[i-1][j-1][c[j]] + v[i], f[i-1][j][k-w[i]] + v[i]\}
$$

最终，答案即为 $f[n][m][0]$。具体实现可以参考以下 Java 代码：

public class Knapsack {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt(); // 物品总数
        int m = scanner.nextInt(); // 行李箱数量
        int[] w = new int[n+1]; // 物品体积
        int[] v = new int[n+1]; // 物品价值
        int[] t = new int[n+1]; // 物品是否可分开带
        int[] c = new int[m+1]; // 行李箱容量
        for (int i = 1; i <= n; i++) {
            w[i] = scanner.nextInt();
            v[i] = scanner.nextInt();
            t[i] = scanner.nextInt();
        }
        for (int i = 1; i <= m; i++) {
            c[i] = scanner.nextInt();
        }
        int[][][] f = new int[n+1][m+1][1001]; // 前 i 个物品，放置在前 j 个行李箱中，第 k 个行李箱还剩余的容量时的最大价值
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                for (int k = 0; k <= c[j]; k++) {
                    f[i][j][k] = f[i-1][j][k];
                    if (k >= w[i]) {
                        f[i][j][k] = Math.max(f[i][j][k], f[i-1][j][k-w[i]] + v[i]);
                    }
                }
                if (t[i] == 1) {
                    for (int k = 0; k <= c[j]; k++) {
                        if (k >= w[i]) {
                            f[i][j][k] = Math.max(f[i][j][k], f[i-1][j-1][c[j]] + v[i]);
                        }
                    }
                }
            }
        }
        System.out.println(f[n][m][0]);
        scanner.close();
    }
}