题目1 难度：一星

将一个压缩的字符串原样输出。连续的3个字符及以上才需要压缩，否则输出!error。

字符串包含'@'和'A-Z'时也输出!error。

示例1:

input:
2fdd

output:
!error

示例2:

input：
3fdd

output:
fffdd

示例3:

input:
4f@A

output:
!error

我的答案

import java.util.*;

/**
 * @author leekari
 * @date 2020/10/15 23:03
 * @description
 */
public class Test {
    public static void main(String[] args){
        Scanner scanner = new Scanner(System.in);
        String s = scanner.nextLine();
        char[] chars = s.toCharArray();
        if (chars.length > 100) {
            System.out.println("!error");
        }else {
            boolean error = false;
            List<Object[]> list = new ArrayList<>();
            for (int i = 0; i < chars.length; i++) {
                Object[] arr = new Object[2];
                if (chars[i] >= '0' && chars[i] <= '9' && i == chars.length - 1) {
                    error = true;
                    System.out.println("!error");
                    break;
                }
                if (chars[i] >= 'A' && chars[i] <= 'Z' || chars[i] == 'a') {
                    error = true;
                    System.out.println("!error");
                    break;
                }
                if (chars[i] >= '0' && chars[i] <= '9') {
                    int count = 0;
                    char c = ' ';
                    if (chars[i + 1] >= '0' && chars[i + 1] <= '9') {
                        count = (chars[i] - 48) * 10 + chars[i + 1] - 48;
                        if (chars[i + 2] > 'a' && chars[i + 2] < 'z') {
                            c = chars[i + 2];
                            i = i + 2;
                        }else {
                            error = true;
                            System.out.println("!error");
                            break;
                        }
                    }else {
                        count = chars[i] - 48;
                        if (chars[i + 1] > 'a' && chars[i + 1] < 'z') {
                            c = chars[i + 1];
                            i = i + 1;
                        }else {
                            error = true;
                            System.out.println("!error");
                            break;
                        }
                    }
                    if (count <= 2) {
                        error = true;
                        System.out.println("!error");
                        break;
                    }
                    arr[0] = count;
                    arr[1] = c;
                    list.add(arr);
                }else if (chars[i] >= 'a' && chars[i] <= 'z'){
                    arr[0] = 1;
                    arr[1] = chars[i];
                    list.add(arr);
                }
            }
            if (!error){
                StringBuilder sb = new StringBuilder();
                for (Object[] o: list) {
                    int count = (int) o[0];
                    char c = (char) o[1];
                    while (count > 0) {
                        sb.append(c);
                        count--;
                    }
                }
                if (sb.length() > 100){
                    System.out.println("!error");
                }else {
                    System.out.println(sb.toString());
                }
            }
        }
    }
}

题目2 难度：一星

存在不等式：

a11 * x1 + a12 * x2 + a13 * x3 + a14 * x4 + a15 * x5 >= b1
a21 * x1 + a22 * x2 + a23 * x3 + a24 * x4 + a25 * x5 >= b2
a31 * x1 + a32 * x2 + a33 * x3 + a34 * x4 + a35 * x5 >= b3

其中a11,a21,...,a35都为int类型

x1,...,x5为double类型，b1,b2,b3为double类型

符号位为'>='或者'<='

判断不等式是否成立，并输出最大差

示例1:

input:
2.36,3,6,7.1,6;1,30,8.6,2.5,21;0.3,69,5.3,6.6,7.8;1,13,2,17,5;340,67,300.6;<=,>=,<=

output:
false 758

示例2:

input:
2.3,3,5.6,7,6;11,3,8.6,25,1;0.3,9,5.3,66,7.8;1,3,2,7,5;340,670,80.6;<=,<=,<=

output:
false 458

我的答案

import java.util.*;

/**
 * @author leekari
 * @date 2020/10/15 23:54
 * @description
 */
public class Test2 {
    public static void main(String[] args){
        Scanner scanner = new Scanner(System.in);
        String str = scanner.nextLine();

        String[] strings = str.split(";");
        String[] a1 = strings[0].split(",");
        String[] a2 = strings[1].split(",");
        String[] a3 = strings[2].split(",");
        String[][] doubleA = new String[][]{a1, a2, a3};
        String[] x = strings[3].split(",");
        String[] b = strings[4].split(",");
        String[] flags = strings[5].split(",");
        boolean ok = true;
        double max = Double.MIN_VALUE;
        for (int i = 0; i < flags.length; i++) {
            String[] a = doubleA[i];
            double result = 0.0;
            for (int j = 0; j < a.length; j++) {
                result += Double.parseDouble(a[j]) * Integer.parseInt(x[j]);
            }
            if (flags[i].equals(">=")) {
                double sub = result - Double.parseDouble(b[i]);
                if (sub < 0){
                    ok = false;
                }
                max = Math.max(max, Math.abs(sub));
            }else {
                double sub = Double.parseDouble(b[i]) - result;
                if (sub < 0){
                    ok = false;
                }
                max = Math.max(max, Math.abs(sub));
            }
        }
        System.out.println(ok + " " + (int)max);
    }
}

题目3 难度：二星

有m*n台计算机，计算机由0，1表示，组成局域网。需要相连的计算机标识为1，就可以组成局域网。计算在m*n计算机中组成的最大的局域网包含几台计算机。

示例：

input:
2 2
1 0
1 1

output:
3

我的错误答案：

import java.util.Scanner;

/**
 * @author leekari
 * @date 2020/10/16 00:19
 * @description
 */
public class Test3 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s = scanner.nextLine();
        String[] strings = s.split(" ");
        int n = Integer.parseInt(strings[0]);
        int m = Integer.parseInt(strings[1]);
        int[][] arr = new int[n][m];
        for (int i = 0; i < m; i++) {
            String line = scanner.nextLine();
            String[] lines = line.split(" ");
            for (int j = 0; j < lines.length; j++) {
                arr[j][i] = Integer.parseInt(lines[j]);
            }
        }

        int[] res = new int[n * m];
        int max = Integer.MIN_VALUE;
        for(int r = 0; r < m; r++) {
            for (int l = 0; l < n; l ++) {
                int count = 0;
                for (int i = r; i < m; i++) {
                    for (int j = l; j < n; j++) {
                        if (arr[j][i] == 1){
                            count++;
                        }else {
                            break;
                        }
                    }
                }
                max = Math.max(count, max);
            }
        }

        System.err.println(max);
    }
}