第二届全国高校计算机技能竞赛——Java赛道
小赛跳高
签到题
import java.util.*;
public class Main{public static void main(String []args) {Scanner sc = new Scanner(System.in);double n = sc.nextDouble();for(int i = 0; i < 4; i++) {n = n * 0.9;}System.out.printf("%.2f", n);}
}找数
签到题
import java.util.Scanner;
public class Main{public static void main(String []args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();for(int i = 1; i <= n; i++) {if((i % 3 == 1) && (i % 5 == 1) && (i % 7 == 1)) {System.out.print(i + " ");}}}
}打分
签到题
注意cnt等于0的情况
最大值取-0x3f3f3f3f
最小值取0x3f3f3f3f
注意当cnt = 0时,说明无人打出大于0的分数。
此时选手的分数为0分，注意保留2位小数
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();int maxsco = -0x3f3f3f3f;int minsco= 0x3f3f3f3f;int sum = 0;int cnt = 0;while (n-->0) {int sco = sc.nextInt();if (sco > 100 || sco < 1) {continue;}sum += sco;cnt++;if (sco > maxsco) {maxsco = sco;}if (sco < minsco) {minsco = sco;}}if(cnt>0) {System.out.printf("%.2f",(double) (sum - maxsco - minsco) / (cnt - 2));}else {System.out.println(0.00);} }
}找子串
不断截取子串，判断后，找出最大串和最小串
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);String str1 = sc.nextLine();String minstr = str1;String maxstr = "";for (int i = 0; i < str1.length(); i++) {for (int j = i + 1; j <= str1.length(); j++) {String substr = str1.substring(i, j);if (substr.compareTo(minstr) < 0) {minstr = substr;}if (substr.compareTo(maxstr) > 0) {maxstr = substr;}}}System.out.println(minstr);System.out.println(maxstr);}
}
矩阵距离
宽搜bfs 队列存点对,往上下左右四个方向搜索
注意初始化所有距离数组dis[][]为-1
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;class Main {static final int[][] dt = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int m = scanner.nextInt();scanner.nextLine(); // 读取换行符char[][] g = new char[n][m];for (int i = 0; i < n; i++) {String line = scanner.nextLine();for (int j = 0; j < m; j++) {g[i][j] = line.charAt(j);}}bfs(g, n, m);}static void bfs(char[][] g, int n, int m) {int[][] dis = new int[n][m];Queue<int[]> q = new LinkedList<>();for(int i = 0; i < n;i++) {for(int j = 0; j < m;j++) {dis[i][j] = -1;}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (g[i][j] == '1') {dis[i][j] = 0;q.offer(new int[]{i, j});}}}while (!q.isEmpty()) {int[] t = q.poll();int x = t[0];int y = t[1];for (int i = 0; i < 4; i++) {int dx = x + dt[i][0];int dy = y + dt[i][1];if (dx >= 0 && dx < n && dy >= 0 && dy < m && dis[dx][dy] == -1) {dis[dx][dy] = dis[x][y] + 1;q.offer(new int[]{dx, dy});}}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {System.out.print(dis[i][j] + " ");}System.out.println();}}
}总结
难度中下,签到题3道,medium2道。
总体难度没去年难,考查基础思维。
注意代码实现细节和边界情况
