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
