2022.3.20米哈游 云游戏

题型一：单选10道

题型二：不定项选择15道

题型三：编程题3道

1. 捡圣遗物
   思路，使用队列，圣遗物放进队列，队列不为空输出圣遗物，为空输出"no reliquaries."

   import java.util.*;
   public class Main {
   public static void main(String[] args) {
       Scanner sc = new Scanner(System.in);
       int q = sc.nextInt();
       int i = 0;
       Queue<Integer> deque = new LinkedList<>();
       while (i<q){
           int p = sc.nextInt();
           if (p == 1){
               deque.add(sc.nextInt());
               deque.add(sc.nextInt());
           }else {
               if (deque.isEmpty()){
                   System.out.println("no reliquaries.");
               }else {
                   System.out.print(deque.poll()+" " +deque.poll());
               }
           }
           i++;
       }
   }
   }

2. 两个排列左移最少步骤，可以一样
   思路：直接暴力查找，从p2中找到与p1的首元素相等的值，然后判断以此为起点，两个队列相不相等，如果相等，去元素下标i和长度n-i中最小的

   import java.util.Scanner;
   public class Main2 {
   public static void main(String[] args) {
       Scanner sc = new Scanner(System.in);
       int t = sc.nextInt();
       for (int j = 0; j < t ; j++) {
           int n = sc.nextInt();
           int[] p1 = new int[n];
           int[] p2 = new int[n];
           for (int i = 0; i < n; i++) {
               p1[i] = sc.nextInt();
           }
           for (int i = 0; i < n; i++) {
               p2[i] = sc.nextInt();
           }
           System.out.println(move(p1,p2));
       }
   }
   private static int move(int[] p1, int[] p2) {
       int n = p1.length;
       for (int i = 0; i < n; i++) {
           int p1index = 0;
           if (p2[i] == p1[p1index]){
               int p2index = i;
               int count = 0;
               while (p1[p1index++] == p2[(p2index++)%n]){
                   count++;
                   if (count==n){
                       return Math.min(i,n-i);
                   }
               }
           }
       }
       return 0;
   }
   }

3. 删除叶子节点，最少删除几个，可以使直径小于给定值
   没思路，不会