解法一：队列+模拟

用N个队列模拟银行窗口的队列，每次找出排在第一位且完成时间最早、窗口编号最小的顾客办理业务，并更然后加入一位未进入队列的顾客，更新队首顾客的完成时间。
需要注意的一个坑：如果一个顾客开始办理业务的时间早于17:00，完成时间晚于17:00，那么银行也需要为他办理业务。

  1. import java.io.*;
  2. import java.util.*;
  4. public class Main {
  6.     public static void main(String[] args) throws IOException {
  7.         StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
  8.         PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
  10.         in.nextToken();
  11.         int N = (int) in.nval;
  12.         in.nextToken();
  13.         int M = (int) in.nval;
  14.         in.nextToken();
  15.         int K = (int) in.nval;
  16.         in.nextToken();
  17.         int Q = (int) in.nval;
  19.         Customer[] customers = new Customer[K];
  20.         for (int i = 0; i < K; ++i) {
  21.             in.nextToken();
  22.             customers[i] = new Customer((int) in.nval);
  23.         }
  25.         List<Queue<Customer>> windows = new ArrayList<>();
  26.         for (int i = 0; i < N; ++i) {
  27.             windows.add(new LinkedList<>());
  28.         }
  29.         int toAdd;
  30.         for (toAdd = 0; (toAdd < K) && (toAdd < N * M); ++toAdd) {
  31.             windows.get(toAdd % N).offer(customers[toAdd]);
  32.         }
  35.         int todo = 0;
  36.         while (todo < K) {
  37.             int min = 0x3f3f3f3f;
  38.             Customer next = null;
  39.             int windowID = -1;
  40.             for (int i = 0; i < windows.size(); ++i) {
  41.                 Customer customer = windows.get(i).peek();
  42.                 if (customer == null) {
  43.                     continue;
  44.                 }
  45.                 // 完成时间最短, 窗口编号小的优先
  46.                 if ((customer.time - customer.cost < 540) && (customer.time < min)) {
  47.                     min = customer.time;
  48.                     next = customer;
  49.                     windowID = i;
  50.                 }
  51.             }
  53.             // 没有需要服务的客户
  54.             if (next == null) {
  55.                 break;
  56.             }
  58.             // 更新自己和队列中下一个人的状态
  59.             next.isFinish = true;
  60.             Queue<Customer> queue = windows.get(windowID);
  61.             queue.poll();
  62.             if (toAdd < K) {
  63.                 queue.offer(customers[toAdd++]);
  64.             }
  65.             if (!queue.isEmpty()) {
  66.                 queue.peek().time += next.time;
  67.             }
  68.             ++todo;
  69.         }
  70.         for (int i = 0; i < Q; ++i) {
  71.             in.nextToken();
  72.             int index = (int) in.nval - 1;
  73.             out.println(customers[index]);
  74.         }
  75.         out.flush();
  76.     }
  77. }
  79. class Customer {
  80.     // 处理时间
  81.     int cost;
  83.     // 完成时间
  84.     int time;
  86.     // 是否能被银行处理
  87.     boolean isFinish;
  89.     public Customer(int cost) {
  90.         this.cost = cost;
  91.         time = cost;
  92.         isFinish = false;
  93.     }
  95.     @Override
  96.     public String toString() {
  97.         if (isFinish) {
  98.             int minute = (time + 480) / 60;
  99.             int second = time % 60;
  100.             return String.format("%02d:%02d", minute, second);
  101.         }
  102.         return "Sorry";
  103.     }
  104. }