HDU1880

        题意就是根据咒语查功能，根据功能查看是否存在相应咒语，题意简单，不过是道不错的练习题。

        下面的都MLE了，听说C++用G++提交才可以AC，否则也MLE；方法很多，不想做了……

        方法一：我用Java的HashMap一直MLE，即便由value反查key减少映射数也一样MLE，听说C++的map可以AC。

        方法二：快排+二分

        方法三：Hash方法，用31求幂。

import java.util.*;

//写俩Map

public class HDU1880 {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str =null;
        Map map1 = new HashMap();
        Map map2 = new HashMap();
        String[] s0;
        while(true) {
            str = sc.nextLine();
            if(str.equals("@END@")) {
                break;
            }
            s0 = str.split(" ", 2);
            str = s0[0].substring(1,s0[0].length()-1);
            map1.put(str,s0[1]);
            map2.put(s0[1], str);
        }
        int n = sc.nextInt();
        for(int i=0; i) {
            String s = sc.next();
            s += sc.nextLine();
            if(s.charAt(0)=='[') {
                s = s.substring(1, s.length()-1);
                boolean tag = map1.containsKey(s);
                if(tag) {
                    System.out.println(map1.get(s));
                }else {
                    System.out.println("what?");
                }
            }else {
                boolean tag = map2.containsKey(s);
                if(tag) {
                    System.out.println(map2.get(s));
                }else {
                    System.out.println("what?");
                }
            }
        }
    }
}

import java.util.*;
//一个Map存两次
public class HDU1880 {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str =null;
        Map map = new HashMap();
        String[] s0;
        while(true) {
            str = sc.nextLine();
            if(str.equals("@END@")) {
                break;
            }
            s0 = str.split(" ", 2);
            str = s0[0].substring(1,s0[0].length()-1);
            map.put(str,s0[1]);
            map.put(s0[1], str);
        }
        int n = sc.nextInt();
        for(int i=0; i) {
            String s = sc.next();
            s += sc.nextLine();
            if(s.charAt(0)=='[') {
                s = s.substring(1, s.length()-1);
            }
            boolean tag = map.containsKey(s);
            if(tag) {
                System.out.println(map.get(s));
            }else {
                System.out.println("what?");
            }
        }
    }
}

import java.util.*;
//map里反查key
public class HDU1880 {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str =null;
        Map map = new HashMap();
        String[] s0;
        while(true) {
            str = sc.nextLine();
            if(str.equals("@END@")) {
                break;
            }
            s0 = str.split(" ", 2);
            str = s0[0].substring(1,s0[0].length()-1);
            map.put(str,s0[1]);
        }
        int n = sc.nextInt();
        for(int i=0; i) {
            String s = sc.next();
            s += sc.nextLine();
            if(s.charAt(0)=='[') {
                s = s.substring(1, s.length()-1);
                boolean tag = map.containsKey(s);
                if(tag) {
                    System.out.println(map.get(s));
                }else {
                    System.out.println("what?");
                }
            }else {
                ArrayList arr = valueGetKey(map,s);
                if(arr.isEmpty()) {
                    System.out.println("what?");
                }else {
                    System.out.println(arr.get(0));
                }
            }
            
        }
    }

    private static ArrayList valueGetKey(Map map,String value) {

        Set set = map.entrySet();
        ArrayList arr = new ArrayList();
        Iterator > it = set.iterator();
        while(it.hasNext()) {
            Map.Entry entry = (Map.Entry)it.next();
            if(entry.getValue().equals(value)) {
                String s = entry.getKey();
                arr.add(s);
            }
        }
        return arr;
    }
}