PAT 1029

感觉已经没啥心情用JAVA和PAT斗智斗勇了，会内存超限，是不是读进来就内存超限了？

二分做的方法有点bug，又写了一个merge sort的，想着百万数量级应该不会超时吧这都超时也太蛋疼了，结果居然内存爆了...

 1 import java.util.*;
 2 import java.io.*;
 3
 4 class FastReader{
 5     BufferedReader reader;
 6     StringTokenizer tokenizer;
 7
 8     public FastReader(InputStream stream){
 9         reader = new BufferedReader(new InputStreamReader(stream), 1 << 10);
10         tokenizer = null;
11     }
12
13     public String next(){
14         while (tokenizer == null || !tokenizer.hasMoreTokens()){
15             try{
16                 tokenizer = new StringTokenizer(reader.readLine());
17             } catch (Exception e){
18                 throw new RuntimeException(e);
19             }
20         }
21
22         return tokenizer.nextToken();
23     }
24
25     public long next_long(){
26         return Long.parseLong(next());
27     }
28
29     public int next_int(){
30         return Integer.parseInt(next());
31     }
32 }
33
34 public class Main{
35     static long[] seq_a;
36     static long[] seq_b;
37
38     public static void main(String[] args){
39         FastReader reader = new FastReader(System.in);
40         int N;
41
42         N = reader.next_int();
43         seq_a = new long[(int) N];
44         for (int i = 0; i < N; i++)
45             seq_a[i] = reader.next_long();
46
47         N = reader.next_int();
48         seq_b = new long[(int) N];
49         for (int i = 0; i < N; i++)
50             seq_b[i] = reader.next_long();
51
52         int total = seq_a.length + seq_b.length;
53         int target = ((total & 0x1) != 0) ? total / 2 + 1 : total / 2;
54
55         int pa = 0, pb = 0;
56         int cnt = 1;
57         while (cnt < target){
58             if (seq_a[pa] < seq_b[pb])
59                 pa++;
60             else
61                 pb++;
62
63             cnt++;
64         }
65
66         long ans = Math.min(seq_a[pa], seq_b[pb]);
67         System.out.println(ans);
68     }
69 }

PAT，想说爱你不容易... 既然这么不待见JAVA为什么还给JAVA这么严苛的条件，给C++倒是很宽嘛，不用二分做都能过的时间...