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...

import java.util.*;
import java.io.*;

class FastReader{
    BufferedReader reader;
    StringTokenizer tokenizer;
    
    public FastReader(InputStream stream){
        reader = new BufferedReader(new InputStreamReader(stream), 1 << 10);
        tokenizer = null;
    }
    
    public String next(){
        while (tokenizer == null || !tokenizer.hasMoreTokens()){
            try{
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (Exception e){
                throw new RuntimeException(e);
            }
        }
        
        return tokenizer.nextToken();
    }
    
    public long next_long(){
        return Long.parseLong(next());
    }
    
    public int next_int(){
        return Integer.parseInt(next());
    }
}

public class Main{
    static long[] seq_a;
    static long[] seq_b;
    
    public static void main(String[] args){
        FastReader reader = new FastReader(System.in);
        int N;
        
        N = reader.next_int();
        seq_a = new long[(int) N];
        for (int i = 0; i < N; i++)
            seq_a[i] = reader.next_long();
        
        N = reader.next_int();
        seq_b = new long[(int) N];
        for (int i = 0; i < N; i++)
            seq_b[i] = reader.next_long();
        
        int total = seq_a.length + seq_b.length;
        int target = ((total & 0x1) != 0) ? total / 2 + 1 : total / 2;
        
        int pa = 0, pb = 0;
        int cnt = 1;
        while (cnt < target){
            if (seq_a[pa] < seq_b[pb])
                pa++;
            else
                pb++;
            
            cnt++;
        }
        
        long ans = Math.min(seq_a[pa], seq_b[pb]);
        System.out.println(ans);
    }
}

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...