此题有大坑

• 坑一：由于字符串不知道多长，因此不能用数组存储（亲自试验，3000不够用，就算够用，开辟一个大数组的耗时也会导致超时）

• 坑二 ：由于字符串过长，使用Scanner读字符串也会导致超时

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Scanner;

public class 损坏的RAID5{
    
    static int N = 1010;
    static String data[] = new String[N];
    static int len[] = new int[N];
    public static void main(String[] args) throws NumberFormatException, IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        String[] split = br.readLine().split(" ");
        int n = Integer.parseInt(split[0]);//磁盘
        int s = Integer.parseInt(split[1]);//条带
        int l = Integer.parseInt(split[2]);//现有磁盘
        for(int i = 0 ; i < l ; i ++) {
            String nextLine[] = br.readLine().split(" ");
            int diskNum =  Integer.parseInt(nextLine[0]);
            data[diskNum] = nextLine[1];
            len[diskNum] = data[diskNum].length();
        }
        int m = Integer.parseInt(br.readLine());
        while(m-->0) {
            int x = Integer.parseInt(br.readLine());
            int i = x / ((n - 1) * s);// x在第i行(编号)
            int j = x % ((n - 1) * s);// 多了j个
            int h = j % s + i * s;// x的高度
            int c = (n - 1 - i % n + j / s + 1) % n;// m所在磁盘号
            String res = find(h,c,n);
            bw.write(res+"\n");
        }
        bw.flush();
    }
    private static String find(int h, int c, int n) {
        if(len[c]/8-1<h) {//读取不到
            return Eor(h,c,n);
        }
        return data[c].substring(8*h,8*h+8);//直接读
    }
    private static String Eor(int h, int c,int n) {
        int res[] = new int[4];
        for(int i = 0 ; i < n ;i ++) {//除了c列，若再出现错误列，返回“-”
            if(i==c) continue;
            if(len[i]/8-1<h) return "-";
            String ss = data[i].substring(8*h,8*h+8);//求异或值
            for(int j = 0 ; j < 4; j ++) {
                int a = Integer.valueOf(ss.substring(j*2,j*2+2),16);
                res[j] ^= a;
            }
        }
        String ans = "";
        for(int i = 0 ; i < 4 ; i ++) {
            ans+= String.format("%02X", res[i]);
        }
        return ans;
    }
}