ccf损坏的RAID5

提交了只得了60分，应该是最后的大数据样例没过

先放在博客里

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
 
using namespace std;
char disk[1000][85000];

int main()
{
    int n, s, l;
    cin >> n >> s >> l;
    for (int i = 0; i < l; ++i)
    {
        int k;
        cin >> k;
        scanf("%s", disk[k]);
    }

    //计算block所在的位置
    int m;
    cin >> m;
    for (int i = 0; i < m; ++i)
    {
        int block,level,diskn,index;
        int level_block_number = (n - 1) * s;
        cin >> block;
        level = block / level_block_number;
        diskn = (n-level+(block % level_block_number) / s ) % n;
        index = block % s;            //index为要求的block在条带中的位置
        //在数组中找到这一段数据
        //disp是在字符数组中的下标 
        //读取一块的大小(4字节),也就是8个字符
        int disp = (level * s + index) * 8;
        //如果要读取数据的盘未损坏且长度够 
        if (disk[diskn][disp]!= '\0')
        {
            char tmp[9]="";
            strncpy(tmp,disk[diskn]+disp,8);
            printf("%s\n",tmp);
        }
        else if (disk[diskn][0]=='\0'&&disk[(diskn+1)%n][disp]!='\0'&& n - 1 == l)
        {
            char iter[9]="";
            int x=0,t;
            for (int j = 0; j < n; ++j)
            {
                if (j != diskn)
                {
                    strncpy(iter,disk[j]+disp,8);
                    sscanf(iter,"%x",&t);
                    x^=t;
                }
            }
            printf("%08X\n", x);
        }
        else printf("-\n");

    }

    return 0;
}