CCF-CSP题解 201903-3 损坏的RAID5

先吐槽先吐槽！因为输入太大，需要用fgets，读n个字符或读到回车终止。

char *fgets(char *str, int n, FILE *stream)

因为scanf模拟考试T了10+次。因为IO超时的题目都是乐色！（x

用数组存磁盘信息。\(char [1000][40960*2]\)不会爆。静态数据区一般2G大小。

然后就是模拟。计算每个询问块号的条带号、单个磁盘条带号、磁盘号。分情况讨论。

%x小写16进制，%X大写。%d十，%o八，%b二。

#include <bits/stdc++.h>
const int maxn = 1000;
const int maxm = 40960;

using namespace std;

char disk[maxn+10][maxm*2+10];
int vis[maxn+10];
int len, got = 0;

int main()
{
    int n, s, l;
    scanf("%d%d%d", &n, &s, &l);

    memset(vis, 0, sizeof(vis));
    for (int i = 1, x; i <= l; i++)
    {
        scanf("%d ", &x);
        vis[x] = 1;
        fgets(disk[x], maxm*2 + 10, stdin);
        if (!got)
        {
            len = strlen(disk[x]) / 8 / s - 1;
            got = 1;
        }
    }

    int m;
    scanf("%d", &m);
    while (m--)
    {
        int b;
        scanf("%d", &b);
        int stripe = b / s;
        int k = stripe / (n-1);
        int diskNum = (n - k%n + stripe % (n-1)) % n;
        if (k > len || (n-l >= 2 && !vis[diskNum]))
        {
            printf("-\n");
        }
        else
        {
            int st = (k*s + b%s) * 8, en = st + 7;
            if (vis[diskNum])
            {
                for (int i = st; i <= en; i++)
                    printf("%c", disk[diskNum][i]);
                printf("\n");
            }
            else
            {
                for (int i = st; i <= en; i++)
                {
                    int ans = 0;
                    for (int j = 0, tmp; j <= n-1; j++)
                    {
                        if (j != diskNum)
                        {
                            if (disk[j][i] >= '0' && disk[j][i] <= '9')
                            {
                                tmp = disk[j][i] - '0';
                            }
                            else
                            {
                                tmp = disk[j][i] - 'A' + 10;
                            }
                            ans ^= tmp;
                        }
                    }
                    printf("%X", ans);
                }
                printf("\n");
            }
        }
    }

    return 0;
}