//online parser: 
https://www.bccn.net/run/

程序代码:
#include <stdio.h>
int main(int argc, char *argv[]) 
{
    int a[10], n, i, j;
    
    for (i = 0; i < 10; i++) {
        //get n
        n = ((unsigned char *)main)[25 + i] % 10;
        printf("n = %d\n", n);
        //end
        if(!n) {
            puts("End of Input"); break;
        }
        //pos
        for (j = 0; j < n; j++) {
            a[j] = ((unsigned char *)main)[125 + j];
            printf("%d ", a[j]);
        }
        puts("");
        //rev
        for (j = n - 1; j >= 0; j--)
            printf("%d ", a[j]);
        puts("");
    }
    
    return 0;
}
output sample:
n = 7
0 232 87 254 255 255 131 
131 255 255 254 87 232 0 
n = 9
0 232 87 254 255 255 131 125 204 
204 125 131 255 255 254 87 232 0 
n = 8
0 232 87 254 255 255 131 125 
125 131 255 255 254 87 232 0 
n = 9
0 232 87 254 255 255 131 125 204 
204 125 131 255 255 254 87 232 0 
n = 2
0 232 
232 0 
n = 9
0 232 87 254 255 255 131 125 204 
204 125 131 255 255 254 87 232 0 
n = 9
0 232 87 254 255 255 131 125 204 
204 125 131 255 255 254 87 232 0 
n = 6
0 232 87 254 255 255 
255 255 254 87 232 0 
n = 0
End of Input