Digital LC-display is widely used in many different fields such as electronic calculator, electronic watch, digital instruments, etc. A simulated example of the numbers represented by digital LC-display device is shown as follows:

Each number represented by LC-display above is composed of s "-" signs for the horizontal segments and s "|" signs for the vertical ones, and each number exactly occupies s+2 columns and 2s+3 rows. Your task is to change the size of the original numbers by changing "s" — the number of signs.

——摘自HDOJ 3575（测试数据不好显示）

这道题题目不难！看了后题目大家都会有想法，是的，我承认不难，但想起自己当初（也才昨天而已）实现的时候真是太痛苦了！都怪自己审题不仔细，忽略了数据的大小。所以一开始时，存储输入数据的数组就只开了100*100， 呃。。写完后本地测试数据轻松过，但提交总是WA！这种情况是很郁闷的！当时改了一下午，空格换行之类的细节想得都快晕了！

到后来是心灰意冷，真没辙了。没办法了，就很无聊的把数据改大了点，草，竟奇迹般地过了。。Accept！！！有木有搞错啊……盼望了一下午啊！可还是高兴不起来，当时真的是哭笑不得。。

注意审题啊，切记……切记……

#include<iostream>
#include<string>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;

int m,n;
char digit[1500][1500];

void f(int row)
{
     int i,j,k;
     int len = strlen(digit[row]);
     for(i=0; i*(m+2)+i-1<len; i++)
     {
              if(i != 0)
              printf(" ");
              printf(" ");
              for(j=0; j<n; j++)
              if(digit[row][(m+2)*i+i+1] == '-')
              printf("-");
              else printf(" ");
              printf(" ");
              }
              printf("\n");
     }
void g(int row)
{
     int i,j,k;
     int len = strlen(digit[row]);
     for(i=0; i*(m+2)+i-1<len; i++)
     {
              if(i != 0)
              printf(" ");
              printf("%c", digit[row][(m+2)*i+i]);
              for(j=0; j<n; j++)
              printf(" ");
              printf("%c", digit[row][i*(m+2)+i+m+1]);
              }
              printf("\n");
     }

int main()
{
    int cases, t;
    int i,j,k;
    
    scanf("%d", &cases);
    t = 1;
    while(t <= cases)
    {
            scanf("%d %d", &m, &n);
            
            getchar();
            for(i=0; i<2*m+3; i++)
             gets(digit[i]);

            
            printf("Case %d:\n", t);
            
            f(0);
            for(i=0; i<n; i++)
            g(1);
            f(m+1);
            for(i=0; i<n; i++)
            g(m+2);
            f(2*m+2);
            t++;
            //if(t != cases)
            printf("\n");
            }   
    return 0;
    }